Chapter 7
INTEGRAL EQUATIONS
Chapter 7 INTEGRAL EQUATIONS
Chapter 7 Integral Equations
7.1 Normed Vector Spaces 1. Euclidian vector space \n 2. Vector space of continuous functions CG( )
3. Vector Space LG2 () 4. Cauchy-Bunyakowski inequality 5. Minkowski inequality
7.2 Linear Operators - continuous operators - bounded operators - Lipschitz condition - contraction operator - successive approximations - Banach fixed point theorem
7.3 Integral Operator
7.4 Integral equations - Fredholm integral equations - Volterra integral equations - integro-differential equations - solution of integral equation
7.5 Solution Methods for Integral Equations 1. Method of successive approximations for Fredholm IE (Neumann series) 2. Method of successive substitutions for Fredholm IE (Resolvent method) 3. Method of successive approximations for Volterra IE
7.6 Connection between integral equations and initial and boundary value problems
1. Reduction of IVP to the Volterra IE 2. Reduction of the Volterra IE to IVP 3. Reduction of BVP to the Fredholm IE
7.7 Exercises
Future Topics:
7.7 Fixed point theorem (see [Hochstadt “Integral equations”, p.25]) (added in 7.2) Elementary existence theorems
7.8 Practical applications (see [Jerri “Introduction to Integral Equations with Applications”])
7.9 Inverse problems (see [ Jerri, p.17])
7.10 Fredholm’s alternatives
Chapter 7 INTEGRAL EQUATIONS
7.1 Normed Vector Spaces
We will start with some definitions and results from the theory of normed vector spaces which will be needed in this chapter (see more details in Chapter 10).
1. Euclidian vector space \n The n-dimensional Euclidian vector space consists of all points n \\=={x() x12 ,x ,...,x n x k ∈} for which the following operations are defined:
n Scalar product ( x,y) =+++ x11 y x 22 y ... x nn y x,y∈ \
22 2 Norm x ==+++()x,x x12 x ... x n Distance ρ ( x,y) = x− y
Convergence lim xk = x if lim x− xk = 0 k →∞ k →∞ \n is a complete vector space (Banach space) relative to defined norm x .
2. Vector space CG() Vector space CG( ) consists of all real valued continuous functions defined on the closed domain G ⊂ \n : C( G) =⊂→{ f() x : D\\n continuous} Norm fmaxfx= ( ) C xG∈
Convergence lim fk = f if lim f− fk = 0 k →∞ k →∞ C CGis a complete vector space (Banach space) relative to defined norm f . ( ) C
3. Vector space LG2 () The space of functions integrable according to Lebesgue (see Section 3.1)
2 \^n LG2 ()= fx:G () ⊂→∫ fx() dx <∞ G Inner product ( f ,g) = ∫ f( x) g( x) dx G 2 Norm f ==f,f f x dx 2 ()∫ () G The following property follows from the definition of the Lebesgue integral
∫ f ()x dx ≤ ∫ f ()x dx G G
LGis a complete normed vectors spaces (Banach spaces) relative to f . 2 ( ) 2
4. Cauchy-Bunyakovsky-Schwarz Inequality (see also Theorem 10.1, p.257)
f,g≤⋅ f g for all f ,g∈ L G ( ) 22 2 ( )
Proof:
If f ,g∈ L2 ( G) , then functions f , g and any combination α f + β g are
also integrable and therefore belong to L2 (G). Consider
f + λ g ∈ L2 (G) , λ ∈ R for which we have Chapter 7 INTEGRAL EQUATIONS
2 2 2 0 ≤ ∫ ()f + λ g dx = ∫ f dx + 2λ∫ fg dx + λ2 ∫ g dx G G G G
The right hand side is a quadratic function of λ . Because this function is non- negative, its discrimenant ( Db=−2 4ac) is non-positive
2 22 4fgdx4fdxgdx0∫∫∫−≤ GGG 2 22 fg dx≤ f dx g dx ∫∫∫ GGG and because ()f ,g = ∫ fgdx ≤ ∫ fg dx ≤ ∫ f g dx , G G G 2 22 f,g≤⋅ f g () 22
from which the claimed inequality yields
f ,g≤⋅ f g () 22 ■
5. Minkowski Inequality (3rd property of the norm “Triangle Inequality”), (see Example 10.7 on p.257)
f + g ≤ f + g for all f ,g∈ L G 2 2 2 2 ()
Proof:
2 Consider f + g = f + g, f + g 2 ( ) = ( f , f )+ ( f ,g)+ (g, f )+ (g, g) 2 2 ≤ f + f ,g + g, f + g 2 ( ) ( ) 2 2 2 ≤ f + 2 f g + g from C-B inequality 2 2 2 2 2 = f + g ( 2 2 )
Then extraction of the square root yields the claimed result. ■
Note that the Minkowski inequality reduces to equality only if functions f and g are equal up to the scalar multiple, f = αg , α ∈ R (why?).
