Homework 7 Solutions

Homework 7 Solutions # (Section.4: The following functions are defined on an interval of length. Sketch the even and odd etensions of each function over the interval [, ]. (a f( =, f ( Even etension of f( Odd etension of f( f o ( f e ( - - (b, < f( =, 3 f ( 3 Even etension of f( Odd etension of f( f e ( f o ( - -3 - - 3 - -3 - - 3

# (Section.4: The following functions are defined on an interval of length. Find the Fourier Sine series and the Fourier Cosine series for each and sketch the functions to which the series converge over the interval [ 3, 3]. (a f( =, f ( Fourier Cosine series Even etension of f( f e ( = b n =, n =,,... - The Fourier coefficients are calculated as follows. a = f(d = a = d = a n = ( nπ f(cos d = cos(nπd = nπ sin(nπ = [sin(nπ sin ] = nπ a n = n =,,...

Then, f( = F C ( = a + ( nπ a n cos n= = + cos(nπ n= for < < f( = F C ( = < < Fourier Cosine series representation of f( F C ( -3 - - 3

Fourier Sine series Odd etension of f( f o ( = a n =, n =,,,... - - The Fourier coefficients are calculated as follows. b n = ( nπ f(sin d = sin(nπd = nπ cos(nπ = [cos(nπ cos] = nπ nπ [ ( n ] Then, f( = F S ( = b n = π ( n n ( nπ b n sin n= n =,,... for < < f( = F S ( = ( n sin(nπ < < π n= n Fourier Sine series representation of f( F S ( - -3 - - 3

(b, < π f( =, π π f ( pi *pi Fourier Cosine series Even etension of f( f e ( = π b n =, n =,,... -*pi -pi pi *pi The Fourier coefficients are calculated as follows. a = f(d = π π f(d = [ π d + π π π ] d = (π + = π a = π a n = ( nπ f(cos d = ( n f(cos d π = [ π ( n π ( n ] cos d + cos d π π = [ ( n π ] π n sin + = [ ( nπ ] sin sin = ( nπ nπ nπ sin a n = ( nπ nπ sin n =,,...

Then, f( = F C ( = a + ( nπ a n cos n= for < < f( = F C ( = + π n= ( ( nπ n n sin cos < < π Fourier Cosine series representation of f( F C ( -6*pi -4*pi -*pi *pi 4*pi 6*pi

Fourier Sine series Odd etension of f( f o ( = π a n =, n =,,,... - -*pi -pi pi *pi The Fourier coefficients are calculated as follows. π b n = ( nπ f(sin d = ( n f(sin d π = [ π ( n π ( n ] sin d + sin d π π = [ ( n π n cos π ] + = [ ( nπ ] cos cos = [ ( nπ ] cos nπ nπ Then, b n = [ ( ] nπ cos nπ f( = F S ( = f( = F S ( = π n= ( nπ b n sin n= n [ cos n =,,... for < < ( ] nπ sin ( n < < Fourier Sine series representation of f( F S ( - -6*pi -4*pi -*pi *pi 4*pi 6*pi

#3 Section.5: (page 6 # 7, 9 (7 th edition: pages 579 58 # 7,9 Problems 7 and 9 both involve solving a heat conduction problem of the form u t = α u, < <, t > ( u(, t =, u(, t =, t > ( u(, = f(,, (3 where α,, and f( are given. It is efficient to first solve the general case, and then substitute the specific quantities given in problems 7 and 9. Separation of Variable Method: et u(, t = X(T(t. Plugging this epression for u into the PDE ( gives Divide by α X(T(t to get X(T (t = α X (T(t. α T (t T(t = X ( X(. Since the left hand side depends on t alone and the right hand side depends on alone, they can be equal only when both are equal to the same constant, call it λ. α T (t T(t = X ( X( = λ. This epression separates into two equations: one for T(t and one for X(, T + α λt =, X + λx =. (4 We can use ( to determine boundary conditions for X(. At =, Since this equality holds for all t, it must be that u(, t = X(T(t =, X( =. (5 At =, and by the same argument we have u(, t = X(T(t =, X( =. (6

Equations (4 (6 suggest the following eigenvalue problem for X(. X + λx = X( =, X( =. We solved this eigenvalue problem in Section. (see lecture notes and found the following eigenvalues and eigenfunctions. λ n = ( nπ X n ( = sin ( n =,,... nπ For each λ n, equation (4 can be used to determine the corresponding solution T n (t, T n + α λ n T n =. The is the familiar first order eponential ODE with a decay rate of α λ n. It is both separable and linear so it can be solved by either the separable variable technique or the integrating factor method. In either case, we get T n (t = C n e α λ nt αnπ = Cn e ( t. We have found infinitely many solutions of the PDE which are given by u n (, t = X n (T n (t = e αnπ ( t ( sin nπ, n =,..... It is easy to check that each solution satisfies the PDE ( and the boundary conditions (. The last step is to satisfy initial condition (3. Note that any linear combination of the u n s also satisfies the PDE and the boundary conditions (principle of superposition. We can attempt to satisfy the initial conditions by considering the linear combination of all the u n s given by u(, t = αnπ B n e ( t sin ( nπ. (7 n= Applying the initial condition (3 yields u(, = ( nπ B n sin n= = f(. By comparing this equation to the Fourier Sine series of f( on the interval (,, it can be seen that the B n s are the Fourier Sine series coefficient, and therefore B n = f( sin ( nπ d. (8 Then the solution given by equations (7 and (8 satisfies the PDE, the boundary conditions, and the initial conditions.

