Section 3.3: Fredholm Integrl Equtions Suppose tht k : [, b] [, b] R nd g : [, b] R re given functions nd tht we wish to find n f : [, b] R tht stisfies f(x) = g(x) + k(x, y) f(y) dy. () Eqution () is known s Fredholm Integrl Eqution (F.I.E.) or Fredholm Integrl Eqution of the second kind. (F.I.E. s of the first kind hve g(x) =.) The function k is referred to s the integrl kernel. The F.I.E. my be written s fixed point eqution T f = f where the opertor T is defined by T f(x) = g(x) + k(x, y) f(y) dy. Theorem: If k : [, b] [, b] R nd g : [, b] R re continuous nd if sup k(x, y) dy <, there exists unique continuous f : [, b] R tht stisfies the Fredholm integrl eqution. Proof: We will show tht the sup condition implies tht T is contrction mpping in C([, b]) (equipped with the usul uniform/sup norm). Then, since C([, b]) is complete, we cn use the Contrction Mpping Theorem to show tht there exists unique fixed point f C([, b]). Note tht T f T f 2 = sup T f (x) T f 2 (x) x b x b = sup k(x, y)(f (y) f 2 (y)) dy So, T is contrction mpping. sup k(x, y) (f (y) f 2 (y)) dy f f 2 sup k(x, y) dy x b } {{ } <
By the Contrction Mpping Theorem, the eqution T f = f, nd therefore the F.I.E., hs unique solution in C([, b]). We now know tht, if the conditions of the previous theorem re stisfied, we my solve (??) by choosing ny f = C([, b]) nd computing f = lim n T n f. The Fredholm Integrl Opertor, denoted by K, is defined s on functions f C([, b]) s Kf := k(x, y) f(y) dy where k is n F.I.E. kernel. Note tht K is liner opertor. The F.I.E. is then written f = g + Kf which cn lso be written T f = g + Kf using the fixed point eqution T f = f. Note tht T f = g + Kf T 2 f = T (T f ) = T (g + Kf ) = g + K(g + Kf ) = g + Kg + K 2 f T 3 f = T (T 2 f ) = g + Kg + K 2 g + K 3 f. T n f = g + Kg + K 2 g 2 + + K n g n + K n f On HW 6, we will see tht lim n K n f =. Thus f = lim n T n f = K n g. n=
Exmple: Solve the Fredholm Integrl Eqution Note tht f(x) = + x f(y) dy. sup k(x, y) dy = sup x dy =. x We need this strictly less thn in order to use our Theorem from pge. To this end, we will bck off of little bit nd consider solving. for < α <. Note tht now f(x) = + sup k(x, y) dy = sup x dy = α 2 <. x α x f(y) dy (2) Furthermore g(x) = nd k(x, y) = x re continuous functions on [, ] so ll of the conditions of the Theorem on pge re stisfied. So, we my strt with ny f C([, ]) nd repetedly pply T where T f(x) = + x f(y) dy. Let f (x) =. Then f (x) = T f (x) = + x f (y) dy = + x dy = + αx, nd nd f 2 (x) = T f (x) = + x f (y) dy = + x( + αy) dy = + x [ α + 2 α3] f 3 (x) = T f 2 (x) = + x f 2(y) dy = + x[ + y(α + 2 α3 )] dy Continuing, we get Therefore, = + x [ α + 2 α3 + 2 2 α 5]. [ f n (x) = + x α + 2 α3 + 2 2 α5 + + ] α2n +. 2 n f(x) = lim n f n (x) = + x n= 2 n α 2n+ = + xα ( ) α 2 n n= 2 = + xα α 2 /2 = + 2α 2 α 2 x
It is esy to check tht this stisfies the given F.I.E. Note tht the sum in tht second to lst line is still convergent for α ( 2, 2) nd furthermore tht the solution stisfies (2) for ny α ± 2! Exmple: Solve the Fredholm Integrl Eqution Note first tht f(x) = sin x + π/2 sin x cos y f(y) dy. π/2 sup k(x, y) dy = sup sin x cos y dy x π/2 π/2 = sup sin x cos y dy = sup sin x = sin(π/2) = x π/2 x π/2 However, in light of the comments t the end of the previous exmple, we re going to try to leve the π/2 in plce. Let f (x) =. Then f (x) = sin x + π/2 sin x cos y f (y) dy = sin x + π/2 sin x cos y dy = sin x + sin x π/2 cos y dy = sin x + sin x = 2 sin x. Now, f 2 (x) = sin x + π/2 sin x cos y f (y) dy = sin x + π/2 sin x cos y 2 sin y dy = sin x + 2 sin x π/2 sin y cos y dy }{{}}{{} u du = sin x + 2 sin x 2 sin2 y π/2 = sin x + 2 sin x 2 = 2 sin x. So, we hve lredy reched our fixed point! Tht is, f n (x) = 2 sin x for n =, 2,.... Thus, we hve f(x) = lim n f n (x) = 2 sin x.
It is esy to see/verify tht this stisfies the given F.I.E. Section 3.4: Boundry Vlue Problems In this section we wish to find solutions to the boundry vlue problem (BVP) given by u (x) = q(x)u(x), < x < u() = u, u() = u When q(x) is constnt, the solution is esy. Recll tht for second order differentil eqution of the form u (x) + bu (x) + cu(x) = one first finds roots r nd r 2 for the uxiliry eqution u 2 + bu + c =. Then If r nd r 2 re rel nd distinct, the solution hs the form u(x) = c e rx + c2e r2x. If the roots re rel nd repeted (r = r 2 = r), the solution hs the form u(x) = c e rx + c 2 xe rx. If the roots re complex (r = + ib, r 2 = ib), the solution hs the form u(x) = c e x cos(bx) + c 2 e x sin(bx). Non-constnt q(x) is more difficult nd is the point of this Section. To solve u (x) = q(x)u(x), < x < u() = u, u() = u
we begin by zeroing out the boundry conditions nd considering the function v(x) := u(x) u + (u u )x. Note tht v (x) = u (x) nd tht q(x)u(x) = q(x)v(x) + q(x)[u + (u u )x]. Our new BVP is given by v (x) = q(x)v(x) + q(x)[u + (u u )x] v() =, v() = On our rod to solution, we first consider something simpler. Theorem: Let f : [, ] R be continuous. The unique solution of the BVP v (x) = f(x), v() =, v() = is given by where g(x, y) = v(x) = g(x, y) f(y) dy { x( y), x y y( x), y x. Proof: Note tht So, which v(x) = x v (s) = f(s) x v (y) = v (y) dy = x f(s) ds dy + c x + c 2. v (s) ds = f(s) ds f(s) ds + c f(s) ds dy + c x
Integrting by prts with u = f(s) ds nd dv = dy (so du = f(y) dy nd v = y) we get for x [, ]. v(x) = { [y Now the boundry conditions give Thus, we hve tht f(s) ds]y=x y= x yf(y) dy} + c x + c 2 = [x x f(s) ds x yf(y) dy] + c x + c 2 = [x x f(y) dy x yf(y) dy] + c x + c 2 = [ x x f(y) dy x yf(y) dy] + c x + c 2 v() = c 2 = v() = yf(y) dy + c = c = yf(y) dy v(x) = x x f(y) dy + x yf(y) dy yf(y) dy = x y( x)f(y) dy + x x( y)f(y) dy = g(x, y)f(y) dy