Chapter 7 INTEGRAL EQUATIONS
7.2 Linear Operators
Let M and N be two complete normed vectors spaces (Banach spaces, see Ch.10) with norms ⋅ and ⋅ , correspondingly. We define an operator L as M N a map (function) from the vector space M to the vector space N :
L : M → N
Introduce the following definitions concerning the operators in the vector spaces:
Operator L : M → N is linear if L(αf + βg) = αLf + βLg for all f , g ∈ M and all α ,β ∈ R
Operator L : M → N is continuous if from f k → f in M
follows Lf k → Lf in N (the image of the convergent sequence in M is a convergent sequence in N )
Operator L : M → N is bounded if there exists c > 0 such that Lf ≤ c f for all f ∈ M N M
The norm of operator L : M → N can be defined as the greatest lower bound of such constant c Lf L = sup N f ≠0 f M
Theorem 7.1 If linear operator L : M → N is bounded then it is continuous
Proof: Let operator L : M → N be bounded, then according to the definition there exists c > 0 such that Lf ≤ c f . N M
Let f k → f in M . That means that lim f k − f = 0 . From the definition of k →∞ M
the limit it follows that for any ε > 0 there exists kNε ∈ such that f − f < ε for all kk≥ . k M ε
To prove the theorem, show now that Lf k → Lf in N or that
lim Lf k − Lf = 0 . We have to show that for any Ε > 0 there exists K Ε ∈ N k →∞ N such that Lf − Lf < Ε for all k ≥ K . k N Ε Ε Choose ε = , then c Ε Lf − Lf = L()f − f ≤ c f − f < c ⋅ = Ε for all kk≥= K. ■ k N k N k M c Ε c Ε
Remark: It is also true that if linear operator is continuous then it is bounded (prove as an exercise). Therefore, for linear operators, properties continuous and bounded are equivalent. Chapter 7 INTEGRAL EQUATIONS
Definition Linear operator L:M→ N satisfies the Lipschitz condition with constant k0≥ if
Lf− Lg≤− k f g for all f,g∈ M
Obviously that if linear operator satisfies the Lipschitz condition (it is called a Lipschitz operator) then it is bounded (take vector g = 0 ) and, therefore, it is continuous.
Definition Linear operator L:M→ N is a contraction if it satisfies the Lipschitz condition with constant k1< .
distance between images becomes smaller
Let S be a closed subset of Banach space M , SM⊂ , and let L:S→ S be an operator.
Definition Solution of operator equation fLf= is called a fixed point of operator L .
Definition Successive approximations is a sequence {}f012,f ,f ,... constructed in the following way:
fS0 ∈ is a starting point
fLf10=
fLf21= #
fLfn1+ = n #
Schematic visualization of successive approximations:
Chapter 7 INTEGRAL EQUATIONS
Successive approximations can be used for solution of operator equation fLf= For example, in this case, the successive approximations converge to the fixed point:
But they do not always converge to the fixed point of operator equation. This example shows that even the choice of the starting point close to the fixed point yields the divergent sequence of successive approximations (apparently they are not very successive / ):
The following theorem establishes the sufficient condition for convergence of successive approximations to the fixed point of operator equation.
Chapter 7 INTEGRAL EQUATIONS
Theorem (Banach Fixed Point Theorem, 1922)
Let S be a non-empty closed subset of Banach space M , SM⊂ , S ≠ ∅ . And let L:S→ S be a contraction operator with constant k1< . Then the sequence of successive approximations
{ fnn f=∈ Lf n10− , f S} converges to the unique fixed point
fS,∈ f=Lf for any starting point fS0 ∈
fn → f and the following estimate is valid 1kn ff−≤ ff − ≤ ff − nnn101k1k+ − Proof:
n • Using mathematical induction, show that ffnn1−≤+ kff 01 − (☼)
0 Verify for n0= ff01−= kff 01 −=− ff 01 true
n Assume for n : ffnn1−≤+ kff 01 −
n1+ Show for n1+ : ffn1++−≤ n2 kff 0 − 1
Indeed, ffn1++− n2 =−Lfnn1 Lf + definition of s.a.
