MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

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1 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions Section 3 Integrl Equtions Integrl Opertors nd Liner Integrl Equtions As we sw in Section on opertor nottion, we work with functions defined in some suitble function spce. For exmple, f (x), g (x) my live in the spce of continuous relvlued functions on, b, i.e. C (, b). We lso sw tht it is possible to define integrl s well s differentil opertors cting on functions. Theorem 2.8 is n exmple of n integrl opertor: u (x) where G (x, y) is Green s function. G (x, y)f (y)dy, Definition 3.: An integrl opertor, K, is n integrl of the form K (x, y)f (y)dy, where K (x, y) is given (rel-vlued) function of two vribles, clled the kernel. The integrl eqution Kf g mps function f to new function g in the intervl, b, e.g. K : C, b C, b, K : f g. Theorem 3.2: The integrl opertor K is liner K (αf + βg) αkf + βkg. Exmple : The Lplce Trnsform is n integrl opertor Lf exp ( sx) f (x) dx f (s), with kernel K(x, s) K(s, x) exp( sx). Let us recp its importnt properties. Theorem 3.3: (iv) Convolution: let then (i) (ii) (iii) ( ) df L s dx f (s) f (), ( ) d 2 f L s 2 f (s) sf () f (), dx ( 2 x ) L f (y)dy s f (s). f (x) g (x) f (y)g (x y)dy L (f (x) g (x)) f (s)ḡ (s).

2 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 2 Recll tht differentil eqution is n eqution contining n unknown function under differentil opertor. Hence, we hve the sme for n integrl opertor. Definition 3.4: An integrl eqution is n eqution contining n unknown function under n integrl opertor. Definition 3.5: () A liner Fredholm integrl eqution of the first kind hs the form Kf g, K (x, y)f (y)dy g (x) ; (b) A liner Fredholm integrl eqution of the second kind hs the form f λkf g, f (x) λ K (x, y)f (y)dy g (x), where the kernel K (x, y) nd forcing (or inhomogeneous) term g (x) re known functions nd f (x) is the unknown function. Also, λ ( R or C) is prmeter. Definition 3.6: If g (x) the integrl eqution is clled homogeneous nd vlue of λ for which λkφ φ possesses non-trivil solution φ is clled n eigenvlue of K corresponding to the eigenfunction φ. Note: λ is the reciprocl of wht we hve previously clled n eigenvlue. Definition 3.7 Volterr integrl equtions of the first nd second kind tke the forms () (b) respectively. f (x) λ K (x, y)f (y)dy g (x), K (x, y)f (y)dy g (x), Note: Volterr equtions my be considered specil cse of Fredholm equtions. Suppose y b nd put { K (x, y) for y x, K (x, y) for x < y b, then ( K (x, y)f (y)dy + x ) K (x, y)f (y)dy K (x, y)f (y)dy +.

3 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 3 Conversion of ODEs to integrl equtions Exmple 2: Boundry vlue problems Let Lu (x) λρ (x) u (x) + g (x) where L is liner ODE opertor s in Section (of t lest second order), ρ (x) > is continuous, g (x) is piecewise continuous nd the unknown u (x) is subject to homogeneous boundry conditions t x, b. Suppose L hs known Green s function under the given boundry conditions, G (x, y) sy, then pplying Theorem 2., u (x) λ λ G (x, y)λρ (y)u(y) + g (y) dy G (x, y)ρ(y)u(y)dy + G (x, y)ρ(y)u(y)dy + h (x). G (x, y)g (y)dy The ltter is Fredholm integrl eqution of the second kind with kernel G (x, y)ρ(y) nd forcing term h (x) G (x, y)g (y)dy. The ODE nd integrl eqution re equivlent: solving the ODE subject to the BCs is equivlent to solving the integrl eqution. Exmple 3: Initil vlue problems Let u (x) + (x) u (x) + b (x) u (x) g (x), where u () α nd u () β nd (x), b (x) nd g (x) re known functions. Chnge dummy vrible from x to y nd then Integrte with respect to y from to z : u (y)dy + (y)u (y)dy + b (y)u(y)dy g (y)dy nd using integrtion by prts, with u, v u, on the second term on the left hnd side u (y) x + (y)u(y)z (y)u(y)dy + b (y)u (y)dy g (y)dy, yields u (z) β + (z) u (z) ()α Integrting now with respect to z over to x gives (y) b (y) u (y)dy g (y)dy. u (z) x βx + (z) u (z) dz () αx (3.) (y) b (y) u (y)dydz g (y)dydz.

