e Versatile Wronskian
Richard Chapling v. November
Given two differentiable functions y1,y2, their Wronskian is the Abel’s eorem determinant eorem (Abel’s Theorem). Suppose that y1 and y2 satisfy B y1 y2 = ′ − ′ . W ′ ′ y1y2 y1y2 ′′ y1 y2 y + py + q = 0, where p,q are also funions of x. en In this handout, we shall explore its uses, both in this course, ′ = − . and in starred sections, in later courses’ material. W pW
Test for Linear Independence Of course it then follows that ∫ ( x ) ( ) = ( ) − . ( , ) W x W a exp p Recall thaty1 andy2 are called linearly independent on a b if there a do not exist α1 and α2 so that α1y1(x) + α2y2(x) = 0 for every x in (a,b) (i.e. α1y1 + α2y2 is the zero function on (a,b)). Proof. We have ′ ′′ ′ ′ ′ ′ ′′ Result . . If y1,y2 are linearly dependent, W = 0. = + − − W y1y2 y1y2 y1y2 y1 y2 W = 0 y (x) , 0 x ∈ (a,b) y y = ′′ − ′′ . If and 1 for any , 1 and 2 are y1y2 y1 y2 linearly dependent. = (− ′ − ) − (− ′ − ) py2 qy2 y1 y2 py1 qy1 = − ( ′ − ′ ) = − , p y1y2 y1y2 pW Proof. . Suppose y1 is linearly dependent on y2. Then either as required. □ y2(x) = 0 for every x, in which case the result follows imme- diately, or we can rewrite the linear dependence condition Corollary . If p is finite and two solutions to the differential equa- = as y1 αy2 for some constant α. But then tion ′′ ′ + + = 0 ′ ′ ′ ′ y py q W = y y − y y = αy y − αy y = 0. 1 2 1 2 2 2 2 2 are linearly independent, they remain linearly independent. ∫ ( )/ ( ) = − x ( ) , Proof. W x W a exp a p , so if W a 0 and p is finite, W (x) , 0. □ . We note that since y2 , 0,
( ) ′ = 2 y2 . Alternative to Reduion of Order W y1 y1 We seek the general solution of the differential equation = , ( / )′ = ′′ ′ Since W 0 and y1 0, we find that y2 y1 0, i.e. y + py + q = 0. () y2/y1 = α is constant, which as before is equivalent to αy1 − y2 = 0, i.e. linear dependence. □ We recall the usual procedure for reduction of order: if y1 is a (nonzero) solution, we substitute y = uy , where u is also a func- ( ) = ∈ ( , ) 1 If we do allow y1 c 0 for some c a b , the latter result is tion of x. Differentiating gives not true. Peano gave the counterexample ( )′ = ′ + ′ uy1 u y1 uy1 2 ( )′′ = ′′ + ′ ′ + ′′, y1(x) B x , y2(x) B x |x |. uy1 u y1 2u y1 uy1 so the () becomes One can check that y and y are linearly independent (there is a 1 2 = ( ′′ + ′ ′ + ′′) + ( ′ + ′ ) + linear combination that vanishes for positive x, and one that is for 0 u y1 2u y1 uy1 p u y1 uy1 quy = ′′ + ( ′ + ) ′ + ( ′′ + + ) negative x, but none for all x). On the other hand, we all know y1u 2y1 py1 u y1 py1 q u ′ ( ) = ′′ ′ ′ that y1 is differentiable with y1 x 2x, and from the definition = y u + (2y + py )u ′ ( ) = | | 1 1 1 of derivative one finds y2 is differentiable with y2 x 2 x , so ′ since y1 satisfies (). This is a first-order linear equation for u , (hence reduion of order) so can be solved using an integrating W = (x|x|)(2x) − (2|x|)(x2) = 0. factor, namely (after∫ initially dividing by y1 to make the leading coefficient 1) y2e p , so 1 ∫ ( x ) ′ A To summarise, the following implications hold: ( ) = exp − , u x 2 p y1 a ¬LI =⇒ W = 0 which can be integrated again to give u as , =⇒ ∫ W 0 LI x ∫ 1 − t ( ) = a p + . (LI and (∀x ∈ (a,b))y1(x) , 0) =⇒ W , 0 u x A 2 e dt B a (y1(t)) (W = 0 and (∀x ∈ (a,b))y1(x) , 0) =⇒ ¬LI