≤−kfnn1 f+ Lipschitz condition n ≤⋅kk f01 − f assumption n1+ =−kff01
• Show that {}fn is a Cauchy sequence, i.e. lim fmn− f= 0 n,m→∞
Consider
fffLfLfLfLffmn−= m − m + m − n + nn − (add and subtract)
Apply Minkowski inequality twice:
fmn− f ≤−fmmLf + Lf mn − Lf + Lf nn − f
≤−fmmLf + k f mn −+− f Lf nn f Lipschitz condition
fLfLff−+− ≤ mm nn 1k− f −+−fff ≤ mm1nn1++ definition of s.a. 1k− kfmn−+ f kf − f ≤ 01 01 equation (☼) 1k−
kkmn+ =−f f → 0 when m,n →∞ 1k− 01 Chapter 7 INTEGRAL EQUATIONS
• Because vector space M is complete, Cauchy sequence { fn } converges to
some fM∈ . And because fSn ∈ and set S is closed (includes all limiting points), fS∈ . Therefore in a limit, equation of successive approximations
fLfn1+ = n ⇒ lim fn1+ = lim Lf n nn→∞ →∞
lim fn1+ = L lim f n nn→∞( →∞ ) converges to fLf= And therefore, fS∈ is a fixed point.
• (Uniqueness) Let f,g∈ S be two fixed points of operator L : fLf= g = Lg Then from f − gLfLgkfg=−≤ − yields (1k− ) f−≤ g 0 Because 1k− > 0 f − g0≤ That is possible only if f − g0= Therefore, fg= So the fixed point is unique. ■
Hugo Steinhaus, the colleague and friend of Stefan Banach, formulated the fixed point theorem in the following way:
“hedgehog cannot be combed”
Chapter 7 INTEGRAL EQUATIONS
7.3 Integral Operator Consider an operator called an integral operator given by the equation
Kf = ∫ K(x, y) f (y)dy x ∈G ⊂ R n G
Obviously, that integral operator is linear.
Function K()x, y is called a kernel of the integral operator. We will consider
kernels K()x, y ∈ L2 (G ×G) , therefore 2 ∫∫ K(x, y) dxdy < ∞ GG In a case of G ⊂ R , the domain G = (a,b), where a,b can be finite or infinite.
Theorem 7.2 Let K be the integral operator with a kernel K()x, y continuous in [a,b]×[a,b]. Then operator K is bounded, and, therefore, continuous. Moreover:
1) K : L a,b → C a,b Kf ≤ M b − a f for f ∈ L a,b 2 () [ ] C 2 2 ()
2) K : L a,b → L a,b Kf ≤ M b − a f for f ∈ L a,b 2 () 2 ( ) 2 ( ) 2 2 ()
3) K : C a,b → C a,b Kf ≤ M b − a f for f ∈C a,b [] [ ] C ( ) C []
Proof: Since K()x, y is continuous in the closed domain [a,b]×[a,b], there exists M > 0 such that M = max K(x, y) . x,y∈[]a,b
1) Let f ∈ L2 (a,b). Then because function K(x, y) is continuous in [][]a,b × a,b , the function (Kf )(x) is continuous in [a,b], and, therefore
K : L2 ()a,b → C[a,b]. Consider
Kf = max (Kf)( x) definition of norm in Ca,b[ ] C xa,b∈[] b = max K()() x, y f y dy definition of integral operator xa,b∈[]∫ a
= max (K( x, y) , f( y)) inner product in La,b2 () xa,b∈[]
≤ max K f Cauchy-Bunyakowski inequality xa,b∈[] 22
b 12 2 ≤ f max K() x, y dy definition of norm in La,b2 () 2 xa,b∈[] ∫ a
b 12 2 ≤ f max M dy replace by M = max K()x, y 2 xa,b∈[] ∫ x,y∈[]a,b a
= M b − a f calculating definite integral 2 Chapter 7 INTEGRAL EQUATIONS
2) Kf = Kf x , Kf x 1 2 definition of norm in La,b 2 (( )( ) ( )( )) 2 ( )
1 2 b 2 = ∫ ()()Kf x dx inner product in La,b2 ( ) a
1 2 2 b b = K()()x, y f y dy dx definition of integral operator ∫∫ a a
1 2 b 2 2 ≤ K f dx Cauchy-Bunyakowski inequality ∫ 2 2 a
1 2 b b 2 = f K()x, y dydx factoring f 2 ∫∫ 2 a a
1 2 b b ≤ f M 2 dydx replace by M = max K(x, y) 2 ∫∫ x,y∈[]a,b a a
1 2 b b ≤ M f dydx calculating definite integral 2 ∫∫ a a
= M f b − a 2 ( )
3) Kf = max (Kf)( x) definition of norm in Ca,b[ ] C xa,b∈[] b = max K()() x, y f y dy definition of integral operator xa,b∈[]∫ a b ≤ max K() x, y f() y dy xa,b∈[]∫ a b ≤ max M f() y dy replace by M = max K(x, y) xa,b∈[]∫ x,y∈[]a,b a b = M ∫ fydy() does not depend on x a b ≤ M max f() y dy ∫ ya,b∈[] a b ≤ M fdy definition of norm in Ca,b ∫ C [ ] a b ≤ M fdy C ∫ a
