Paper XII : Ordinary Differential Equation
Lecture notes for the Post-graduate, sem 3 Course Department of Mathematics Ramakrishna Mission Vidyamandira Belur Math, INDIA Course Instructor : Dr. Arnab Jyoti Das Gupta August, 2020 to January, 2021 2
Syllabus
1. Preliminaries – Initial Value problem and the equivalent integral equation, mth order equation in d-dimensions as a first order system, concepts of local existence, existence in the large and uniqueness of solutions withy examples.
2. Basic Theorems – Ascoli-Arzela Theorem. A Theorem on convergence of solutions of a family of initial-value problems.
3. Picard-Lindelof Theorem – Peano’s existence Theorem and corollary. Maximal intervals of exis- tence. Extension Theorem and corollaries. Kamke’s convergence Theorem. Kneser’s Theorem (Statement only).
4. Differential inequalities and Uniqueness – Gronwall’s inequality. Maximal and minimal solu- tions. Differential inequalities. A Theorem of Winter. Uniqueness Theorems. Nagumo’s and Osgood’s criteria.
5. Egres pointstand Lyapunov functions. Successive approximations.
6. Variation of constants, reduction to smaller sustems. Basic inequalities, constant coefficients. Floquet Theory. Adjoint systems, Higher order equations.
7. Linear second order equations – Preliminaries. Basic facts. Theorems of Sturm. Sturm Liou- vilee Boundary value Problems.
References
1. P. Hartman, Ordinary Differential Equations, John Wiley (1964).
2. E.A. Coddington and N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill, NY (1955).
3. G.F. Simmons : Differential Equaitons.
4. W. E. Boyce and R. C. DiPrima, Elementary Differential Equations and Boundary Value problems.
5. S. L. Ross, Differential Equation Contents
1 Existence and Uniqueness of solutions 5 1.1 Notations ...... 5 1.2 Initial Value problem ...... 6 1.3 Uniqueness of solutions ...... 16 1.3.1 Lipschitz condition ...... 17 1.4 Method of successive spproximations ...... 20 1.5 Continuation of solutions ...... 22 1.6 System of differential equations of first order ...... 24 1.7 Higher order ODEs as system of first order ODEs ...... 25 1.8 Dependency on Initial conditions ...... 28
2 System of first order ordinary differential equations 33 2.1 System of First order ODEs ...... 33 2.2 Systems of linear odes ...... 34 2.3 Uniqueness of the solution of the system of differential equations ...... 37 2.3.1 Existence of Fundamental set of solutions ...... 40 2.3.2 Linear Differential operators (constant coefficients) ...... 40 2.3.3 Linear Differential operators (Variable coefficients) ...... 44 2.3.4 Existence and uniqueness theorem ...... 45 2.4 Inhomogeneous system of first order linear odes ...... 48 2.4.1 n-th order linear ode as a system of first order linear odes ...... 51 2.4.2 n-th order linear ode with constant coefficients ...... 53 2.5 Phase potrait ...... 61
3 Differential Inequalities 65 3.1 Gronwall’s Inequality ...... 65 3.2 Solution of a differential inequality ...... 66
4 Some more Existence and Uniqueness results 71 4.1 Maximal And Minimal solutions ...... 71 4.2 Uniqueness results ...... 73
3 4 CONTENTS
5 Sturm-Liouville Theory 79 5.1 Adjoint of a second order linear ODE ...... 79 5.2 Self-adjoint 2nd order linear ode ...... 80 5.3 Basic results of Sturm theory ...... 82 5.4 Sturm-Liouville Problems ...... 87
6 Variation of parameters 91 6.1 General theory for second order linear odes ...... 93
7 Liapunov functions 97 7.1 Stability of non-linear odes ...... 97 7.1.1 Liapunov’s direct method ...... 98 7.2 Instability theorems ...... 104 Chapter 1 Existence and Uniqueness of solutions
Lecture 1
1.1 Notations
Through out our discussion we will be using the following notations.
• I = (a, b) will denote an open interval in R.
• Ck(I) will denote the set of all complex valued functions having k-continuous derivatives on I.
• When I is an interval other than open interval, we can extend the above definition as follows
– If f has right hand k-th derivative existing at a which is continuous from the right at a, then we will say f ∈ Ck([a, b)).
– If f has left hand k-th derivative existing at b which is continuous from the left at b, then we will say f ∈ Ck((a, b]).
– Analogously, we have the condition for f ∈ Ck([a, b]).
• D will denote the domain, meaning an open connected set in the real (t, x) plane, where t is the independent variable and x will be a solution or the dependent variable.
• Ck(D) will denote the set of all complex valued functions on D such that all k-th order ∂kf partial derivatives ∂tp∂xq , p + q = k, exist and are continuous on D.
• C0(I) or C(I) will denote the set of all continuous functions on I.
• If D is such that it has multiple boundary points, which are also limit points, then one may look at the continuity of the left-hand and / or righ-hand derivatives at each such points to define Ck(D) accordingly.
5 6 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
Our Aim : To solve the following problem :- Find a differentiable function ϕ defined on I such that
1. ∀t ∈ I, (t, ϕ(t)) ∈ D and
2. ϕ0(t) = f(t, ϕ(t)), ∀t ∈ I. where f ∈ C(D) and D is a domain.
Remark. 1. Such a problem is called an ordinary differential equation of the first order.
2. It is also represented as (E) x0(t) = f(t, x), t ∈ I
3. If such a differentiable function ϕ exists then ϕ is called a solution of the differential equation (E) on I.
4. Since, f ∈ C(D), ϕ0 ∈ C(I) =⇒ ϕ ∈ C1(I).
5. From the geometrical point of view, the above problem can be rephrased as finding a solution ϕ ∈ C1(I) whose graph (t, ϕ(t)) has slope f(t, ϕ(t)) at the point (t, ϕ(t)).
1.2 Initial Value problem
To find an interval I containing τ and a solution ϕ of (E) on I satisfying ϕ(τ) = ξ, i.e. satisfying
x0(t) = f(t, x(t)), x(τ) = ξ
Remark. 1. ODEs like x0(t) = 1 have infinitely many solutions x(t) = t + c, where c is a constant.
2. To avoid such situations we try to impose conditions on the solutions to obtain either a unique or a smaller class of solutions.
Coming back to our initial value problem
(IVP ) x0(t) = f(t, x(t)), x(τ) = ξ 1.2. INITIAL VALUE PROBLEM 7
If ϕ is a solution to the above problem, then we should be able to integrate both sides and obtain
Z t ϕ(t) − ϕ(τ) = f(s, ϕ(s))ds τ Z t =⇒ ϕ(t) = ϕ(τ) + f(s, ϕ(s))ds, ∀t ∈ I τ
On the other hand if we start with a function implicitly defined as
Z t (∗) Ψ(t) = ξ + f(s, ψ(s))ds, ∀t ∈ I τ then, we have Ψ ∈ C1(I) as f is continuous. Taking derivatives wrt t we have
Ψ 0(t) = f(t, Ψ(t)), ∀t ∈ I
Additionally, we have Ψ(τ) = ξ
Thus, Ψ is a solution of IVP. This shows we have a one-to-one correspondence between the solutions of IVP and the C1(I) functions of the form (∗). Hence, the above IVP is equivalent to finding the solution of the integral equations (∗).
Remark. 1. Though we have obtained an equivalent problem of the IVP, we have not yet solved it.
2. In fact, we still have not answered the question of whether such a solution exists or not.
3. Even if the solution exists for one particular interval I, will it exist on whole of R or on certain other intervals I0?
dy 2 Example. Consider an example dt = y with the initial condition y(1) = −1. THen, clearly 1 1 y(t) = − t is a solution. But, note that t is undefined when t = 0. THus, it will be a solution only for those intervals I, that do not contain the point t = 0.
Remark. The above example shows the interval plays an important role in answering the question of existence of solutions. 8 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
Definition 1.2.1 (-approximate solutions). Let f ∈ C(D) be real valued. A function ϕ ∈ C(I) is said to be an -approximate solution of (E) on the interval I, if it satisfies the following conditions
1. (t, ϕ(t)) ∈ D, ∀t ∈ I
2. ϕ ∈ C1(I), except at most for finitely many points, where ϕ0 may have simple discontinuities.
3. |ϕ0(t) − f(t, ϕ(t))| ≤ , ∀t ∈ I\S, where S is the set of all simple discontinuities of the function ϕ0.
Remark. 1. Any function ϕ ∈ C(I) having property (2), given above, is said to have piecewise 1 continuous derivative on I and is denoted by ϕ ∈ Cp (I).
2. Recall that a function f has a simple discontinuity at a point c if the left and right limits of f exist at c, but are not equal.
3. If = 0, then it means ϕ ∈ C1(I), in which case S = φ and we have our solution.
Some Notations to be used later :
1. Rectangular regions R will denote the following
|t − τ| ≤ a, |x − ξ| ≤ b, a, b > 0, i.e. R := [τ − a, τ + a] × [ξ − b, ξ + b]
It is a rectangular region having center at (τ, ξ).
2. M = max |f(t, x)|, since, f is continuous and R is compact and M exists. (t,x)∈R
b 3. α = min a, M .
Lecture 2
Theorem 1.2.2 (Existence of solution). Let f ∈ C(R) and > 0. Then, there exists an -approximate solution ϕ of (E) on the interval [τ − α, τ + α] such that ϕ(τ) = ξ.
This can be rephrased as 1.2. INITIAL VALUE PROBLEM 9
Let us consider an initial value problem
x0(t) = f(t, x), t ∈ [τ − a, τ + a] (1) and x(τ) = ξ,
Further, consider b, ∈ R+ such that f ∈ C(R), where R = [τ − a, τ + a] × [ξ − b, ξ + b]. Then, there exists an α ∈ R+ and and -approximate solution of (1) on the interval [τ − α, τ + α], where α = min a, b and M = max |f(t, x)|. M (t,x)∈R
Proof. Since, R is compact and f ∈ C(R), f is uniformly continuous on R. Thus, for the given
> 0, ∃δ > 0 such that |f(t, x) − f(t,˜ x˜)| ≤ , whenever ||(t, x) − (t,˜ x˜)|| ≤ δ. (See figure (1.1)).
Figure 1.1: Diagramatic representation of the neighborhoods
Also, we now have the existence of M and α as defined before. Clearly, α ≤ a and hence we won’t run into an problem of going outside the domain, if we divide the interval [τ, τ + α] into n-equal parts t0 = τ < t1 < t2 < ··· < tn = τ + α in such a way that δ max |tk − tk−1| ≤ min δ, 1≤k≤n M
In the interval t ∈ [t0, t1], construct a line segment passing through (t0, ξ) with slope f(t0, ξ). Let this line segment meet the boundary t = t1 at the point (t1, ξ1). 10 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
Figure 1.2: Diagramatic representation of the approximate solution.
Then, construct another line segment passint through (t1, ξ1) with slope f(t1, ξ1) in the interval
[t1, t2]. Follow this process, till the last interval [tn−1, tn]. (see figure (1.2)).
Note :-
1. ξk = ξk−1 + (t − tk−1)f(tk−1, ξk−1), ∀k ≥ 1.
2. By the definition of M, the above line segments will remain in the region T .
Thus, if we construct a function ϕ defined as
ξ , if t = τ ϕ(t) = ϕ(tk−1) + f(tk−1, ϕ(tk−1))(t − tk−1) , t ∈ [tk−1, tk], 1 ≤ k ≤ n then ϕ forms the required -approximate solution where ϕ is extended to [τ − α, τ] in the same way as in [τ, τ + α].
Remark. 1. ϕ has the exact graph, that we had constructed using the previous line segments.
2. Clearly,
(a) ∀t ∈ [τ, τ + α], (t, ϕ(t)) ∈ T ⊆ R.
1 (b) ϕ ∈ Cp ([τ, τ + α]) as the points t1, t2, ··· , tn may pose problems. 1.2. INITIAL VALUE PROBLEM 11
0 (c) |ϕ (t) − f(t, ϕ(t))| = |f(tk−1, ϕ(tk−1)) − f(t, ϕ(t))| ≤ , ∀t ∈ (tk−1, tk).
Definition 1.2.3 (Equicontinuous collection of functions). Let F = {f : I → R}, where I ⊆ R is an interval. THis collection of functions F is said to be equicontinuous on I if for each > 0, ∃δ > 0 such that
|f(t) − f(t˜)| < , whenever f ∈ F; t, t˜∈ I; satisfy |t − t˜| < δ.
Figure 1.3: Graphical illustration of equicontinuous collection of functions.
Remark. With respect to the above figure note the following
1. Within [a, b], {f1, f2} forms an equicontinuous class of functions but within [b, β], they don’t.
2. Graphically, one may understand the collection of equicontinuous functions as follows :- Given any horizontal strip of height > 0, we will be able to obtain a δ > 0, such that whenever we consider a vertical strip of width δ, within the given interval I, we will have the graph of all the functions of this collection enclosed within a rectangle of horizontal dimension δ and vertical dimension . (see figure (1.4)).
3. One of the most important properties of the set of all equicontinuous functions is given by the following lemma (Ascoli or often referred to as Arzella-Ascoli’s theorem). 12 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
Figure 1.4: Graphical illustration of two equicontinuous functions.
Lemma 1.2.4. Let I ⊆ R be a bounded interval and let F be an infinite collection of uniformly bounded, equicontinuous functions defined on I. Then, F contains a sequence {fn} which is uniformly convergent on I.
Proof. As Q is countable, ∃ a bijection ϕ : N → Q ∪ I. Let {rk}, k ∈ N be the enumerated collection of the rational numbers in I.
Since, the collection F is uniformly bounded, the collection of real numbers {f(r1)|f ∈ F} is a bounded set and hence, has a convergent sequence.
Thus, ∃ a sequence of functions {fn,1} of F such that {fn,1(r1)} is convergent.
Similarly for the collection {fn,1(r2)}, we will find a subsequence of functions {fn,2} ⊆ F such that
{fn,2(r2)} is convergent.
Thus, for each k ∈ N, we will be able to obtain a subsequence {fn,k} ⊆ {fn,k−1} ⊆ · · · ⊆ {fn,1} ⊆ F, such that {fn,k(rk)} is convergent. [Note, uniformly boundedness is required to insure that
{f(rn)|f ∈ F} is bounded for each n ∈ N.] Note that for each fixed k ∈ N, the sequence of functions {fn,k} are convergent at the points r1, r2, ··· , rk .
Now, construct a new sequence of functions {gn} ⊆ F defined as gn = fn,n.
Claim : {gn} is the required uniformly convergent subsequence on I . 1.2. INITIAL VALUE PROBLEM 13
Verification : for any fixed k ∈ N, the tail sequence {gn}n≥k is a subsequence of {fn,k} and hence, the sequence of real numbers {gn(rk)} converges.
As k was arbitrarily chosen from N, we have for every k ∈ N, {gn(rk)} is convergent.
Since, {rk} was an enumeration of rationals in I, we have {gn(r)} Is convergent for all rationals r ∈ I.
Hence, for every > 0, rk ∈ I ∩ Q, ∃N(rk) such that
|gn(rk) − gm(rk)| < , ∀n, m ≥ N(rk)
Again as the given collection F is equicontinuous, for the above chosen > 0, ∃δ > 0 such that
|f(t) − f(t˜)| < , ∀f ∈ F; t, t˜∈ I; |t − t˜| < δ
Since, I is a bounded interval in R, we can split I into finitely many subintervals Ij, j = 1, 2, ··· , p, such that the length of the largest sub interval is less than δ . (see figure (1.5).) Now, as the set of
Figure 1.5: Division of intervals.
all rational in I is dense in I, therefore for each of the above subintervals Ij, we can find a rational rk(j) ∈ Ij (this rational depends on the enumeration as well as the sub interval Ij).
Finally, to show uniform convergence of the sequence of functions {gn} over the interval I, we need to show that for each > 0, ∃M ∈ N such that for every t ∈ I, |gn(t) − gm(t)|, whenever n, m ≥ M.
Now, t ∈ I =⇒ ∃j ∈ {1, 2, ··· p} such that t ∈ Ij. Hence, we have
|gn(t) − gm(t)| ≤ |gn(t) − gn(rk(j))| + |gn(rk(j)) − gm(rk(j))| + |gm(rk(j)) − gm(t)|. 14 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
Now, t, rk(j) ∈ Ij =⇒ |t − rk(j)| < δ, which implies |gn(t) − gn(rk(j))| < and |gm(rk(j)) − gm(t)| < .
Also, convergence of {gn} over I ∩ Q, gives |gn(rk(j)) − gm(rk(j))| < , ∀n, m ≥ N(rk(j)).
Thus, we have |gn(t) − gm(t)| < 3, ∀n, m ≥ N(rk(j)).
Now, we have finitely many subintervals Ij and for each subinterval we have fixed a rational number rk(j). Thus, we have chosen only finitely many rationals rk(j), j = 1, 2, ··· , p, representing each of the finitely many intervals Ij.
Let M = max {N(rk(j))}. Then, for n, m ≥ M we have 1≤j≤p
|gn(t) − gm(t)| < 3, ∀t ∈ I
Hence, {gn} forms an uniformly convergent sequence on I.
Lecture 3
Theorem 1.2.5 (Cauchy-Peano existence theorem). Let R be the rectangular region as defined earlier. Let f ∈ C(R), then ∃ϕ ∈ C1([τ − α, τ + α]), which satisfies the IVP
ϕ0(t) = f(t, ϕ(t)) on [τ − α, τ + α], ϕ(τ) = ξ
b Here, α = min{a, M }, M and a, b are defined as earlier.
1 Proof. Let n = n , n ∈ N. Then, by theorem 1.1, for each n ∈ N∃ an -approximate solution of the IVP, say ϕn such that ϕn(τ) = ξ∀n ∈ N, on the interval I = [τ − α, τ + α], where α is defined as b min{a M }.
By construction of each ϕn, as per theorem 1.1, we have
ϕn(t) − ϕn(t˜)| ≤ M|t − t˜|, ∀t, t˜∈ I (i) 1.2. INITIAL VALUE PROBLEM 15
Thus, for t˜= τ, we have
|ϕn(t) − ϕn(τ)| = M|t − τ| b =⇒ |ϕ (t) − ξ| ≤ b as |t − τ| ≤ α ≤ n M
=⇒ |ϕn(t)| ≤ |ϕn(t) − ξ| + |ξ| ≤ b + |ξ| (ii)
This is true ∀n ∈ N. Thus, the sequence {ϕn} is uniformly bounded by (b + |ξ|).
Again, (i) suggests that {ϕn} are equicontinuous. Hence, by the Ascoli’s lemma, there exists a subsequence {ϕnk } uniformly convergent on [τ − α, τ + α].
Let ϕnk → ϕ as k → ∞. As ϕnk ’s are continuous and {ϕnk } uniformly converges to ϕ, the limit ϕ must also be continuous on [τ − α, τ + α].
