AMATH 731: Applied Functional Analysis Fall Some basics of integral equations

Transcription

1 AMATH 731: Applied Functionl Anlysis Fll Introduction Some bsics of integrl equtions An integrl eqution is n eqution in which the unknown function u(t) ppers under n integrl sign, e.g., K(t, s)u(s) ds. (1) (There re lso integrodifferentil equtions, the function ppers both under n integrl sign nd in derivtive.) The function K(t, s) is known s the kernel of the integrl eqution. The mthemticl representtion of mny problems in science or engineering ppers in nturl wy in terms of integrl equtions. For exmple, it is conceivble tht prticulr stte vrible u(t) t ny time t depends upon its vlue for ll previous times, i.e., u(t τ) for τ > 0. In such cse we my hve to sum over ll such vlues (or functions of such vlues), which implies tht n integrtion over u(t τ) must be mde. The result is n integrl eqution. For more detils on these nd other pplictions, the reder is referred to the book, Introduction to Integrl Equtions with Applictions, by A.B. Jerri (Mrcel Dekker, 1985): popultion dynmics nd surges in birth rtes, mortlity of equipment nd the rte of replcement, biologicl species living together, the torsion of wire or rod, the control of rotting shft, the propgtion of nerve impulse, the smoke filtrtion in cigrette, the chnce of crossing dense trffic, the shpe of hnging chin, the deflection of rotting rod, shpe of wire tht llows bed to descent on it with predetermined time (Abel s problem). Problems tht re nturlly formulted in terms of differentil equtions, my be reformulted in terms of integrl equtions. This includes initil vlue problems s well s boundry vlue problems. As n exmple, you will recll tht the initil vlue problem is equivlent to the integrl eqution x (t) = f(x(t), t), x(t 0 ) = x 0, (2) x(t) = x 0 + f(x(s), s) ds. t 0 (3) Becuse of the dditive nture of the integrl, such integrl representtions lend themselves much better to nlysis, i.e., bounds nd estimtes. 1

2 2 Clssifiction The most generl liner integrl eqution in u(t) cn be presented s follows, h(t) (t) When b(t) = t, this eqution is clled Volterr integrl eqution, h(t) When h = 0, it is clled Volterr eqution of the first kind, f(t) + When h(t) = 1, it is clled Volterr eqution of the second kind, K(t, s)u(s) ds. (4) K(t, s)u(s) ds. (5) K(t, s)u(s) ds = 0. (6) K(t, s)u(s) ds. (7) When b(t) = b, constnt, Eq. (4) is clled Fredholm integrl eqution, h(t) When h = 0, it is clled Fredholm eqution of the first kind, f(t) + When h(t) = 1, it is clled Fredholm eqution of the second kind, K(t, s)u(s) ds. (8) K(t, s)u(s) ds = 0. (9) K(t, s)u(s) ds. (10) Note: In mny presenttions, n dditionl constnt λ ppers in front of the integrl, e.g., λ becuse of the ppernce of such constnt in mny pplictions. K(t, s)u(s) ds, (11) Let us rewrite the Fredholm eqution of the first kind, Eq. (9) s follows: f(t) = H(t, s)u(s) ds. (12) For specil cses of the intervl of integrtion [, b], nd the kernel H(t, s), the function f(t) my be seen to be prticulr integrl trnsform of the function u(s). For exmple, when = 0, b = nd H(t, s) = e s, then f(t) is the Lplce trnsform of u(s). This is lso the form of the Green s function solution to boundry vlue problem (BVP) for liner second-order inhomogeneous DE (see below). 2

3 The Volterr nd Fredholm integrl equtions obviously look lmost like, with the exception of the upper bounds in their integrls. However, different methods of solution re required. Initil vlue problems cn be represented s Volterr integrl equtions, boundry vlue problems s Fredholm integrl equtions. We mention here tht nonliner counterprts of the bove equtions exist when the term u(s) in the integrls is replced by nonliner function of u, i.e., f(u(s), s). In the liner cse, it is esy to see tht if u 1 nd u 2 re solutions to n integrl eqution, then ny liner combintion c 1 u 1 + c 2 u 2 is lso solution. 3 Formultion of IVPs nd BVPs in terms of integrl equtions In wht follows, we simply stte results without ny derivtion. 3.1 Generl second-order liner initil vlue problem The solution u(t) of the generl initil vlue problem u (t) + A(t)u (t) + B(t)u(t) = g(t), u() = c 1, u () = c 2, (13) my be expressed in terms of Volterr integrl eqution of the second kind, K(t, s) ds, (14) nd f(t) = (t s)g(s) ds + (t )[A()c 1 + c 2 ] (15) K(t, s) = (s t)[b(s) A (s)] A(s). (16) 3.2 Generl second-order liner boundry vlue problem The solution u(t) of the generl boundry vlue problem u (t) + A(t)u (t) + B(t)u(t) = g(t), u() = c 1, u(b) = c 2, (17) my be expressed in terms of Fredholm integrl eqution of the second kind, K(t, s) ds, (18) f(t) = c 1 + (t s)g(s) ds + [ (t ) c 2 c 1 (s ) (b s)g(s) ds ]. (19) nd K(t, s) = [ (t b) (b ) [A(s) ( s)[a (s) B(s)]], t > s, (t ) (b ) [A(s) (b s)[a (s) B(s)]], s < t. ]. (20) 3