=−M ba f calculating definite integral ( ) C
■
Chapter 7 INTEGRAL EQUATIONS
7.4 Integral Equations Integral equations are equations in which the unknown function is under the integral sign. The typical integral equations for unknown function u()x , xG∈⊂\n (in this chapter , we consider xa,b∈( ) ⊂ \ ) include integral term in the form of integral operator with the kernel K(x, y) bx() KuKx,yuydy= ∫ ()() a
The main types of integral equations are the following:
I Fredholm integral equation 1) Fredholm’s integral equation of the 1st kind:
b ∫ K ()()x,y u y dy= f () x Ku = f non-homogeneous eqn a b ∫ K ()()x,y u y dy= 0 Ku = 0 homogeneous eqn a
2) Fredholm’s integral equation of the 2nd kind: λ ∈C is a parameter
b ux()=+λ∫ Kx,yuydyf ( )() () x u = λKu + f non-homogeneous eqn a b ux()= λ∫ Kx,yuydy ( )() u = λKu homogeneous eqn a
II Volterra integral equation 1) Volterra’s integral equation of the 1st kind:
x ∫ K()()x, y u y dy = f ()x 0
2) Volterra’s integral equation of the 2nd kind:
x u()x = λ∫ K (x, y )()u y dy + f ()x 0
Note that Volterra’s equations can be viewed as a special case of Fredholm’s equations with K(x, y) = 0 for 0 < x < y < a (it is called a Volterra kernel). y
a
0 a x
Chapter 7 INTEGRAL EQUATIONS
III Integro-Differential Equation includes an unknown function under the integral sign and also any derivative of the unknown function. For example: du = u()x + ∫ K (x, y )()u y dy + f ()x dx G An important representation of the integro-differential equation is a Radiative Transfer Equation describing energy transport in the absorbing, emitting and scattering media (analogous equations appear in the theory of neutron transport).
Solution of integral equation is any function u(x)satisfying this equation:
u = λKu + f non-homogeneous equation
u = λKu homogeneous equation
The value of the parameter λ ∈C for which the homogeneous integral equation
has a non-trivial solution u ∈ L2 which is called an eigenvalue of the kernel K(x, y), and the corresponding solution is called an eigenfunction of this kernel.
Eigenvalue problem We will distinguish eigenvalue problems for the integral kernel (integral equation): u = λKu
and for the integral operator 1 Ku = u λ The eigenvalues of the integral operator are recipical to eigenvalues of the integral kernel, and eigenfunctions are the same in both cases.
Existence of the solution of Fredholm’s integral equation
Consider Fredholm’s integral equation of the 2nd kind: u = λKu + f (◊) with bounded integral operator K which also satisfies the Lipschitz condition:
KuKukuu12−≤− 12, k0≥ Rewrite integral equation in the form uTu= (◊◊) where operator T is defined by Tu= λ Ku+ f Note that operator T is not linear. Obviously, if u is a fixed point of operator equation (◊◊), then u is a solution of integral equation (◊). Consider
Tu12− Tu =+−+λλKuf12() Kuf
=−λλKuKu12
=−λ K (uu12)
≤−λ ku12 u If λ k1< , then operator T is a contraction and according to Banach fixed point theorem , there exist a unique fixed point of equation (◊◊) . This unique fixed point is also a solution of Fredholm’s equation (◊). Therefore, the following conclusion can be made: Fredholm’s integral equation of the 2nd kind with bounded kernel has a unique solution for sufficiently small λ , in fact λ < 1k. Chapter 7 INTEGRAL EQUATIONS
7.5 Solution Methods for Integral Equations
1. The Method of Successive Approximations for Fredholm’s Integral Equation
For the integral equation
u = λKu + f
the following iterations of the method of successive approximations are set by:
u0 (x) = f (x)
u n (x) = λKu n−1 + f n = 1,2,...