4 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 4 This cn be simplified by ppeling to the following theorem. Theorem 3.8: For ny f (x), f (y)dydz (x y)f (y)dy. Proof: Interchnging the order of integrtion over the tringulr domin in the yz-plne revels tht the integrl equls f (y)dydz f (y) dzdy y which is trivilly integrted over z to give (x y)f (y)dy s required. Alterntively, integrting the repeted integrl by prts with u (z) f (y)dy, v (z), i.e. u (z) f (z), v (z) z, gives f (y)dydz x z f (y)dy xf (y)dy zx f (y)dy zf (z) dz z Returning to eqution (3.), it my now be written s zf (z) dz yf (y)dy. Thus, u (x) α βx () αx + (x y) (y) b (y) u (y)dy (x y)g (y)dy. (y)u (y)dy u (x) + { (y) (x y) (y) b (y)}u(y)dy (x y)g (y)dy + β + ()αx + α. This is Volterr integrl eqution of the second kind. Notes: () BVPs correspond to Fredholm equtions. (2) Initil vlue problems (IVPs) correspond to Volterr equtions. (3) The integrl eqution formultion implicitly contins the boundry/initil conditions; in the ODE they re imposed seprtely.

5 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 5 Exmple 4 Write the Initil Vlue Problem u (y) + yu (y) + 2u(y) subject to u() α, u () β s Volterr integrl eqution. Integrte with respect to y between nd z: u (z) β + Integrte second term by prts: nd substitute bck to get yu (y)dy + 2 yu (y)dy yu(y) z zu(z) u (z) β + zu(z) + Integrte gin, with respect to z between nd x: u(x) α βx + zu(z)dz + u(y)dy u(y)dy, y(y)dy u(y)dy. u(y)dydz. Chnge dummy vrible in first integrl term to y nd use Theorem 3.8 in lst term to get Simplifiction gives u(x) α βx + yu(y)dy + u(x) + x u(y)dy α + βx. (x y)u(y)

6 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 6 Fredholm Equtions with Degenerte Kernels We hve seen tht Fredholm integrl eqution of the second kind is defined s f (x) λ K (x, y)f (y)dy + g (x). (3.2) Definition 3.9: The kernel K (x, y) is sid to be degenerte (seprble) if it cn be written s sum of terms, ech being product of function of x nd function of y. Thus, K (x, y) u j (x) v j (y) u T (x)v(y) u (x) v (y) u,v, (3.3) j where the ltter nottion is the inner product for finite vector spces (i.e. the dot product). Eqution (3.2) my be solved by reduction to set of simultneous liner lgebric equtions s we shll now show. Substituting (3.3) into (3.2) gives b f (x) λ u j (x) v j (y) f (y)dy + g (x) nd letting then λ j c j j u j (x) f (x) λ v i (y)f (y)dy + g (x) v j (y)f (y)dy v j, f, (3.4) c j u j (x) + g (x). (3.5) j For this clss of kernel, it is sufficient to find the c j in order to obtin the solution to the integrl eqution. Eliminting f between equtions (3.4) nd (3.5) (i.e. tke inner product of both sides with v i ) gives b c i v i (y) λ c j u j (y) + g (y) dy, j or interchnging the summtion nd integrtion, c i λ c j v i (y)u j (y)dy + j v i (y)g (y)dy. (3.6) Writing nd ij g i v i (y)u j (y)dy v i, u j, (3.7) v i (y)g (y)dy v i, g, (3.8)

7 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 7 then (3.6) becomes By defining the mtrices c i λ ij c j + g i. (3.9) j A ( ij ), c c c 2. c n this eqution my be written in mtrix nottion s i.e. c λac + g, g g g 2. g n (I λa)c g (3.) where I is the identity. This is just simple liner system of equtions for c. We therefore need to understnd how we solve the cnonicl system Ax b where A is given mtrix, b is the given forcing vector nd x is the vector to be determined. Let s stte n importnt theorem from Liner Algebr: Theorem 3.: (Fredholm Alterntive) Consider the liner system Ax b (3.) where A is n n n mtrix, x is n unknown n column vector, nd b is specified n column vector. We lso introduce the relted (djoint) homogeneous problem with p n rnk (A) non-trivil linerly independent solutions A T ˆx (3.2) ˆx, ˆx 2,..., ˆx p. Reminder, rnk(a) is the number of linerly independent rows (or columns) of the mtrix A. Then the following lterntives hold: either (i) DetA, so tht there exists unique solution to (3.) given by x A b for ech given b. (And b x ) or (ii) DetA nd then () If b is such tht b, ˆx j for ll j then there re infinitely mny solutions to eqution (3.). (b) If b is such tht b, ˆx j for ny j then there is no solution to eqution (3.).