© R Chapling ′ ′ We can obtain an equation for a second solution a different way. But these are linear equations for u and u , and so can be solved ′ 1 2 Since we know that W satisfies W + pW = 0, which we can al- in exactly the same way as usual, which gives ways solve, we know W up to a constant factor. Therefore we can ′ −y2 f ′ y1 f re-interpret the formula for the Wronskian as u = , u = , 1 W 2 W ′ − ′ = , y1y2 y1y2 W whereW is once again the Wronskian. We can now find an expres- a first-order equation for y2. Again, this is easy to solve using an sion for y with a single integration: the complementary function integrating factor, so naturally emerges as the constants of integration in the ui , ( ) ′ ∫ ∫ y2 = W y2 f y1 f y = −y1 + y2 . y1 y2 1 ∫ W W y2 W = C + y1 y2 It is worth noting that this can be written using one integral, as 1∫ W ∫ = + , x y (t)y (x) − y (x)y (t) y2 Cy1 y1 2 dt ( ) = 1 2 1 2 ( ) . y y x ( ) f t dt 1 a W t where we have written the constant separately (rather than leaving This will reappear in IB Mehod in the theory of Green’s funions. it implicit in the integral) to emphasise that we may modify y2 by adding any constant multiple of y1. We notice that this is identical to the reduction of order solution, but the algebra was easier. * Higher order differential equations So our new procedure is If y ,...,y are differentiable functions, their Wronskian is the ′ + = 1 n . Solve the Abel equation W pW 0 to find W . determinant = ′ − ′ . Solve the expression used to define W , W y1y2 y1y2, as y1 y2 ··· yn a first-order equation in y2. ′ ′ ′ y y ··· y 1 2 n W B ...... Variation of Parameters (n−1) (n−1) ··· (n−1) y y yn The next stage is to produce a general way to obtain solutions to 1 2 inhomogenious equations. Suppose we have found all of the solu- This may be used to aid in the solution of higher-order differential tions to the homogeneous equation (i.e. the complementary func- equations, although its uses are less straightforward. + tion). The big leap is that, since Ay1 By2 satisfies the equation The general linear homogeneous ordinary differential equation with A and B constant, replacing A and B by functions may give of order n can be written as us enough extra freedom to satisfy the inhomogeneous equation. Fortunately, this does turn out to be the case. ∑n−1 (n) + (k) = We want to solve a differential equation in the standard form y pky 0 () = ′′ ′ k 0 y + py + qy = f , () where f is a function of x. Let y1 and y2 be two linearly independ- ent solutions to (). Then we expect that we can write . Abel’s theorem
y = u1y1 + u2y2, () Abel’s theorem carries over with little alteration: one can show from the definition of determinant that if and we wish to determine ui . There is a certain amount of freedom ( ) ( ) ··· ( ) in choosing u1 and u2 since if we have uy1y2, it is unclear which a11 t a12 t a1n t ( ) ( ) ··· ( ) ∑ ∏n term to put it in: we have two unknown functions, but only one a21 t a22 t a2n t A(t) = . . . . = sgn(σ) a ( )(t), equation. Therefore we specify the extra condition . . . . . iσ i . . . ∈ = σ Sn i 1 ′ + ′ = a (t) a (t) ··· a (t) u1y1 u2y2 0 () n1 n2 nn to make the problem more definite; it turns out that this will be then by the product rule, sufficient. Differentiating y, we find ′ ′ ′ a (t) a (t) ··· a (t) ′ ′ ′ ′ ′ ′ ′ 11 12 1n y = (u y1 + u y2) + u1y + u2y = u1y + u2y , ( ) ( ) ··· ( ) 1 2 1 2 1 2 ′ a21 t a22 t a2n t ′′ = ′ ′ + ′ ′ + ′′ + ′′. A (t) = . . . . y u1y1 u2y2 u1y1 u2y2 ...... ( ) ( ) ··· ( ) Inserting this into the differential equation (), we find an1 t an2 t ann t ( ) ( ) ··· ( ) = ′′+ ′+ = ( ′′+ ′ + )+ ( ′′+ ′ + )+ ′ ′ + ′ ′ , a11 t a12 t a1n t f y py qy u1 y1 py1 qy1 u2 y2 py2 qy2 u1y1 u2y2 ( ) ( ) ··· ( ) a21 t a22 t a2n t + ··· + . . . . , and the first two terms cancel since yi solve the homogeneous ...... equation. We are left with two equations, ′ ( ) ′ ( ) ··· ′ ( ) an1 t an2 t ann t ′ ′ u y1 + u y2 = 0 1 2 i.e. one may write the derivative as the sum of the determinants ′ ′ + ′ ′ = . u1y1 u2y2 f where the jth row has been replaced by its derivative. We might expect this, based on the simple examples we’ve done using heuristics like “multiply by another x” for constant-coefficient equations. This particular condition also turns out to be a major simplification: it will avoid us having to worry about terms containing second derivatives of the ui .
