Problem (p.613) Determine all solutions, if any, to the boundary value problem. y + 9y = 0; 0 < x < π, y(0) = 0, y (π) = 6,

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1 Problem (p.613) Determine all solutions, if any, to the boundary value problem y + 9y = 0; 0 < x < π, y(0) = 0, y (π) = 6, by first finding a general solution to the differential equation. Solution. The characteristic equation is r = 0 with roots ±3i, so the general solution is y(x) = C 1 cos 3x + C 2 sin 3x with y (x) = 3C 1 sin 3x + 3C 2 cos 3x. Using the boundary conditions, we get C 1 = 0 and 3C 2 = 6, thus y(x) = 2 sin 3x is the only solution of the boundary value problem. 1

2 Problem (p.613) Determine all solutions, if any, to the boundary value problem y 2y + y = 0; 1 < x < 1, y( 1) = 0, y(1) = 2, by first finding a general solution to the differential equation. Solution. The characteristic equation is r 2 2r + 1 = 0 with a double root 1, so the general solution is y(x) = C 1 e x + C 2 xe x. Using the boundary conditions, we get C 1 e 1 C 2 e 1 = 0 and C 1 e + C 2 e = 2, thus C 1 = C 2 = e 1 and y(x) = (1 + x)e x 1 is the only solution of the boundary value problem. 2

3 Problem (p.613) Find the eigenvalues and the corresponding eigenfunctions for the boundary value problem y + λy = 0; 0 < x < π, y (0) = 0, y(π) = 0. Solution. The characteristic equation is r 2 + λ = 0. We will consider three cases: 1. If λ < 0, there are two real roots ± λ and the general solution is y(x) = C 1 e λx + C 2 e λx with y (x) = C 1 λe λx C 2 λe λx. Using boundary conditions, we get C 1 λ C2 λ = 0 and C1 e λπ λπ + C 2 e = 0. The only solution is C 1 = C 2 = 0, so there are no negative eigenvalues. 2. If λ = 0, there is one double root 0 and the general solution is y(x) = C 1 + C 2 x with y (x) = C 2. Using boundary conditions, we get C 2 = 0 and C 1 + C 2 π = 0. The only solution is C 1 = C 2 = 0, so 0 is not an eigenvalue. 3. If λ > 0, there are two complex roots ± λi and the general solution is y(x) = C 1 cos λx + C 2 sin λx with y (x) = C 1 λ sin λx + C2 λ cos λx. Using boundary conditions, we get C 2 λ = 0 and C1 cos λπ + C 2 sin λπ = 0. The first relation implies C 2 = 0, so in order to get non-trivial solutions we need to require cos λπ = 0. Then λπ = π/2 + nπ for non-negative integer n (since λ > 0 by assumption), i.e. eigenvalues are λ n = (2n+1)2 and corresponding eigenfunctions are c 4 n cos ( 2n+1x ). 2 Answer. The eigenvalues are λ n = (2n+1)2 with eigenfunctions c 4 n cos ( 2n+1x ), where n = 2 0, 1, 2,... and c n are arbitrary constants. 3

4 Problem (p.613) Find the eigenvalues and the corresponding eigenfunctions for the boundary value problem y + λy = 0; 0 < x < π/2, y (0) = 0, y (π/2) = 0. Solution. The characteristic equation is r 2 + λ = 0. We will consider three cases: 1. If λ < 0, there are two real roots ± λ and the general solution is y(x) = C 1 e λx + C 2 e λx with y (x) = C 1 λe λx C 2 λe λx. Using boundary conditions, we get C 1 λ C2 λ = 0 and C1 λe λπ/2 C 2 λe λπ/2 = 0. The only solution is C 1 = C 2 = 0, so there are no negative eigenvalues. 2. If λ = 0, there is one double root 0 and the general solution is y(x) = C 1 + C 2 x with y (x) = C 2. Using boundary conditions, we get C 2 = 0 and C 2 = 0. Therefore, λ = 0 is an eigenvalue with corresponding eigenfunctions being constants. 3. If λ > 0, there are two complex roots ± λi and the general solution is y(x) = C 1 cos λx + C 2 sin λx with y (x) = C 1 λ sin λx + C2 λ cos λx. Using boundary conditions, we get C 2 λ = 0 and C1 λ sin( λπ/2) + C2 λ cos( λπ/2) = 0. The first relation implies C 2 = 0, so in order to get non-trivial solutions we need to require sin( λπ/2) = 0. Then λπ/2 = nπ for positive integer n (since λ > 0 by assumption), i.e. eigenvalues are λ n = 4n 2 and corresponding eigenfunctions are c n cos 2nx. Note that we can combine the second case with the third one by allowing n = 0, since cos 0 = 1. Answer. The eigenvalues are λ n = 4n 2 with eigenfunctions c n cos 2nx, where n = 0, 1, 2,... and c n are arbitrary constants. 4