n Lemma 7.1 u x = λk K k f where K k = K K " K n () ∑ ( () ) k =0 k times
Proof by mathematical induction (assume that the formula for n is true):
0 0 n = 0 u0 (x) = λ K f = f (x) confirmed
n = n + 1 uxn1+ ( ) = λKu n + f by definition
n = λK ∑ λk K k f + f by assumption k =0 n =+f ∑ λ k1++Kf k1 linearity k0= n+1 = f + ∑ λ p K p f change of index p = k + 1 p=1 n+1 = λ0 K 0 f + ∑ λ p K p f p=1 n+1 = ∑ λ p K p f p=0 n+1 = ∑ λk K k f change of index p = k ■ k =0
∞ Neumann Series ∑ λk K k f is called to be the Neumann Series k =0
Estimation of iterations K n f = K(K n−1 f ) C C
≤ M (b − a) K n−1 f Theorem 7.2 (3) C ≤ M 2 (b − a)2 K n−2 f C … ≤ M n b − a n f ( ) C Chapter 7 INTEGRAL EQUATIONS
∞ ∞ λk K k f ≤ λ kkK f Cauchy-Bunyakovsky inequality ∑ ∑ C k =0 C k0= ∞ k ≤ f λ M k b − a k Theorem 7.2 3) C ∑ () k =0 ∞ k = f λ M b − a geometric series C ∑ []() k =0 1 converges if λ < M ()b − a f = C 1− λ M ()b − a
∞ 1 Therefore, the Neumann series ∑ λk K k f converges for λ < . k =0 M ()b − a Denote the sum of the Neumann series as a function u()x : ∞ u()x = ∑ λk K k f k =0 Show that this function satisfies the integral equation u = λKu + f . Consider iterations
u n (x) = λKu n−1 + f then
u(x) = lim u n (x) n→∞
= λK lim u n−1 (x)+ f n→∞ b
= λ K()x, y lim u n−1 ()y dy + f ∫ n→∞ a b = λ∫ K()()x, y u y dy + f a f And , recalling estimation, u()x ≤ C C 1− λ M ()b − a
show that this solution is unique. For that, it is enough to show that the homogeneous equation u = λKu has only a trivial solution. Indeed, if
u0 = λKu0 , then u0 ∈C[a,b] and , according to Theorem 6.2 3), u ≤ λ M b − a u , then 0 C ( ) 0 C 1− λ M b − a u ≤ 0 [ ( )] 0 C 1 Because λ < , [1− λ M (b − a)]> 0 and, therefore, u = 0 . That M ()b − a 0 C yields, that u (x) = 0 for all x ∈[a,b]. So, only the trivial solution exists for the homogeneous equation.
The non-homogeneous equation u = λKu + f can be rewritten in the form (I − λK )u = f where I is an identity operator Then solution of this equation can be treated as an inversion of the operator u = (I − λK )−1 f 1 Therefore, if λ < , then there exists an inverse operator (I − λK )−1 . M ()b − a
The above mentioned results can be formulated in the following theorem: Chapter 7 INTEGRAL EQUATIONS
Theorem 7.3 Fredholm’s integral equation u = λKu + f 1 with λ < and continuous kernel K()x, y has a M ()b − a unique solution u(x)∈C[a,b] for any f (x)∈C[]a,b . This solution is given by a convergent Neumann series ∞ u()x = ∑ λk K k f k =0 and satisfies f u()x ≤ C . C 1− λ M ()b − a 1 If λ < , then there exists an inverse operator M ()b − a (I − λK )−1 .
Conditions of Theorem 7.3 are only just sufficient conditions; if these conditions are not satisfied, solution of the integral equation still can exists and the Neumann series can be convergent.
Example 7.1 Find the solution of the integral equation 1 1 u()x = e x + ∫ u()y dy e 0 by the method of successive approximations and in the form of the Neumann series.
Identify: K(x, y) = 1 f (x) = e x b − a = 1 1 M = 1 λ = e
1 1 1 Check condition: λ < < < 1 M ()b − a e 1⋅1
1) iterations:
x u0 ()x = e 1 1 x 1 x 1 x x 1 x 1 x 1 u1 ()x = e + ∫ u0 ()y dy = e + ∫ e dy = e + [e ]0 = e + 1− e 0 e 0 e e 1 1 1 1 1 1 u x = e x + u y dy = e x + e x + 1− dy = e x + 1− 2 () ∫ 1 () ∫ 2 e 0 e 0 e e … 1 1 1 u x = e x + u y dy = e x + 1− n () ∫ n−1 () n e 0 e
Then solution of the integral equation is a limit of iterations
x 1 x u()x = lim u n (x) = lime + 1− = e + 1 n→∞ n→∞ e n
This result can be validated by a direct substitution into the given integral equation. Chapter 7 INTEGRAL EQUATIONS
2) Neumann series:
∞ u()x = ∑ λk K k f = f (x)+ λ1 K 1 f + λ2 K 2 f +" k =0
f (x) = e x 1 Kf = ∫ e x dy = e −1 0 1 K 2 f = ∫ ()e − 1 dy = e −1 0 … K n f = e −1
Then the Neumann series is
1 1 1 u(x) = e x + ()e −1 + ()e −1 +"+ ()e −1 +" e e 2 e n
11 1 = ee1e1e1x −−+−+()()() −+() e1 −++""() e1 −+ e ee2n
∞ 1 = e x − e −1 + e −1 ()()∑ n n=0 e
(e − 1) = e x − e + 1+ 1 1− e
= e x − e + 1+ e
= e x + 1
So, the Neumann series approach produces the same solution.