© R Chapling If one applies this to the Wronskian, one finds that in all but the Differentiating the expression fory n times and applying these con- last term, the determinant has two rows equal and so is equal to ditions gives zero. Therefore ∑n (k) (k) ··· y = ujy k ∈ {0, 1,...,n − 1} y1 y2 yn j ′ ′ ··· ′ j=1 y1 y2 yn n ′ . . . . . ∑ W = . . . . . (n) ′ (n−1) (n) y = u y + ujy , ( −2) ( −2) ( −2) j j j n n ··· n = y1 y2 yn j 1 (n) (n) (n) y y ··· y 1 2 n and substituting into the differential equation and using that yj are solutions to the homogeneous equation causes this last to become But if∑all the yi satisfy (), each term in the last row can be written − ( ) as − n 1 p y k . Applying linearity of the determinant in the ∑n k=0 k ′ (n−1) = . last row, we find that we again obtain n determinants, all but one ujyj f of which have two rows equal and so are zero. This leaves j=1 Hence we have a linear system of n equations y1 y2 ··· yn ′ ′ ′ y y ··· y ∑n 1 2 n ′ (k) = ∈ { , ,..., − }, . . . . ujy f δkn k 0 1 n 1 ′ = . . . . . = − , j W . . . pn−1W j=1 ( −2) ( −2) ( −2) y n y n ··· y n 1 2 n which we can invert in the usual way: for example, Cramer’s rule (n−1) (n−1) (n−1) ′ −pn−1y −pn−1y ··· −pn−1y = / 1 2 n gives uj Wj W , where Wi is the Wronskianwith the jth column of the matrix replaced by (0, 0,..., f ). A single integration then which is of exactly the same form as Abel’s theorem for second- gives the result. order equations. Thus we can again solve for W as before, ( ∫ ) x * Integrals of solutions to Sturm–Liouville W (x) = W (a) exp − pn−1 . a equations
e following seions will make much more sense aer the second part of IB M. . Reduion of order A Sturm–Liouville equation is a second-order differential equa- tion of the form Reduction of order works in a similar, but rather messier, man- ′ ′ −(pu ) + qu = λwu, ner: if we know y1 solves (), we insert uy1 into this equation. We again find that the coefficient of u vanishes, and we have a linear where p > 0, λ is called the eigenvalue, and w > 0 the weight func- ′ equation of order n − 1 for u . But this may be even harder to solve tion. than the original equation! In the order 2 case we were lucky that Such equations (with particular boundary conditions that force the equation was reduced to a linear first-order one, which we can the solutions to have mani interesting furry animals many inter- always solve. But even a third-order equation may reduce to a hor- esting properties) are studied in IB Mehod. We are interested∫ rible second-order equation that we can’t deal with. Exactly the in two results here: firstly, in finding an expression for u1u2w, same problem occurs with the Wronskian: it gives an order n − 1 where u1 and u2 are solutions with corresponding distinct eigen- equation for yn in terms of y1,y2,...,yn, which is useless unless values λ1 and λ2 respectively. Perhaps surprisingly, this is possible we know the rest of the yi , and not even easy to solve. Thus the to do in closed form. second-order case does not usefully generalise. We begin with the equations ′ ′ −(pu ) + qu1 = λ1wu1 1 () . Variation of parameters with Cramer’s rule −( ′ )′ + = w pu2 qu2 λ2 u2 Contrastingly, variation of parameters extends simply to higher- satisfied by the functions. We can obtain the integral we are inter- order inhomogeneous differential equations: given ested in in two ways: by multiplying the first equation by u2 and ∑n−1 integrating, or the second by u1 and integrating. These give ( ) ( ) ∫ ∫ ∫ y n + p y k = f , n − ( ′ )′ + = w = pu1 u2 qu1u2 λ1 u1u2 k 0 ∫ ∫ ∫ − ( ′ )′ + = w . we look for a solution of the form pu2 u1 qu1u2 λ2 u1u2 ∑n = , Subtracting, we notice the q terms cancel, so y ujyj ∫ ∫ = j 1 ( − ) w = − ( ′ )′ − ( ′ )′ . λ1 λ2 u1u2 pu1 u2 pu2 u1 where yj are a linearly independent set of solutions to the homo- geneous equation. Now impose the extra conditions Integrating by parts gives ∫ ∫ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ∑n − (pu ) u − (pu ) u = −pu u + pu u − (pu u − pu u ) ′ ( ) 1 2 2 1 1 2 2 1 1 2 2 1 u y k = 0, k ∈ {0, 1,...,n − 2} j j = ( ′ − ′ ). j=1 p u1u2 u1u2 Again, we see that just beyond the corners of your waking mind the material in this course there lie equations that cannot be solved explicitly. Most (but not all!) of the equations arising from physics are second-order, which somewhat mitigates this problem, but even a general second-order equation may not have a solution that can be written down as a simple integral. A general approach will be discussed in II Fhe Comple Mehod, but still only applies to polynomial coefficients.