4 3.3 Green s functions Not much will be sid here the topic is too vst. Given the following BVP ssocited with nonhomogeneous second-order DE, u (t) + A(t)u (t) + B(t)u(t) = g(t), < t < b, α 1 u() + α 2 u () = 0, β 1 u(b) + β 2 u (b) = 0. (21) The Green s function method of solving (21) consists in expressing the solution in the form u(t) = G(t, s)g(s) ds, (22) G(t, s) is the Green s function ssocited with the BVP. It my be constructed from two linerly independent solutions u 1 (t) nd u 2 (t) of the ssocited homogeneous DE, u (t) + A(t)u (t) + B(t)u(t) = 0. (23) Note tht Eq. (22) is Fredholm integrl eqution of the second kind, cf. Eq. (9). Exmple: Problem Set 2, Question 5, it is stted tht the unique solution to the BVP u = h(t), u(0) = u(1) = 0, (24) is given by u(t) = the Green s function G(t, s) is given by 1 0 G(t, s)h(s) ds, (25) G(t, s) = s(t 1), 0 s t 1, t(s 1), 0 t s 1. (26) Note tht G(t, s) = G(s, t). 4 Liner Volterr equtions In wht follows, we exmine the liner Volterr eqution of the second kind, Method 1: Method of Successive Substitution K(t, s)u(s) ds. (27) From (27), define the integrl opertor T : C[, b] C[, b], s follows: v = T u, v(t) = (Tu)(t) = f(t) + K(t, s)u(s) ds. (28) Clerly solution to (27) is fixed point of T, i.e., Tu = u. We could proceed s ws done in the course notes ( the nonliner problem ws treted) to estblish sufficient conditions for T to be contrctive on some intervl [, b]. However, it is possible to show tht, for sufficiently lrge n, the composition T n is contrction. This cn be done by defining the sequence of functions u 0, u 1 = Tu 0 = Tf nd u n+1 = Tu n, n = 1, 2,. (29) 4

5 It is not too hrd to show tht u n = T n u 0 cn be written s u n = T n (u 0 ) = f(t) + K(t, s)f(s) ds + K n (t, s) is the nth iterted kernel defined s follows, K n+1 (t, s) = It then follows tht for u, v C[, b], (T n u)(t) (T n v)(t) s K 2 (t, s)f(s) ds + + K n (t, s)f(s) ds, (30) K(t, τ)k n (τ, s)dτ, K 1 (t, s) = K(t, s). (31) K n (t, s) u(s) v(s) ds. (32) We ssume tht K(t, s) is continuous on the bounded squre domin D = [, b] [, b]. Therefore it is bounded, i.e., K(t, s) M on D. An induction rgument estblishes tht K n (t, τ) Mn (n 1)! (t τ)n 1, τ t. (33) Using the sup metric on C[, b], we cn then estblish tht for u, v C[, b], d(t n u, T n v) α n d(u, v), (34) α = Mn (b ) n (35) n! For sufficiently lrge n, sy n = p, α p < 1, implying tht T p is contrctive. It then follows tht T p hs unique fixed point u. This function u is lso the unique fixed point of T (exercise) nd therefore stisfies Eq. (27). Method 2: Neumnn series We shll write Eq. (27) in the form (I L)u(t) = f(t), (36) L represents the liner integrl opertor. Formlly, the solution to this eqution would be given by the series [I + L + L 2 + L 3 + ]u(t) (37) This is ctully forml representtion of the series f(t) = f(t) + n=1 K n (t, s)f(s) ds, (38) the K n (t, s) re the iterted kernels defined erlier. It now remins to prove the convergence of this series. With some work, it cn be shown tht [ ] K n K n 1 f(t) k(t) f, (39) (n 1! nd K = [ 1/2 b k(t) = K(t, s) ds] 2 (40) [ K(t, s) 2 ds dt] 1/2. (41) It follows from (39) tht the series in (38) is uniformly convergent. Since every term is continuous function of t, it follows tht the function defined by this sum is continuous. 5