5 Problem (p.613) Find the eigenvalues and the corresponding eigenfunctions for the boundary value problem y 2y + λy = 0; 0 < x < π, y(0) = 0, y(π) = 0. Solution. The characteristic equation is r 2 2r + λ = 0. Its roots are λ 1,2 = 2± 4 4λ 2 = 1 ± 1 λ. We will consider three cases: 1. If λ < 1, roots are real and distinct, the general solution is y(x) = C 1 e λ 1x + C 2 e λ 2x. Using boundary conditions, we get C 1 + C 2 = 0 and C 1 e λ 1π + C 2 e λ 2π = 0. The only solution is C 1 = C 2 = 0, so there are no such eigenvalues. 2. If λ = 1, there is one double root 1 and the general solution is y(x) = C 1 e x + C 2 xe x. Using boundary conditions, we get C 1 = 0 and C 1 e π + C 2 πe π = 0. The only solution is C 1 = C 2 = 0, so λ = 1 is not an eigenvalue. 3. If λ > 1, roots are complex and the general solution is y(x) = C 1 e x cos λ 1x + C 2 e x sin λ 1x. Using boundary conditions, we get C 1 = 0 and C 1 e π cos λ 1π + C 2 e π sin λ 1π = 0, so in order to get non-trivial solutions we need to require sin λ 1π = 0. Then λ 1π = nπ for positive integer n (since λ 1 > 0 by assumption), i.e. eigenvalues are λ n = n and corresponding eigenfunctions are c n e x sin nx. Answer. The eigenvalues are λ n = n with eigenfunctions c n e x sin nx, where n = 1, 2, 3,... and c n are arbitrary constants. 5

6 Problem (p.613) Solve the heat flow problem u t (x, t) = u 3 2 (x, t), x2 0 < x < π, t > 0, u(0, t) = u(π, t) = 0, t > 0, u(x, 0) = sin 4x + 3 sin 6x sin 10x, 0 < x < π. Solution. Using the method of separation of variables (see Problems 28 and 30 for examples), we show that u(x, t) = X(x)T (t) is a solution of the partial differential equation if X (x) + λx(x) = 0, T (t) + 3λT (t) = 0. Combining the first ODE with boundary conditions X(0) = X(π) = 0 and solving this boundary value problem (see Problems 10, 12, and 14 for examples) we see that eigenvalues are λ n = n 2 with eigenfunctions X n (x) = c n sin nx for n = 1, 2, 3,.... Solving the equation for T (t) with these values for λ gives T n (t) = a n e 3n2t, where a n are any constants. Thus functions u n (x, t) = e 3n2t sin nx satisfy the differential equation and boundary conditions. Now we take into account the initial conditions. Note that u n (x, 0) = sin nx, the equation is linear, and the boundary conditions are homogeneous, so any linear combination of solutions is again a solution. Then u(x, t) = u 4 (x, t) + 3u 6 (x, t) u 10 (x, t) is a solution that satisfies the initial condition as well. Answer. u(x, t) = e 48t sin 4x + 3e 108t sin 6x e 300t sin 10x. 6

7 Problem (p.614) Solve the vibrating string problem 2 u t (x, t) = 2 u 2 32 (x, t), x2 0 < x < π, t > 0, u(0, t) = u(π, t) = 0, t > 0, u(x, 0) = 0, 0 < x < π, u (x, 0) = 2 sin 3x + 9 sin 7x sin 10x, t 0 < x < π. Solution. Using the method of separation of variables (see Problems 28 and 30 for examples), we show that u(x, t) = X(x)T (t) is a solution of the partial differential equation if X (x) + λx(x) = 0, T (t) λt (t) = 0. Combining the first ODE with boundary conditions X(0) = X(π) = 0 and solving this boundary value problem (see Problems 10, 12, and 14 for examples) we see that eigenvalues are λ n = n 2 with eigenfunctions X n (x) = c n sin nx for n = 1, 2, 3,.... Solving the equation for T (t) with these values for λ gives T n (t) = a n cos 3nt + b n sin 3nt, where a n and b n are any constants. Thus functions u n (x, t) = (a n cos 3nt + b n sin 3nt) sin nx satisfy the differential equation and boundary conditions. Now we take into account the initial conditions. Since the equation is linear and the boundary conditions are homogeneous, any linear combination of solutions is again a solution. Our goal is to pick a linear combination of u n (x, t) that will satisfy initial conditions. Note that u n (x, 0) = a n sin nx. Since we want u(x, 0) = 0 for all x, we let a n = 0. Then u n (x, t) = 3nb t n cos 3nt sin nx and un (x, 0) = 3nb t n sin nx. We see that the combination of u 3, u 7, and u 10 with b 3 = 2, b 9 7 = 3, and b 7 10 = 1 is the required solution. 30 Answer. u(x, t) = 2 9 sin 9t sin 3x sin 21t sin 7x sin 30t sin 10x. 30 7

