Strauss PDEs 2e: Section 5.3 - Exercise 4 Page of 6 Exercise 4 Consider the problem u t = ku xx for < x < l, with the boundary conditions u(, t) = U, u x (l, t) =, and the initial condition u(x, ) =, where U is a constant. (a) Find the solution in series form. (Hint: Consider u(x, t) U.) (b) Using a direct argument, show that the series converges for t >. (c) If ɛ is a given margin of error, estimate how long a time is required for the value u(l, t) at the endpoint to be approximated by the constant U within error ɛ. (Hint: It is an alternating series with first term U, so that the error is less than the next term.) Solution Part (a) Since the boundary conditions are not homogeneous, the method of separation of variables cannot be applied. Make the change of variables, v(x, t) = u(x, t) U, in order to make them so. Find the derivatives of u in terms of this new variable. v t = u t v tt = u tt v x = u x v xx = u xx As a result, v satisfies the same PDE as u. u t = ku xx v t = kv xx The initial and boundary conditions associated with it are as follows. v(, t) = u(, t) U = U U = v x (l, t) = u x (l, t) = v(x, ) = u(x, ) U = U = U The method of separation of variables can be applied to solve for v because the PDE and its boundary conditions are linear and homogeneous. Assume a product solution of the form v(x, t) = X(x)T (t) and plug it into the PDE and the boundary conditions. v t = kv xx XT = kx T v(, t) = X()T (t) = X() = v x (l, t) = X (l)t (t) = X (l) =
Strauss PDEs 2e: Section 5.3 - Exercise 4 Page 2 of 6 Now separate variables in the PDE: divide both sides by kxt to bring all constants and functions of t to the left side and all functions of x to the right side. T kt = X X The only way a function of t can be equal to a function of x is if both are equal to a constant. T kt = X X = λ Values of λ for which X() = and X (l) = are satisfied are called the eigenvalues, and the nontrivial functions X(x) associated with them are called the eigenfunctions. Determination of Positive Eigenvalues: λ = µ 2 Assuming λ is positive, the differential equation for X becomes Multiply both sides by X. X X = µ2. X = µ 2 X The general solution can be written in terms of hyperbolic sine and hyperbolic cosine. X(x) = C cosh µx + C 2 sinh µx Apply the boundary conditions to determine C and C 2. X() = C = X (l) = µ(c sinh µl + C 2 cosh µl) = The second equation reduces to C 2 cosh µl =. Since hyperbolic cosine is not oscillatory, the only way this equation is satisfied is if C 2 =. The trivial solution is obtained, so there are no positive eigenvalues. Determination of the Zero Eigenvalue: λ = Assuming λ is zero, the differential equation for X becomes Multiply both sides by X. X X =. X = The general solution is obtained by integrating both sides with respect to x twice. After the first integration, we have X (x) = C 3. Apply the boundary condition x (l) = to determine C 3. X (l) = C 3 =
Strauss PDEs 2e: Section 5.3 - Exercise 4 Page 3 of 6 Integrate both sides of the previous equation with respect to x once more. X(x) = C 4 Apply the boundary condition x() = to determine C 4. X() = C 4 = The trivial solution X(x) = is obtained again, so zero is not an eigenvalue. Determination of Negative Eigenvalues: λ = γ 2 Assuming λ is negative, the differential equation for X becomes Multiply both sides by X. X X = γ2. X = γ 2 X The general solution can be written in terms of hyperbolic sine and hyperbolic cosine. X(x) = C 5 cos γx + C 6 sin γx Apply the boundary conditions to determine C 5 and C 6. The second equation reduces to X() = C 5 = X (l) = γ( C 5 sin γl + C 6 cos γl) = C 6 γ cos γl = cos γl = γl = π (2n ), n =, 2,... 2 γ n = π (2n ), n =, 2,.... 2l The eigenfunctions associated with the eigenvalues λ = γ 2 are X(x) = C 6 sin µx X n (x) = sin 2l (2n )x, n =, 2,.... Because there are negative eigenvalues, the ODE for T will now be solved. Multiply both sides by kt. T kt = γ2 T = kγ 2 T The general solution is written in terms of the exponential function. T (t) = C 7 e kγ2 t T n (t) = exp 4l 2 (2n )2 t, n =, 2,...