6 5 Liner Fredholm equtions The liner Fredholm eqution of the second kind is given by K(t, s)u(s) ds. (42) Two methods of solving this problem, nmely (i) successive substitution nd (ii) Neumnn series nd their reltionship were discussed erlier in this course (see relevnt hndouts). Becuse of the fixed intervl of integrtion, the restrictions on the liner integrl opertor to ensure convergence re much tighter. (For exmple, we require L < 1 for the convergence of geometric series in this cse, in contrst to the bsolute convergence of the Neumnn series for ny bounded L for the Volterr problem.) 6 Nonliner integrl equtions 6.1 Nonliner Volterr eqution of the second kind The nonliner Volterr eqution of the second kind will hve the form, F(t, s, u(s)) ds. (43) Assume tht f(t) is continuous on [, b] nd F(t, s, u) is continuous in with respect to its three vribles for t b, s t nd c u d. Note tht this cse includes the stndrd integrl eqution solution (3) for the initil vlue problem in Eq. (2). As for the liner cse, we consider the nonliner opertor T given by v(t) = (Tu)(t) = f(t) + F(t, s, u(s)) ds. (44) A fixed point of T corresponds to solution of (43). There re couple of technicl bounds involving, b nd the bounds of f(t) which we omit here. We simply outline the min ides behind estblishing contrctivity of T. We once gin ssume tht F is Lipschitz with respect to u, i.e., F(t, s, u(s))f(t, s, v(s) K u(s) v(s). (45) Then (Tu)(t) (Tv)(t) = K [F(t, s, u(s) F(t, s, v(s)] ds F(t, s, u(s) F(t, s, v(s) ds u(s) v(s) ds. (46) From here, one could proceed to estblish the Lipschitz nture of T with respect to the sup metric, with pproprite constrints on nd b: d (Tu, Tv) K(b )d (u, v). (47) This is quite restrictive result. We must restrict the size of the intervl [, b] so tht K(b ) < 1. However, in simily wy s for liner Volterr equtions, we my estblish tht T n is contrctive for sufficiently lrge n. We sketch the ide below. Consider the inequlity in Eq. (46) for T(Tu) nd T(Tv), 6

7 (T 2 u)(t) (T 2 v)(t) = K [F(t, s, Tu(s) F(t, s, Tv(s)] ds F(t, s, Tu(s) F(t, s, Tv(s) ds K 2 Tu(s) Tv(s) ds s u(w) v(w) dw ds (48) from which it follows tht d (T 2 u, T 2 v) K 2(b )2 d (u, v). (49) 2 Iterting this result (which my lso be proved by induction), d (T n u, T n v) c n d (u, v), (50) c n = Kn (b ) n. (51) n! For sufficiently lrge n, c n < 1, from which it follows tht T n is contrction. This implies tht T n, hence T, hs unique fixed point u, the solution of the integrl eqution (43). 6.2 Nonliner Fredholm eqution of the second kind The nonliner Fredhom eqution of the second kind will hve the form, F(t, s, u(s)) ds. (52) Once gin, we ssume tht f(t) is continuous on [, b] nd F(t, s, u) is continuous in with respect to its three vribles for t b, s t nd c u d. As for the liner cse, we consider the nonliner opertor T given by v(t) = (Tu)(t) = f(t) + F(t, s, u(s)) ds. (53) A fixed point of T corresponds to solution of (52). We hve the sme technicl bounds involving, b nd the bounds of f(t) s for the nonliner Volterr cse, which we omit here. And once gin we ssume tht F is Lipschitz with respect to u, i.e., F(t, s, u(s))f(t, s, v(s) K u(s) v(s). (54) Then (Tu)(t) (Tv)(t) = L [F(t, s, u(s) F(t, s, v(s)] ds F(t, s, u(s) F(t, s, v(s) ds u(s) v(s) ds (55) 7

8 From here, one cn estblish the Lipschitz nture of T with respect to the sup metric, i.e., d (Tu, Tv) K(b )d (u, v). (56) For b such tht K(b ) < 1, it follows tht T is contrctive. Unfortuntely, we my not iterte this procedure s ws done for the nonliner Volterr eqution. This is limittion tht is cused by the itertion over the fixed intervl [, b] s opposed to the intervl [, t] for the Volterr cse. 8