Chapter 7 INTEGRAL EQUATIONS
2. The Method of Successive Substitutions for Fredholm’s Integral Equation (the Resolvent Method)
Iterated kernel Let integral operator K has a continuous kernel K(x, y), then define:
Repeated operator K n = K(K n−1 )= (K n−1 )K n = 2,3,...
′ ′ ′ It has a kernel K n (x, y) = ∫ K()()x, y K n−1 y , y dy G
Indeed, K 1 fx = K x,yfydy ( )( ) ∫ () () G Kx,y1 ()
(K 2 f )(x) = [K(Kf )](x) = ∫∫K()()()x, y′ K y′, y f y dydy′ GG = ∫∫K ()()x,y′′′ K y ,y dy f() y dy GG
Kx,y2 () …
′ ′ ′ Kernel K n (x, y) = ∫ K()()x, y K n−1 y , y dy G
′ ′ ′ = ∫ K n−1 ()()x, y K y , y dy G
is called an iterated kernel. Kernels K n (x, y) are continuous, and if domain G = ()a,b , then n n−1 K n (x, y) ≤ M (b − a)
Resolvent Function defined by the infinite series ∞ k R()x, y,λ = ∑ λ K k +1 ()x, y k =0 is called a resolvent.
Theorem 7.4 Solution of integral equation u = λKu + f with continuous 1 kernel K(x, y) is unique in C[a,b] for λ < , and for M ()b − a any f ∈C[a,b] is given by
b u()x = f ()x + λ∫ R (x, y,λ )()f y dy a
i.e. there exists inverse operator 1 (I − λK )−1 = I + λR , λ < M ()b − a
Chapter 7 INTEGRAL EQUATIONS
Example 7.2 Find solution of integral equation 23 1 1 u()x = x + ∫ xyu()y dy 6 8 0 by the resolvent method .
23 Identify: K(x, y) = xy f ()x = x b − a = 1 6 1 M = 1 λ = 8
1 1 1 Check condition: λ < < < 1 M ()b − a 8 1⋅1 Iterated kernels:
K 1 (x, y) = xy 1 1 1 y′3 xy K 2 (x, y) = K ()()x,y′ K1 y′′ ,y dy = xy′y′ydy′ = xy = ∫ ∫ 3 3 0 0 0 1 1 1 xy′ xy y′3 xy K 3 (x, y) = K22()()x,y′ K y′′ ,y dy = y′ydy′ = = ∫ ∫ 3 3 3 2 0 0 0 3 …
xy K (x, y) = n 3 n−1
Resolvent: ∞ k R(x, y,λ) = ∑ λ K k +1 ()x, y k =0 1 xy 1 xy 1 xy 1 xy = xy + + + + ...+ + ... 8 3 8 2 3 2 8 3 3 3 8 n 3 n 1 1 1 1 1 1 1 1 = xy 1+ + + + ...+ + ... 2 2 3 3 n n 8 3 8 3 8 3 8 3 1 = xy 1 1− 24 24 = xy 23
Solution: b u(x) = f ()x + λ∫ R (x, y,λ )()f y dy a 23 1 1 24 23 = x + ∫ xy ydy 6 8 0 23 6 23 1 1 = x + x∫ y 2 dy 6 2 0 1 23 1 y 3 = x + x 6 2 3 0
= 4x
Chapter 7 INTEGRAL EQUATIONS
3. The Method of Successive Approximations for the Volterra Integral Equation of the 2nd kind
Consider the Volterra integral equation of the 2nd kind x ux()=+λ∫ Kx,yuydyf ( ) () () x 0 where K()x, y is a continuous kernel, K ( x,y) ∈× C([ a,b] [ a,b]) .
The method of successive approximation is defined by the following iterations:
u0 ()x = f ()x n n k u n ()x = ∑ λ K f = λKu n−1 + f k =0
Theorem 7.5 The Volterra integral equation of the 2nd kind x ux()=+λ∫ Kx,yuydyf ( ) () () x 0 with continuous kernel K(x, y) and with any λ ∈ R
has a unique solution u(x)∈C[0,a] for any f ()x ∈C[]0,a . This solution is given by a uniformly convergent Neumann series ∞ u(x) = ∑ λn ()K k f ()x k =0 and its norm satisfies u x ≤ f e λ Ma ( ) C C
Example 7.3 Find solution of integral equation x u()x = 1+ ∫ u ()y dy 0 by the method of successive approximations.