Z t Now, ϕn(t) = ξ + [f(s, ϕn(s)) + ∆n(s)] ds (iii) τ
0 0 where ∆n(s) = ϕn(s) − f(s, ϕn(s)) defined only on those points where ϕn exists. We define 0 ∆n(s) = 0, ∀s for which ϕn does not exist.
Since, ϕn is an n-approximate solution on the [τ − α, τ + α], |∆n(s)| ≤ n. Again, as f is uniformly continuous on R and ϕnk → ϕ uniformly on [τ − α, τ + α], it follows that lim f(t, ϕn (t)) = f t, lim ϕn (t) = f(t, ϕ(t)) (iv) k→∞ k k→∞ k
The convergence is uniform. Thus, (iii) and (iv) gives
Z t Z t
ϕ(t) = lim ϕnk (t) = lim ξ + {f(s, ϕnk (s)) + ∆nk (s)}ds = ξ + f(s, ϕ(s))ds (v) k→∞ k→∞ τ τ
Now, (v) suggests ϕ(τ) = ξ and ϕ0(t) = f(t, ϕ(t)) (As f is a continuous function). Thus, ϕ is a solution of the given IVP.
Question 1.2.6. 1. Is the choice of the subsequence in the above proof necessary? Justify.
2. If uniqueness is assumed, will the choice be unnecessary? [Ref: Theory of ODEs by Coddington and Levinson, Chapter 1.] 16 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
Lecture 4
Theorem 1.2.7. Let f ∈ C(D) and (τ, ξ) ∈ D. Then, ∃ a solution ϕ of the IVP on some t-interval containing τ in its interior.
Proof. D is a domain and as per our conventions D is open. Therefore, there exists r > 0 such that B((τ, ξ), r) ⊆ D as (τ, ξ) ∈ D. Let R be a closed rectangle centered at (τ, ξ) and contained in B((τ, ξ), r). Thus, f ∈ C(R). Hence, 1 by Cauch-Peano’s theorem, ∃ϕ ∈ C (I1) which solves the given IVP. Here,
I1 = [τ − α, τ + α] ⊆ [τ − a, τ + a].
1.3 Uniqueness of solutions √ Example. Consider the ode x0 = 3 x with initial condition x(0) = 0. Let the interval for this problem be [0, 1]. For c ∈ [0, 1] define
0 , ∀t ∈ [0, c] ϕc(t) = 3 n 2(t−c) o 2 3 , otherwise.
0 √ Then, note that ϕc(t) = ϕc and ϕc(0) = 0. But, ϕc1 6= ϕc2 for c1 6= c2. (see figure (1.6).) Thus,
Figure 1.6: ϕc are different for different values of c. uniqueness is not guaranteed, even though the right hand side is continuous. Hence, we require something more than continuity of f(t, x). One such sufficient condition is the Lipschitz condition. 1.3. UNIQUENESS OF SOLUTIONS 17
1.3.1 Lipschitz condition
Definition 1.3.1. One variable : Let f : D → R,D ⊆ R. Then f is said to be Lipschitz continuous over D if ∃ a constant k > 0 such that
|f(x1) − f(x2)| ≤ k|x1 − x2|, ∀x1, x2 ∈ D
Two variables : Let f : D → R,D ⊆ R2. Then f is said to be Lipschitz continuous in second variable if ∃ a constant k > 0 such that
|f(t, x1) − f(t, x2)| ≤ k|x1 − x2|, ∀(t, x1), (t, x2) ∈ D
( Note, here the first variable is fixed.)
Remark. 1. Similarly one may define Lipschitz continuity with respect to any particular variable.
2. Note that one may also define Lipschitz continuity with respect to all the variables together as f : D → R,D ⊆ Rm is said to be Lipschitz continuous if ∃ a constant k > 0 such that
|f(x1) − f(x2)| ≤ k||x1 − x2||, ∀x1, x2 ∈ D
Example. 1. f(x) = sin x is Lipschitz continuous on R. √ 2. g(x) = x is not Lipschitz continuous on [0, 1].
√ 3. f(t, x) = sin x t is Lipschitz continuous in x but not in t.
Notation : We will say f ∈ (C, Lip) on D to state that the two variable function f is continuous on D and lipschitz continuous in the second variable.
Theorem 1.3.2. Let us consider the IVP
ϕ0(t) = f(t, x) on I, ϕ(τ) = ξ 18 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
1 Suppose f ∈ (C, Lip) on D, with Lipschitz constant k. Let ϕ1, ϕ2 ∈ Cp on some interval (a, b), containing τ be 1 and 2 approximate solutions of the above IVP. Further assume that
|ϕ1(τ) − ϕ2(τ)| ≤ δ for some fixed τ ∈ (a, b) and a fixed δ ≥ 0. If = 1 + 2, then for all t ∈ (a, b),
|ϕ (t) − ϕ (t)| ≤ δek|t−τ| + ek|t−τ| − 1 1 2 k
Proof. Consider t ∈ [τ, b). Since, ϕ1, ϕ2 are 1, 2 approximate solutions, we have
0 |ϕi(s) − f(s, ϕi(s))| ≤ i (i = 1, 2) (1.1) at all but finitely many points on [τ, b). Integrating from τ to t we get
Z t Z t 0 |ϕi(t) − ϕi(τ) − f(s, ϕi(s))ds| ≤ |ϕi(s) − f(s, ϕi(s))|ds ≤ i(t − τ) (1.2) τ τ
Summing up, we get
Z t
[ϕ1(t) − ϕ2(t)] − [ϕ1(τ) − ϕ2(τ)] − [f(s, ϕ1(s)) − f(s, ϕ2(s))]ds ≤ (t − τ) (1.3) τ where = 1 + 2. Let r(t) = |ϕ1(t) − ϕ2(t)|. Thus, from (1.2) and (1.3) we have
Z t r(t) ≤ r(τ) + |f(s, ϕ1(s)) − f(s, ϕ2(s))|ds + (t − τ). τ
As f is Lipschitz on D, Z t r(t) ≤ r(τ) + k r(s)ds + (t − τ) (1.4) τ R t Now, define R(t) = τ r(s)ds, t ∈ [τ, b). Thus, (1.4) becomes
0 R (t) − kR(t) ≤ δ + (t − τ) ∵ r(τ) ≤ δ
Multiplying both sides by e−k(t−τ) and integrating from τ to t we get
δ e−k(t−τ)R(t) ≤ 1 − e−k(t−τ) − e−k(t−τ)[1 + k(t − τ)] + k k2 k2 1.3. UNIQUENESS OF SOLUTIONS 19
Thus, we have δ R(t) ≤ ek(t−τ) − 1 − [1 + k(t − τ)] + ek(t−τ) (1.5) k k2 k2 Combining (1.4) and (1.5) we get
r(t) ≤ δek(t−τ) + ek(t−τ) − 1 , ∀t ∈ [τ, b) k
Similar results will be obtained for (a, τ]. Hence, we have
|ϕ (t) − ϕ (t)| ≤ δek(t−τ) + ek(t−τ) − 1 , ∀t ∈ (a, b) 1 2 k
Remark. 1. If ϕ1 = ϕ (actual solution), then for any 2 approximate solution ϕ2, we have
|ϕ(t) − ϕ2 (t)| ≤ δek(t−τ) + 2 ek(t−τ) − 1 , ∀t ∈ (a, b) 2 k
2 implies that ϕ2 → ϕ as 2 → 0, δ → 0.
2. We can better this result by taking ϕ2 such that the initial value is satisfied at τ. Then, δ = 0, which implies |ϕ(t) − ϕ2 (t)| ≤ 2 ek(t−τ) − 1 , ∀t ∈ (a, b) 2 k
3. If δ = = 0, i.e. two exact solutions ϕ1, ϕ2 passing through the same initial point at t = τ, then
|ϕ1(t) − ϕ2(t)| ≤ 0 =⇒ ϕ1 = ϕ2
Thus, we have uniqueness of solution of IVP.
Lecture 5
Theorem 1.3.3. Let f ∈ (C, Lip) in D and (τ, ξ) ∈ D. If ϕ1 and ϕ2 are any two solutions of 0 ϕ (t) = f(t, ϕ(t)) on (a, b), a < τ < b, such that ϕ1(τ) = ϕ2(τ) = ξ, then ϕ1 = ϕ2.
Proof. Since, ϕ1(τ) = ϕ2(tau) = ξ, we can take δ = 0 in theorem (1.3.2). 20 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
0 Also, as ϕ1, ϕ2 are solutions of ϕ (t) = f(t, ϕ(t)), 1 = 2 = 0.
∴ |ϕ1(t) − ϕ2(t)| ≤ 0 =⇒ ϕ1(t) = ϕ2(t), ∀t ∈ (a, b) =⇒ ϕ1 = ϕ2
1.4 Method of successive spproximations
Now, we will see a constructive proof of existence and uniqueness of the solution of IVP.
Theorem 1.4.1 (Picard-Lindel¨of). If f ∈ (C, Lip) on a rectangular region R, then ∃ successive approximations ϕk on |t − τ| ≤ α ascontinuous functions and converge uniformly on this interval to the unique solution ϕ of (E) such that ϕ(τ) = ξ.
Proof. Consider the interval I1 = [τ − α, τ], similar arguments will hold for I2 = [τ, τ + α]. Let’s define
Z t ϕ0(t) = ξ and ϕk+1(t) = ξ + f(s, ϕk(s))ds, ∀k ∈ N ∪ {0}, ∀t ∈ [τ − α, τ + α] = I. (1.6) τ
1 Then, ϕ0 ∈ C (I1) and |ϕ0(t) − ξ| ≤ M(τ − t), ∀t ∈ I1, where I1 = [τ − α, τ] and M = max |f(t, ξ)|. (t,ξ)∈R 1 Let’s assume that ϕk ∈ C (I1) and |ϕk(t) − ξ| ≤ M(τ − t), ∀t ∈ I1. 1 ∴ ϕk+1 ∈ C (I1) (by definition of ϕk+1 and f ∈ C(R). Now,
Z t |ϕk+1(t) − ξ| = | f(s, ϕk(s))ds| τ Z τ ≤ |f(s, ϕk(s))|ds t
≤ M(τ − t), ∀t ∈ I1
1 Thus, by the principle of Mathematical Induction ∀k ∈ N ∪ {0}, ϕk ∈ C (I1) and
|ϕk(t) − ξ| ≤ M(τ − t), ∀t ∈ I1 (1.7) 1.4. METHOD OF SUCCESSIVE SPPROXIMATIONS 21
Let ∆k(t) = |ϕk+1(t) − ϕk(t)|, t ∈ I1. Then, we have
Z t Z t
∆k(t) = f(s, ϕk(s))ds − f(s, ϕk−1(s))ds τ τ Z τ ≤ |f(s, ϕk(s)) − f(s, ϕk−1(s))|ds t Z τ ≤ c |ϕk(s) − ϕk−1(s)|ds, where c is the Lipschitz constant of f on R. t Z τ = c ∆k−1(s)ds (1.8) t
Again, by (1.7) we have
∆0(t) = |ϕ1(t) − ϕ0(t)| ≤ M(τ − t) (1.9)
Thus, proceeding inductively we have
Z τ Z τ k ∆k(t) ≤ c ··· ∆0(s)ds t t | {z } k−times (τ − t)k+1 M ck+1(τ − t)k+1 = ckM = , ∀t ∈ I (1.10) (k + 1)! k (k + 1)! 1 ∞ ∞ X M X ck+1(τ − t)k+1 ∆ (t) ≤ , ∀t ∈ I ∴ k c (k + 1)! 1 k=0 k=0 M ≤ ecα as |τ − t| ≤ α c
∞ n−1 P P ∴ ∆k(t) is uniformly convergent for t ∈ I1. This implies ϕn(t) = ϕ0(t) + [ϕk+1(t) − ϕk(t)] k=0 k=0 converges uniformly to continuous limit function ϕ on I1.
Since, (t, ϕk(t)) ∈ R, ∀k ∈ N ∪ {0} and t ∈ I1, we have (t, ϕ(t)) ∈ R, ∀t ∈ I1. Hence, f(s, ϕ(s)) is defined ∀s ∈ I1.
Z τ Z τ Z τ
∴ [f(s, ϕ(s)) − f(s, ϕk(s))]ds ≤ |f(s, ϕ(s)) − f(s, ϕk(s))|ds ≤ c |ϕ(s) − ϕk(s)|ds t t t
R t which tends to 0 uniformly as k → ∞. By (1.6), ϕ(t) = ξ + τ f(s, ϕ(s))ds, t ∈ I1. Similarly, we 22 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
have the result for I2 and hence we have
Z t ϕ(t) = ξ + f(s, ϕ(s))ds, ∀t ∈ I τ
Further, uniqueness follows from theorem (1.3.3).
Lecture 6
1.5 Continuation of solutions
Recall what we have obtained till now. We have an IVP ϕ0(t) = f(t, ϕ(t)), ∀t ∈ [τ − a, τ + a], ϕ(τ) = ξ. The solution curve lies within the shaded region given below. Picture will come Thus, even though we start with the domain [τ − a, τ + a], we obtain the solution in the interval [τ − α, τ + α] subseteq[τ − a, τ + a] which is generally a proper subset. Infact in many cases the solution might exist in a very small neighborhood of the initial point.
Example. Consider the ode
dy = −cosec y =⇒ y = cos−1(t + c), dt where c ∈ R is the integration constant. Note, that for the solution to be well-defined |t + c| ≤ 1, i.e. even though the differential equation can be defined on whole of R, the solution will exist only in the interval t ∈ [−(1 + c), (1 − c)]. π π Thus, if y(0) = 2 is the initial condition, then c = cos 2 = 0, which implies the solution will exist on [−1, 1].
dy t2 Example. On the other hand the ode dt = 2ty has the general solution y = ce , which exists ∀t ∈ R.
Thus, we need to see when can we extend the region of existence of the solution and up to how far can we extend it. 1.5. CONTINUATION OF SOLUTIONS 23
Theorem 1.5.1. Let D be a domain in the (t, x) plane and f ∈ C(D) is bounded on D. If ϕ is a solution of the ode ϕ0(t) = f(t, ϕ(t)) on an interval (a, b), then the limits ϕ(a + 0) = lim ϕ(a + h) h→0+ and ϕ(b − 0) = lim ϕ(b − h) exist. Further, if (a, ϕ(a + 0)) or (b, ϕ(b − 0)) is in D, then the h→0+ solution ϕ may be continued to the left of a or right of b respectively.
Proof. Let M ∈ R+ such that |f(t, x)| ≤ M, ∀(t, x) ∈ D. Also, assume ϕ passes through (τ, ξ) ∈ D and τ ∈ (a, b). Then, Z t ϕ(t) = ξ + f(s, ϕ(s))ds, ∀t ∈ (a, b) τ
R t2 Thus, ∀t1, t2 ∈ (a, b), |ϕ(t1) − ϕ(t2)| ≤ |f(s, ϕ(s))|ds ≤ M|t2 − t1|. t1 Thus, as t1, t2 → a + 0 we have |ϕ(t1) − ϕ(t2)| → 0. Hence, by the Cauchy criterion of convergence that ϕ(a + 0) exists. Similarly, ϕ(b − 0) exists. ϕ(t) , ∀t ∈ (a, b) Now, if (a, ϕ(a + 0)) ∈ D, defineϕ ˜(t) = . Then,ϕ ˜ is a solution of the given ϕ(a + 0) , t = a ode of class C1 on [a, b). Infact
Z t ϕ˜(t) = ξ + f(s, ϕ˜(s))ds τ 0 0 ϕ˜+(a) =ϕ ˜ (a + 0) = f(a, ϕ˜(a))
Thisϕ ˜ is called a continuation of the solution ϕ to [a, b). Similarly, ϕ can be extended to (a, b] if (b, ϕ(b − 0)) ∈ D. Now, taking τ = a and ξ =ϕ ˜(a), we have by the existence theorem a solution ϕ∗ ∈ C1 on some interval [a − α, a], α > 0 such that ϕ∗0 (t) = f(t, ϕ∗(t)), ∀t ∈ [a − α, a] and ϕ∗(a) =ϕ ˜(a). ϕ(t) , if t ∈ (a, b) Thus, definingϕ ˆ(t) on [a − α, b) asϕ ˆ(t) = we have a solution for the ϕ∗(t) , if t ∈ [a − α, a] given ode on [a − α, b). Similarly, one can proceed on the right end point. Hence, we can extend the solution continuously on the left of a and right of b if (a, ϕ(a + 0)) ∈ D and (b, ϕ(b − 0)) ∈ D respectively.
dy Remark. 1. In the previous example of dt = −cosec(y), the solution could be extended to the left and right end points of [−1, 1]. But, that may not be the case always. 24 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
dy 2 −1 2. For example dt = y has solution ϕ(t) = −t defined on (−1, 1). But, it can not be extended to its right end pont as ϕ does not stay in the region D where f is bounded.
Lecture 7
1.6 System of differential equations of first order
Consider a first order ordinary differential equation given by
0 ϕ1(t) = f1(t, ϕ1(t)), ∀t ∈ I where I ⊆ R is an interval. Now, if we consider n-number of such first order odes,
0 ϕi(t) = fi(t, ϕ1(t), ··· , ϕn(t)), ∀t ∈ I
1+n where fi ∈ C(D), i = 1, 2, ··· , n, where D is a domain in R and each fi is a function of
(t, x1, x2, ··· , xn), then we have a system of n-ordinary differential equations of first order.
1+n Question 1.6.1. Let D ⊆ R , be an open conneccted set and fi ∈ C(D), 1 ≤ i ≤ n, i ∈ N. The problem is to find n-differentiable functions ϕ1, ϕ2, ··· , ϕn defined on a real t interval I such that
1. (t, ϕ1(t), ··· , ϕn(t)) ∈ D, ∀t ∈ I.
0 2. ϕi(t) = fi(t, ϕ1(t), ··· , ϕn(t)), ∀t ∈ I, ∀1 ≤ i ≤ n.
Remark. 1. In compact notation this can be written as
(a) ∀t ∈ I, (t, ϕ(t)) ∈ D, where ϕ(t) = (ϕ1(t), ··· , ϕn(t)).
dϕ(t) t (b) dt = F(t, ϕ(t)), where F(t, ϕ(t)) = (f1(t, ϕ(t)), f2(t, ϕ(t)), ··· , fn(t, ϕ(t)))
2. This problem is called a system of n-ordinary differential equations of the first order.
3. If such an interval I and functions (ϕ1(t), ··· , ϕn(t)) exist then the set of functions
(ϕ1(t), ··· , ϕn(t)) is called a solution of the system on I. 1.7. HIGHER ORDER ODES AS SYSTEM OF FIRST ORDER ODES 25
4. Let (τ, ξ1, ··· , ξn) ∈ D. The initial value problem consists of finding a solution (ϕ1, ··· , ϕn) of
the system on an interval I containing τ such that ϕi(τ) = ξi, ∀1 ≤ i ≤ n.
