MATH 412 Fourier Series and PDE- Spring 2010 SOLUTIONS to HOMEWORK 6
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1 MATH 412 Fourier Series and PDE- Spring 21 SOLUTIONS to HOMEWORK 6 Problem 1. Solve the following problem u t u xx =1+xcos t <x<1, t> u x (,t)=u x (1,t)=sint t u(x, ) = 1 + cos(2πx) <x<1, t>. Solution: We look for the solution u of the form u = v + w where w solves the Neumann boundary conditions. we take w(x, t) =x sin t. Then v satisfies v t v xx =1+xcos t x cos t =1 <x<1, t> v x (,t)=v x (1,t)= t v(x, ) = 1 + cos(2πx) <x<1, t>. We look for eigenvalues and eigenfunctions of the problem X + λx =, X () = X (1) =. <x<1 The eigenvalues and the corresponding eigenfunctions of this problem are λ n =() 2, X n (x) =cosx, n. Then we look for the solution u in the form of an infinite series v(x, t) = T n (t)x n (x) = T n (t)cosx. n n Differentiating and substituting into the equation we find that v t v xx = n [T n (t)+()2 T n (t)] cos x =1 (1) and at t = 1+cos2πx = v(x, ) = n T n ()] cos x. (2) The equation (1) gives T n(t)+() 2 T n (t) = 1 n = n = (3) 1
2 and (2) gives T n () = 1 n =and n =2 otherwise. From (3) and (4), if n =,thent (t) =1implying that T (t) =t +1.Ifn 1, then T n (t) =A n e n2 π 2t which, in view of (4), is equal to T 2 (t) =e 4π2t and T n (t) =for n 1 and different than 2. So, and v(x, t) =1+t + e 4π2t cos(2πx), u(x, t) =1+t + e 4π2t cos(2πx)+x sin t. Problem 2. Solve the following problem u tt c 2 u xx = <x<1, t> u(,t)=1,u(1,t)=2π t u(x, ) = x + π, u t (x, ) = x 1. Solution: Write a solution u as the sum u = v + w where w solves the Dirichlet boundary conditions u(,t)=1,u(1,t)=2π, t. wetake w(x, t) =1+x[2π 1]. Since v = u w, we have v(x, ) = (x+π) x(2π 1) 1 = 2(1 π)(x 1/2) and v t (x, ) = so that v solves the following wave equation with homogenoeous Dirichlet boundary conditions, v tt c 2 v xx =, v(,t)=v(1,t)=, t <x<1, t>, v(x, ) = 2(1 π)(x 1/2), v t (x, ) =, x 1. We seek a solution v using separation of variables. Take v(x) =X(x)T (t). The eigenvalue problem for X is as follows, (4) X + λx =, X() = X(1) =. <x<1 The eigenvalues and the corresponding eigenfunctions are as follows, For λ = λ n, T is a solution of λ n =() 2, X n (x) =sinx, n 1. T n + λ n c 2 T n = T n (t) =A n cos(ct)+b n sin(ct). 2
3 The proposed solution is an infinite series v(x, t) = n 1[A n cos(ct)+b n sin(ct)] sin x. At t =, 2(1 π)(x 1/2) = v(x, ) = n 1 A n sin x, (5) and since v t (x, t) = n 1 c[ A n sin(ct)+b n cos(ct)] sin x, v t (x, ) = n 1 cb n sin x. (6) From (6), From (5), A n =4(1 π) So, =4(1 π) B n =, n 1. ( x 1 ) sin(x) dx 2 x sin(x) dx 2(1 π) 4(1 π) [ ] 1 4(1 π) = x cos x + 4(1 π)( 1)n = +, n is odd = 4(1 π), n is even. u(x, t) =1+x(2π 1) 2(1 π) [( 1) n 1] = 4(1 π) π sin(x) dx 2(1 π) [ ] 1 cos x dx + cos x 2(1 π) [( 1) n +1] n 1 cos(2ct)sin2x. 2n Problem 3. Find the solution u of the reduced Helmholtz equation, u ku = <x<,y<π u(,y)=1,u(π, y) = y π u(x, ) = u(x, π) = x π. where k is a positive constant. Solution: Using separation of variables write u(x, y) =X(x)Y (y). Then X X = Y Y + k. 3
4 Since the left-hand side depends on x and the right-hand side, both sides are equal to some constant λ. wegettwoequations X (λ + k)λx = and Y + λy =. The boundary condition along the sides [, 1] } and [, 1] π} imply that Y () = Y (π) =.weconsidertheeigenvalueproblem Y + λy =, Y () + Y (π) =. The eigenvalues and the eigenfunctions are <y<π (7) For λ n = n 2,weconsider The general solution is given by λ n = n 2, Y n (y) =sinnx, n 1. X n (n 2 + k)x n =. (8) X n (x) =A n cosh n 2 + kx + B n sinh n 2 + kx. At x =,wehavex(π, y) =(recall that u(π, y) =)sothat A n cosh n 2 + kπ + B n sinh n 2 + kπ =, i.e., A n = B n tanh n 2 + kπ. u(x, y) = n 1 B n [ tanh n 2 + kπ cosh n 2 + kx +sinh n 2 + kx]sinny. To find B n s we use the inhomogeneous boundary condition at x =, 1=u(,y)= n 1( B n )tanh n 2 + kπ sin ny from which B n tanh n 2 + kπ = 2 π u(x, y) π sin ny dy = 2, n is even [1 cos ] = n is odd. = X 4[tanh p (2n 1) 2 + kπ cosh p (2n 1) 2 + kx sinh p (2n 1) 2 + kx] (2n 1)π tanh p sin(2n 1)y. (2n 1) n kπ 4, 4