Identify: K(x, y) = 1 f (x) = 1 M = 1 λ = 1
K 0 f = f ()x = 1 x x 1 0 x K f = ∫ K()x, y ()K f ()y dy = ∫ 1⋅1dy = [y]0 = x 0 0 x x x 2 2 2 1 y x K f = K()x, y ()K f ()y dy = 1⋅ydy = = ∫ ∫ 2 2 0 0 0 x x x 2 3 3 3 2 y 1 y x K f = K()x, y ()K f ()y dy = 1⋅ dy = = ∫ ∫ 2 2 3 2 ⋅3 0 0 0 … x n K n f = n!
∞ ∞ x k Solution: u(x) = ∑ λn ()K k f ()x = ∑ = e x k =0 k =0 k! Chapter 7 INTEGRAL EQUATIONS
7.6 Connection between integral equations and initial and boundary value problems
1. Reduction of IVP to the Volterra integral equation
Example 7.4 Reduce IVP u′ − 3x 2 u = 0 u(0) = 1 to the Volterra integral equation.
Integrate the differential equation from 0 to x :
x 2 ∫ uy′()− 3yuydy0 () = 0 xx ∫∫()udy′ − () 3yudy02 = 00 x ux()− u0 ()−= 3yuydy0∫ 2 () use the initial condition u(0) = 1 0 x ux()=+ 1 3yuydy∫ 2 () is a Volterra equation with K ( x,y) = y2 0
2. Reduction of the Volterra integral equation to IVP
Recall the Liebnitz rule for differentiation of expressions with integrals:
d b(x) bx()∂g ()x,y db() x da() x ∫ g()x, y dy =+∫ dy g x,b() x − g x,a() x dx a()x ax() ∂x dx dx
In particularly,
d x ∫ g()y dy = g(x) dx 0
d x x g(x, y) ∫ g()x, y dy = ∫ dy + g()x,x dx 0 0 ∂x
Reduction of the Volterra integral equation to IVP is performed by consecutive differentiation of the integral equation with respect to variable x and substitution x = 0 for setting of the initial conditions.
Example 7.5 Reduce the Volterra integral equation x u()x = x 3 + ∫ (x − y )2 u ()y dy 0 initial value problem.
substitute x = 0 to get initial condition
x 0 u()x = x 3 + ∫ (x − y )2 u ()y dy u0()=+ 03 ∫ ( x − y )2 u () ydy u(0) = 0 0 0 0 Chapter 7 INTEGRAL EQUATIONS
x 0 u′()x = 3x 2 + ∫ 2 (x − y )()u y dy u′()0 = 30 2 + ∫ 2 (x − y )()u y dy u′()0 = 0 0 0 x x u′′()x = 3x 2 + 2∫ u ()y dy u′′()0 = 30 2 + 2∫ u ()y dy u′′()0 = 0 0 0
u′′′()x = 6 x + 2u ()x
Therefore, the integral equation is reduced to IVP for 3rd order ODE:
u2u6x′′′ −= u(0) = 0 u′(0) = 0 u′′(0) = 0
3. Reduction of BVP to the Fredholm integral equation
Recall repeated integration formula:
xxttn3t2 1 n1− "" ∫∫ ∫∫f() t112 dt dt dt n1n− dt=− ∫() x t f () t dt 00 00()n1!− 0
Example 7.6 Reduce the boundary value problem y′′(x)+ y(x) = x x ∈ (0,π ) y(0) = 1 y(π ) = π − 1 to the Fredholm integral equation.