8 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 8 In the cse of (ii)(), then there re infinitely mny solutions becuse the theorem sttes tht we cn find prticulr solution x PS nd furthermore, the homogeneous system Ax (3.3) hs p n rnk (A) > non-trivil linerly independent solutions x,x 2,...,x p. so tht there re infinitely mny solutions becuse we cn write p x x PS + α jˆx j where α j re rbitrry constnts (nd hence there re infinitely mny solutions). No proof of this theorem is given. To illustrte this theorem consider the following simple 2 2 mtrix exmple: Exmple 5 Determine the solution structure of the liner system Ax b when ( ) ( ) 2 (I) A (II) A 2 2 nd in the cse of (II) when ( ) b 2 j b ( ) (3.4) (3.5) (I) Since Det(A) the solution exists for ny b, given by x A b. (II) Here Det(A) so we hve to consider solutions to the djoint homogeneous system, i.e. i.e. ( 2 2 A T ˆx (3.6) ) ˆx. (3.7) This hs the non-trivil linerly independent solution ˆx (2 ) T. It is cler tht there should be such solution, i.e. p n rnk(a) 2. Note lso tht the homogeneous system i.e. ( 2 2 Aˆx (3.8) ) ˆx (3.9) hs the non-trivil linerly independent solution x ( ) T. If solutions do exist they will therefore hve the form x x PS + α x. A solution to the problem Ax b will exist if ˆx b. This condition does hold for b ( 2) T nd so the theorem predicts tht solution will exist. Indeed it does, note tht x PS (/2 /2) T nd so x x PS + α x is the infinite set of solutions. The orthogonlity condition does not hold for b ( ) T nd so the theorem predicts tht solution will not exist. This is cler from looking t the system.

9 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 9 Now let us pply the Fredholm Alterntive theorem to eqution (3.) in order to solve the problem of degenerte kernels in generl. Cse (i) if det (I λa) (3.2) then the Fredholm Alterntive theorem tells us tht (3.) hs unique solution for c: c (I λa) g. (3.2) Hence (3.2), with degenerte kernel (3.3), hs the solution (3.5): or from (3.2) f (x) λ c i u i (x) + g (x) λ(u (x)) T c + g (x) i which my be expressed, from (3.8), s f(x) λ f(x) λ(u (x)) T (I λa) g + g (x), (u (x)) T (I λa) v (y) g (y)dy + g (x). Definition 3.: The resolvent kernel R (λ, x, y) is such tht the integrl representtion for the solution holds. f (x) λ R (λ, x, y)g (y)dy + g (x) Theorem 3.2: For degenerte kernel, the resolvent kernel is given by R (λ, x, y) (u (x)) T (I λa) v (y). Cse (i) covered the simple cse when there is unique solution. Let us know concern ourselves with the cse when the determinnt of the mtrix on the left hnd side of the liner system is zero. Cse (ii) suppose det (I λa), (3.22) nd tht the homogeneous eqution hs p non-trivil linerly independent solutions (I λa)c (3.23) c,c 2,...,c p. Then, the homogeneous form of the integrl eqution (3.2), i.e. f (x) λ K (x, y)f (y)dy, (3.24)

10 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions with degenerte kernel (3.3), hs p solutions, from (3.5): f j (x) λ c j i u i (x) (3.25) with j, 2,..., p. Turning to the inhomogenous eqution, (3.), it hs solution if nd only if the forcing term g is orthogonl to every solution of or Hence (3.8) yields which is equivlent to Thus, writing then i (I λa) T h i.e. h T g (3.26) h i g i. i h i v i (y)g (y)dy i ( ) h i v i (y) g (y)dy. i h (y) h i v i (y) (3.27) i h (y)g (y)dy, which mens tht g (x) must be orthogonl to h (x) on, b. Let us explore the function h(x) little; we strt be expressing (3.26) s h i λ ji h j. j Without loss of generlity ssume tht ll the v i (x) in (3.3) re linerly independent (since if one is dependent on the others, eliminte it nd obtin seprble kernel with n replced by (n )). Multiply the ith eqution in (3.26) by v i (x) nd sum over ll i from to n: h i v i (x) λ i i ji h j v i (x), j i.e. from (3.7) i h i v i (x) λ i h j v j (y)u i (y)v i (x) dy. j Using (3.27) nd (3.3) we see tht this reduces to the integrl eqution: h (x) λ K (y, x)h(y)dy. (3.28)