5 Problem 4. Consider the Laplace equation u = in the domain < x,y < πwith the boundary condition u y(x, π) =x 2 a, u y(x, ) = x π u x(,y)=u x(π, y) = y π. Find all values for which the problem is solvable. Solve the problem for these values of a. Solution: If nu := u u =(u x,u y) (n 1,n 2), thenalong[, 1] }, wehave (u x,u y)=(u x, ) and (n 1,n 2)=(, 1) so that nu =.Alongπ} [, 1], (u x,u y)= (,u y) and (n 1,n 2)=(1, ), and nu =. Along [, 1] π}, (u x,u y)=(u x,x 2 a) and (n 1,n 2)=(, 1), and nu = x 2 a. Finally,Along} [, 1], (u x,u y)=(,u y) and (n 1,n 2)=( 1, ) so that nu =.Consequently, Z «π nuds= (x 2 3 a) dx = 3 aπ. R Since the necessary condition for the existence of solution of the Neumann problem is R R nu ds=,wherer = (x, y) <x,y<π}, itfollowsthata = π2 /3. We look for the product solutions u(x, y) =X(x)Y (y). Then So, X X = Y Y = λ. X + λx = and Y λy = (9) Note that along the sides } [, 1] and π} [, 1], u satisfies the homogoneous boundary conditions. This implies that X () = X (π) =. The eigenvalues and eigenfunctions of the problem X + λx =, <x<π X () = X (π) =. are λ n = n 2, X n(x) =cosnx, n. With λ = λ n = n 2,thegeneralsolutionforY in (9) is given by (1) Y n(y) =A n cosh ny + B n sinh ny, n. Along the side [, 1] }, u y(x, ) = so that Y n() =. Since Y n(y) =na n sinh ny + nb n cosh ny, =Y n() = nb n, n 1. Y n(y) =A n cosh ny for n. Then the product solutions are u n(x, y) =A n cosh ny cos nx and u(x, y) = X n A n cosh ny cos nx. Along the side [,π] π}, u y(x, π) =x 2 π 2 /3. x 2 π 2 /3=u y(x, π) = X n 1 na n sinh cos nx. 5
6 This gives for n 1, na n sinh = 2 π (x 2 π 2 /3) cos nx dx = 2 π = 2 ˆx2 sin nx π 4 = 4 n 2 π [x cos nx]π 4 n 2 π x sin nx dx x 2 cos nx dx 2π 3 cos nx dx = 4 n 2 ( 1)n cos nx dx u(x, y) =A +4 X ( 1) n cosh ny cos nx. n 2 n 1 Problem 5. Let D = (x, y) <x,y<π} and let D be the boundary of D. (a) Assume that v satisfies v xx + v yy + xv x + yv y > in D. Showthatv has no local maximum in D. (b) Consider the problem u xx + u yy + xu x + yu y = u(x, y) =f(x, y) (x, y) D (x, y) D, (11) where f is a given continuous function. Show that if u is a solution, then the maximum is achieved on the boundary D. Hint: Use the function v(x, y) = u(x, y)+εx 2. (c) Show that the problem (11) has at most one solution. Solution: (a) Assume that v has a local maximum at (x,y ) D. Then v(x,y )= (, ) and v(x,y ). Then u xx(x,y )+u yy(x,y )+x u x(x,y )+y u x(x,y )=u xx(x,y )+u yy(x,y ) contradicting v xx + v yy + xv x + yv y > in D. v doesn t have a local maximum in D. (b) Since u is continuous in D it attains its maximum in D. Consider v(x, y) = u(x, y)+εx 2.Then v + xv x + yv y = u + xu x + yu y + ε(2 + x 2 )=ε(2 + x 2 ) > So v attains its maximum at some boundary point. Let M =maxu(x, y) (x, y) D}. Thenforany(x, y) D, v(x, y) =u(x, y)+εx 2 M + επ 2. Since v takes its maximum at the boundary point, So, v(x, y) M + επ 2, for all (x, y) D. u(x, y) =v(x, y) εx 2 v(x, y) M + επ 2, for all (x, y) D. Taking ε +,wegetv(x, y) M for all or all (x, y) D showing the claim. Assume that there are two solutions u and v of (11). Set w = u v. Thenw solves w xx + w yy + xw x + yw y =, w(x, y) =, (x, y) D (x, y) D, By (b), w(x, y). Replace w by w. Then w satisfies the same equation (12). Then w. Consequently,w, i.e.,u 1 u 2. (12) 6
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