MT Integral equations

Transcription

1 MT58 - Integrl equtions Introduction Integrl equtions occur in vriety of pplictions, often eing otined from differentil eqution. The reson for doing this is tht it my mke solution of the prolem esier or, sometimes, enle us to prove fundmentl results on the eistence nd uniqueness of the solution. Denoting the unknown function y φ we consider liner integrl equtions which involve n integrl of the form K(,s)φ(s)ds or K(,s)φ(s)ds The type with integrtion over fied intervl is clled Fredholm eqution, while if the upper limit is, vrile, it is Volterr eqution. The other fundmentl division of these equtions is into first nd second kinds. For Fredholm eqution these re f () = φ() = f () + λ K(,s)φ(s)ds K(,s)φ(s)ds The corresponding Volterr equtions hve the upper limit replced with. The numericl prmeter λ is introduced in front of the integrl for resons tht will ecome pprent in due course. We shll minly del with equtions of the second kind. Series solutions One firly ovious thing to try for the equtions of the second kind is to mke n epnsion in λ nd hope tht, t lest for smll enough vlues, this might converge. To illustrte the method let us egin with simple Volterr eqution, φ() = + λ φ(s)ds. For smll λ, φ () = is first pproimtion. Insert this in the integrl term to get etter pproimtion φ () = + λ sds = + λ. Agin put this into the integrl to get φ () = + s + λs ds = + λ + 6 λ 3. Continuing this process, we get

2 φ n () = + λ n! λn n nd, s we let n, the series converges to ( λ eλ ). Sustituting into the eqution verifies tht this is the correct solution. So, for this Volterr eqution the technique of epnsion in series gives result which is convergent for ll λ. We now show tht this works for ll Volterr equtions, suject to some firly generl conditions. Suppose tht we look for solution with in some finite intervl [,]nd tht on this intervl f ()is ounded with f () < m. We lso suppose tht, on the intervl [,] [,], K(,s) < M. Then φ () = f () < m φ () = f () + λ K(, s)f (s)ds ( ) < m + λ mm φ () = f () + K(, s)φ () < m + M(m + mm(s ) )ds = m + λ M( ) + Crrying on like this we get φ n () < m + λ M( ) + λ M ( ) n! λ n M n ( ) n λ M ( ) Since the series here is of eponentil type we get convergence for ll vlues of λ. Now see whether the sme thing works for Fredholm equtions. Agin we look t n emple - φ() = + λ φ(s)ds similr to the previous one, ecept for the fied rnge of integrtion. Now we get φ () = nd, continuing in this wy φ () = + λ sds = + λ φ () = + λ s + λ ds = + λ + λ φ n () = + ( λ + λ λ n ).

3 Now we get geometric series which only converges if λ <. In this simple prolem the series cn e summed nd letting n we get the solution λ φ() = + λ ( ). Since we hve mnged to sum the series, we get solution vlid for ll λ. In more generl prolems, the series will not e esily summle nd will only e vlid for restricted rnge of vlues of λ. The series otined in this wy, for either type of eqution is known s the Neumnn series. Emple: Find the first two terms of the Neumnn series for the eqution We get Seprle kernels φ () = sin φ () = sin + λ π/ φ() = sin + λ cos( s)φ(s)ds. π/ π/ cos( s)sinsds = sin + λ sin(( + )s = sin + λ cos(( + ) π ) cos(( ) π + ) + ( ) + sin(( )s)ds Suppose the kernel of Fredholm eqution is product of function of nd function of s, to the eqution is of the form φ() = f () + λ u() v(s)φ(s)ds Clerly the u()cn come outside the integrl nd if we write c = v(s)φ(s)ds then we hve φ() = f () + λcu(). All we need to find is the constnt c nd this cn e done y multiplying the lst eqution y v()nd integrting from to. This gives

4 so c = f ()v()d + λc u()v()d c = λ v()f ()d u()v()d This leds to the solution, so long s the vlue of λ is not such s to mke the denomintor vnish. Emple: Solve The solution is with from the lst two equtions we get So, unless λ = 6 we hve φ() = + λ. 3 s φ(s)ds. φ() = + λ 3 c c = s φ(s)ds. c = 4 5 d + λc d = λc. φ() = + 5 6λ 3 6 λ. The vlue of λ for which the solution reks down is n eigenvlue of the prolem. Now, go ck to the generl form of the eqution nd look wht hppens t this vlue. Define the homogeneous eqution corresponding to our originl eqution y φ() = λ u()v(s)φ(s)ds (ie put f=). Multiplying this y v()nd integrting gives which implies tht v()φ()d = λ u()v()d v(s)φ(s)ds λ v()u()d =.