11 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions Note tht this is the homogeneous form of the trnspose of the Fredholm integrl eqution (3.2), i.e. it hs no forcing term on the right hnd side nd the kernel is written K(y, x) rther thn K(x, y). In conclusion for Cse (ii), the integrl eqution (3.2) with seprble kernel of the form (3.3) hs solution if nd only if g (x) is orthogonl to every solution h (x) of the homogeneous eqution (3.28). The generl solution is then f (x) f () (x) + p α j f (j) (x) (3.29) where f () (x) is prticulr solution of (3.2), (3.3) nd the α j re rbitrry constnts. Exmple 6: Consider the integrl eqution f (x) λ j Find (i) the vlues of λ for which it hs unique solution, (ii) the solution in this cse, (iii) the resolvent kernel. sin (x y)f (y)dy + g (x). (3.3) For those vlues of λ for which the solution is not unique, find (iv) condition which g (x) must stisfy in order for solution to exist, (v) the generl solution in this cse. The solution proceeds s follows. Expnd the kernel: f (x) λ nd hence it is cler tht is seprble. Thus nd so write which gives f (x) λ λ sin (x y)f (y)dy + g (x) (sin x cosy cosxsin y)f (y)dy + g (x) K (x, y) sin x cosy cosxsin y sin x f (y)cosydy cosx f (y)sin ydy + g (x), c c 2 Substituting this vlue of f(x) into (3.3) gives c f (y) cos ydy, (3.3) f (y) sin ydy, (3.32) f (x) λ c sin x c 2 cosx + g (x). (3.33) {λ c sin y c 2 cosy + g (y)} cosydy

12 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 2 Defining λc sin y cos ydy λc 2 cos 2 ydy + g g (y)cosydy, g (y)cosydy. nd noting the vlues of the integrls sin y cos ydy sin 2ydy π 2 4 cos 2y, cos 2 ydy ( + cos 2y)dy y + π sin 2y 2 π, yields c 2 πλc 2 + g. Repeting this procedure, putting f(x) from (3.33) into (3.32), gives Observing tht c 2 {λ c sin y c 2 cosy + g (y)}sin ydy λc sin 2 ydy λc 2 cosy sin ydy + sin 2 ydy 2 ( cos 2y)dy 2 g (y)sin ydy. y 2 sin 2y π 2 π, nd writing then we obtin g 2 g (y)sin ydy, c 2 2 πλc + g 2. Thus, there is pir of simultneous equtions for c, c 2 : or in mtrix nottion c + πλc 2 2 g, πλc 2 + c 2 g 2, 2 πλ 2 πλ c c 2 g Cse (i): These equtions hve unique solution provided det πλ 2 πλ, 2 g 2. (3.34) i.e. + 4 π2 λ 2 or λ ± 2i π.

13 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 3 In this cse or Hence, c c 2 c c 2 2 πλ πλ 2 g πλ g g π2 λ 2 πλ πλg g π2 λ 2 πλg 2 + g 2 π g (y)cosydy πλ g (y)sin ydy π2 λ 2 πλ g (y)cos ydy + g (y)sin ydy 2 π cos y πλ sin y + 2 g (y)dy. 4 π2 λ 2 πλ cosy + sin y 2 g 2 or f (x) λ c sin x c 2 cosx + g (x) λ + 4 π2 λ 2 sin x (cosy 2 ) πλ sin y f(x) +g (x) λ + 4 π2 λ 2 ( ) cos x πλ cosy + sin y g (y)dy 2 sin (x y) 2 πλ cos (x y) g (y)dy + g (x). This is the required solution, nd we cn observe tht the resolvent kernel is R (x, y, λ) sin (x y) πλ cos (x y) π2 λ 2 Cse (ii): If det 2 πλ 2 πλ i.e. λ + 2i 2i or π π then there is either no solution or infinitely mny solutions. With this, (3.34) becomes ±i c g. i Solving these equtions using row opertions, R2 R2 ± ir, gives ±i c g c 2 g 2 ± ig or c 2 c ± ic 2 g, g 2 ± ig. g 2

14 MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions 4 The second eqution plces restriction on g (x), which by definition of the g i, is (sin y ± i cosy)g (y)dy. (3.35) This is the condition tht g (x) must stisfy for the integrl eqution to be soluble, i.e. if g(x) does not stisfy this, then (3.3) does not hve solution. Suppose this condition holds then we cn set c 2 to tke ny rbitrry constnt vlue, c 2 α, sy. Thus, c iα + g, nd hence from (3.33), the solution of (3.3) is, when λ ±2i/π, f (x) ± 2i π ( iα + g )sin x α cosx + g (x) 2α 2i (sin x i cosx) ± π π g sin x + g (x) for rbitrry α, with constrint g 2 ig or equivlently (3.35). We rrived t the bove conclusions vi simple row opertions. Fredholm s theorem would hve lso told us the sme informtion regrding constrints on g nd g 2.