0 2 x1 = t + x1 + x2 + sin x3 0 2 Example. Let x2 = cos(tx1 − x2x3) ∀t ∈ (−10, 0). 0 2 x3 = t Then, these 3 odes represent a system of first order odes. If we introduce the vector notations, then we have
X0 = F(t, X), ∀t ∈ (−10, 10), where
0 2 x1 x1 f1(t, X) t + x1 + x2 + sin x3 0 0 X = x2 ,X = x , F(t, X) = f2(t, X) = cos(tx1 − x2x3) 2 0 2 x3 x3 f3(t, X) t
Remark. 1. We will use |.| to denote the l1 norm and ||.|| to denote the l2 norm for the R1+n.
2. Note, using these two norms or any other equivalent norm one may obtain equivalent definitions of Lipschitz continuity and -approximate solutions of a system of odes. Hence, all the theorems already proved for one variable ode is valid for a system of n-equations also.
3. A special case arises in the study of the system of odes when we consider the right hand side functions as linear functions.
4. We will study this extensively later on.
1.7 Higher order ODEs as system of first order ODEs
One interesting fact is that a m-th order ode can be expressed as a system of m- first order odes.
Example. Consider the equation
d3y d2y dy + cos t + ety + sin ty + y2 + t = 0 dt3 dt2 dt 26 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
This is a third order, first degree ode. Let us introduce a new set of dependent variables. Let
y1 = y dy dy y = 1 = 2 dt dt dy d2y y = 2 = 3 dt dt2 dy d3y d2y dy =⇒ 3 = = − cos t − ety − (sin(ty) + y2 + t) dt dt3 dt2 dt dy3 =⇒ = − cos ty − ety1 y − (sin(ty ) + y2 + t) dt 3 2 1 1
Thus, we can express this as a system
y1 y2 d y2 = y3 dt ty1 2 y3 − cos ty3 − e y2 − (sin(ty1) + y1 + t)
Example. Consider the equation
d4y d2y dy t2 + 2t + cos t + et = 0 dt4 dt2 dt
Then, introducing the variables
dy dy dy y = y, t = 1 , y = 2 , and y = 3 1 2 dt 3 dt 4 dt we get dy4 1 t = − 2 2ty3 + cos ty2 + e dt4 t and the system can be written as
y1 y2 0 1 0 0 y1 0 d y2 y3 0 0 1 0 y2 0 = = + dt y y 0 0 0 1 y 0 3 4 3 1 t cos t 2 1 t y4 t2 [2ty3 + cos ty2 + e ] 0 − t2 − t 0 y4 − t2 e 1.7. HIGHER ORDER ODES AS SYSTEM OF FIRST ORDER ODES 27
In vector notation, this becomes dY (t) = A(t)Y (t) + b(t) dt where y1 0 1 0 0 0 y2 0 0 1 0 0 Y (t) = ,A(t) = and b(t) = y 0 0 0 1 0 3 cos t 2 1 t y4 0 − t2 − t 0 − t2 e
Example. Consider another ode
d3y d2y dy 4 − 6 + 7 + 6y + 8t = 0 dt3 dt2 dt
Then, we have the system
y1 0 1 0 0 dY = AY + b, where Y = y2 ,A = 0 0 1 and b = 0 dt 2 7 2 y3 − 3 − 4 − 3 −2t
Remark. 1. This is a linear system of first order ode with constant corfficients. These type of systems are by far the easiest to solve.
2. Previous example was also a linear system but, with variable coefficients.
Thus, we have found out a way to relate the solutions of a higher order ode with that of a system of first order odes, theory of which will be covered in the chapter 4 later on.
Lecture 8
Before we move on to the vast theory of linear systems of odes, let’s look in to this important area of how the initial conditions of an IVP influence the solutions. For linear systems it is easier to visualize these dependencies. 28 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
1.8 Dependency on Initial conditions
Consider an ode with Lipschitz continuous (in 2nd variable) f.
ϕ0(t) = f(t, ϕ(t)) on some interval I with initial condition
ϕ(t0) = ξ
Then, the solution must satisfy Z t ϕ(t) = ξ + f(s, ϕ(s))ds t0
Thus, the solution depends on the initial parameters ξ and t0.
0 t−t0 Example. y = y with initial condition y(t0) = y0 has a solution y(t) = y0e .
Remark. 1. Thus, we will consider the solution not only as a function of t, but also as a
function of two more variables, i.e. t0 and ξ (initial conditions).
2. We are interested in how the solution behaves with respect to all three or any of these variables, in particular whether the solution varies continuously depending on the variables or not.
Let D be the domain in R1+n[(t, x) space] and f ∈ (C, Lip) in D. Let Ψ be a solution of the ode
x0(t) = f(t, x(t)) on I
Thus, we have (t, Ψ(t)) ∈ D, ∀t ∈ I. By uniqueness theorem ∃ a unique solution passing through any fixed point (τ, ξ) ∈ D close enough to the given solution. Here, ξ0 = Ψ(τ). The blue curve, lying within the butterfly region on both sides of (τ, ξ0) is the solution curve Ψ(t). Now, consider the star (τ, ξ), very close to (τ, ξ0) as a new initial condition.
Claim : We will have a unique solution on the interval I for this initial condition (τ, ξ).
Remark. 1. This is different from the extension of solutions on I. 1.8. DEPENDENCY ON INITIAL CONDITIONS 29
Figure 1.7: Existence of solutions of an IVP
2. In the previous extension theorems we extended the solutions to the left of (τ − α) and right of (τ + α), but did not consider outside the butterfly region in [τ − α, τ + α].
Lecture 9
Theorem 1.8.1. Let f ∈ (C, Lip) in a domain D in R1+n[(t, x) space] and let ψ be a solution of the ode x0(t) = f(t, x(t)) on I = [a, b]
Then, ∃δ > 0, such that for a fixed (τ, ξ) ∈ U, where U is an open δ-rectangular (l1) neighborhood of (τ, ψ(τ) and ∃ a unique solution ϕ is the above ode on I with ϕ(t = τ, τ, ξ) = ξ. Moreover ϕ ∈ C on V := t ∈ (a, b) and (τ, ξ) ∈ U, i.e. V := (a, b) × U. [Here ϕ is a function of t, τ and ξ]
Remark. Refer to the previous picture with ξ0 = ψ(τ). Then, this theorem guarantees the existence of a rectangular neighborhood U :(a, b) × (ψ(τ) − δ, ψ(τ) + δ) such that for every initial condition (τ, ξ) ∈ U, ∃ unique solution.
Proof. Let δ > 0 be such that the region U1 defined by
U1 : t ∈ I, |x − ψ(t)| ≤ δ1, i.e.U1 = {(t, x)|t ∈ I, |x − ψ(t)| < δ1} 30 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS
−k(b−a) is a subset of D. Then, let δ > 0 be such that δ < e δ1, where k is the Lipschitz constant of f. Let U be the set U = {(τ, ξ)|a < τ < b, |ξ − ψ(τ)| < δ}. Now, for any (τ, ξ) ∈ U, ∃ψ, satisfying the system of odes locally and passing through the initial point (τ, ξ). Since, locally ϕ is a solution of the system of odes,
Z t ∴ ϕ(t, τ, ξ) = ξ + f(s, ϕ(s, τ, ξ))ds, ∀t for which ϕ exists . τ
For t ∈ I Z t ψ(t) = ψ(τ) + f(s, ψ(s))ds, ∵ ψ is a solution. (1.11) τ As proven earlier, two solutions of same system of ode satisfy the inequality
k|t−τ| k|t−τ| k|b−a| |ϕ(t, τ, ξ) − ψ(t)| ≤ |ξ − ψ(τ)|e < δe ≤ δe < δ1
This shows that (t, ϕ(t)) ∈ U1, ∀t such that it is defined. Thus, we can extend the solutions to the whole interval I, using the idea of extension theorem. Construct a sequence of function defined by
ϕ0(t, τ, ξ) = ψ(t) + ξ − ψ(τ) (1.12) Z t ϕj+1(t, τ, ξ) = ξ + f(s, ϕj(s, τ, ξ))ds, j ∈ N ∪ {0} (1.13) τ
Then, for (τ, ξ) ∈ U, |ϕ0(t, τ, ξ) − ψ(t)| = |ξ − ψ(τ)| < δ1, which shows that (t, ϕ0(t, τ, ξ)) ∈ U1 for t ∈ I. Clearly, ϕ0 ∈ C(V ) and from (1.11), (1.12) and (1.13) we have
Z t
|ϕ1(t, τ, ξ) − ϕ0(t, τ, ξ)| = {f(s, ϕ0(s, τ, ξ)) − f(s, ψ(s))}ds τ ≤ k|ξ − ψ(τ)||t − τ|
k|t−τ| ∴ |ϕ(t, τ, ξ) − ψ(t)| ≤ (1 + k|t − τ|) |ξ − ψ(τ)| < e |ξ − ψ(τ)| < δ1
provided t ∈ I, (τ, ξ) ∈ U. Thus, (t, ϕ1(t, τ, ξ)) ∈ U1 and ϕ1 ∈ C(V ). Using mathematical induction 1.8. DEPENDENCY ON INITIAL CONDITIONS 31
one can show that if ψ0, ψ1, ··· , ψj are all in U1 and continuous on V , then
kj+1|t − τ|j+1 |ϕ (t, τ, ξ) − ϕ (t, τ, ξ)| ≤ |ξ − ψ(τ)| if t ∈ I and (τ, ξ) ∈ U. (1.14) j+1 j (j + 1)!
Thus, we have
j X |ϕj+1(t, τ, ξ) − ψ(t)| ≤ |ϕα+1(t, τ, ξ) − ϕα(t, τ, ξ)| + |ϕ0(t, τ, ξ) − ψ(t)| α=0 " j # X kα+1|t − τ|α+1 ≤ 1 + |ξ − ψ(τ)| < δ (k + 1)! 1 α=0
=⇒ (t, ϕj+1(t, τ, ξ)) ∈ U1.
Also, by (1.12) and (1.13), ϕj+1 ∈ C(V ). Thus, by mathematical induction, we have
∀j ∈ N ∪ {0}, (t, ϕj(t, τ, ξ)) ∈ U1 and ϕj ∈ C(V ). Hence, by (1.14) ϕj converges uniformly on V to ϕ, which implies ϕ is continuous on V . 32 CHAPTER 1. EXISTENCE AND UNIQUENESS OF SOLUTIONS Chapter 2 System of first order ordinary differential equations
Lecture 10
References : This chapter is mainly based on the following two references.
1. Calculus II by Tom Apostol, Chapter : Systems of Differential Equations.
2. Elementary Differential Equations and Boundary Value problems by Boyce and Diprima, Chapter 7 : Systems of first order linear equations, Ninth Edition, Wiley Publications.
2.1 System of First order ODEs
A general system of n-first order odes is of the form
dy1 = f (t, y , ··· , y ) dt 1 1 n dy2 dt = f2(t, y1, ··· , yn) (S1) . . . . dyn dt = fn(t, y1, ··· , yn)
We will assume fi, i = 1, 2, ··· , n to be constinuous in all the variables, i.e. t, yi, i = 1, 2, ··· , n; and fis are Lipschitz continuous with respect to the dependent variables yjs.
Remark. • This does not mean that f1 is Lipschitz in only y1 variable.
• This means that each fi is Lipschitz in all the variables yj, j = 1, ··· , n.
• It is very difficult to solve systems like (S1). Hence, we will restrict ourselves to a very
specific case where fis are linear functions in yjs. Thus, we will be working (for now only) with systems of n-linear ordinary differential equations of first order.
33 34 CHAPTER 2. FIRST ORDER ODE SYSTEMS
2.2 Systems of linear odes
We are going to consider the following system
0 y1 = p11(t)y1 + p12(t)y2 + ··· + p1n(t)yn + q1(t) 0 y2 = p21(t)y1 + p22(t)y2 + ··· + p2n(t)yn + q2(t) ......
0 yn = pn1(t)y1 + pn2(t)y2 + ··· + pnn(t)yn + qn(t)
where yi s are dependent unknown functions, pijs and q − is are the given functions of t, defined on some interval J.
Recall that a linear n-th order ode can be converted to a system on n linear first order odes. Thus, we can treat n-th order linear odes as special cases of system of linear first order odes. Before we do that, we would like to introduce some abstract concepts.
Matrix functions
Let J ⊆ R be an interval. We define a function P : J → Mnm(R) as p11(t) p12(t) ··· p1m(t) p (t) p (t) ··· p (t) 21 22 2m P (t) = . . . . . . . . pn1(t) pn2(t) ··· pnm(t) where n, m ∈ N and pij : R → R are functions of t. P is called a matrix function in 1 variable. 2.2. SYSTEMS OF LINEAR ODES 35
Integral of a matrix function
If P (t) = [pij(t)]n×m be a matrix function defined on an interval J, then P is said to be integrable over J iff each pij is integrable over J, i, j = 1, 2, ··· , n. The integral is given by
Z Z P (t) = pij(t)ft J J n×m
Derivatives of a matrix function
We define it in a similar way as P 0(t) = p0 (t) ij n×m
Note :- All basic differential rules for sums and products of differentiable functions hold for matrix functions also. Further, if P and Q are two square matrix functions of the same size, then (PQ)0 = PQ0 + P 0Q.
Exponential of a matrix
For this we will work with square matrices only. Let A = [aij be an n × n matrix over R or C. We define the exponential of A as ∞ X 1 eA = Ak k! k=0
Note : For this we require that the power series of matrix (rhs) converges.
Norm of a matrix
Let A = [aij be an n × n matrix over R or C. We will consider the following norm
n n X X ||A|| = |aij| j=1 i=1
Remark. This is the l1 norm for matrices. There are different norms for matrices, but the results that we will prove here will also hold for all the other equivalent norms. 36 CHAPTER 2. FIRST ORDER ODE SYSTEMS
Fundamental properties of norms
1. ||A + B|| ≤ ||A|| + ||B||, triangle inequality.
2. ||AB|| ≤ ||A||||B||, (this is slightly different from scalar norms).
3. ||cA|| ≤ |c|||A||
Convergence of series of matrices
(k) Let {Ak} be an infinite sequence of matrices of order m × n. Let aij be the (ij)-th entry of Ak. Then we will say the series of matrices ∞ X Ak k=1 is convergent if all the mn series
∞ X (k) aij , 1 ≤ i ≤ m; 1 ≤ j ≤ n k=1 converges and we write ∞ " ∞ # X X (k) Ak = aij k=1 k=1 m×n
Note : This is just the component wise convergence of the series.
An easy test for convergence of series of matrices
P∞ P∞ If k=1 ||Ak|| converges, then so does k=1 Ak.
∞ k P Ak Remark. Defining A = I when k = 0, we have a well defined series k! for every square matrix k=0 A. Further, we have the inequality
∞ X Ak 1 ≤ ||A||k, ∀k ∈ k! k! N k=0 2.3. UNIQUENESS OF THE SOLUTION OF THE SYSTEM OF DIFFERENTIAL EQUATIONS37
Differential equation satisfied by etA
Claim : Let E(t) = etA. Then E satisfies the matrix differential equation E0(t) = E(t)A = AE(t).
0 d tA Verification : Note E (t) = dt e . As the power series on the rhs is convergent ∀t ∈ R, we can do term by term differentiation to obtain E0(t) = E(t)A = AE(t).
Question 2.2.1. 1. Show that A commutes with E(t).
2. Let D be a diagonal matrix. Show that eD is also a diagonal matrix. What about etD?
3. Let A be a diagonalisable matrix, i.e. for there exist a diagonal matrix D and an invertible matrix P such that D = P AP −1. Show that eA = P −1eDP . What is the relationship between eAt and eDt?
Lecture 11
2.3 Uniqueness of the solution of the system of differential equations
Statement :- Let A and B be given n]timesn constant matrices. Then the only n × n matrix function F satisfying the initial value problem
0 tA F (t) = AF (t),F (0) = B, for t ∈ R is F (t) = e B
tA d tA tA Proof :- To prove that e B is a solution, note that dt (e B) = A(e )B since matrix multiplication is associative. Therefore, etAB is a solution of the given ode. To prove the uniqueness, consider F be any solution of the given ode system and G(t) = e−tAF (t) which implies G0(t) = 0 =⇒ G(t) = G(0) = B =⇒ F (t) = etAB.
Question 2.3.1. 1. For any two square matrices A, B of same order such that they commute, show that eA+B = eAeB. Will the relation hold if they don’t commute? 38 CHAPTER 2. FIRST ORDER ODE SYSTEMS
2. Let A be an n × n matrix such that Am+1 = 0 for some m ∈ . Then, ∞ m N A P 1 k P 1 k e = I + k! A = I + k! A . k=1 k=1
3. Let A be an n × n strictly upper triangular matrix, i.e. aij = 0, ∀i ≥ j. Then, ∃m ∈ N such that Am+1 = 0.
4. For a general square matrix A, it is difficult to obtain eAt. Using Cay;ey-Hamilton theorem,Putzer gave a procedure to obtain eA. This can be seen in the book Calculus II by Tom Apostol, Chapter 7, pg 206.
Remark. Solving the system of first order odes F 0(t) = AF (t),F (0) = B directly, using the exponential form F (t) = eAtB is difficult for a general n × n matrix as calculating the exponential might be a tough task. So we explore the properties of the system a bit more and see whether we can use those results.
Theorem 2.3.2 (Principle of superposition). Consider a general homogeneous system of n linear first order odes given by
X0(t) = P (t)X(t) (2.1) p11(t) p12(t) ··· p1m(t) p (t) p (t) ··· p (t) 21 22 2m where P (t) = . . . . . . . . pn1(t) pn2(t) ··· pnm(t) is a matrix function of t. If X and Y are two solutions of (2.1) then any linear combination of X and Y will again be a solution of (2.1).
Proof. Clearly, X0(t) = P (t)X(t) and Y 0(t) = P (t)Y (t) which implies 0 (c1X + c2Y ) (t) = P (t)(c1X + c2Y )(t) for c1, c2 ∈ R.
Theorem 2.3.3. If x(i), i = 1, 2, ··· , n are linearly independent solutions of (2.1), then any solution Y of (2.1) can be uniquely expressed as a linear combination of X(i)s. In other words, The set of all solutions of (2.1) forms a vector space with respect to function addition and scalar multiplication. The dimension of S is atmost n for an n × n system. 2.3. UNIQUENESS OF THE SOLUTION OF THE SYSTEM OF DIFFERENTIAL EQUATIONS39
Proof. Using the principle of superposition, it is very easy to show that S will be a vector space. To prove that it dimension is at most n, we require the concepts of Wronskian. The proof is done in the upcoming sections for linear homogeneous systems with constant and variable coefficients separately.
Definition 2.3.4 (Wronskian). Let’s consider the homogeneous system of first order linear ordinary differential equations given by (2.1). Let x(i), i = 1, 2, ··· , n are linearly independent solutions of (2.1). Then we define the Wronskian of the n-solutions as the determinant given by
(1) (2) (n) X1 (t) X1 (t) ··· X1 (t) (1) (2) (n) X (t) X (t) ··· X (t) (1) (2) (n) 2 2 2 W X ,X , ··· ,X (t) = ......