5 There cn e no solution of the homogeneous eqution unless this condition is stisfied. If it is stisfied then it is esily verified tht φ() = ku() is solution for ritrry k. Going ck to the inhomogeneous eqution, it cn e seen tht if λ is equl to the eigenvlue, nd solution eists, then the condition v()f ()d = must e stisfied. Thus the solution cn only eist for certin functions f. If such solution eists then ny multiple of the solution of the homogeneous eqution cn e dded to it. To summrise, we hve, () λ eigenvlue - unique solution eists, or () λ = eigenvlue - solution my or my not eist nd if it does it is not unique. Homogeneous eqution hs solution, ny multiple of which cn e dded. There re cler prllels etween this nd the mtri eqution X = F + λax where X nd F re column vectors nd A mtri. This is discrete nlogue of the Fredholm eqution. This hs unique solution X = (I λa) X if the inverse eists, which it does ecept for certin vlues of λ (the inverses of the mtri eigenvlues). At these vlues of λ the homogeneous eqution (F=) hs nontrivil solution while the inhomogeneous eqution my or my not hve solution, nd if it does ritrry multiples of the solution of the homogeneous eqution cn e dded to it. Degenerte Kernels The method outlined ove cn e etended to the sitution where K(,y) = u i ()v i (y) over some finite rnge of i. To see this we look t n emple - If we let then the solution is i φ() = + λ (y + y )φ(y)dy. c = c = yφ(y) y φ(y) φ() = + λ(c +c ). If we multiply this y then y nd integrte from to in ech cse we otin

6 c = 3 + λ( 3 c + 4 c ) c = 4 + λ( 4 c + 5 c ) or 3 λ 4 λ 4 λ 5 λ c c = 4 5 This gives c c = 48(λ 5) 6λ / 3 4 8λ + λ 6λ 8(λ 3) / 4 nd hence uniquely defined solution unless λ 8λ + 4 = ie λ = 64 ± 4 4. These two vlues re the eigenvlues for this prolem. If λ tkes one of these vlues then the eqution 3 λ 4 λ c 4 λ 5 λ c = hs non-trivil solution. This gives non-trivil solution to the homogeneous integrl eqution (which my e multiplied y n ritrry constnt). If we consider kernel of the generl form K(,y) = n i= u i ()v i (y) then we cn follow the sme procedure nd will end up with n n n mtri to invert. The determinnt will e n nth degree polynomil in λ with n solutions (possily comple nd possily repeted) which re the eigenvlues. If λ is not n eigenvlue then the mtri cn e inverted nd unique solution found. If it is n eigenvlue then the homogeneous eqution hs non-trivil solution. In this cse, the originl eqution my or my not hve solution. To otin the condition under which solution eists we introduce the trnsposed eqution φ() = f () + K(y,)φ(y)dy where the rguments in K re swpped over. This produces mtri which is the trnspose of the originl nd hs the sme eigenvlues. If ψ() is solution of the

7 homogeneous trnsposed eqution for the eigenvlue λ then if we multiply the orginl eqution y ψ() nd integrte from to we get φ()ψ() = ψ()f () + λ K(,y)ψ()φ(y)ddy. From the definition of ψ() it stisfies ψ() = λ K(t,)ψ(t)dt from which it cn e seen tht the integrl in the lst term ove just gives ψ(y) nd the whole term is φ(y)ψ(y)dy which, of course, just cncels the left hnd side. So, we conclude tht if solution eists we must hve Emple: Solve ψ()f () =. π/ φ() = sin() + λ cos( y)φ(y)dy. The first thing to note is tht using the formul for cos( y) rings this into the form we wnt, nmely With we get or π/ φ() = sin + λ (cos cosy + sin siny)φ(y)dy. c = c = π/ π/ cos()φ() d sin()φ() d c = + λ c π 4 +c c = π 4 + λ c +c π 4