© R Chapling Oh hey, it’s the Wronskian again! Even better, the remaining in- the uniqueness theorem for linear differential equations: if v is not ′ tegrals have disappeared, so we have found that identically zero but v(a) = 0, v (a) must be nonzero). Integrating ∫ () over (z , z ) gives ′ − ′ ( , ) 1 2 u1u2 u1u2 W u1 u2 ∫ u1u2w = p = p . () [ ] z ( ) − − u ′ ′ z2 2 ′ p λ1 λ2 λ1 λ2 (vPu −upv ) = (Q − q)u2 + (P − p)u 2 + (W (u, v))2 . v v2 z1 z1 This is also a consequence of the identity Whether or not v(z1) = 0 or v(z2) = 0, the left-hand side evaluates ′ u1Lu2 − u2Lu1 = −(pW (u1,u2)) , to zero at both endpoints. But each term on the right is nonnegat- ive, and the last term is strictly positive unless v ∝ u. So if v ̸∝ u where L = −DpD + q is the Sturm–Liouville operator, which holds we obtain a contradiction, which implies that v has a zero, the first for any u1,u2; it is known as Lagrange’s identity. possibility. ′ A natural∫ question is whether we can extract any information If v ∝ u, since u2,u 2 > 0 are continuous and p, P,q,Q are con- 2w 7→ ( ) about u1 from this. Suppose that x uλ x solves the equa- tinuous the only way to ensure that the right-hand side vanishes tion is that p = P and q = Q. □ −( ′ )′ + = w puλ quλ λ uλ > ⩾ and is differentiable in λ (and therefore also continuous). Then ap- Corollary . Suppose P 0, Q 0. en any nonzero solution −( ′)′ + = plying (), we find for eigenvalues λ and λ + h to Pu Qu 0 has at most one root. ∫ ′ ′ + − Proof. We apply the Sturm comparison theorem using p = P and uλ huλ uλ+huλ u + u w = p = v( ) = −( v′)′ = λ h λ h q 0. x 1 is a nonzero solution of P 0, and has no ′ ′ ′ ′ roots. For any interval (z1, z2), if there is a solution u with roots at + − + − uλ huλ uλuλ uλuλ uλ+huλ v ( , ) □ = p z1 and z2, must have a zero in z1 z2 , a contradiction. h ′ − ′ ′ u + − u u + u Corollary (Sturm Comparison Theorem). Let P,Q,q : [a,b] ⇒ R = pu λ h λ − pu λ h λ . λ h λ h be continuous, with q ⩽ Q and consider the two differential equations → ′ ′ Taking the limit as h 0 gives −(Py ) + Qy = 0 () ∫ −( ′)′ + = . 2w = ( ′ ∂ − ∂ ′ ) = (∂ , ), Py qy 0 () uλ p uλ λuλ uλ λuλ pW λuλ uλ If u is a nonzero solution of the former with successive zeros z1 and which may or may not be easy to compute. z2 and v a solution of the laer, then either . v has a zero in (z , z ), or * Comparison eorems 1 2 . there is α ∈ R so that v = αu, and then q = Q. In this section we shall discuss something a bit more exciting: when do solutions of Sturm–Liouville equations have zeros? This is the casep = P of the Sturm–Picone Comparison Theorem. eorem (Sturm–Picone Comparison Theorem). Let Corollary (Sturm Separation Theorem). If u, v are linearly inde- P,p,Q,q : [a,b] ⇒ R be continuous, with pendent solutions of ′ ′ −(Py ) + Qy = 0 0 < p ⩽ P, q ⩽ Q and u(z1) = u(z2) = 0, then v has a zero in (z1, z2). and consider the two differential equations = −( ′)′ + = Proof. This is the q Q case of Sturm–Picone Comparison The- Py Qy 0 () v , ′ ′ orem. We know that λu, so the result follows from the other −(py ) + qy = 0. () possibility in the theorem. □
