MATH 551 HOMEWORK II SOLUTIONS Graded problems: 1,3,4,6,7. Solutions to ungraded problems abridged. Let D d dx and xf, gy ş b fpxqgpxqdx. Integration by parts is then expressed as a xdf, gy fg b a xf, Dgy. 1. Problem 1 (a) We have the following operator and domain. L D 2 3yp0q ` y 1 p0q 0 S yp1q ` 2y 1 p1q 0 We compute the adjoint. xly, vy xd 2 y, vy y 1 v 1 0 xdy, Dvy ry 1 v yv 1 s 1 0 ` xy, D 2 vy Since L L, L is formally self adjoint. The boundary conditions are 0 ry 1 p1qvp1q yp1qv 1 p1qs ry 1 p0qvp0q yp0qv 1 p0qs rvp1q ` 2v 1 p1qsy 1 p1q ` r3vp0q ` v 1 p0qsyp0q The domain of the adjoint is then vp1q ` 2v S 1 p1q 0 3vp0q ` v 1 p0q 0 Since S S, we have that L is self adjoint. (b) We have the following operator and domain $ & yp0q 7y 1 p0q 0 L D 3 ` D 2 D S yp1q 0 % y 1 p1q 4y 2 p1q 0 We compute the adjoint of the summands, citing Part (a) for D 2. xdy, vy yv 1 0 xy, Dvy yv 1 0 ` xy, Dvy To compute the adjoint of D 3 we use the first steps of Part (a): xd 3 y, vy ry 2 v y 1 v 1 s 1 0 ` xdy, D 2 vy ry 2 v y 1 v 1 ` yv 2 s 1 0 ` xy, D 3 vy 1
2 MATH 551 HOMEWORK II SOLUTIONS Combining these yields xly, vy xpd 3 ` D 2 Dqy, vy xd 3 y, vy ` xd 2 y, vy xdy, vy ry 2 v y 1 v 1 ` yv 2 ` y 1 v yv 1 yvs 1 0 xy, D 3 vy ` xy, D 2 vy ` xy, Dvy ry 2 v y 1 v 1 ` yv 2 ` y 1 v yv 1 yvs 1 0 ` xy, p D 3 ` D 2 ` Dqvy Thus L D 3 ` D 2 ` D. The boundary conditions are 0 y 2 p1qvp1q y 1 p1qv 1 p1q ` y 1 p1qvp1q y 2 p1qr5vp1q 4v 1 p1qs 0 y 2 p0qvp0q y 1 p0qv 1 p0q ` yp0qv 2 p0q ` y 1 p0qvp0q yp0qv 1 p0q yp0qvp0q y 2 p0qvp0q ` y 1 p0qr7v 2 p0q 8v 1 p0q 6vp0qs The domain of the adjoint is then $ & vp0q 0 S 8v % 1 p0q 7v 2 p0q 0 5vp1q 4v 1 p1q 0 Neither operator is self adjoint since L L and S S. (c) We have the following operator and domain. α 1 yp0q ` α 2 y 1 p0q 0 L DpD ` q S β 1 yp1q ` β 2 y 1 p1q 0 It is straightforward to see that We then compute xqy, vy xy, qvy. xdpdy, vy ppxqry 1 vs 1 0 xpdy, Dvy ppxqry 1 vs 1 0 xdy, pdvy ppxqry 1 v yv 1 s 1 0 ` xy, DpDvy Summing yields L DpD ` q. For nonzero p, the boundary conditions are 0 y 1 p1qvp1q yp1qv 1 p1q y 1 p0qvp0q ` yp0qv 1 p0q y 1 p1qrvp1q ` β2 β 1 v 1 p1qs y 1 p0qrvp0q ` α2 α 1 v 1 p0qs The domain of the adjoint is then α 1 vp0q ` α 2 v S 1 p0q 0 β 1 vp1q ` β 2 v 1 p1q 0 When p, q are real and hence p p, q q, the formal operator is self adjoint. When additionally p 0, we have S S, and hence that L is self adjoint.
MATH 551 HOMEWORK II SOLUTIONS 3 2. Problem 2 (a) We have the following boundary value problem. u 2 pxq 9e 4x up0q 5 up1q 7 We have that u B satisfies u 2 B 0 and hence u B c 1 x ` c 2. Imposing boundary conditions yields the solution u B 12x 5. We have that u F satisfies u 2 F 9e4x with homogeneous boundary conditions. We find the eigenvalues and their corresponding eigenfunctions to be λ n pnπq 2 φ n pxq sinpnπxq, yielding the eigenfunction expansion 8ÿ x9e 4x, sinpnπxqy u F a n φ n a n λ n xsinpnπxq, sinpnπxqy 18rp 1qn e 4 1s nπp16 ` pnπq 2. n 1 Alternatively, we can solve directly to find u F 9 16 e4x ` b 1 x ` b 2. The boundary conditions then induce the system 9 16 ` b 2 0 9 16 e4 ` b 1 ` b 2 0 This yields the following solution. u F 9 16 pe4x ` p1 e 4 qx 1q Summing gives us the solution to the boundary value problem u 9 16 e4x ` r12 ` 9 16 p1 e4 qsx 89 16 (b) We have the following boundary value problem u 2 ` 8u 1 ` 12u 9 u 1 p0q ` 4up0q 5 u 1 pπq ` 4u 1 pπq 3 Letting wpxq e 4x upxq, we have a Sturm-Liouville problem for w: w 2 pxq 4wpxq e 4x pu 2 ` 8u 1 ` 12uq 9e 4x with the following boundary conditions. w 1 p0q u 1 p0q ` 4up0q 5 w 1 p1q e 4π ru 1 p1q ` 4up1qs 3e 4π The homogeneous equation has general solution w B c 1 e 2x ` c 2 e 2x. Applying the boundary conditions forces the solution w B pxq 2pe4π ` 1q 3e 6π 5 re2x ` e 4π e 2x s. We want w F satisfying w 2 F w F 9e 4x with homogeneous boundary conditions. We find the eigenvalues and their corresponding eigenfunctions to be λ n 4 ` n 2 n 0 φ n cospnxq, yielding an eigenfunction expansion 8ÿ 1 p 1q n e 4π w F a n φ n a n πpn 2 ` 16qpn ` 4q 2.