(1) (2) (n) Xn (t) Xn (t) ··· Xn (t)
Remark. The Wronskian maps n-solutions for each t to a scalar given by the determinant value. If we fix n-solutions and vary t over an interval I, then the Wronskian of these n-functions can be considered as a function of t.
(1) (2) (n) Definition 2.3.5 (Linearly independent solutions). If W X ,X , ··· ,X (t0) 6= 0 for some
fixed t0 then we say the solutions are linearly independent at t0. If the Wronskian is non-zero for all t ∈ I, then we say the solutions are linearly independent on whole I.
Definition 2.3.6 (Fundamental set of solutions). Any collection of n-solutions of the n × n system (2.1), which are linearly independent over an interval I is said to be a fundamental set of solutions for the system over the interval I.
Theorem 2.3.7 (Abel’s theorem). If x(i), i = 1, 2, ··· , n are linearly independent solutions of (2.1) on an interval I = (α, β), then the Wronskian W X(1),X(2), ··· ,X(n) (t) is either identically zero or else never vanishes on I.
Proof. Exapanding the Wronskian determinant using the first row and differentiating it we get
n (i) n dW X dy X (j) dYj = 1 Y + y dt dt i 1 dt i=1 j=1 40 CHAPTER 2. FIRST ORDER ODE SYSTEMS
dYj Using the equation (2.1) to substitute for dt we obtain
" n # dW X Z = p (t) W (t) =⇒ W (t) = c exp trace(P (t))dt dt ii i=1 where c is the constant of integration. If for some t0 ∈ I,W (t0) = 0, then c = 0, i.e. W (t) = 0, ∀t ∈ I.
2.3.1 Existence of Fundamental set of solutions
Theorem 2.3.8. If X(i), i = 1, 2, ··· , n are n solutions of (2.1) on an interval I = (α, β) (i) (i) (1) (2) (n) corresponding to the initial conditions y (t0) = e , where {e , e , ··· , e } form a fundamental set of solutions of (2.1).
Proof. As the Wronskian for a homogeneous linear system either vanishes everywhere or never vanishes on the interval I, therefore the given solutions X(i), i = 1, 2, ··· , n are linearly independent on I iff they are linearly independent for t = t0. Now,
1 ··· 0
0 ··· 0
W (t0) = . . . = |In| = 1 6= 0 . . .
0 ··· 1
Thus, X(i), i = 1, 2, ··· , n forms a fundamental set of solutions.
Lecture 12
2.3.2 Linear Differential operators (constant coefficients)
Linear Differential operators (constant coefficients) Let us define a linear differential operator with constant coefficients as follows
dn dn−1 d L ≡ + a + ... + a + a c dxn 1 dxn−1 n−1 dx n 2.3. UNIQUENESS OF THE SOLUTION OF THE SYSTEM OF DIFFERENTIAL EQUATIONS41
Then any linear ode dny dn−1y dy + a + ... + a + a y = 0 dxn 1 dxn−1 n−1 dx n can be written as dny dn−1y dy L (y) = + a + ... + a + a y = 0 c dxn 1 dxn−1 n−1 dx n
Question 2.3.9. Does this linear differential operator Lc has any relation with the linear transformations on vector spaces?
Definition 2.3.10 (Linear operator). Let V and W be two vector spaces over a field F and
L : V → W be a function. L is said to be a linear operator from V to W if for every v1, v2 ∈ V and any c ∈ F ,
L(cv1 + v2) = cL(v1) + L(v2).
Coming back to the differential operator
dn dn−1 d L ≡ + a + ... + a + a c dxn 1 dxn−1 n−1 dx n
n 0 Considering Lc : C → C , we have
n Lc(cf + g) = cLc(f) + g, ∀f, g ∈ C , ∀c ∈ F
n Thus, Lc is a linear operator on C . The kernel of Lc is the solution set of the equation Lc(y) = 0, i.e. the solution set of the homogeneous differential equation
dny dn−1y dy L (y) = + a + ... + a + a y = 0 c dxn 1 dxn−1 n−1 dx n
n Dimension of kernel : Note that, V = ker(Lc) is a vector subspace of C . Hence, it has a basis.
Is the kernel of the linear operator Lc finite dimensional? Here, we need the help of Wronskian. If ker(Lc) is infinite dimensional, then there exists linearly independent functions yi ∈ ker(Lc) for 42 CHAPTER 2. FIRST ORDER ODE SYSTEMS
i = 1, 2, ..., (n + 1). Now, we have the Wronskian of these functions W (y1, ..., yn+1) =
y y ... y y 1 2 n n+1
y0 y0 ... y0 y0 1 2 n n+1
......
y(n−1) y(n−1) ... y(n−1) y(n−1) 1 2 n n+1 (n) (n) (n) (n) y1 y2 ... yn yn+1
0 Pn Using row operations Rn+1 = Rn+1 + i=1 aiRi, we have
y y ... y y 1 2 n n+1
y0 y0 ... y0 y0 1 2 n n+1
W (y1, ..., yn+1) = ......
y(n−1) y(n−1) ... y(n−1) y(n−1) 1 2 n n+1
Lc(y1) Lc(y2) ... Lc(yn) Lc(yn+1)
But, we know Lc(yi) = 0, ∀i = 1, 2, ..., n + 1. Hence, W (y1, ..., yn+1) = 0.
Thus, ker(Lc) is finite dimensional and dim(ker(Lc)) ≤ n. dy mx Now, taking the hint from dx + cy = 0, we look for solutions of the form y = e for our linear n Pn n−i mx homogeneous ode. This gives (m + i=1 aim ) e = 0, which in turn gives us the auxilliary equation n n X n−i m + aim = 0 i=1 This gives us exactly n linearly independent solutions (after taking care of the repeated roots of the auxilliary equation.)
Hence, dim(ker(Lc)) = n and the basis of the kernel of the linear differential operator Lc contributes to the complementary function. In fact, the complementary function is the linear combination of the basis elements of the ker(Lc). This takes care of the homogeneous linear ode with constant coefficients.
Question 2.3.11. What about the non-homogeneous linear odes with constant coefficients?
Answer. For this we have the Particular integral. But, to find out from where does this particular integral come, we need some more linear algebra. 2.3. UNIQUENESS OF THE SOLUTION OF THE SYSTEM OF DIFFERENTIAL EQUATIONS43
Figure 2.1: Diagramatic representation of Quotient spaces
Quotient spaces
Definition 2.3.12 (Cosets). Let W be a subspace of a vector space V . Let α ∈ V , then the set
α + W := {α + w|w ∈ W } is called a coset of W in V .
Remark. Note that, even if α 6= β, it is possible that the cosets α + W = β + W . For example, take β = α + w0, where w0 ∈ W is a non-null vector.
Definition 2.3.13 (Quotient spaces). The set of all distinct cosets of W in V , denoted by V/W is called a Quotient space.
Question 2.3.14. Let W be the kernel of the linear transformation L : V → S. Further, let s ∈ S and w1, w2 ∈ V are two solutions of L(y) = s. Does there exists an α ∈ V such that w1, w2 ∈ α + W ? 44 CHAPTER 2. FIRST ORDER ODE SYSTEMS
Answer. Now, L(w1 − w2) = L(w1) − L(w2) = 0 =⇒ (w1 − w2) ∈ W .
If w2 ∈ α + W , then w2 = α + w for some w ∈ W .
But, then w1 = w2 + (w1 − w2) = α + [w + (w1 − w2)] ∈ α + W .
Infact, we can express this coset of the solutions to the non-homogeneous linear equation as w1 + W .
Particular integral : Coming back to the non-homogeneous linear ode with constant coefficients
Lc(y) = f(x), where f ∈ Lip(R)
Let y1 be one of its solution. Then note that
∀y ∈ ker(Lc), Lc(y + y1) = Lc(y1) = f
Hence, Y = y + y1 is a solution of the non-homogeneous problem. Also, by previous observation, any other solution y2 of the non-homogeneous problem will always lie in the same coset as y1, i.e. y2 ∈ y1 + ker(Lc). Thus, any solution Y of the non-homogeneous equation is given by
Y = y + y1, where y ∈ ker(Lc) and y1 ∈ y1 + ker(Lc) which can be reframed as Y = CF + PI, where CF is the complementary function (belonging to the kernel of Lc and PI is the particular integral belonging in the coset y1 + ker(Lc).
Lecture 13
2.3.3 Linear Differential operators (Variable coefficients)
Linear Differential operators (Variable coefficients) Now, let’s consider a homogeneous linear ode with variable coefficients, (the coefficients can be functions of x) given by
dny dn−1y dy L (y) = + a (x) + ... + a (x) + a (x)y = 0 v dxn 1 dxn−1 n−1 dx n 2.3. UNIQUENESS OF THE SOLUTION OF THE SYSTEM OF DIFFERENTIAL EQUATIONS45
n 0 Again, we can easily show that Lv : C → C is a linear transformation. And one can similarly show that the solution space of the above homogeneous equation (=ker(Lv))forms a vector subspace of Cn.
Now, one may think that we can similarly show that the ker(Lv) is finite dimensional using the Wronskian argument. But, the problem is - here the coefficients are functions of x and not scalars. Hence, row operations won’t help.
2.3.4 Existence and uniqueness theorem
We will be using the following theorem (without proof) in our quest to prove that the kernel of Lv is finite dimensional. This theorem can be derived from the general existence and uniqueness theorem of the ordinary differential equations.
Theorem 2.3.15. Let us consider an n−th order linear ordinary differential operator
dn dn−1 d L ≡ + a (x) + ... + a (x) + a (x), v dxn 1 dxn−1 n−1 dx n where ai, i = 1, 2, ..., n, are continuous functions on some open interval J ⊂ R. If x0 ∈ J and if k0, k1, ..., kn−1 are n given real numbers, then there exists a unique solution y = f(x), to the homogeneous linear ode Lv(y) = 0 on J and which also satisfies the initial conditions
0 (n−1) f(x0) = k0, f (x0) = k1, ..., f (x0) = kn−1.
n Consider the equation Lv(y) = 0. Then, given any (k0, k1, ...., kn−1) ∈ R and x0 ∈ J, there will exist a unique solution y = f(x) of Lv(y) = 0, such that
f(x0) k0 0 f (x0) k1 . = . . . (n−1) f (x0) kn−1
Consider a general ode F(x, y, y0, ..., yn) = g(x) with initial conditions 0 n (y, y , ..., y )(x0) = (k0, k1, ..., kn). Then, existence theorem says that there exists a solution 46 CHAPTER 2. FIRST ORDER ODE SYSTEMS
0 n y = f(x) such that F(y) = g and (y, y , ..., y )(x0) = (k0, k1, ..., kn). Whereas, the uniqueness theorem guarantees that no other f exists.
Kernel is finite dimensional
n 0 Theorem 2.3.16. Let Lv : C → C be a linear differential operator of order n given by
dn dn−1 d L = + a (x) + ... + a (x) + a (x) v dxn 1 dxn−1 n−1 dx n
Then the solution space of the equation Lv(y) = 0 has dimension n.
n Sketch of the proof. • Consider T : ker(Lv) → F defined by
dy dyn−1 T (y) = y(x ), (x ), ..., (x ) , 0 dx 0 dxn−1 0
where y ∈ ker(Lv) and x0 ∈ F .
• Now, by the uniqueness theorem of odes we have T (y) = 0 =⇒ y = 0, as we are in the homogeneous case.
• So, ker(T ) is trivial. Which means T is one-one.
• Also, by existence theorem, for any n-tuple α ∈ F n, there exists a solution of T (y) = α.
n • Thus, T is a bijection, which implies dim(ker(Lv)) = dim(F ) = n.
Remark. • Even though we have obtained the result that the kernel of Lv is finite
dimensional, it is very difficult to find a basis of ker(Lv).
Question 2.3.17. Why is it difficult to find a basis of ker(Lv), when it was so easy to find a basis for ker(Lc)?
Answer. • Lc involved only constant coefficients, that helped us to reduce the problem of ode to finding the roots of a polynomial.
• But, Lv has functions as coefficients and it is very difficult to manage so many variants of functions together. 2.3. UNIQUENESS OF THE SOLUTION OF THE SYSTEM OF DIFFERENTIAL EQUATIONS47
What to do?
• Not only the basis, it is difficult to find the particular integral for the non-homogeneous linear
ode Lv(y) = f.
• But, if we have knowledge of at least one solution explicitly, then it becomes easier to find the complete primitive.
• We use different methods based on
– the coefficients of Lv, – the informations given along with the problem.
• If we go back to 1st order linear odes with function coefficients, then we know how to solve them using integrating factors.
• Thus, if we can somehow reduce the order of the equation and involve some 1st order linear odes, then we can hope for solutions.
• This idea leads us to different methods like
1. Change of dependent variable.
2. Change of independent variable.
3. Factorisation of operators.
4. Power series method if the coefficients are analytic.
Eigen values of a linear transformation Let V be a vector space over the field F (R or C) and L : V → V be a linear transformation. We say λ ∈ F is an eigen value of L if ∃v ∈ V \{0}, such that L(v) = λv. v is then called an eigen vector of L corresponding to the eigen value λ.
Special Case : Matrices
• Let A be an n × n matrix.
• Then, it can be considered as a linear transformation. 48 CHAPTER 2. FIRST ORDER ODE SYSTEMS
• To obtain an eigen value of A, we must find a non-zero vector v such that Av = λv.
• But, Av = λv iff (A − λI)v = 0.
• Thus, (A − λI) is singular matrix and hence det(A − λI) = 0, which is known as the characteristic equation of A.
• Once, the eigen values are obtained, one can find the eigen vectors v by finding the kernel of (A − λI) or directly solving Av = λv.
Question 2.3.18. 1. It is repeatedly mentioned that the roots of the auxilliary equation are the eigen values of some Linear transformation.
2. Which linear transformation?
3. Similarly, emx are the eigen functions for which linear operator?
4. Are they eigen values and eigen vectors of Lc?
mx mx 5. Clearly, NO. As Lc(e ) = 0 and not m(e ).
6. To obtain the answers we need to change the set-up. But, before that we need to look into something else.
Lecture 14
2.4 Inhomogeneous system of first order linear odes
With variable coefficients, there was not much to do. Let’s visit another section where again we have a lot to do - the system of linear equations of first order with constant coefficients. Consider 2.4. INHOMOGENEOUS SYSTEM 49 the following system of linear odes
dy 1 = P y + P y + ... + P y + q (t) dt 11 1 12 2 1n n 1 dy 2 = P y + P y + ... + P y + q (t) dt 21 1 22 2 2n n 2 ...... dy n = P y + P y + ... + P y + q (t) dt n1 1 n2 2 nn n n where Pij are constants and qi(t) are functions of t for i, j ∈ 1, 2, ..., n.
t Taking Y = (y1, ..., yn) ,P = (Pij)n×n, the above system can be written as
dY = PY + Q(t), where and Q(t) = (q (t), ..., q (t))t dt 1 n
Observations : As in the case of a single linear ode with constant coefficients, this case can be dealt very smoothly. One can also verify that the solutions of this system will form a vector space and the linear transformation
d L ≡ I − P :(C1)n → (C0)n sc dt is a linear transformation. Thus, again if Q(t) ≡ 0, we have the solution of the homogeneous system
Lsc = 0 is the kernel of the linear operator Lsc.
The advantage of having a constant coefficients homogeneous system is that, one can row-reduce the matrix P to obtain an upper triangular matrix and then easily solve the system. Let’s restrict ourselves to a setup where we can obtain a very simple matrix - reducing n−th order linear ode to a system of n first order linear odes. 50 CHAPTER 2. FIRST ORDER ODE SYSTEMS
P is a Diagonal matrix : The system L(Y ) = 0 becomes an autonomous system, i.e. each of the dependent variables depends only on itself :
dy 1 = P y dt 11 1 dy 2 = P y dt 22 2 ...... dy n = P y dt nn n
Each equation can be solved easily, and we have the solutions as
Piit yi(t) = Cie , i = 1, 2, ..., n.
P is diagonalizable : Then, there is a diagonal matrix D, such that D = AP A−1. Using change of variables Z = AY , we have
Z0 = AY 0 = A(PY ) = AP A−1Z = DZ.
Then we can solve for the new system Z0 = DZ. Finally, transform the solutions back to Y .
General P matrix : To solve the system Lsc(Y ) = 0, we need to do the following :
1. Find the eigen values and their corresponding eigen vectors of the matrix P, i.e. find the roots of det(P − rI) = 0
2. There are three possibilities for the eigenvalues of P
(a) All eigenvalues are real and different from each other.
• Then associated with each eigenvalue ri is a real eigenvector vi.
• The set of n eigenvectors vi, i = 1, 2, ..., n, is linearly independent. 2.4. INHOMOGENEOUS SYSTEM 51
• The corresponding solutions of the differential system are
rit yi(t) = vie
• and the general solution becomes
n X rit Y (t) = civie . i=1
3. Some eigenvalues occur in complex conjugate pairs.
• Then there are still n linearly independent solutions, provided that all the eigenvalues are different.
• And we have the general solution of the form
n X rit Y (t) = civie . i=1
4. Some eigenvalues are repeated.
• Number of corresponding linearly independent eigenvectors may be smaller than the algebraic multiplicity of the eigenvalue.
• We need to seek additional solutions of another form. • Why does this look familiar? • This is similar to the n−th order linear odes with constant coefficients. • Thus, a repeated eigen value will give rise to solutions of the form erit, terit, t2erit and so on.
2.4.1 n-th order linear ode as a system of first order linear odes n-th order linear ode as a system of n equations Let us go back to the n-th order linear ode with constant coefficients
dny dn−1y dy L (y) = 0, i.e. + a + ... + a + a y = 0 c dtn 1 dtn−1 n−1 dt n 52 CHAPTER 2. FIRST ORDER ODE SYSTEMS
Define the following for i = 1, 2, ..., n − 1
0 0 0 y1 = y, y2 = y1, ..., yi+1 = yi, ..., yn = yn−1
Then we have our system
y1 0 1 0 ... 0 0 y1 y 0 0 1 ... 0 0 y 2 2 d . ...... . = dt . ...... . yn−1 0 0 0 ... 0 1 yn−1 yn −an −an−1 −an−2 ... −a2 −a1 yn
Example. Given a linear ode, say d2y dy − 3 + 2y = 0 dt2 dt Form an auxilliary equation m2 − 3m + 2 = 0 and find its solutions m = 1, 2. Converting it to the system we have ! ! ! d y 0 1 y 1 = 1 dt y2 −2 3 y2 The characterestic equation of the coefficient matrix P is
−r 1 2 = −r(3 − r) + 2 = r − 3r + 2 = 0 −2 3 − r
The characterestic equation of the coefficient matrix of the system arising from the 2nd order linear ode is same as the auxilliary equation of the original 2nd order equation. 2.4. INHOMOGENEOUS SYSTEM 53
Lecture 15
2.4.2 n-th order linear ode with constant coefficients
To find the solution of the system, we need to find the eigen values of the coefficient matrix P, i.e.
find the roots of
−r 1 0 ... 0 0
0 −r 1 ... 0 0
......