Set y′′(x) = u(x)
x x integrate ∫ y′′()t dt = ∫ u ()t dt 0 0 x y′()x − y′ ()0 = ∫ u ()t dt 0
t x x 2 ′ ′ integrate ∫ []y ()t 2 − y ()0 dt 2 = ∫∫ u()t1 dt1 dt 2 0 0 0 t x 2 ′ y()x − y ()0 − y ()0 x = ∫∫ u (t1 )dt1 dt 2 0 0 x y()x − y ()0 − y′ ()0 x = ∫ (x − t )()u t dt repeated integration 0
Use the first boundary condition x y()x = 1+ y′ ()0 x + ∫ (x − t )()u t dt 0
In this expression, y′(0) is not known. Substitute x = π and apply the second boundary condition Chapter 7 INTEGRAL EQUATIONS
π y()π = 1+ y′ ()0 π + ∫ (π − t )()u t dt 0 π π −1 = 1+ y′()0 π + ∫ (π − t )()u t dt 0 Solve for y′(0) 2 1 π y′()0 = 1− − ∫ ()()π − t u t dt π π 0 Then 2 1 π x y(x) = 1+ 1− − ∫ ()()()()π − t u t dtx + ∫ x − t u t dt π π 0 0 2 x π x = 1+ x − x − ∫ (π − t )()u t dt + ∫ (x − t )()u t dt π π 0 0 Now substitute this expression for y(x) and y′′()x = u ()x into the original differential equation
2 x π x u + 1+ x − x − ∫ ()()()()π − t u t dt + ∫ x − t u t dt = x π π 0 0 2 x π x u + 1− x − ∫ (π − t )()u t dt + ∫ (x − t )()u t dt = 0 π π 0 0 2 x π x u = −1+ x + ∫ (π − t )()u t dt − ∫ (x − t )()u t dt π π 0 0 2 x x x π x u = −1+ x + ∫ ()()π − t u t dt + ∫ ()()()()π − t u t dt − ∫ x − t u t dt π π 0 π x 0 2 x x x x π u = x − 1+ ∫ (π − t )()u t dt − ∫ (x − t )()u t dt + ∫ ()()π − t u t dt π π 0 0 π x 2 x x x π x()π − t u = x − 1+ ∫ (π − t )()u t dt − ∫ (x − t )()u t dt + ∫ u()t dt π 0 π 0 x π 2 x x π x()π − t u = x − 1+ ∫ ()()()π − t − x − t u t dt + ∫ u()t dt π 0 π x π 2 x t(π − x) π x(π − t) u = x − 1+ ∫ u()t dt + ∫ u()t dt π 0 π x π
It yields a Fredholm integral equation
2 π ux1Kx,tutdt=−+∫ ()() π 0
with a kernel
t(π − x) 0 ≤ t ≤ x K()x,t = π x()π − t x ≤ t ≤ π π
Chapter 7 INTEGRAL EQUATIONS
7.7 Exercises
1. Prove part 3) of the Theorem 7.2.
2. Classify each of the following integral equations as Fredholm or Volterra integral equation, linear or non-linear, homogeneous or non-homogeneous, identify the parameter λ and the kernel K ( x,y) : 1 a) u() x=+ x∫ xyu( y )dy 0 x b) u() x=+ 1 x2 +∫ ( x − y ) u( y )dy 0 x c) ux()=+ ex2∫ yu(y)dy 0 1 2 d) u() x=−∫ ( x y ) u( y )dy 0 x11 1 e) ux()=+ 1∫ dy 4xyu(y)0 +
3. Reduce the following integral equation to an initial value problem
x ux()=+ x∫ ( y − xuydy )() 0
4. Find the equivalent Volterra integral equation to the following initial value problem
y′′ ( xyxcosx) +=( ) y (00) = y′()01=
5. Derive the equivalent Fredholm integral equation for the following boundary value problem
yyx′′ + = x ∈(0,1) y (01) = y ()10=
6. Solve the following integral equations by using the successive approximation method and the resolvent method:
1 a) u() x=+ xλ∫ xyu () y dy 0 π 1 2 b) ux()=+ x∫ cosxuydy() 4 0
7. Solve the following integral equation by using the successive approximations method
x ux()=− 1∫ ( y − xuydy )() 0
Chapter 7 INTEGRAL EQUATIONS
8. Solve the following integral equations:
x a) u() x=+− sin 2x∫ u ( x s ) sin () s ds 0 x b) ux()=+ x2as∫ u′ ( x − se )− ds u0( ) = 0 0
9. Using mathematical induction prove identity for iterated kernel (7.5 2):
′ ′ ′ K n (x, y) = ∫ K()()x, y K n−1 y , y dy G
10. Using mathematical induction verify the following estimate for iterated kernels (7.5 2):
n n−1 K n (x, y) ≤ M (b − a)
11. Verify result of Example 7.4 by solving both IVP and derived integral equation.
Chapter 7 INTEGRAL EQUATIONS
Stefan Banach (1892 -1945) Scottish Café Lvov
The Scottish café in Lvov (Ukraine) was a meeting place for many mathematicians including Banach, Steinhaus, Ulam, Mazur, Kac, Schauder, Kaczmarz and others. Problems were written in a book kept by the landlord. A collection of these problems appeared later as the Scottish Book. R D Mauldin, The Scottish Book, Mathematics from the Scottish Café (1981) contains the problems as well as some solutions and commentaries.
Ivar Fredholm (1866 – 1927)
Fredholm is best remembered for his work on integral equations and spectral theory. Find out more at: http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Fredholm.html
Vito Volterra (1860 - 1940)
Volterra published papers on partial differential equations, particularly the equation of cylindrical waves. His most famous work was done on integral equations. He published many papers on what is now called 'an integral equation of Volterra type'.
Find out more at: http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Volterra.html