4 MATH 551 HOMEWORK II SOLUTIONS (a) By direct computation: 3. Problem 3 Lpe rx q pe rx q 2 ` 6pe rx q 1 ` 9e rx r 2 e rx ` 6re rx ` 9e rx pr ` 3q 2 e rx (b) Letting r 3, Lpe 3x q 0. Hence Ce 3x is a solution to Ly 0. (c) By the commuting of derivatives: (d) By direct computation: B r Lpzq B r pb 2 xz ` 6B x z ` 9zq pb 2 x ` 6B x ` 9qB r z LpB r zq LpB r pe rx qq B r Lpe rx q B r rpr ` 3q 2 e rx s pr ` 3qr2 ` xpr ` 3qse rx (e) Letting r 3 we see that Lpxe 3x q 0, giving us solutions y Cxe 3x. 4. Problem 4 (a) Let ϕ uv 3 u 1 v 2 ` u 2 v 1 u 3 v. Then ϕ 1 u 1 v 3 ` uv p4q u 2 v 2 u 1 v 3 ` u 3 v 1 ` u 2 v 2 u p4q v u 3 v 1 uv p4q vu p4q ulv vlu Therefore ulv vlu is an exact differential. (b) By part (a) and the FTC: ş 1 0 pulv vluqdx ş 1 0 ϕ1 dx ϕ 1 0 ruv 3 u 1 v 2 ` u 2 v 1 u 3 vs 1 0 (c) Since upbq 0 vpbq for b 0, 1, we are reduced to ϕ 1 0 ru 2 v 1 u 1 v 2 s 1 0. By other boundary conditions, ϕp1q 0 ϕp0q, and thus ş 1 pulv vluqdx 0. 0
MATH 551 HOMEWORK II SOLUTIONS 5 5. Problem 5 (b) Variation of parameters yields upxq ş 1 0 fpx 0qGpx, x 0 qdx 0 with x x ă x 0 Gpx, x 0 q x 0 x ą x 0 (c) We must ensure that Gpx, x 0 q satisfies G xx px, x 0 q δpx x 0 q Gp0, x 0 q 0 G x pl, x 0 q 0 Enforcing the boundary and jump conditions yields x x ă x 0 Gpx, x 0 q x 0 x ą x 0 (d) As usual upxq ş 1 0 fpx 0qGpx, x 0 qdx 0, except we now express Gpx, x 0 q as Gpx, x 0 q 2 8ÿ sinpnπx{lq sinpnπx 0 {Lq L pnπ{lq 2 n 1 6. Problem 6 (a) We have the following boundary value problem u 2 pxq fpxq up0q A u 1 plq B By 5(c), the homogenoeus BC solution is u F pxq ş L 0 fpx 0qGpx, x 0 qdx with x x ă x 0 Gpx, x 0 q x 0 x ą x 0 This yields the following conditions Gp0, x 0 q 0 G x p0, x 0 q 1 GpL, x 0 q x 0 G x pl, x 0 q 0 We find u B satisfying u u F ` u B in two ways. By Green s Formula: ş L 0 rupxqg xxpx, x 0 q u 2 pxqgpx, x 0 qsdx rug x px, x 0 q Gpx, x 0 qu 1 pxqs L 0 ş L 0 upxqδpx, x 0qdx şl 0 fpxqgpx, x 0qdx up0qg x p0, x 0 q GpL, x 0 qu 1 plq upxq u F pxq up0q ` xu 1 plq upxq u F pxq ` A ` BX Alternatively, u B solves u 2 B 0 with given boundary conditions, forcing u B pxq Ax ` B.
6 MATH 551 HOMEWORK II SOLUTIONS 7. Problem 7 (a) We have the boundary value problem u 2 ` u sin x up0q upπq 0. The general homogeneous solution is upxq c 1 sin x ` c 2 cos x. The boundary conditions force c 2 0, yielding φ h sin x. Then ş π 0 φ hpxqfpxqdx ş π 0 sin2 xdx ą 0. By the Fredholm alternative, the nonhomogeneous problem has no solutions. (c) We have the boundary value problem u 2 u sin x up0q upπq 0. The general homogeneous solution is upxq c 1 e x ` c 2 e x. The boundary conditions force c 1 c 2 0, yielding φ h 0. By the Fredholm alternative, there exists a unique nonhomogeneous solution.