......
0 0 0 ... −r 1
−an −an−1 −an−2 ... −a2 −a1 − r Which will give rise to the same polynomial equation as the auxilliary equation of the n−th order linear ode. Thus, the auxilliary equation is basically the characterestic equations of the system of linear first order odes arising from the n-th order linear ode. Consider an n-th order linear homogeneous ode with constant coefficients
dny dn−1y dy L (y) = + a + ... + a + a y = 0 c dtn 1 dtn−1 n−1 dt n
We can transform it into a system of n number of first order linear odes with constant coefficients dY dx = AY , where y 0 1 0 ... 0 0 y0 0 0 1 ... 0 0 00 y ...... Y = ,A = . ...... . 0 0 0 ... 0 1 n−1 y −an −an−1 −an−2 ... −a2 −a1
Roots of the auxilliary equation of the n-th order linear ode are precisely the eigen values of the matrix A. The complimentary function and hence the basis of the kernel of L are formed by the eigen functions corresponding to these eigen values. 54 CHAPTER 2. FIRST ORDER ODE SYSTEMS
Distinct roots means distinct eigen values and distinct eigen functions. Problems arise for repeated eigen values - eigen functions can not be repeated.
Example. Consider the 3rd order linear ode
d3y d2y dy − 7 + 16 − 12y = 0 dx3 dx2 dx
dY Corresponding system dx = AY , where y 0 1 0 Y = y0 ,A = 0 0 1 y00 12 −16 7
The characterestic equation is m3 − 7m2 + 16m − 12 = 0 having the roots 2, 2 and 3. Clearly, e2x and e3x are two linearly independent solutions of the n-th order ode. Are they solutions of the system?
Answer is No. e2x ∈ C3 but A : C1 × C1 × C1 → C0 × C0 × C0. So, A can not act on e2x. We need to find an eigen vector of A. What about a 3-tuple with all the entries as e2x?
e2x e2x A e2x = e2x e2x (12 + 7 − 16)e2x
Clearly, this idea is not working. Now, observe that, as far as A is concerned, e2x does not have a very important role. A(ξe2x) = (Aξ)e2x, ∀ξ ∈ F 3
To get A(ξe2x) = 2(ξe2x), we thus, need
A(ξe2x) = (Aξ)e2x = 2(ξe2x) =⇒ Aξ = 2ξ
Thus, we need to find an eigen vector of A itself. t Aξ = 2ξ =⇒ ξ2 = 2ξ1, ξ3 = 4ξ1. Thus, ξ = (1, 2, 4) is an eigen vector of A with respect to eigen 2.4. INHOMOGENEOUS SYSTEM 55 value 2. Thus, an eigen function of the system of odes is given by
e2x 2x ξe = 2e2x 4e2x
Now, the problem is we don’t have another independent eigen vector of A wrt 2. Wrt the eigen value 3, we have easier solution. Following is an eigen function corresponding to 3.
1 3x 3 e 9
To obtain another linearly independent eigen vector of A corresponding to the eigen value 2, we need to introduce another concept called generalized eigen vectors.
Generalized eigen vectors : Let’s consider the matrix
! ! 5 0 5 1 A = and B = 0 5 0 5
Then, A and B both have characteristic equations (5 − r)2 = 0 and repeated eigen value 5 with multiplicity 2. Now, let’s find eigen vectors for A. Then,
! ! 1 0 Aξ = 5ξ =⇒ and are the linearly independent eigen vectors. 0 1
As for the eigen vectors for B,
! ! 5ξ + ξ 5ξ Bξ = 5ξ =⇒ 1 2 = 1 5ξ2 5ξ2
! 1 which implies ξ2 = 0. Thus, is an eigen vector of B. No matter how you try, you can not find 0 another linearly independent eigen vector of B. 56 CHAPTER 2. FIRST ORDER ODE SYSTEMS
Remark. • Previous two examples show that (A − 5I)2 = 0 and (B − 5I)2 = 0.
• But, the kernel of (A − 5I) is 2 dimensional, whereas the kernel of (B − 5I) is 1 dimensional.
• Now, (X − 5I)2 = 0 implies (X − 5I){(X − 5I)ξ} = 0, ∀ξ ∈ R2.
• Also, det(X − 5I) = 0 implies ker(X − 5I) is non-trivial. Now, two cases arise
– (X − 5I) = 0, same as A above. – (X − 5I) 6= 0 but (X − 5I)2 = 0, same as B above.
• Now, the kernel of (A − 5I) is 2 dimensional implies (A − 5I) = 0.
• But, the kernel of (B − 5I) is 1 dimensional, which implies (B − 5I) 6= 0.
• As (B − 5I) 6= 0 but, (B − 5I)2 ≡ 0, we have at least one vector v s.t.
(B − 5I)v 6= 0 but (B − 5I)2v = 0
• We need to find such vectors and call them as the generlized eigen vector of B wrt the eigen value 5. ! 0 1 • Now, (B − 5I)v 6= 0 =⇒ v2 6= 0, as (B − 5I) = . 0 0 ! 0 • Thus, is a generalized eigen vector wrt eigen value 5. 1
Claim : For 2 x 2 matrices, r is repeated eigen value, v if generalized e-vector
1. If you take w = (B − rI)v, then w is an eigen vector of B wrt r.
2. We may say that the generalized eigen vectors are preimages of the eigen vectors of B under the transformation (B − rI).
Definition 2.4.1 (Generalized eigen vector of nxn matrix wrt an eigen value r with algebraic multiplicity p). v is called a generalized eigen vector of B if
(B − rI)v 6= 0, but ∃1 < q ≤ p, s.t.(B − rI)qv = 0 even though (B − rI)q−1v 6= 0 2.4. INHOMOGENEOUS SYSTEM 57
In other words, ∃1 < q ≤ p, v ∈ ker((B − rI)q) ∩ range((B − rI)q−1)
Example. 5 1 0 5 0 0 A = 0 5 1 ,B = 0 5 1 0 0 5 0 0 5
3 Let e1, e2, e3 be the standard basis vectors of F . Then, e1 is an eigen vector of A. Kernel of (A-5I) is 1 dimensional, hence it has two linearly independent generalized eigen vectors. Note that 2 2 (A − 5I) e2 = 0 but (A − 5I) e3 6= 0.
On the other hand, e1 and e2 are both eigen vectors of B. Kernel of (B-5I) is 2 dimensional, hence it 2 has only one linearly independent generalized eigen vectors. Note that (B − 5I) e3 = 0. Again, consider the matrix 5 1 0 0 0 5 0 0 D = 0 0 5 0 0 0 0 5
It has 3 linearly independent e-vectors and 1- generalized e-vector. Hence, kernel of (D − 5I) is 3dimensional. Also, (D − 5I)3v = 0, ∀v.
Example. Consider the 3rd order linear ode
d3y d2y dy − 7 + 16 − 12y = 0 dx3 dx2 dx
dY Corresponding system dx = AY , where y 0 1 0 Y = y0 ,A = 0 0 1 y00 12 −16 7
The characterestic equation is m3 − 7m2 + 16m − 12 = 0 having the roots 2, 2 and 3. Following are 58 CHAPTER 2. FIRST ORDER ODE SYSTEMS its two linearly independent solutions
1 1 3x 2x Y1 = 3 e ,Y2 = 2 e , 9 4
We need to search for the third linearly independent solution as we already know the kernel has dimension 3.
generalized e-vector of A corresponding to the e-value 2 : We need to find a v s.t. 2 (A − 2I) v = 0. The above equation is easier to solve if we try to find v s.t. (A − 2I)v = ve, where 1 ve = 2 4
Solving it we obtain a generalised eigen value
1 v = 3 8
Verification : 1 1 1 3x 2x 2x Y = c1 3 e + c2 2 e + c3 3 e 9 4 8
dY is the general solution of the ode system dx = AY . Now, 1 1 1 dY 3x 2x 2x = 3c1 3 e + 2c2 2 e + 2c3 3 e dx 9 4 8 2.4. INHOMOGENEOUS SYSTEM 59
But,
1 1 1 1 3x 2x 2x 2x AY = 3c1 3 e + 2c2 2 e + 2c3 3 e + c3 2 e as (A − 2I)v = ve =⇒ Av = 2v + ve. 9 4 8 4
There is an extra term and hence it can not be a solution of the system.
Remark. • Revist the reduction of the problem of finding eigen functions for the linear system d ( dx − A) to the problem of finding eigen vectors of A.
d rx rx • Our argument was that ( dx − A)ξe = 0 ⇔ (rI − A)ξe = 0.
• Hence, as erx is never zero, we must have (rI − A)ξ = 0.
• Carefully, look at the equivalence in second point.
d rx rx • The equivalence was possible as the operator dx acting on e produced a similar function re .
• If we can replace this function by some other function f(x), such that the action of the d operator dx on it produces functions similar to f(x) then we may be able to obtain a new set of linearly independent solutions.
Important observations : (A − rI)Y = 0 has infinitely many linearly independent solutions of the form Y = vf(x), where v remains the fixed eigen vector and f(x) varies over linearly 1 d independent C functions. But, we have proved that the linear operator ( dx − A) has kernel of dimension 3. So, we won’t be able to find more than 3 linearly independent solutions by this process.
d Suitable candidate : Looking at the relation dx = AY , we find our suitable choice should be
1. exponentials.
2. polynomials times exponentials. 60 CHAPTER 2. FIRST ORDER ODE SYSTEMS
Exponentials gave us one solution. Hence, we need something of form 2 to get the other linearly 2x d independent solution. Let Y = ξxe be a solution of the ode system ( dx − A)Y = 0. This will give ξ1 + 2ξ1x ξ2x 2x 2x ξ2 + 2ξ2x e = ξ3x e ξ3 + 2ξ3x (12ξ1 − 16ξ2 + 7ξ3)x which gives ξ1 = ξ2 = ξ3 = 0. Hence, we need to search for a different type of solution. The other option left is to combine the forms 1 and 2. Look for solutions of the form Y = ξe2x + ηxe2x.
d Solution : Now, ( dx − A)Y = 0 will give
2ξe2x + ηe2x + 2ηxe2x = Aξe2x + Aηxe2x =⇒ (2ξ + η − Aξ)e2x = (A − 2I)ηxe2x
As e2x and xe2x are linearly independent, lhs = rhs = 0. Also, as e2x never vanishes, we must have
(2ξ + η − Aξ) = (A − 2I)η = 0 =⇒ (A − 2I)η = 0 and Aξ = 2ξ + η. i.e. η and ξ are respectively an eigen vector and a generalized eigen vector of A. Thus, taking
1 1 2x 2x η = 2 and ξ = 3 we have our 3rd linearly independent solution Y = ξe + ηxe 4 8
Remark. 1. Thus, the linearly independent eigen functions are
1 1 1 1 3x 2x 2x 2x Y1 = 3 e ,Y2 = 2 e ,Y3 = 3 e + 2 xe 9 4 8 4 2.5. PHASE POTRAIT 61
2. Thus, the general solution or complimentary function is
1 1 1 1 3x 2x 2x 2x Y = c1 3 e + c2 2 e + c3 3 e + 2 xe 9 4 8 4
Which can also be written as 1 1 1 1 3x 2x 2x Y = c1 3 e + c2 2 + c3 3 e + c3 2 xe 9 4 8 4
Lecture 16
Let’s conclude this chapter with a discussion on the phase potrait for 2 × 2 linear systems.
2.5 Phase potrait
Consider the homogeneous system of linear first order odes given by
X0(t) = PX(t) where P is an 2 × 2 constant matrix.
Remark. Though, the idea developed here can be used for a general n × n system, we will restrict ourselves to the 2 × 2 systems only as it is easy to visualise.
Procedure :
0 0 1. We will plot the direction/gradient fields given by grad(X(t)) = (x1(t), x2(t)) at the points X(t). This 2d plane will be called the phase plane and the diagram that we will obtain will be called the phase potrait.
2. Evaluating PX for a large collection of values of t ∈ I, we can draw a plot of the direction
fields of the tangent vectors in the x1 − x2 plane. 62 CHAPTER 2. FIRST ORDER ODE SYSTEMS
0 0 Figure 2.2: Phase potrait for the x1(t) = 2x1(t), x2(t) = −3x2(t)
3. A plot that shows a representative sample of trajectories for a given system is called a phase potrait.
Example. ! 2 0 X0(t) = X 0 −3
Since, it is a diagonal matrix (this type of systems are called uncoupled systems as the equations are independent of each other) we can straight away write
0 x1(t) = 2x1(t) 0 x2(t) = −3x2(t)
The phase potrait for the above system is given by (2.2).
Example. ! 1 1 X0(t) = X 4 1 Plot the direction field and determine the qualitative behaviour of solutions. Then, find the general solution and draw a phase potrait showing several trajectories.
Solution. This is a coupled system unlike the previous example. Hence, directly we can not work with solutions. But, as the directional fields work with the derivatives of the solutions, we can evaluate the derivatives at each point and obtain the tangent vectors at each point. 2.5. PHASE POTRAIT 63
Figure 2.3: Phase potrait : Pic Courtesy : Elementary differential equations and boundary value problems, pg 311
For x1 = 1, x2 = 0 we have ! ! ! ! x0 1 1 1 1 1 = = 0 x2 4 1 0 4
Similarly, for x1 = 0, x2 = 1, we have ! ! x0 1 1 = 0 x2 1
This means at (0,1), the tangent vector is (1,1), which makes an equal angle with the x1 and x2 directions. Similarly, at (1,0), the tangent vector is (1,4), which will be shifted more towards x2 direction. So the phase potrait will look like figure (2.3) Similarly, at (-1,0) and (0,-1) the tangent vectors will be parallel to those of (1,0) and (0,1) but in opposite directions respectively.
Also, note that for x1 = 0, x2 arbitrary, the tangent vectors are all parallel and make equal angles with the x1 and x2 axis, but they are in opposite directions for x2 > 0 and x2 < 0.
Similarly, for x1 6= 0, x2 = 0, we have a set of parallel tangent vectors reversing directions for x1 > 0 and x1 < 0.
Now, for x1 = x2 = 1 we have ! ! x0 2 1 = 0 x2 5 Going on this way we will be able to plot the phase potraits. 64 CHAPTER 2. FIRST ORDER ODE SYSTEMS Chapter 3 Differential Inequalities
3.1 Gronwall’s Inequality
Theorem 3.1.1. Let λ(t) be a real valued continuous function and µ(t) is a non-negative continuous function on I = [a, b]. If any continuous function y(t) satisfies
Z t y(t) ≤ λ(t) + µ(s)y(s)ds, ∀t ∈ I a then we have Z t R t µ(σ)dσ y(t) ≤ λ(t) + λ(s)µ(s)e s ds, ∀t ∈ I a In particular, if λ is constant, then R t µ(σ)dσ y(t) ≤ λe a
R t Proof. Let z(t) = a µ(s)y(s)ds, t ∈ I. Since, µ, y are both continuous, z is differentiable. Also, z(a) = 0.
Z t ∴ z(t) − z(a) = µ(s)y(s)ds a =⇒ z0(t) = µ(t)y(t) Z t =⇒ z0(t) − µ(t)z(t) = µ(t) y(t) − µ(s)y(s)ds a =⇒ z0(t) − µ(t)z(t) ≤ µ(t)λ(t) by the given inequality.
0 − R t µ(σ)dσ − R t µ(σ)dσ =⇒ [z (t) − µ(t)z(t)]e a ≤ µ(t)λ(t)e a
d h − R t µ(σ)dσi − R t µ(σ)dσ =⇒ z(t)e a ≤ µ(t)λ(t)e a dt
Integrating from a to t
Z t − R t µ(σ)dσ − R s µ(σ)dσ z(t)e a ≤ µ(s)λ(s)e a ds a Z t R t µ(σ)dσ =⇒ z(t) ≤ µ(s)λ(s)e s ds a
65 66 CHAPTER 3. DIFFERENTIAL INEQUALITIES
By the given inequality
Z t Z t R t µ(σ)dσ y(t) ≤ λ(t) + µ(s)y(s)ds = λ(t) + z(t) ≤ λ(t) + µ(s)λ(s)e s a a
Further, if λ(t) = λ (a constant) we have
Z t R t µ(σ)dσ R t µ(σ)dσ z(t) ≤ λ µ(s)e s ds =⇒ z(t) ≤ −λ + λe a a
R t µ(σ)dσ Therefore, we have the required result y(t) ≤ λe s .
Lecture 17
3.2 Solution of a differential inequality
dy We have dealt with the problems of finding solutions for the differential equations dt = f(t, y). dy dy What about the functions y(t) = ψ(t), that satisfies the inequalities dt ≤ f(t, y) or dt ≥ f(t, y) or dy dy dt < f(t, y) or dt > f(t, y)? We would like to explore the possibilities of finding solutions for these problems.
Definition 3.2.1 (Solution of a differential inequality). Let f(t, x) be continuous on a region D ⊂ R × R. A function x(t) is said to be a solution of the differential inequality
dx > f(t, x), t ∈ I = [t , t + α) dt 0 0 if the following conditions hold
1. x0(t) exists ∀t ∈ I,
2. (t, x(t)) ∈ D, ∀t ∈ I and
3. x0(t) > f(t, x(t)), ∀t ∈ I.
Remark. 1. The interval I can be any type of interval.
2. Equivalent definitions for systems can be made. 3.2. SOLUTION OF A DIFFERENTIAL INEQUALITY 67
3. Analogous definitions hold for x0(t) ≥ f(t, x), [≤ f(t, x) or < f(t, x)].
Example. Consider the differential inequality
dy < −{y(t)}2 on (0, π) dt
Verify that y(t) = cot(t) is a solution. Further note that
1 z(t) = −ct, 0 < c < is a constant π2 is also a solution.
2 Theorem 3.2.2. Let f(t, x) be continuous on a region D ⊆ R and y1, y2 be two solutions of the differential inequalities.
0 y1(t) ≤ f(t, y1(t)) 0 y2(t) > f(t, y2(t))
over the interval I = [t0, t0 + α). If y1(t0) < y2(t0), then y1(t) < y2(t), ∀t ∈ I.
Proof. Let us assume the set A = {t ∈ I|y1(t) ≥ y2(t)}= 6 φ. Since, A ⊆ I,A is bounded. Infact t0 is a lower bound. ∗ ∗ ∴ It has a greatest lower bound, let it be t . Hence, t ≥ t0.
Now, y1, y2 both being solutions of some differential inequalities implies they are continuous.
∗ ∗ Claim : y(t ) = y2(t ). ∗ If the claim is false, then either (y1 − y2)(t ) > (or <)0.
∗ ∗ Case 1 (> 0): There exists a neighborhood of t where y1 − y2 > 0, which implies ∃t1 < t such ∗ that y1(t1) > y2(t1) =⇒ t1 ∈ A. This contradicts that t = inf(A).
