SECTION 2.6 THE SECOND ALTERNATIVE

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1 54 SECTION 2.6 THE SECOND ALTERNATIVE We ow discuss the probles where the Secod Alterative holds. The suppositio is that there is a otrivial solutio for L(y) =, B (y) = B 2 (y) =. The Fredhol Theores assure us that, if f is cotiuous, the there is a solutio for L(y) = f, with B (y) = B 2 (y) = provided < f, w > = f(x) w(x) dx = for all solutios w of the equatio L*(w) =, B *(w) = B 2 *(w) =. As before, we will costruct Gree's fuctios G such that, i case f satisfies the above requireet, the y(x) = G(x,t) f(t) dt provides a solutio for L(y) = f. I this secod alterative, there ay be ay solutios for the equatio L(y) = f. Cosequetly, we expect there ay be ay Gree's fuctios. I the techique developed below, G(,t) is always i M= {y: B (y) = B (y) = }. This is ot ecessarily true for Gree's fuctios costructed by other ethods: see for exaple the costructio foud by Do Joes while a graduate research assistat at GEORGIA TECH ad give i a appedix. We agai divide the probles ito three cases accordig to the ature of the boudary coditios. We shall illustrate ethods of costructio. The first case to cosider is where the boudary coditios arise as iitial coditios. This case is ot pertiet for the iitial value proble has a uique solutio. Thus, case oe is always i the first alterative. EXAMPLE:(Secod Alterative,uixed, two poit boudary coditios) Suppose that L(y) = y'' + y' - 2y, B (y) = y() - y'(), ad B 2 (y) = y() - y'(). It is the purpose of this exaple to show that there is o fuctio G such that L(G(,t))(x) = δ(x,t). Note that L*(z) = z'' - z' -2z ad M* = {y: 2z() = z'(), 2z() = z'()}. Notrivial fuctios i the ullspace of {L*, B *, B 2 *} are ultiples of e 2t. Hece, we are i the secod alterative. The Fredhol Alterative theore suggests that there will be o fuctio G such that, if t is i (,), the the distributio equatio L(G(,t)) (x) = δ(x,t) holds uless δ(x,t) e 2t dt =.

2 55 Of course, the value of this itegral is ot zero. For this situatio, we ust odify the costructio of the Gree's fuctio. CONSTRUCTION OF G IN THE SECOND ALTERNATIVE, th ORDER Step () Fid the ullspace of { L*, M*} Step (2) Fid a orthooral basis for this ullspace. Call this basis v,v 2,...v, <. Step (3) Costruct u p such that L(u p ) = v p, p =, 2,... Step (4) Costruct G such that L(G(,t))(x) = δ(x,t) - v p (x)v p (t). p= THEOREM If < t <, the there is G(,t) such that L(G(,t))(x) = δ(x,t) - v p (x)v p (t). p= INDICATION OF PROOF. By the Fredhol Alterative Theores, there will be such a fuctio G provided = < δ(x,t) - v p (x)v p (t), w(t) > [δ(x,t) - v p (x)v p (t)] w(t) dt p= p= for all w i the ullspace of L*. This ca be verified by writig w i ters of this orthooral basis, w(x) = α p v p (x), ad evaluatig the dot p= product. HOW TO CONSTRUCT G SUCH THAT L(G(,t))(x) = δ(x,t) - v p (x)v p (t). p= First, fid liearly idepedet solutios {y p } p= of the hoogeeous equatio L(y) =. The, fid solutios {u p } p=, <, for the equatios