If u is a nonzero solution of the former with successive zeros z1 and z2 and v a solution of the laer, then either Exercise Prove this directly by considering the Wronskian. An extremely simple example of this theorem is that between v ( , ) . has a zero in z1 z2 , or any two roots of sin ax there is a root of cos ax. A less silly ex- ample is that the same is true of α cos ax + β sin ax and γ cos ax + . there is α ∈ R so that v = αu, and then p = P and q = Q. δ sin ax for any α, β,γ, δ with αδ − βγ , 0. Proof. We start by quoting the following rather mysterious (but Finally, we prove easily verified) identity due to Picone: Proposition . Let λ1 < λ2, and suppose that u1,u2 solve the ( ) ′ 2 Sturm–Liouville equations u ′ ′ ′ ′ u ′ ′ ′ p (vPu −upv ) = u(Pu ) − (pv ) +(P −p)u 2 + (W (u, v))2, v v v2 −( ′)′ + = w () pui qui λi ui which holds for any differentiable functions with v , 0. If we now on (a,b) with the same boundary conditions u (a) = u (b) = 0. en impose that u and v solve () and (), this simplifies to i i u2 has more zeros in (a,b) than u1. ( ) ′ u ′ ′ ′ p (vPu − upv ) = (Q − q)u2 + (P − p)u 2 + (W (u, v))2 () We first note that this result does make sense: if u had infin- v v2 1 itely many zeros, there would be a point z where they accumulate, v ( , ) v( ) = ( ) = ′ ( ) = Suppose that has no zeros in z1 z2 . Even if z1 0, the and one can then show that u1 z u1 z 0, which again by limit of u/v exists at z1 (this follows from L’Hôpital’s Rule and uniqueness would force u1 to be exactly zero. But you knew that. Although you may also have known that.
© R Chapling ′ ′ Proof. We apply the Sturm Comparison Theorem to −(py ) + (q − ′ ′ λ1w)y = 0 and −(py ) + (q − λ2w)y = 0. Since λ1 < λ2 and w > 0,
q − λ1w > q − λ2w,
and so if u1 has n zeros in the interior of the interval, and we write them as z0 = a, z1, z2,..., zn+1 = b, the SCT gives at least one zero in (zi−1, zi ) for i ∈ {1, 2,...,n + 1}, i.e. at least n + 1 in the interior. □
To go further would require us to use actual Sturm–Liouville theory, so we shall stop here. The theorem to remember for later is:
eorem . Let λ0 < λ1 < λ2 < ··· be the eigenvalues of the Sturm–Liouville problem ′ ′ −(pu ) + qu = λwu u(a) = u(b) = 0.
with corresponding eigenfunions un. en λn has n zeros in (a,b).
. ** Alternative explanation for Sturm–Liouville zeros The following approach requires more theory, but provides per- haps a more intuitive picture: let u(x, λ) be the solution to
−∂1(p∂1u) + qu = λu u(a, λ) = 0, ∂1(a, λ) = 1
(as the equation is linear, the value of the derivative is immaterial provided it is nonzero, so we can choose the latter for definiteness). The position of a zero z of u depends on λ: it satisfies the equation
u(z(λ), λ) = 0.
Hence we can take differentiate implicitly with respect to λ: ′ 0 = z (λ)∂1u(z(λ), λ) + ∂2u(z(λ), λ).
Since zeros are simple, at a zero z0 and eigenvalue λ0, we have u(z0, λ0) = 0 and ∂1u(z0, λ0) , 0, so ∂ ( , ) ′ ( ) = − 2u z0 λ0 z0 λ0 ∂1u(z0, λ0) Also, differentiating the differential equation with respect to λ at λ0 gives −∂1(p∂1∂2u) + q∂2u = w(u + λ0∂2u) Multiplying by u and integrating with respect to x gives ∫ z0 2 w(u(x, λ0)) dx a ∫ z0 = − u∂1(p∂1∂2u) + u(q − λ0w)∂2u dx a ∫ z0 = [− ∂ ∂ ]z0 + (∂ )(∂ ∂ ) + ( − w)∂ pu 1 2u a p 1u 1 2u u q λ0 2u dx a ∫ z0 = + [(∂ ) (∂ )]z0 + (∂ ) − ∂ ( ∂ ) + ( − w) 0 2u p 1u a 2u 1 p 1u q λ0 u dx a = p(z0)(∂2u(z0, λ0))(∂1u(z0, λ0)) + 0. ′ ( ) Plugging this into the equation for z0 λ0 gives ∫ z ′ 1 0 ( ) = − ( ( , ))2w( ) < 0, z0 λ0 2 u x λ0 x dx (∂1u(z0, λ0)) a so as λ increases, the zeros of u move towards a.
© R Chapling