∗ Case 2 (< 0): There exists a neighborhood N of t where y1 − y2 < 0. Now, choosing t2 from ∗ this neighborhood, we should have τ ∈ A such that t ≤ τ < t2, (by the property of the infimum of a set). 68 CHAPTER 3. DIFFERENTIAL INEQUALITIES
But, then τ ∈ N =⇒ y1(τ) < y2(τ), which contradicts τ ∈ A. ∗ ∗ Hence, the only possibility is that our claim y1(t ) = y2(t ) is true. ∗ Also, as y1(t0) < y2(t0) and t ≥ t0, therefore t∗ > t0. ∗ ∗ ∗ ∗ Now, t = inf(A) =⇒ y1(t − h) < y2(t − h), ∀h > 0, such that t − h ∈ I. ∗ ∗ 0 ∗ y1(t −h)−y(t ) ∴ y1(t − 0) = limh→0+ −h ∗ ∗ ∗ ∗ ∗ ∗ y1(t −h)−y1(t ) y2(t −h)−y2(t ) Now, y1(t − h) < y2(t − h) =⇒ −h > −h . Thus, we have 0 ∗ 0 ∗ y1(t − 0) ≥ y2(t − 0). 0 ∗ 0 ∗ Since, y1(t ) and y2(t ) exists, we have
0 ∗ 0 ∗ 0 ∗ 0 ∗ y1(t − 0) = y1(t ), y2(t − 0) = y2(t )
0 ∗ 0 ∗ Therefore, we have y1(t ) ≥ y2(t ). 0 0 Again, by the given hypothesis y1(t) ≤ f(t, y1(t)) and y2(t) > f(t, y2(t)), ∀t ∈ I. Therefore, we have 0 ∗ ∗ ∗ 0 ∗ y1(t ) ≤ f(t, y1(t )) = f(t, y2(t )) < y2(t ), which is a contradiction.
Therefore, the initial assumption that A 6= φ is not possible. Hence, y1(t) < y2(t), ∀t ∈ I.
Lecture 18
dy Definition 3.2.3 (Sub-solution). Solutions of the differential inequalities of the form dt ≤ f(t, y) are called sub-solutions.
dy Definition 3.2.4 (Super-solution). Solutions of the differential inequalities of the form dt ≥ f(t, y) are called super-solutions.
Theorem 3.2.5. Let f(t, x) be continuous on D ⊆ R2. Further assume
dx 1. dt = f(t, x), x(t0) = x0, where (t0, x0) ∈ D,I = [t0, t0 + α).
dx1 dx2 2. x1(t) and x2(t) are solutions of dt < f(t, x1), dt > f(t, x2) in I.
3. x1(t0) ≤ x0 ≤ x2(t0) where x0 = x(t0).
0 Then ∀t ∈ I , we have x1(t) < x(t) < x2(t). 3.2. SOLUTION OF A DIFFERENTIAL INEQUALITY 69
Proof. If x0 = x(t0) < x2(t0), then x(t) and x1(t) satisfy the hypothesis of the previous theorem. 0 Therefore, we have x(t) < x2(t), ∀t ∈ I .
If x0 = x2(t0), define z(t) = x2(t) − x(t). Therefore, 0 0 0 0 z (t) = x2(t) − x (t) =⇒ z (t) > f(t0, x2(t0)) − f(t0, x(t0)) = 0. Hence, Z is strictly increasing in a neighborhood of t0 in I. Let N = [t0, t0 + δ] be such a neighborhood of t0. This implies z(t0 + δ) > z(t0) = 0.
∴ In the interval Iδ = [t0 + δ, t0 + α), z(t0 + δ) > 0 =⇒ x2(t0 + δ) > x(t0 + δ).
Then, x and x2 satisfy the hypothesis of the previous theorem on the interval
Iδ =⇒ x(t) < x2(t), ∀t ∈ Iδ. Now, δ can be chosen arbitrarily small. Hence, 0 ∀δ > 0, x(t) < x2(t), ∀t ≥ t0 + δ =⇒ x(t) < x2(t), ∀t > t0, t ∈ I . For the sub-solution and exact solution part, one may proceed as follows :
Take y1(t) = −x1(t), y(t) = −x(t) and proceed as before.
Remark. The above result suggests that if the sub-solution is less than or equal to the exact solution or less than or equal to the super solution initially, then the inequality is maintained strictly in the interior of the interval. 70 CHAPTER 3. DIFFERENTIAL INEQUALITIES Chapter 4 Some more Existence and Uniqueness results
Lecture 19
4.1 Maximal And Minimal solutions
Definition 4.1.1 (Maximal Solution). Let D ⊆ R2 be open and f : D → R be continuous. A solution r(t) of the IVP dx = f(t, x), x(t ) = x dt 0 0 where (t0, x0) ∈ D, is said to be a maximal solution if for any arbitrary solution y(t) of the above IVP, we have y(t) ≤ r(t) for every t in the common domain of existence of r and y.
Definition 4.1.2 (Minimal Solution). Let D ⊆ R2 be open and f : D → R be continuous. A solution s(t) of the IVP dx = f(t, x), x(t ) = x dt 0 0 where (t0, x0) ∈ D, is said to be a minimal solution if for any arbitrary solution y(t) of the above IVP, we have y(t) ≥ s(t) for every t in the common domain of existence of s and y.
Remark. 1. Maximal solution is the one which dominates all other solutions in their common region of existence.
2. This does not mean that a maximal solution has the maximal interval of existence.
Theorem 4.1.3. Let f be a continuous function on the region
S = {(x, y)|x0 ≤ x ≤ x0 + α, |y − y0| ≤ b} contained inside the domain D of f. Then, ∃ a maximal b and a minimal solution of the IVP in the interval [x0, x0 + α], where α = min{a, 2M+b } and M ≥ max |f(x, y)| on D.
b Proof. Existence of maximal solution : Let 0 < ≤ 2 . Consider the IVPs
dy = f(x, y), y(x ) = y (4.1) dx 0 0 dy = f(x, y) + , y(x ) = y + (4.2) dx 0 0
71 72 CHAPTER 4. SOME MORE EXISTENCE AND UNIQUENESS RESULTS
Now, f is continuous on S, which implies f = f + is continuous on b S = {(x, y)|x0 ≤ x ≤ x0 + a, |y − (y0 + )| ≤ 2 } and S ⊆ S. b Also, we have |f(x, y)| ≤ |f(x, y)| + ≤ M + 2 , ∀(x, y) ∈ S. ∴ by Peano’s existence theorem we have the existence of a solution y(x) of the IVP- in the b interval [x0, x0 + α] where α = min{a, 2M+b }.
Let 0 < 2 < 1 ≤ and y1 , y2 be the solutions of the IVP-1, IVP-2, respectively.
Note that y2 (x0) = y0 + 2 < y0 + 1 = y1 (x0). 0 0 Again, y2 (x) = f(x, y2(x)) + 2 and y1 (x) = f(x, y1(x)) + 1 > f(x, y1(x)) + 2. 0 0 Now, we have the following set-up y2 (x) = f2 (x, y2(x)) and y1 (x) > f2 (x, y1(x)).
∴ By previous corollary we have y2 (x) < y1 (x), ∀x ∈ [x0, x0 + α].
Varying → 0, we have a family of equicontinuous and uniformly bounded functions on [x0, x0 + α].
∴ By Ascoli-Arzela theorem ∃ a decreasing sequence {n} ↓ 0 such that {yn } converges uniformly on [x0, x0 + α].
Let r(x) = limn→∞ yn (x). Thus, yn → r uniformly on [x0, x0 + α].
∴ r(x0) = y0. Now, f is continuous on S which is compact. Hence, f is uniformly continuous on S. R x y = (y0 + n) + [f(t, y (t)) + n]dt. ∴ n x0 n ∴ Taking limits as n → ∞ Z x r(x) = y0 + f(t, r(t))dt =⇒ r is a solution of IVP. x0
To show r(x) is a maximal solution : Let y(x) be any other solution of the IVP in [x0, x0 + α]. 0 Then, for > 0, y(x0) = y0 < y0 + = y(x0) = f(x, y). Again, y (x) = f(x, y) < f(x, y) + and 0 y(x) = f(x, y) + = f(x, y).
By previous corollary, y(x) < y(x), ∀x ∈ [x0, x0 + α]. Since, > 0 was arbitrary, varying over ns and taking limits as n → ∞, we have
y(x) ≤ r(x), ∀x ∈ [x0, x0 + α] =⇒ r is a maximal solution.
Uniqueness of maximal solution over the fixed interval [x0, x0 + α]: The uniqueness is ensured by the fact that we have uniform convergence yn (x) → r(x), ∀x ∈ [x0, x0 + α]. 4.2. UNIQUENESS RESULTS 73
For Minimal solution : Consider the IVP-
dy = f(x, y) − , y(x ) = y − dx 0 0 and proceed in the similar way as in the case of the maximal solution to obtain the existence and uniqueness of the minimal solution.
Lecture 20
4.2 Uniqueness results
Till now, we have seen existence and uniqueness results together. Also, such uniqueness results du depended on the conditions satisfied by the right hand side function f ov the IVP dt = f(t, u). Now, we will be exploring some more uniqueness results that will depend on the conditions satisfied by u.
Lemma 4.2.1. Let w(z) be an increasing continuous function on [0, α), α ∈ R+ and w(0) = 0 and R α dz w(z) > 0 for z > 0 and lim→0+ w(z) = +∞. Further, assume u(x) to be a continuous function on [0, α] satisfying Z x u(x) ≤ w(u(t))dt, 0 < x ≤ α 0 then u(x) = 0, ∀x ∈ [0, α].
Proof. Let v(x) = max{u(t): t ∈ [0, x]}. Since, u is a continuous function, it attains its maximum in [0, x], which implies v(x) is well-defined ∀x ∈ [0, α]. Now, assume v(x) > 0, ∀0 < x ≤ α. By the definition of v(x), u(x) ≤ v(x) = max{u(t) : 0 ≤ t ≤ x}, ∀x ∈ [0, α].
Again, as u is continuous, ∃yx such that u(yx) = v(x) and 0 ≤ yx ≤ x.
Z yx Z x ∴ v(x) = u(yx) ≤ w(u(t))dt ≤ w(u(t))dt 0 0 Now, as u(t) ≤ v(t) and w ↑, we have
Z yx Z x Z x v(x) = u(yx) ≤ w(u(t))dt ≤ w(u(t))dt ≤ w(v(t))dt 0 0 0 74 CHAPTER 4. SOME MORE EXISTENCE AND UNIQUENESS RESULTS
R x 0 Letv ¯(x) = 0 w(v(t))dt. Thenv ¯(0) = 0, v(x) ≤ v¯(x) andv ¯ (x) = w(v(x)) ≤ w(¯v(x)). Now, for x 6= 0, v(x) > 0 =⇒ v¯(x) > 0 =⇒ w(¯v(x)) > 0.
v¯0(x) ≤ 1 ∴ w(¯v(x)) and for 0 < δ < a, Z a v¯0(t) Z a dt ≤ dt = a − δ δ w(¯v(t) δ But, substituting z =v ¯(t), we have
Z a v¯0(t) Z α dz dt = δ w(¯v(t) w(z) wherev ¯(δ) = andv ¯(a) = α. But, → 0+ =⇒ RHS → ∞ Z a v¯0(t) ∴ dt ≤ a − δ δ w(¯v(t) is not possible. Thus, our assumption that v(x) > 0 for 0 < x ≤ α is false. As u was non-negative, v(x) ≥ 0. But, v can not be positive. Hence,
v(x) = 0 on [0, α] =⇒ u(x) = 0 on [0, α]
Theorem 4.2.2 (Osgood’s uniqueness theorem). Let f(x, y) be continuous in
S = {(x, y): |x − x0| ≤ a, |y − y0| ≤ b} and ∀(x, y1), (x, y2) ∈ S it satisfies
|f(x, y1) − f(x, y2)| ≤ w(|y1 − y2|) where w(z) is as in lemma (4.2.1). Then
dy = f(x, y), y(x ) = y dx 0 0 has at most one solution in |x − x0| ≤ a. 4.2. UNIQUENESS RESULTS 75
Proof. Let y1, y2 be two solutions of the IVP
dy = f(x, y), y(x ) = y dx 0 0
As, |f(x, y1) − f(x, y2)| ≤ w(|y1 − y2|), therefore
Z x y1(x) − y2(x) = (f(t, y1) − f(t, y2)) dt x0 Z x =⇒ |y1(x) − y2(x)| ≤ |f(t, y1) − f(t, y2)| dt x0 Z x ≤ w (|f(t, y1) − f(t, y2)|) dt x0
Thus, u(x) = |y1(x) − y2(x)| satisfies the hypothesis of lemma (4.2.1) and hence u(x) = 0 on the given interval.
∴ y1(x) = y2(x) in the given interval. Thus, the IVP has a unique solution. Remark. 1. This is a uniqueness theorem. Hence, existence of solution is not guaranteed.
2. It may happen that f satisfies the hypothesis of Osgood’s uniqueness theorem even though the IVP has no solution!
Lemma 4.2.3. Let u(x) be a non-negative continuous function in |x − x0| ≤ a and u(x0) = 0. 0 Further, if u is differentiable at x = x0 with u (x0) = 0, then the inequality
Z x u(t) u(x) ≤ dt =⇒ u(x) ≡ 0 in |x − x0| ≤ a. x0 t − x0
Proof. case 1 : (x0 ≤ x ≤ x0 + a) Define
Z x u(t) v(x) = dt x0 t − x0
Note, this integral exists as u(x) 0 lim = u (x0) = 0 x→x0 x − x0 Again, u(x) v(x) v0(x) = ≤ x − x0 x − x0 76 CHAPTER 4. SOME MORE EXISTENCE AND UNIQUENESS RESULTS
Now, 0 d v(x) v (x) v(x) 1 0 v(x) = − 2 = v (x) − < 0, ∵ x ≥ x0 dx x − x0 x − x0 (x − x0) x − x0 x − x0 d v(x) v(x) ∴ < 0 =⇒ is not increasing. dx x − x0 x − x0
Also, as v(x0) = 0, we have v(x) ≤ 0, which contradicts that v(x) > 0.
∴ v(x) ≡ 0 =⇒ u(x) ≡ 0 on [x0, x0 + a]
case 2 : (x0 − a ≤ x ≤ x0) This case can be done similarly and is left as an homework.
Lecture 21
Theorem 4.2.4 (Nagumo’s Uniqueness theorem). Let f(x, y) be continuous on the rectangular region S = [x0 − a, x0 + a] × [y0 − b, y) + b]. Also, assume that for any (x, y1), (x, y2) ∈ S, we have
y1 − y2 |f(x, y1) − f(x, y2)| ≤ k , where x 6= x0, 0 < k ≤ 1 x − x0
Then, the IVP dy = f(x, y) on |x − x | ≤ a and y(x ) = y dx 0 0 0 has at most one solution in [x0 − a, x0 + a].
Proof. Let y1 and y2 be two solutions of the given IVP and defines
u(x) = |y1(x) − y2(x)|, |x − x0| ≤ a
Then, u(x) ≥ 0 and u is continuous. Now,
Z x0 Z
u(x) = |y1(x) − y2(x)| = f(s, y1)ds − x0xf(s, y2)ds x Z x Z x |y1(s) − y2(s)| ≤ |f(s, y1) − f(s, y2)|ds ≤ k ds x0 x0 |s − x0| Z x Z x |y1(s) − y2(s)| u(s) ≤ ds = ds x0 |s − x0| x0 s − x0 4.2. UNIQUENESS RESULTS 77
u(x0 + h) − u(x0) 1 ∴ lim = lim |y1(x0 + h) − y2(x0 + h)| , ∵ u(x0) = 0 h→0 h h→0 h 1 0 0 = lim |y1(x0) + hy1(x0 + θh) − y2(x0) − hy2(x0 + θ2h)| h→0 h
But, y1(x0) = y2(x0) = y0. Hence, by the mean value theorem
1 0 0 u(x) ≤ lim |h||y1(x0 + θh) − y2(x0 + θ2h)| h→0 h |h| = lim |f(x0 + θ1h, y1(x0 + θ1h)) − f(x0 + θ2h, y1(x0 + θ2h))| h→0 h |h| = 0 as is a bounded function and rest of the part tends to 0 h u(x0 + h) − u(x0) 0 ∴ lim = 0 =⇒ u is differentiable at x0 and u (x0) = 0. h→0 h
∴ By lemma (4.2.3), u(x) ≡ 0 on |x − x0| ≤ a, which means that IVP has atmost one solution. 78 CHAPTER 4. SOME MORE EXISTENCE AND UNIQUENESS RESULTS Chapter 5 Sturm-Liouville Theory
Lecture 21 (contd.)
Reference : Differential Equations by S. L. Ross, 3rd edition, chapters 11 and 12.
5.1 Adjoint of a second order linear ODE
d2 d Consider the second order linear differential operator L ≡ a0(t) dt2 + a1(t) dt + a2(t), where ai(t) are differentiable and a0(t) 6= 0
2−i Definition 5.1.1 (Adjoint of a second order linear ODE). Let L(x) = 0 where ai ∈ C on [a, b].
Also, a0(t) 6= 0 on [a, b]. The adjoint of the above 2nd order linear differential equation is
d2 d [a (t)x] − [a (t)x] + a (t)x = 0 dt2 0 dt 1 2
On simplifiction this gives
d2x dx a (t) + [2a0 (t) − a (t)] + [a00(t) − a0 (t) + a (t)]x = 0 0 dt2 0 1 dt 0 1 2
Example. Consider the second order linear ode
d2x dx t2 + 7t + 8x = 0 dt2 dt
Find the adjoint.
2 Solution. Here, a0(t) = t , a1(t) = 7t and a2(t) = 8. Therefore the adjoint equation is given by
d2x dx t2 − 3t + 3x = 0 dt2 dt
79 80 CHAPTER 5. STURM-LIOUVILLE THEORY
Lecture 22
5.2 Self-adjoint 2nd order linear ode
Definition 5.2.1 (Self-adjoint of a second order linear ODE). A second order linear ode L(x) = 0 d2 d 2−i where L ≡ a0(t) dt2 + a1(t) dt + a2(t) and ai ∈ C on [a, b] with a0(t) 6= 0, is said to be self-adjoint if the adjoint of the above 2nd order linear differential equation, given by
d2 d [a (t)x] − [a (t)x] + a (t)x = 0 dt2 0 dt 1 2 is same as itself.
Theorem 5.2.2. Let us consider a second order linear ode
d2x dx a (t) + a (t) + a (t)x = 0 0 dt2 1 dt 2
2−i where ai ∈ C on [a, b] with a0(t) 6= 0. Then, the necessary and sufficient condition for this ode to be self-adjoint is
0 a1(t) = a0(t), ∀t ∈ I
Proof. Equating the respective coefficients will give the result.