Now equations (7 and (8 can be used to solve problems 7 and 9. #7 In this problem, equations ( (3 are given with =, α =, f( = sin(π sin(5π. Then by equation (7, the solution is u(, t = B n e (nπ t sin(nπ, n= where the B n s are the coefficients of the Fourier Sine series of f( on the interval (,. Though the B n s can be calculated using equation (8, the initial condition has a special form and it is more convenient to apply the initial condition directly to get u(, = B n sin(nπ = sin(π sin(5π n= The B n s can be determined by equating the coefficients of like terms (sin(nπ, n =,,..., which yields B =, B 5 =, and B n = otherwise. Therefore all terms of the series are zero ecept for the second (n = and fifth (n = 5 terms, so the solution is u(, t = e (πt sin(π e (5πt sin(5π. #9 In this problem, equations ( (3 are given with Then by equation (7, the solution is = 4, α =, f( = 5. u(, t = B n e n π t 6 nπ sin( 4, (9 n= where the B n s are the coefficients of the Fourier Sine series of f( on the interval (,4. The B n s are determined from equation (8. B n = ( nπ f( sin = 5 4 ( nπ nπ cos 4 4 d = 4 = nπ 4 ( nπ 5 sin 4 d [cos(nπ cos] = nπ [ ( n ] B n = nπ [ ( n ], n =,,... ( The solution is given by equations (9 and ( and can be written as u(, t = π ( n e n π t 6 sin ( nπ n= n 4.

#4 Section.6: Consider the heat flow problem u t = u, < <, t > ( u(, t = 4, u(, t =, t > ( u(, =,. (3 (a Find the steady state (time independent solution, call it v(, by setting u t = in the PDE and solving the resulting boundary value problem. With u t = and u(, t = v(, equations ( and ( reduce to v =, < <, (4 v( = 4, v( =. (5 The general solution of equation (4 is v( = C +C. The boundary conditions (5 determine the constants to be C = 4 and C =, and therefore v( = (. (b et w(, t = u(, t v(, where v( is the steady state solution found in part (a. Find the resulting PDE, boundary conditions, and initial condition for w(, t. The temperature u(, t can be written as u(, t = w(, t + v( and substituted into equation ( to give w t + v t = w + v, and since v t = v =, we get w t = w. Equation ( determines the boundary conditions for w. w(, t = u(, t v( = 4 4 = w(, t = w(, t = u(, t v( = = Equation (3 determines the initial condition for w. w(, t = w(, = u(, v( = ( = 4 w(, = 4

(c et w(, t = X( T(t and find the separated equations for X( and T(t. Solve the eigenvalue problem for X(, and then determine T(t. Note that w satisfies equations ( (3 with =, α =, f( = 4. The separated equations were already determined and solved above. By equation (4, the separated equations are T + λt =, X + λx =, and the boundary conditions for X( are The eigenvalues and eigenfunctions are given by X( =, X( =. λ n = ( nπ X n ( = sin ( nπ and the corresponding solutions for T(t were found to be n =,,..., T n (t = B n e n π 4 t, n =,,.... (d Complete the solution by solving the intial-boundary value problem for w(, t and writing the solution for u(, t. It follows from equations (7 and (8 that and Therefore, B n = w(, t = ( nπ f( sin d = = 4 ( nπ nπ cos B n e n π t 4 sin ( nπ n= ( nπ 4 sin d = 8 8 [cos(nπ cos] = nπ nπ [ ( n ] B n = 8 nπ [ ( n ], n =,,.... w(, t = 8 ( n e n π t 4 sin ( nπ π n= n, and since u(, t = w(, t + v(, the solution is u(, t = ( 8 ( n e n π t 4 sin ( nπ π n= n.

#5 Section.6: Suppose that a metal rod of unit length is completely insulated and that the temperature u(, t of the rod satisfies u t = 3u, < <, t > u (, t = u (, t =, where f( is the initial temperature of the rod. u(, = f( (a et u(, t = X(T(t. Find the separated equations, solve the eigenvalue problem for X( (don t forget to check λ =, and determine T(t. The equations are separated as before. T + 3λT = X + λx = The boundary condition for X( are found as follows. u (, t = X (T(t = X ( = u (, t = X (T(t = X ( = The eigenvalue problem for X( was solved on homework 6 (problem #(c with =. The eigenvalues and eigenfunctions were found to be λ n = (nπ X n ( = cos(nπ n =,,,..., The corresponding solutions for T(t are found by solving T n + 3λ nt n =, which gives T n (t = C n e 3λnt = C n e 3n π t, n =,,,....

(b Find the general solution and show that u(, t approaches a constant as t. How is the constant related to the initial temperature f(? We have infinitely many solutions given by u n (, t = X n (T n (t = e 3n π t cos(nπ, n =,,..... The initial conditions can be satisfied by considering the following linear combination of the u n s. Applying the initial condition yields u(, t = A + n= A n e 3n π t cos(nπ u(, t = A + A n cos(nπ = f( n= By comparing this equation to the Fourier Cosine series of f( on the interval (,, it can be seen that the A n s are the Fourier Cosine series coefficient, and therefore A n = f( cos(nπd, n =,,,... As t, all terms of the series approach zero ecept for the constant term A, so that u(, t A as t. Recall from calculus that the integral A = f(d gives the average value of f( over the interval (,. It makes sense that the temperature in the rod approaches this value as t. Since the rod is completely insulated on all sides, it neither gains nor loses heat, so the total heat in the rod is always equal to the initial amount. As t, the heat diffuses through the rod until it is uniformly distributed and thermal equilibrium (constant temperature is attained. Since the rod is insulated, this constant temperature must be the average of the initial temperature distribution f(.