3 L(u p )(x) = v p (x). It is ot required that these solutios should satisfy ay special boudary coditios. The proble of fidig G is ow a proble of fidig costats {C p } p= ad {D p } p= such that p= C p y p (x) - v p (t) u p (x) if x < t p= G(x,t) =. D p y p (x) - v p (t) u p (x) if t < x p= p= where {C p } p= ad {D p } p= are deteried by these 2 equatios: 56 (a) B p (G(,t)) =, p=,2,..., (b) = P G(x,t)/ x p x=t+ - P G(x,t)/ x p x=t-, < p < -2, (c) /a (t) = - G(x,t)/ x - x=t+ - - G(x,t)/ x - x=t-. CONTINUATION OF THE PREVIOUS EXAMPLE Recall that L(y) = y'' +y' -2y, B (y) = y() - y'(), ad B 2 (y) = y() - y'(). Liearly idepedet solutios for L(y) = are e -2x ad e x. A oralized basis for the oe-diesioal ullspace of {L*,B *,B 2 *} is αe 2x where α is the positive uber give by α2 = / (e 2x ) 2 dx = 4/(e 4 -). A solutio u for the equatio y'' +y' -2y = αe 2x is u(x) = αe 2x /4. Now, G is give by: G(x,t) = Ae-2x +Be x - α 2 e 2(x+t) /4 if x < t Ce -2x +De x - α 2 e 2(x+t) /4 if t < x. The four costats - A,B,C, ad D - ca be solved by these four equatios: () = B (G(,t)) = G(,t) - G(x,t)/ x x= = 3A + α 2 e t /4 (2) = B2(G(,t)) = G(,t) - G(x,t)/ x x= = 3Ce -2 + a 2 e 2(+t) /4, (3) = G(t +,t) - G(t -,t) = (C-A)e -2t + (D-B)e t, ad (4) = G(x,t)/ x x=t+ - G(x,t)/ x x=t- = -2(C-A)e -2t + (D-B)e t. Upo solvig this syste of four equatios ad four ukows, a ifiity of solutios will be foud deteried by these three equatios: A = -α 2 e 2t /2, C = Ae 4, D-B = e -t /3. EXERCISE: FOR EACH OF THE FOLLOWING, GIVE L*,B *, B 2 *, ad G.

4 (a) L(y) = y'' +y' -2y, B (y) = y()-y'(),b 2 (y) = y() - y'(). (b) L(y) =4y'' - y, B (y) = y() - 2y'(), B 2 (y) = y() - 2y'(). (c) L(y) = y'' - 2y' - 3y, B (y) = 3y() - y'(), B 2 (y) = 3y() - y'(). 57 EXAMPLE(Secod Alterative, ixed, two poit boudary coditios.) Suppose that L(y) = y'', B (y) = y() + y(), B 2 (y) = y'() - y'(). The L*(z) = z'', B *(z) = z()- z(), B 2 *(z) = z'() + z'(). All solutios of {L,B, B 2 } are ultiples of 2x -. A otrivial solutio of [L*,B *, B 2 *} is the costat fuctio. Also, the fuctio V(x) = fors a basis for the ull space of = L*(z) i M*. The fuctio u(x) = x2/2 satisfies L(u) =. Thus G(x,t) = A + Bx - x2/2 if x < t C + Dx - x2/2 if t < x. We have four ukows; we have the followig four equatios: () = G(,t) +G(,t) = A + C + D - /2 (2) = dg(x,t)/dx x= - dg(x,t)/dx x= = B - (D-) (3) = G(t +,t) - G(t -,t) = C-A + (D-B)t (4) = dg(x,t)/dx x=t+ - dg(x,t)/dx x=t- = D-B. As expected, there is a ifiity of solutios to these equatios which ay be foud by choosig D ad the B = D - 2C = -(t+d) + /2 A = C + t. EXERCISE: I Verify that each of these probles is secod alterative ad fid L*,B *, B 2 *,ad G. () L(y) = y'', B (y) = y() - y(), B 2 (y) = y'() - y'(), (2) L(y) = y'' + 9π 2 y, B (y) = y() - y(), B 2 (y) = y'() + y'(), (3) L(y) = y'' + y' - 2y, B (y) = e y() - y(), B 2 (y) = e y'() - y'(). II. Costruct L*, B* ad G for each of the followig L's ad with periodic boudary coditios y() = y(), y'() = y'(): (a) L(y) = y'', (b) L(y) = y'' + π 2 y (c) L(y) = 2y'' + y' - y, NONHOMOGENEOUS BOUNDARY CONDITIONS To solve the equatios L(y) = f, B (y) = α, B 2 (y) = β, first costruct G for the proble L(y) = f, B (y)=, B 2 (y) =. The costruct fuctios z ad z 2 such that B (z ) =, B 2 (z ), ad B (z 2 ), B 2 (z 2 ) =. The solutio for the origial proble is

5 y(x) = G(x,t)f(t) dt + β z (x) B 2 (z ) + α z 2(x) B (z 2 ) 58