Example (Legendre’s Equation).
d2x dx (1 − t2) − 2t + n(n+)x = 0 dt2 dt
2 0 Here, a0(t) = 1 − t , a1(t) = −2t. Thus, a0(t) = −2t = a1(t). Hence, this is a self-adjoint ode.
Remark. If a second order linear ode
d2x dx a (t) + a (t) + a (t)x = 0 0 dt2 1 dt 2 5.2. SELF-ADJOINT 2ND ORDER LINEAR ODE 81 is self-adjoint, then it can be written as
d dx a (t) + a (t)x = 0 dt 0 dt 2
Question 5.2.3. Can we convert a general second order linear ode in to a self-adjoint second order linear ode?
Let the second order linear ode given by
d2x dx a (t) + a (t) + a (t)x = 0 0 dt2 1 dt 2 is not self-adjoint. Let v(t) be a non-trivial function such that
d2x dx v(t)[a (t) + a (t) + a (t)x] = 0 0 dt2 1 dt 2 is self-adjoint. That would require
d [v(t)a (t)] = v(t)a (t) dt 0 1 0 0 ⇔ v (t)a0(t) + v(t)a0(t) = v(t)a1(t) v0(t) a (t) − a0 (t) ⇔ = 1 0 v(t) a0(t) Z v0(t) Z a0 (t) Z a (t) ⇔ + 0 dt = 1 dt v(t) a0(t) a0(t) 1 R a1(t) dt ⇔ v(t) = e a0(t) , where k is a constant of integration. ka0(t)
R a (t) 1 1 dt Neglecting the arbitrary constant we have v(t) = e a0(t) , which is the required multiplier. a0(t)
d2x dx Theorem 5.2.4. Let a0(t) dt2 + a1(t) dt + a2(t)x = 0 be a second order linear differential equation defined on I = [a, b].Then, it can be transformed into a self-adjoint ode
d dx P (t) + Q(t)x = 0 on I, dt dt 82 CHAPTER 5. STURM-LIOUVILLE THEORY where R a1(t) dt a2(t) P (t) = e a0(t) , and Q(t) = P (t) a0(t)
Proof. (Justification of the proof is already given in the discussion prior to theorem. Here, well-definedness will be taken care of.)
Since, a0, a1 are continuous on I and a0(t) 6= 0∀t ∈ I, P (t) exists and is differentiable on I.
R a (t) 0 1 dt a1(t) d a1(t) P (t) = e a0(t) a0(t) dt a0(t)
d dx Hence, dt P (t) dt + Q(t)x = 0 is well-defined and self-adjoint.
5.3 Basic results of Sturm theory
d dx Theorem 5.3.1. Let f be a solution of dt P (t) dt + Q(t)x = 0 on I = [a, b]. If f has infinitely many zeros on I, then f ≡ 0 on I.
Proof. f has infinitely many zeros in I. Let S = {t ∈ I|f(t) = 0}. As S ⊂ I and is infinite, there exists a sequence {sn}, with sn ∈ S, ∀n ∈ N, such that it converges to some limit s0 (Bolzano-Weirstrass’ theorem).
Also, as I is closed, s0 ∈ I. Now, f is a solution of the given self-adjoint ode, f is differentiable on
I. This implies f(s0) = lim f(sn) = 0. n→∞ 0 f(s)−f(s0) 0 Again, f (s0) = lim . Since, f (s0) exists as f is differentiable, we have s−s0 s→s0
f(sn) − f(s0) f(s) − f(s0) 0 lim = lim = f (s0) n→∞ sn − s0 s→s0 s − s0
0 But, f(sn) = 0 = f(s0). Hence, f (s0) = 0. Thus, f is a solution of the given self-adjoint 2nd order ode with initial conditions 0 f(s0) = 0 = f (s0). 0 Since, the given ode is linear, it has a unique solution satisfying the conditions x(s0) = x (s0) = 0, where s0 ∈ I. ∴ f has to be the trivial solution on I. 5.3. BASIC RESULTS OF STURM THEORY 83
Lecture 23
d dx Theorem 5.3.2 (Abel’s formula). Let f and g be two solutions of dt P (t) dt + Q(t)x = 0 on I = [a, b], then ∀t ∈ I, P (t)[f(t)g0(t) − f 0(t)g(t)] = k where k is a constant.
Proof. We have
d [P (t)f 0(t)] + Q(t)f(t) = 0 (5.1) dt d [P (t)g0(t)] + Q(t)g(t) = 0 (5.2) dt
Combining the two we get
d d g(t) [P (t)f 0(t)] − f(t) [P (t)g0(t)] = 0 dt dt
Integrating from a to t ∈ I, we have
Z t d Z t d g(t) [P (s)f 0(s)] ds = f(s) [P (s)g0(s)] ds a dt a dt Z t Z t 0 t 0 0 0 t 0 0 =⇒ [g(s)P (s)f (s)]a − g (s)P (s)f (s)ds = [f(s)P (s)g (s)]a − f (s)P (s)g (s)ds a a =⇒ P (t)g(t)f 0(t) − P (a)g(a)f 0(a) = P (t)f(t)g0(t) − P (a)f(a)g0(a) =⇒ P (t)[f(t)g0(t) − g(t)f 0(t)] = P (a)[f(a)g0(a) − g(a)f 0(a)]
Now, RHS is a constant. Let it be k. Hence,
P (t)[f(t)g0(t) − g(t)f 0(t)] = k, ∀t ∈ I
Remark. Abel’s formula states that any two solutions f and g of a second order linear self-adjoint d dx ode dt P (t) dt + Q(t)x = 0, satisfy the condition P (t)W (f, g)(t) = constant. 84 CHAPTER 5. STURM-LIOUVILLE THEORY
d dx Theorem 5.3.3. Let f and g be two solutions of dt P (t) dt + Q(t)x = 0 on I = [a, b], such that they have a common zero at t0 ∈ I. Then f and g are linearly dependent on I.
Proof. By Abel’s formula
P (t)W (f, g)(t) = P (t0)W (f, g)(t0), ∀t ∈ I =⇒ P (t)W (f, g)(t) = 0∀t ∈ I
Now, P (t) 6= 0, ∀t ∈ I implies
W (f, g)(t) = 0, ∀t ∈ I =⇒ f and g are linearly dependent on I.
Theorem 5.3.4. Let f and g be two non-trivial linearly dependent solutions of d dx dt P (t) dt + Q(t)x = 0 on I = [a, b]. Then, f(t0) = 0 for some t0 ∈ I will imply g(t0) = 0.
Proof. ∵ f and g are linearly dependent on I, ∃ constants c1, c2 not both zero such that c1f(t) + c2g(t) = 0∀t ∈ I. As f and g are non-trivial solutions, neither of them are identically zero on I. Thus, 2 cases arise. case 1 : If c1 = 0. Then, c2 6= 0 =⇒ c2g(t) = 0, ∀t ∈ I =⇒ g ≡ 0, which is a contradiction. case 2 : If c2 = 0, then c1 6= 0, which implies f(t) ≡ 0 on I, which is again a contradiction.
Thus, none of c1 or c2 is zero and neither of f and g is identically 0 on I.
Now, f(t0) = 0 and c1f(t0) + c2g(t0) = 0 implies g(t0) = 0.
Theorem 5.3.5 (Sturm seperation theorem). Let f and g be two solutions of d dx dt P (t) dt + Q(t)x = 0 on I = [a, b]. Then, between any two consecutive zeros of f, there is precisely one zero of g.
Proof. Let t0 and t1(> t0) be the consecutive zeros of the function f in I. Further, as f is continuous on I, f(t) has same sign on (t0, t1). Now, as g is linearly independent of f,
∴ by previous theorems, g(t0) 6= 0 and g(t1) 6= 0. f If possible let g(t) 6= 0 on (t0, t1). Then, g is a differentiable function on [t0, t1]. f f Now, g (t0) = g (t1) = 0. d f ∴ By Rolle’s theorem ∃t2 ∈ (t0, t1) such that dt g (t2) = 0. 5.3. BASIC RESULTS OF STURM THEORY 85
0 0 d f g(t2)f (t2) − f(t2)g (t2) W (f, g)(t2) But, (t2) = 2 = 2 dt g {g(t2)} {g(t2)}
Thus, we have W (f, g)(t2) = 0. Since, f and g are linearly independent solutions of a linear ode,
W (f, g)(t) 6= 0∀t ∈ I in particular on (t0, t1)
Thus, we arrive at a contradiction.
Hence, our assumption that g never vanishes on (t0, t1) is false.
∴ g vanishes at least once on (t0, t1). If possible let t2, t3 ∈ (t0, t1) be two consecutive zeros of g, with t2 < t3.
Now, as per the first part of the proof, f will vanish at least once in (t2, t3).
But, that will contradict that t0, t1 were consecutive zeros of f.
∴ g vanishes exactly once in (t0, t1).
Lecture 24
Theorem 5.3.6 (Sturm’s comparison theorem). Let P (t) be a differentiable function and 0 Q1(t),Q2(t) be continuous functions on I = [a, b]. Further, assume P (t) is continuous, P (t) > 0 and Q2(t) > Q1(t) on I. d dx d dx Let φ1 and φ2 be the real valued solutions of dt P (t) dt + Q1(t)x = 0 and dt P (t) dt + Q2(t)x = 0 respectively.
Further, if t1 and t2 are consecutive zeros of φ1 in I, then φ2 has at least one zero in (t1, t2).
Proof. If possible, let φ2(t) 6= 0 on (t1, t2). Without any loss of generality assume
φ1(t), φ2(t) > 0∀t ∈ (t1, t2). By hypothesis
d [P (t)φ0 (t)] + Q (t)φ (t) = 0 (5.3) dt 1 1 1 d [P (t)φ0 (t)] + Q (t)φ (t) = 0 (5.4) dt 2 2 2 86 CHAPTER 5. STURM-LIOUVILLE THEORY
Multiplying (5.3) by φ2(t) and (5.4) by −φ1(t) and summing up we get
d [P (t){φ0 (t)φ (t) − φ (t)φ0 (t)}] = {Q (t) − Q (t)}φ (t)φ (t) dt 1 2 1 2 2 1 1 2 Z t2 [P (t){φ0 (t)φ (t) − φ (t)φ0 (t)}]t2 = {Q (t) − Q (t)}φ (t)φ (t)dt 1 2 1 2 t=t1 2 1 1 2 t1
Now, φ1(t1) = φ1(t2) = 0. This implies
Z t2 0 0 P (t2)φ1(t2)φ2(t2) − P (t1)φ1(t1)φ2(t1) = {Q2(t) − Q1(t)}φ1(t)φ2(t)dt t1
0 But, by hypothesis P (t2) > 0, φ1(t2) = 0, φ1(t) > 0 on (t1, t2). This implies φ1(t2) < 0. 0 0 Also, φ2(t2) > 0. Therefore P (t2)φ1(t2)φ2(t2) < 0. 0 0 Similarly, P (t1) > 0, φ1(t1) > 0, φ2(t1) > 0 implies P (t1)φ1(t1)φ2(t1) > 0. 0 0 Thus, P (t2)φ1(t2)φ2(t2) − P (t1)φ1(t1)φ2(t1) < 0.
But, Q2(t) > Q1(t) on I, in particular on [t1, t2]. Also, φ1(t)φ2(t) > 0 there. This implies
Z t2 {Q2(t) − Q1(t)}φ1(t)φ2(t)dt > 0 t1 which becomes a contradiction.
Thus, our initial assumption that φ2 never vanishes on (t1, t2) is false. Hence, ∃t3 ∈ (t1, t2) such that φ2(t3) = 0.
Example. Consider
d2x + A2x = 0 (5.5) dt2 d2x + B2x = 0 (5.6) dt2 where A, B are fixed real numbers with B > A > 0. Then, φ1(t) = sin(At) and φ2(t) = sin(Bt) solves (5.5) and (5.6). nπ (n+1)π The consecutive zeros of φ1 are A and A , n ∈ N. nπ (n+1)π The Sturm comparison theorem suggests that ∃ξn ∈ A , A such that φ2(ξn) = 0.
Remark. Uniqueness of zeros of φ2 in between two consecutive zeros of φ1 is not there, but 5.4. STURM-LIOUVILLE PROBLEMS 87 existence is there.
5.4 Sturm-Liouville Problems
Definition 5.4.1. Consider a boundary value problem consisting of 1. second order homogeneous linear ode of the form
d dy p(x) + [q(x) + λr(x)]y = 0 dx dx
where p, q, r are real valued functions of x with p(x) > 0, having continuous derivative and q, r are just continuous and r(x) > 0 on I = [a, b]. Also, λ is a parameter independent of x.
2. Two supplementary conditions of the form
0 A1y(a) + A2y (a) = 0 boundary conditions 0 B1y(b) + B2y (b) = 0
where A1,A2,B1,B2 ∈ R and A1,A2 are not both zero and B1,B2 are not both zero. We are interested in finding the values of λ so that the above system admits non-trivial solutions. This type of problems are called Sturm-Liouville problem (system).
d2y Example. Consider the ode dx2 + λy = 0 on [0, π], with y(0) = 0, y(π) = 0. This is an example of Sturm-Liouville problem.
Lecture 25
Let L : C2(R) → C0(R) be given by
d df L(f)(x) = p(x) + q(x)f(x) dx dx
d dy Then the Sturm-Liouville problem dx p(x) dx + [q(x) + λr(x)]y = 0 can be re-written as
L(y) = −λr(x)y 88 CHAPTER 5. STURM-LIOUVILLE THEORY
If r(x) = 1, then we say −λ is an eigen value of L is the equation admits a non-trivial solution. For a general r(x) we say (−λ) is an eigen value of L with respect to the weight function r(x).
Remark. 1. Using standard notations used for the Sturm-Liouville problems, we will say λ and not (−λ) as the eigen values of L.
2. In fact we will say that λ (for which y 6= 0 solution exists) is an eigen value (characteristic value) of the ode.
d2y Example. dx2 + λy = 0, y(0) = 0, y(π) = 0. d2y Case 1 : (λ = 0) The equation becomes dx2 = 0, which has only trivial solutions. Case 2 : (λ > 0) In this case the general solution of the problem is given by √ √ 2 y(x) = c1 cos( λx) + c2 sin( λx). We find that non-trivial solutions exist for λ = n , n ∈ N. Case 3 : (λ < 0) In this case the general solution of the problem is given by √ √ y(x) = c1 exp( λx) + c2 exp(− λx). Imposing the boundary conditions gives that no non-trivial solution exists. Thus, the given problem has only positive eigen values given by λ = n2, n ∈ N.
Remark. The boundary / initial conditions play a vital role in determining the eigen values.
π Example. Consider the previous problem with boundary conditions y(0) = 0, y 2 = 0. Case 1 : (λ = 0) Only trivial solution exists. Case 2 : (λ > 0) In this case the general solution of the problem is given by √ √ y(x) = c1 cos( λx) + c2 sin( λx). Imposing the boundary conditions we find that the non-trivial solutions exist for λ = 4n2, n ∈ N. Case 3 : (λ > 0) In this case the general solution of the problem is given by √ √ y(x) = c1 exp( λx) + c2 exp(− λx). Imposing the boundary conditions gives that no non-trivial solution exists.
Question 5.4.2. 1. Is it possible for some boundary conditions the previous problem has no eigen values?
2. Does there exist infinitely many eigen values of any Sturm-Liouville problem?
3. Will every Sturm-Liouville problem have isolated eigen values? 5.4. STURM-LIOUVILLE PROBLEMS 89
Theorem 5.4.3. Let’s consider a general Sturm-Liouville problem
d dy p(x) + [q(x) + λr(x)]y = 0 (5.7) dx dx with boundary conditions
0 A1y(α) + A2y (α) = 0 0 B1y(β) + B2y (β) = 0
where A1,A2 are not both 0 and B1,B2 are not both 0. Then,
1. There exists infinitely many eigen values λn, n ∈ N of the given problem. In particular, they
can be arranged in the increasing order as λ1 < λ2 < ··· < λn < λn+1 < ··· such that λn → ∞ as n → +∞.
2. For each eigen value λn, ∃ an one parameter family of eigen functions φn.[∵ constant multiple of eigen functions are also eigen functions.]
0 3. Each eigen space is one dimensional. Two eigen functions φn and φn corresponding to the
same eigen value λn, vary only by a constant multiple.
4. Each eigen function φn corresponding to the eigen value λn has exactly (n − 1) zeros in the open interval (α, β).
Proof is omitted for this theorem.
d2y Example. Consider dx2 + λy = 0, y(0) = y(π) = 0. 2 Then, the eigen values are λn = n , n ∈ N. Hence, we have
2 2 λ1 = 1 < 4 < 9 < ··· < n < (n + 1) < ···
For λ1 = 1 we have the eigen function y(x) = sin x = φ1(x).
For λ2 = 4 we have the eigen function y(x) = sin(2x) = φ2(x).
For λ5 = 25 we have the eigen function y(x) = sin(5x) = φ5(x). 90 CHAPTER 5. STURM-LIOUVILLE THEORY
Remark. Recall from linear algebra that the eigen vectors corresponding to the different eigen values are linearly independent.
Definition 5.4.4 (Orthogonal functions). Let f and g be two continuous functions of x ∈ [a, b]. Let r(x) be a continuous, then f and g are said to be orthogonal with respect to r(x) iff
Z b f(x)g(x)r(x)dx = 0 a
Observation 5.4.5. Cn([a, b]) is an inner product space with respect to the inner product
Z b hf, gi = f(x)g(x)r(x)dx z
Definition 5.4.6. Let {fn}n∈N, be an infinite collection of functions on [a, b]. {fn} is said to be an orthogonal system with respect to the weight function r(x) on [a, b] if for m 6= n, φm is orthogonal to φn, i.e. Z b φm(x)φn(x)r(x)dx = 0, for m 6= n a
d dy Theorem 5.4.7. Consider the Sturm-Liouville problem dx p(x) dx + [q(x) + λr(x)]y = 0. Let {λn}, n ∈ N be the eigen values with φn being the corresponding eigen functions. Then the set
{φn}n∈N is an orthogonal set of functions with respect to r(x) over [a, b].
[Proof can be found in the book of S. L. Ross (given in the reference)]. Chapter 6 Variation of parameters
Lecture 26
Reference : Differential Equations by S. L. Ross, 3rd edition. Consider the general n-th order linear inhomogeneous ode with constant coefficients given by
dny dn−1y dy + c + ··· + c + c y = F (x) dxn 1 dxn−1 n−1 dx n
Reduced Equation dny dn−1y dy + c + ··· + c + c y = 0 dxn 1 dxn−1 n−1 dx n
Auxilliary Equation n n−1 m + c1m + ··· + cn−1m + cn = 0
Let m1, m2, ··· , mn be n-roots of auxilliary equation. Then the complimentary function comprises of emix, xemix, ··· , etc.
Particular Integral F (x) P(D)y = F (x) =⇒ y = P(D)
d where P(D) is polynomial of D ≡ dx .
General Solution y = C.F. + P.I.