8 Problem (p.614) Find the formal solution to the vibrating string problem 2 u t (x, t) = 2 u 2 42 (x, t), x2 0 < x < π, t > 0, u(0, t) = u(π, t) = 0, t > 0, 1 u(x, 0) = sin nx, n2 0 < x < π, u t (x, 0) = ( 1) n+1 sin nx, 0 < x < π. n Solution. Using the method of separation of variables (see Problems 28 and 30 for examples), we show that u(x, t) = X(x)T (t) is a solution of the partial differential equation if X (x) + λx(x) = 0, T (t) λt (t) = 0. Combining the first ODE with boundary conditions X(0) = X(π) = 0 and solving this boundary value problem (see Problems 10, 12, and 14 for examples) we see that eigenvalues are λ n = n 2 with eigenfunctions X n (x) = c n sin nx for n = 1, 2, 3,.... Solving the equation for T (t) with these values for λ gives T n (t) = a n cos 4nt + b n sin 4nt, where a n and b n are any constants. Thus functions u n (x, t) = (a n cos 4nt + b n sin 4nt) sin nx satisfy the differential equation and boundary conditions. Since the equation is linear and the boundary conditions are homogeneous, any (finite) linear combination of solutions is again a solution. Since we are asked to find a formal solution, we assume that the infinite linear combination of all u n (x, t) is also a solution. Then u(x, t) = (a n cos 4nt + b n sin 4nt) sin nx and our goal is to determine a n and b n using initial conditions. From our formula we have u(x, 0) = a n sin nx. Comparing this expression with the condition for u(x, t) we see that a n = 1 n 2. Using our formula again we compute u t (x, t) = 4n( a n sin 4nt + b n cos 4nt) sin nx, and we see that b n = ( 1)n+1 Answer. u(x, t) = 4n 2. ( ) 1 ( 1)n+1 cos 4nt + sin 4nt sin nx. n2 4n 2 u t (x, 0) = 4nb n sin nx, 8

9 Problem (p.614) Show that if u(x, t) = X(x)T (t) is a solution of the partial differential equation 2 u t 2 + u t + u = α2 2 u x 2, then X(x) and T (t) must satisfy the following ordinary differential equations: where λ is a constant. X (x) λx(x) = 0, T (t) + T (t) + (1 λα 2 )T (t) = 0, Solution. Substituting u(x, t) = X(x)T (t) into the equation, we get X(x)T (t) + X(x)T (t) + X(x)T (t) = α 2 X (x)t (t). Assuming that u(x, t) 0, we can divide by X(x)T (t): T (t) T (t) + T (t) T (t) + 1 = α2 X (x) X(x). Now the left hand side is a function of t only, while the right hand side is the function of x only. Since they are equal, both must be constant. Let s denote this constant by λα 2. Then T (t) T (t) + T (t) T (t) + 1 = λα2, α 2 X (x) X(x) = λα2. Clearing denominators and moving all terms to the left gives us the desired equations for X(x) and T (t). 9

10 Problem (p.614) Show that if u(r, θ, z) = R(r)T (θ)z(z) is a solution of the partial differential equation 2 u r + 1 u 2 r r u r 2 θ + 2 u 2 z = 0, 2 then R(r), T (θ), and Z(z) must satisfy the following ordinary differential equations: where λ and µ are constants. T (θ) + µt (θ) = 0, (1) Z (z) + λz(z) = 0, (2) r 2 R (r) + rr (r) (r 2 λ + µ)r(r) = 0, (3) Solution. Substituting u(r, θ, z) = R(r)T (θ)z(z) into the equation, we get R (r)t (θ)z(z) + 1 r R (r)t (θ)z(z) + 1 r 2 R(r)T (θ)z(z) + R(r)T (θ)z (z) = 0. Assuming that u(r, θ, z) 0, we can divide by R(r)T (θ)z(z): R (r) R(r) + 1 R (r) r R(r) + 1 T (θ) r 2 T (θ) + Z (z) Z(z) = 0, R (r) R(r) + 1 R (r) r R(r) + 1 T (θ) r 2 T (θ) = (z) Z Z(z). Now the left hand side is a function of r and θ, while the right hand side is the function of z. Since they are equal, both must be constant. Let s denote this constant by λ. Then R (r) R(r) + 1 R (r) r R(r) + 1 T (θ) r 2 T (θ) = λ, Z (z) Z(z) = λ. It is easy to transform the last equation into (2). In the equation for R(r) and T (θ) we separate variables one more time: r 2 R (r) R(r) + r R (r) R(r) + T (θ) T (θ) = r2 λ, r 2 R (r) R(r) + r R (r) R(r) r2 λ = T (θ) T (θ). Now the left hand side is a function of r, while the right hand side is the function of θ. Since they are equal, both must be constant. Let s denote this constant by µ. Then r 2 R (r) R(r) + r R (r) R(r) r2 λ = µ, T (θ) T (θ) = µ. Clearing denominators and moving all terms to the left, we get (1) and (3). 10