8 λ π 4 λ λ λ π 4 c c =. π 4 The mtri here cn e inverted unless λ = π 4 ±, the result eing 8 c c = 6 8λπ + λ π 4λ. 4λ + 4π + λπ 6 8λπ + λ π 4λ Looking t wht hppens t the eigenvlues, with the + sign we get c = c if we put the right hnd side equl to zero, giving the solution cos + sin (or ny multiple of this) for the homogeneous eqution. This is lso the solution of the homogeneous trnsposed eqution (which is the sme s the originl). For our eqution to hve solution for this eigenvlue we need π/ sin sin + cos = which is not true, so no solution eists. This cn lso e seen from the ove mtri eqution, the two equtions eing incomptile if λ hs this vlue. The condition for the eistence of solution with the inhomogeneous term replced with generl function f() is π/ f () cos + sin =. It cn e seen tht wht this does is ensure tht the vector on the right hnd side of the mtri eqution is such s to e comptile with the left hnd side, the two equtions then eing identicl. A similr nlysis holds for the other eigenvlue. Resolvent kernel Consider the Fredholm eqution in the usul form nd suppose for the moment tht the kernel is degenerte, Then, s we hve seen K(,y) = n i= φ() = f () + λ c i = u i () v i (y). v()φ()d c i u i ().

9 Multiplying the integrl eqution y ech v i () in turn, we get the system of liner equtions, the solution of which is c i = f i + λ f i = v i j ij c j ()f ()d ij = v i ()u j () c i = with the mtri B given y B = (I λa). This produces solution of the form with φ() = f () + λ j ij f j R(,y;λ)f (y)dy R(,y;λ) = ij u i ()v j (y). i,j R is clled the resolvent kernel nd is uniquely defined ecept when λ is n eigenvlue, ie A λi is singulr. If λ is n eigenvlue then the homogeneous solution hs non-trivil solution nd the full eqution my or my not hve solution, s discussed previously. If solution eists it is not unique, since ny multiple of the solution of the homogeneous eqution cn e dded to it. Fredholm theory Fredholm otined generl epression for the resolvent kernel, vlid even if the kernel is not degenerte. We shll not go into the detils, ut the theory sys tht R(,y;λ) = D(,y;λ) d(λ) where the numertor nd denomintor cn oth e epressed s infinite series D(,y;λ) = n= ( ) n D n! n (,y)λ n ( ) n d(λ) = d n= n! n λ n. We hve lredy seen how to construct single Neumnn series for the solution. It, however, is only convergent for sufficiently smll λ, while the series here re convergent for ll vlues. The solution only reks down when d(λ) =, condition which determines the eigenvlues. The functions D n (,y) nd the constnts d n re found from the recurrence reltions

10 D (,y) = K(,y) d = d n = D n (,)d D n (,y) = K(,y)d n n To illustrte how this works we look t the eqution φ() = + λ K(,z)D n (z,y)dz. 3 y φ(y)dy which we hve lredy solved s n emple of n eqution with seprle kernel. Here we get D (,y) = 3 y d = 5 d = 6 D (,y) = 6 3 y ( 3 z )(z 3 y )dz = 6 3 y 6 3 z =. nd ll susequent terms vnish. The resolvent kernel is thus The solution is R(,y;λ) = 3 y 6 λ = 6 3 y 6 λ. 3 y φ() = + λ y dy = + 6λ 3 6 λ 5 6 λ, the sme s found efore y different method. Solution of Volterr eqution y differentition. A Volterr eqution with simple seprle kernel cn e solved y reducing it to differentil eqution. To illustrte this consider the emple φ() = 5 + s φ(s)ds. We cn divide this through y, so tht the integrl term does not depend on, getting φ() = 4 + s φ(s)ds.