Remark. This method of finding the particular integral is helpful only if F (x) has any of the following forms
1. polynomial in x (including constants),
2. exponentials,
3. trignometric functions,
91 92 CHAPTER 6. VARIATION OF PARAMETERS
4. combination of above 3 types of functions.
We look for some other methods to find the particular integral for a broader categories of F (x). One such method is the Variation of parameters.
d2y Example. Let us start with a second order linear ode dx2 + y = tan x.
Then the complimentary function is yc = c1 cos x + c2 sin x. We now replace the arbitrary constants in the complimentary function by some arbitrary functions (twice differentiable). Thus, we consider a function
f(x) = v1(x) cos(x) + v2(x) sin(x)
If f solves the given ode then
f 00(x) + f(x) = tan x
0 0 0 f (x) = v1(x) cos x − v1(x) sin x + v2(x) sin x + v2(x) cos x
0 0 We impose the condition v1(x) cos x + v2(x) sin x = 0. Thus, we get
0 f (x) = −v1(x) sin x + v2(x) cos x 00 0 0 =⇒ f (x) = − (v1(x) + v2(x)) sin x + (−v1(x) + v2(x)) cos x
Since, we want f to be a solution of the given ode, we have
f 00(x) + f(x) = tan x
0 0 =⇒ − v1(x) sin x + v2(x) cos x = tan x
Thus, we have the following 2 × 2 system
! " #−1 ! ! v0 (x) cos x sin x 0 − sin x tan x 1 = = 0 v2(x) − sin x cos x tan x sin x
∴ v2(x) = − cos x + c3 and v1(x) = sin x − ln(sec x + tan x) + c4
Thus, the particular integral is given by yp = c3 sin x + c4 cos x − cos x ln(sec x + tan x). Choosing 6.1. GENERAL THEORY FOR SECOND ORDER LINEAR ODES 93
c3 = c4 = 0 we get the particular integral as yp = − cos x ln(sec x + tan x).
Remark. Since, c3 sin x + c4 cos x is the complimentary function, hence this part won’t contribute to the ode. Hence, we can neglect c3, c4 in the particular integral itself.
6.1 General theory for second order linear odes
dy dy Let a0(x) dx2 + a1(x) dx + a2(x)y = F (x) be the general second order linear ode, where a0(x) 6= 0.
Let yc = c1y1(x) + c2y2(x) be the known complimentary function. Replace c1, c2 in the 1 complimentary function by two arbitrary C functions v1(x), v2(x) respectively.
Assume yp(x) = v1(x)y1(x) + v2(x)y2(x) be a particular integral. Then,
0 0 0 0 0 yp(x) = {v1(x)y1(x) + v2(x)y2(x)} + {v1(x)y1(x) + v2(x)y2(x)} 0 0 = v1(x)y1(x) + v2(x)y2(x)
0 0 if we impose the condition v1(x)y1(x) + v2(x)y2(x) = 0. Then, we obtain
00 0 0 0 0 00 00 yp (x) = {v1(x)y1(x) + v2(x)y2(x)} + {v1(x)y1 (x) + v2(x)y2 (x)}
Since, yp(x) is a particular integral, it must satisfy the given ode. Hence,
00 0 a0(x)yp (x) + a1(x)yp(x) + yp(x) = F (x) 00 0 00 0 0 0 0 0 =⇒ {a0(x)y1 (x) + a1(x)y1(x) + y1(x)}v1(x) + {a0(x)y2 (x) + a1(x)y2(x) + y2(x)}v2(x) + a0(x){v1(x)y1(x) + v2(x)y2(x)} = F (x)
Since, y1, y2 are solutions of the homogeneous ode, the first two terms in the above expression will vanish. Thus, we are left with
0 0 0 0 F (x) v1(x)y1(x) + v2(x)y2(x) = a0(x) 94 CHAPTER 6. VARIATION OF PARAMETERS
Thus, we have two conditions
0 0 v1(x)y1(x) + v2(x)y2(x) = 0
0 0 0 0 F (x) v1(x)y1(x) + v2(x)y2(x) = a0(x)
As the coefficient matrix is the Wronskian matrix of the linearly independent solutions y1, y2 of the reduced equation, it is invertible. Hence, we have unique solutions
0 F (x)y2(x) v1(x) = a0(x)W [y1, y2](x)
0 F (x)y1(x) v2(x) = a0(x)W [y1, y2](x)
Thus, the particular integral of the given problem is yp(x) = v1(x)y1(x) + v2(x)y2(x), where
Z x F (t)y2(t) v1(x) = dt a a0(t)W [y1, y2](t) Z x F (t)y1(t) v2(x) = dt a a0(t)W [y1, y2](t) where a < b are the end points of the interval I and x ∈ I.
Remark. This approach is equally valid for the m − th order linear odes.
Example. d3y d2y dy − 6 + 11 − 6y = ex dx3 dx2 dx Auxilliary equation : m3 − 6m2 + 11m − 6 = 0, whose roots are m = 1, 2, 3. x 2x 3x Complimentary function : yc = c1e + c2e + c3e . x 2x 3x Let yp = v1(x)e + v2(x)e + v3(x)e be a particular integral of the given ode, where 1 v1(x), v2(x), v3(x) are C functions of x
0 0 x 0 2x 0 3x x 2x 3x yp(x) = (v1(x)e + v2(x)e + v3(x)e ) + (v1e + 2v2e + 3v3e )
0 x 0 2x 0 3x Condition 1 : v1(x)e + v2(x)e + v3(x)e = 0, which implies 6.1. GENERAL THEORY FOR SECOND ORDER LINEAR ODES 95
00 x 2x 3x yp (x) = v1(x)e + 4v2(x)e + 9v3(x)e . Therefore
000 0 x 0 2x 0 3x x 2x 3x yp (x) = (v1(x)e + 4v2(x)e + 9v3(x)e ) + (v1(x)e + 8v2e + 27v3e )
Thus, from the given equation we have
00 00 0 x yp (x) − 6yp (x) + 11yp(x) − 6yp(x) = e 0 x 0 2x 0 3x x v1(x)e + 4v2(x)e + 9v3(x)e = e
Thus, we have the 3 × 3 system
x 2x 3x 0 e e e v1(x) 0 ex 2e2x 3e3x v0 (x) = 0 2 x 2x 3x 0 x e 4e 9e v3(x) e
x 2x 3x ∴ W [e , e , e ] 6= 0. This has unique solution given by
1 v (x) = x 1 2 −x v2(x) = e 1 v (x) = − e−2x 3 4
1 x 3 x ∴ yp(x) = 2 xe + 4 e and the general solution is given by
1 y(x) = c ex + c e2x + c e3x + xex 1 2 3 2
3 x Note, we have neglected the term 4 e as it can be absorbed in the complimentary function itself. 96 CHAPTER 6. VARIATION OF PARAMETERS Chapter 7 Liapunov functions
Lecture 27
Reference : Differential Equations by S. L. Ross, 3rd edition. Let us consider a linear system
dx 1 = 2x + x dt 1 2 dx 2 = x + x dt 1 2 " # " #" # d x 2 1 x 1 = 1 dt x2 1 1 x2
√ 3± 5 The eigen values of the coefficient matrices are λ = 2 . Therefore the solution of the system is λ1t λ2t x1(t) = c1e , x2(t) = c2e .
Further, the equilibrium points of the system are x1 = 0 = x2.
Now, as t ↑, ||(x1(t), x2(t))|| ↑, i.e. the solution curve moves away from the equilibrium point (0, 0) as t increases. Hence, the origin is an unstable equilibrium point.
Remark. We know, how to check the stability of the equilibrium points of a linear system. What about the nonlinear systems?
7.1 Stability of non-linear odes
Consider a nonlinear system given by
dx 1 = P (x , x ) dt 1 2 dx 2 = Q(x , x ) dt 1 2
Equilibrium points for this system satisfy P (x1, x2) = Q(x1, x2) = 0.
97 98 CHAPTER 7. LIAPUNOV FUNCTIONS
Example. Consider a nonlinear system given by
dx 1 = P (x , x ) dt 1 2 dx 2 = Q(x , x ) dt 1 2
2 2 where P (x1, x2) = x1 + x2 and Q(x1, x2) = sin(x1x2). Thus, origin is the only equilibrium point.
2 2 2 2 Example. Taking P (x1, x2) = x1 + x2 and Q(x1, x2) = x1 + x2 − 2 in the above example, then this system does not have any equilibrium point.
In this chapter we will be concerned about the nature of the equilibrium point rather than finding one.
7.1.1 Liapunov’s direct method
We are interested to see whether a given equilibrium point of a first order ode system is stable or not.
Definition 7.1.1. Let E(x, y) be a differentiable function of (x, y) on some domain D ⊂ R2 containing the origin. Then, E is said to be
1. Positive definite if : E(0, 0) = 0 and E(x, y) > 0, ∀(x, y) 6= (0, 0).
2. Positive semi-definite if E(x, y) ≥ 0, ∀(x, y) ∈ D.
3. Negative definite if : E(x, y) = 0 iff (x, y) = (0, 0) and E(x, y) < 0 otherwise.
4. Negative semi-definite if E(x, y) ≤ 0, ∀(x, y) ∈ D.
Setup : Let us consider a system of two non-linear first order odes
dx 1 = P (x , x ) dt 1 2 dx 2 = Q(x , x ) dt 1 2 7.1. STABILITY OF NON-LINEAR ODES 99
Let E(x, y) be a differentiable function defined on the domain D containing the origin and the range of the solution (x1, x2). Then, if x1(t), x2(t) are solutions of the system (1) then E(x1, x2) can be considered as a differentiable function of t.
dE ∂E dx1 ∂E dx2 ∴ = + dt ∂x1 dt ∂x2 dt ∂E ∂E = P (x1, x2) + Q(x1, x2) ∂x1 ∂x2
Definition 7.1.2. Let E(x, y) be any differentiable function on D, as defined above. Then, the derivative of E with respect to the system
dx 1 = P (x , x ) dt 1 2 dx 2 = Q(x , x ) dt 1 2 is given by
dE ∂E ∂E ∴ = P (x1, x2) + Q(x1, x2) dt ∂x1 ∂x2
Definition 7.1.3. Consider the system
dx 1 = P (x , x ) dt 1 2 dx 2 = Q(x , x ) dt 1 2 and E(x, y) as given in the setup. If E is positive definite on D and the derivative of E wrt the system is negative semi-definite, then we say that E is a Liapunov function for the given system.In other words, A differentiable function E(x, y) on D is said to be a Liapunov function for the given system if
1. E is positive definite and
dE 2. dt is negative semi-definite. 100 CHAPTER 7. LIAPUNOV FUNCTIONS
Example. Consider
dx 1 = −x + x2 dt 1 2 dx 2 = −x + x2 dt 2 1
Then, (0, 0) is an equilibrium point. Consider E(x, y) = x2 + y2, which implies the derivative of E dE 2 2 wrt the given system is dt = −2(x1 + x2) + 2x1x2(x1 + x2). dE 1 1 Now, E(x, y) is positive definite and dt is negative semidefinite on |x1| < 2 , |x2| < 2 . Therefore, E is a Liapunov function for the given system.
Remark. 1. We will be considering systems for which the origin (0, 0) is an equilibrium point.
2. In case (0, 0) is not an equilibrium point, we will translate the equilibrium point to (0, 0) and work with an equivalent system.
Theorem 7.1.4. Consider a system of first order odes
dx dx 1 = P (x , x ), 2 = Q(x , x ) (7.1) dt 1 2 dt 1 2
Assume (7.1) has an isolated equilibrium or critical point at the origin and P,Q have continuous first order partial derivatives. If ∃ a Liapunov function for (7.1) in some neighborhood of the origin, then the origin is a stable equilibrium point.
Proof. Since, the Liapunov function E exists in a neighborhood of origin, ∃ > 0 such that E is 2 2 2 2 defined on B2(0, 0). Then, define K = {(x, y) ∈ R |x + y = }.
Again, as E is continuous and K is compact, therefore E attains its minimum on K, i.e.
∃(x, y) ∈ K such that
E(x, y) = inf E(x, y) = m (say) (7.2) (x,y)∈K
As E is a Liapunov function, E is positive definite. Also, as it is continuous on K ⊆ B2(0, 0), E must be uniformly continuous on B(0, 0).
∴ ∃ > δ > 0 such that E(x, y) < m, ∀(x, y) ∈ Kδ
2 2 2 2 where Kδ = {(x, y) ∈ R |x + y ≤ δ }. 7.1. STABILITY OF NON-LINEAR ODES 101
Figure 7.1: Diagramatic representation for the neighborhoods
Consider C be any path satisfying the system (7.1), i.e. C(t) = (f(t), g(t)) where f and g solve 0 2 2 2 (7.1), starting within the region (Kδ) , i.e. C(t0) ∈ Bδ(0, 0), i.e. f(t0) + g(t0) < δ . This implies
E(f(t0), g(t0)) < m. dE dE Since, dt is negative semi-definite and dt [f(t), g(t)] ≤ 0, ∀t such that (f(t), g(t)) ∈ B2(0, 0), E must be non-increasing along C(t).
This implies E[f(t), g(t)] ≤ E[f(t0), g(t0)] < m, which implies the curve C(t) has to lie inside the region bounded by K. As otherwise, ∃ at least one point on the curve C(t) for which E[C(t)] ≥ m, which implies that the origin is a stable equilibrium point.
Lecture 28
dX Definition 7.1.5 (Stable critical point). Let P be a critical point of the system dt = F (X). The critical point P is called a stable critical point if for any > 0, ∃δ > 0 such that for any initial condition X(t0) = X0 ∈ Bδ(P ), we will have the solution of the given system X(t) ∈ B(P ), ∀t ≥ t0. (see figure (7.1.1).)
dX Definition 7.1.6 (Asymptotically stable). A critical point P of the system dt = F (X) is said to be an asymptotically stable critical point if P is a stable critical point and as t → ∞,X(t) → P . (see figure (7.1.1).)
dx dy Theorem 7.1.7. Let dt = P (x, y), dt = Q(x, y) and the origin be a critical point. Further assume P and Q have continuous first order partial derivatives ∀(x, y) ∈ D. If there exists a Liapunov 102 CHAPTER 7. LIAPUNOV FUNCTIONS
Figure 7.2: Diagramatic representation of a stable critical point
Figure 7.3: Diagramatic representation asymptotically stable critical point. 7.1. STABILITY OF NON-LINEAR ODES 103
Figure 7.4: Diagramatic representation of the neighborhoods function E(x, y) in the region D, containing (0,0) and E˙ is negative definite there, then (0, 0) is asymptotically stable.
Proof. Since, E˙ is negative definite, it is also negative semi-definite. Hence, (0, 0) is a stable critical point.
Let C(t) = (f(t), g(t)) be a solution curve of the given system with initial condition (f(t0), g(t0)) belonging to Kδ (for some fixed > 0, as obtained in the previous theorem). (see figure (7.1.1).) R t dE ˙ Now, E(f(t), g(t)) − E(f(t0), g(t0)) = (f(s), g(s))ds. Also, as E is negative definite, E has to t0 ds be strictly decreasing except at the origin.
If possible let C(t) 9 (0, 0) as t → ∞. But, C(t) ∈ K (due to stability), which means it is bounded. ˙ Case 1 : If C(t) converges to some point γ = (γ1, γ2) 6= (0, 0), then E(γ) > 0 and E(γ) < 0.
˙ ∴ E(f(t), g(t)) − E(f(t0), g(t0)) ≤ −k(t − t0), where − k = max {E(f(t), g(t))} t∈[t0,t] Further, as E˙ (f(t), g(t)) → E˙ (γ) < 0 =⇒ −k < 0 =⇒ E(f(t), g(t)) → −∞, as t → ∞, which contradicts the fact that E is positive definite.
Case 2 : If C(t) remains within K but does not converge. As C(t) is bounded, ∃λ > 0 and a sequence {tn} → ∞ such that ||C(tn) − (0, 0)|| ≥ λ. R tn ˙ Then, E(C(tn)) − E(C(t0)) ≤ E(C(s))ds. t0 ˙ ˙ Now, E is continuous on K. This implies the maximum of E exists on K\Bλ(0, 0). (see figure ˙ ˙ (7.1.1).) Let −k be the maximum of E on K\Bλ(0, 0). Again, as E is negtive definite, ˙ −k < E(x, y), ∀(x, y) ∈ Bλ(0, 0) and −k < 0. 104 CHAPTER 7. LIAPUNOV FUNCTIONS
Figure 7.5: Diagramatic representation of the solution curve
Therefore E(C(tn)) − E(C(t0)) ≤ −k(tn − t0) =⇒ E(C(tn)) → −∞ as n → ∞, which again contradicts that E is positive definite. Thus, our assumption that C(t) 9 (0, 0) as t → ∞ is false. Hence, origin is an asymptotically stable critical point.
Example. Consider the system
dx = x + x2 − 3xy dt dy = −2x + y + 3y2 dt
Origin is a critical point. We find that it is an unstable critical point.
To determine the instability of the critical points we need the following theorems (without proof).
Lecture 29
7.2 Instability theorems
Theorem 7.2.1 (Liapunov instability theorem). Let origin be a critical point of the system dx dy dt = P (x, y) and dt = Q(x, y). Suppose there exists a continuously differentiable function E(x, y) such that
1. E is positive definite and
dE 2. dt > 0 in a neighborhood of the origin. 7.2. INSTABILITY THEOREMS 105
Figure 7.6: Diagramatic representation of Liapunov instability condition
Then the origin is an unstable equilibrium point. (see figure (7.6)).
Theorem 7.2.2 (Chataev instability theorem). Let the origin be a critical point of the system dx dy dt = P (x, y) and dt = Q(x, y). Suppose there exists a continuously differentiable function E(x, y) on a neighborhood U of the origin and a non-empty set U1 ⊆ U such that
1. (0, 0) ∈ U1,
2. E(x, y) > 0, ∀(x, y) ∈ U1 {(0, 0)},
˙ ∂E ∂E 3. E = ∂x P (x, y) + ∂y Q(x, y) > 0, ∀(x, y) ∈ U1\{(0, 0)},
4. E(x, y) = 0, ∀(x, y) ∈ ∂U1.
Then, (0, 0) is an unstable critical point. (See figure (7.7).)
Example.
dx = x + x2 − 3xy = P (x, y) dt dy = −2x + y + 3y2 = Q(x, y) dt
For the postive definite function E(x, y) = x2 + y2, E˙ (x, y) = 2[x2 + y2 + (x − 3y)(x2 − y2) + xy(y − 2)] Now, E˙ (0, y) = 2[y2 + 3y3]. If y is sufficiently small in magnitude, then E˙ (0, y) > 0. Thus, by Chataev instability theorem, origin is unstable. 106 CHAPTER 7. LIAPUNOV FUNCTIONS
Figure 7.7: Diagramatic representation of Chataev instability condition
Remark. 1. There are a lot of stability and instability theorems to check the nature of stability of the critical points of a non-linear system.
2. The list of such theorems is non-exhaustive.
3. These are all sufficient conditions and not necessary.