Strauss PDEs 2e: Section 5.3 - Exercise 4 Page 4 of 6 According to the principle of linear superposition, the solution to the PDE for v(x, t) is a linear combination of all products T n (t)x n (x) over all the eigenvalues. v(x, t) = B n exp 4l 2 (2n )2 t sin 2l (2n )x The initial condition will now be used to determine the coefficients B n. v(x, ) = B n sin 2l (2n )x = U To solve for B n, multiply both sides by sin γ m x, where m is an integer, B n sin 2l (2n )x sin 2l (2m )x = U sin 2l (2m )x, and then integrate both sides with respect to x from to l. B n B n sin 2l (2n )x sin 2l (2m )x dx = sin 2l (2n )x sin 2l (2m )x ( U) sin 2l (2m )x dx 2l l dx = U π(2m ) cos 2l (2m )x Because the sine functions are orthogonal, the integral on the left side is zero if n m. All the terms in the infinite series vanish as a result except for one, the n = m term. The coefficients are B n [ sin 2 π 2l (2n )x 2l dx = U π(2n ) ( ) ( ) l 2l B n = U 2 π(2n ) 4 B n = U π(2n ), which means the solution for v is 4 v(x, t) = ( U) π(2n ) exp 4l 2 (2n )2 t sin 2l (2n )x = U 4 π 2n exp 4l 2 (2n )2 t sin 2l (2n )x. Therefore, since u(x, t) = v(x, t) + U, { u(x, t) = U 4 π 2n exp } 4l 2 (2n )2 t sin 2l (2n )x.
Strauss PDEs 2e: Section 5.3 - Exercise 4 Page 5 of 6 Part (b) If we can show that the series in the solution converges absolutely, then it will converge. 2n exp 4l 2 (2n )2 t sin 2l (2n )x sin = 2n exp 4l 2 (2n )2 t 2l (2n )x 2n exp 4l 2 (2n )2 t Apply the ratio test to show that this bigger series converges. [ exp k (2n + ) 2 t lim 2n+ 4l 2 n [ 2n exp k (2n ) 4l 2 t = lim 2n [k n 2n + exp 4l 2 [(2n )2 (2n + ) 2 t 2 2 n = lim n 2 + exp [k 4l n 2 [(2n )2 (2n + ) 2 t = lim [k exp n 4l 2 ( 8n)t ) = lim ( 2kn exp n l 2 t = The limit is zero because t is positive. Since the limit is less than one, the bigger series 2n exp 4l 2 (2n )2 t converges by the ratio test. And by the comparison test, 2n exp 4l 2 (2n )2 t sin 2l (2n )x converges as well. Therefore, converges. Part (c) 2n exp 4l 2 (2n )2 t sin 2l (2n )x Plug in x = l into the solution. u(l, t) = U 4U π The sine evaluates to ( ) n. = U + 4U π 2n exp 4l 2 (2n )2 t sin 2 (2n ) ( ) n 2n exp 4l 2 (2n )2 t
Strauss PDEs 2e: Section 5.3 - Exercise 4 Page 6 of 6 U ɛ serves as a lower bound, so set ɛ equal in magnitude to the first term in the infinite series, which is the lowest value in the sequence. ɛ = 4 U ) ( k π exp 4l 2 t Solve this equation for t. ) ɛπ ( k 4 U = exp 4l 2 t ln ɛπ = k 4 U 4l 2 t 4l 2 ɛπ ln k 4 U = t Therefore, the time required to reach U ɛ at the endpoint is t = 4l 2 ɛπ ln k 4 U = 4l2 ɛπ kπ 2 ln 4 U. Figure : This figure illustrates the solution u(x, t) versus x at various times for k =, l =, and U = using only the first 3 terms in the infinite series. The curves in red, orange, yellow, green, blue, and purple correspond to the times t =, t =., t =.4, t =.7, t =, and t = 2, respectively.