11 Differentiting with respect to gives d φ() d = 4 φ() 3 + φ() = which is simple liner differentil eqution. We get d φ() d e 4 4 = 4 3 e 4 4 nd so φ() = 4 +Ce 4 4. This involves n ritrry constnt, wheres Volterr integrl eqution hs unique solution. We cn evlute the constnt y going ck to the integrl eqution. The condition tht φ() = when = tells us nothing, ut if we use the fct tht φ() / s we see tht the solution is Integrl trnsform methods φ() = 4e Recll tht the Lplce trnsform of function f () is defined y Let us now consider n integrl of the form nd look t its LT. %f (p) = f ()e p d. F() = g( y)f (y)dy %F(p) = e p g( y)f (y)dyd. The integrl is over the shded re in the digrm elow, y

12 nd if we chnge the order of integrtion we get Now let u = y nd get %F(p) = y g( y)f (y)e p ddy. %F(p) = g(u)f (y)e pu py dydu = g(u)e pu du f (y)e py dy = %g(p) f % (p). Our originl integrl is clled the convolution of the functions f nd g, so we hve rrived t the conclusion tht the LT of the convolution of two functions is the product of the LT s of the individul functions. This cn sometimes e used to solve n integrl eqution if the integrl term tkes the form of convolution. Emple: Solve Here g() = nd φ() = + ( y)φ(y)dy. %g(p) = e p d = p. Tking the LT trnsform of the integrl eqution we get %φ(p) = p + p % φ(p) leding to %φ(p) = p = p p + φ() = (e e ) = sinh. Ael s Integrl Eqution: This is the following Volterr eqution of the first find - f () = ( φ(s)ds ( < ). Note tht the integrnd is singulr t the upper limit ut tht the integrl eists. Now g() nd %g(p) = e d = e u du = The gmm function which is used here is defined y

13 So, from the convolution theorem we get Γ() = e u u du. %φ(p) = % f It is now convenient to introduce function ψ() such tht φ() = ψ () nd ψ() =. From the properties of Lplce trnsforms we then hve the result % φ)(p) = p % ψ(p), so tht %ψ(p) = % f But, since the LT is the LT / Γ(@) nd so it follows from the convolution theorem tht If we use the result { } ( f (s)ds ψ() = = π sin π@ then we get the solution in the form Numericl methods sin π@ d φ() = π d. ( f (s)ds. Just s is the cse for differentil equtions, there re mny integrl equtions for which simple nlytic solution is not possile. In this cse numericl techniques cn e used. It is quite esy to see how numericl method cn e implemented. If we let φ i e the vlue of the solution t series of points i spced out long the intervl of interest, then, supposing we re deling with Fredholm eqution of the second kind, we hve φ i = φ( i ) = f ( i ) + K( i,s)φ(s)ds. If we now use some numericl pproimtion for the integrl, using the vlues t our finite set of points φ i = f i + c ij φ j. j This is now mtri eqution for the set of function vlues nd we cn use numericl mtri inversion pckge to get the required result. To illustrte this, nd see how well it works we consider the eqution

14 π/ φ() = sin() + cos( y)φ(y)dy. This, of course, is seprle eqution nd we hve lredy found its solution with n ritrry multiplier λ in front of the integrl. We tke it so tht we cn compre our eventul pproimtion with the ect nswer. If we divide the intervl, π into su-intervls nd let φ i = φ( iπ ) f i = sin( iπ ) K ij = cos( iπ jπ ) for i =,...,, then with Simpson s rule used to pproimte the integrl we get φ i = f i + π 6 (K i φ + 4K i φ + K i φ + 4K i3 φ K i φ ). To find the unknown function vlues we hve to invert n mtri which is redily done using numericl mtri inversion routine..5 R i φ() π i.

15 The ove grph shows the solution otined in this wy. On the scle of this grph, the difference etween this numericl solution nd the true solution is lmost indistinguishle, even though we hve only tken intervls. The mimum error is out.4 4. In the cse of Volterr eqution which we wnt to solve over some rnge we cn use the sme procedure. Suppose, for emple, we tke the eqution φ() = + ( y)φ(y)dy (whose solution we hve lredy seen to e sinh()) nd we wnt to solve it over the rnge [,]. Then we write it s φ() = + K(,y)φ(y)dy K(,y) = y > y otherwise nd tret it in the sme wy s the Fredholm eqution. With [,] divided into intervls we get the following grph of the solution. 4 3 φ() R i.5.5 i

16 Agin, with only smll numer of points we get good pproimtion, the mimum error over this rnge eing round.. Further reding Integrl Equtions - A Short Course - Ll, G. Chmers Integrl equtions B L Moiseiwitsch