First Midterm Exam Solutions

First Midterm Exam Solutions 1. Solve the diffusion equation ∂u/∂t = α∂2u/∂x2 for u(x,t) with the boundary conditions that u(0,t) = 0 and ∂u/∂x = 0 at x = L. a) Find the general solution and the expression for the expansion coefficient for any initial condition, u(x,0) = f(x). Applying separation of variables to the diffusion equation gives the following result. u( x, t ) = e − λ αt [C1 sin(λx ) + C2 cos(λx )] 2 [1] Taking the first derivative of this equation with respect to x gives. 2 ∂u( x, t ) = e − λ αt λ [C1 cos(λx ) − C2 sin(λx )] ∂x [2] The boundary condition that u = 0 at x = 0 gives u ( x, t ) = e − λ αt [C1 sin(0) + C2 ] = 0 2 [3] This requires that C2 = 0. The boundary condition that the ∂u/∂x = 0 at x = L gives the following result. ∂u ( x, t ) ∂x 2 x= L = 0 = e −λ αt λ[C1 cos(λL)] = 0 [4] We can satisfy this condition by having cos(λL) = 0 which requires λL to be an odd integer times π/2. We can express this condition as follows. λL = (2n + 1)π or 2 λ= (2n + 1)π 2L integer n [5] We can now write the general solution for our problem as the sum of all solutions that satisfy the differential equation and the boundary conditions. Each solution can have a different constant which we call Cn in the equation below. ∞ u ( x, t ) = ∑ Cn e −λnαt sin(λn x) n =0 2 λn = (2n + 1)π 2L [6] As usual we find the coefficients, Cn, by using an eigenfunction expansion for the initial condition. First midterm solutions Page 2 ∞ u ( x,0) = f ( x) = ∑ Cn sin (2n + 1)πx [7] L n =0 If we multiply both sides of this equation by sin[(2m+1)x/2L, and integrate the result from 0 to L, we obtain. L ∫ f (x )sin 0 (2m + 1)πx dx = L sin (2m + 1)πx  ∞ ∫ 2L ∑ Cn sin  n =0 2L 0 (2m + 1)πx  dx 2L   [8] Because of the orthogonality of the sine eigenfunctions, there is only one term in the summation on the right side of this equation. L ∫ 0 2L L L 2 x 2L 2(2m + 1)πx   (2m + 1)πx  = − sin dx C m sin m    2L 2L   2 4(2m + 1)π  0 [9] 0 L (2m + 1)πL − 0C = LC m L = − sin  m 2 L  2 2(2m + 1)π  (2m + 1)πx dx = C f ( x )sin ∫ Solving for Cm gives 2 Cm = L L ∫ f (x )sin 0 (2m + 1)πx dx [10] 2L b) Write the solution for the specific initial condition that u(x,0) = f(x) = U, a constant. In this case, equation [10] for Cn becomes (after replacing m by n) L (2n + 1)πx dx = 2U 2 L − cos (2n + 1)πx  2 Cn = U sin  L L (2n + 1)π  2L 2L x =0 x= L ∫ 0 (2n + 1)πL + cos(0) = 4U 4U  = − cos   (2n + 1)π (2n + 1)π  2L [11] Substituting this result into equation [6] gives. u ( x, t ) = 4U ∞ ∑ n =0 e − (2 n+1)2 π2 αt 4 L2 sin (2n + 1)πx (2n + 1)π 2L [12] c) What are the first three terms in your equation for part b)? 9 π 2 αt 25π2αt   − π 2 αt − − x x 1 1 3 1 5πx π π 2 2 2  4 L 4 L 4 L u ( x, t ) = 4U e e sin sin sin + e + +   π 2 L 5π 2L 2 L 3π   [13] First midterm solutions Page 3 d) Find the answer to part a) change if the original boundary conditions were replaced by the boundary conditions that u(0,t) = A and ∂u/∂x = B at x = L, where A and B are constants? (Continue to use the initial condition that u(x,0) = f(x).) In this case our solution for u(x,t) would be written as the sum of two functions. u ( x, t ) = u ( x, t ) + w( x ) [14] Here, v(x,t) is the solution to the original problem, with zero boundary conditions. The solution for v(x,t) is the same as the solution for u(x,t) found in equation [6]. The differential equation and boundary conditions for w are found from the requirements that u has to satisfy the diffusion equation and boundary conditions, while v satisfies the diffusion equation with zero boundary conditions. These requirements can only be satisfied if w(x)is given by the following differential equation and boundary conditions. d 2w dx 2 =0 dw dx w(0 ) = A x= L =B [15] This gives the following result for w. w = Bx + A [16] (Note that dw/dx =B so that for long times the gradient is a constant for all x values.) The solution for u = v + w would then be given by a combination of equations [6] and [16]. ∞ u ( x, t ) = ∑ Cn e −λnαt sin(λn x) + Bx + A 2 n =0 λn = (2n + 1)π [17] 2L Here, the equation for Cn, given by equation [10] would have to be modified to include the Bx + A term. 2 Cm = L L ∫ [ f (x ) − Bx − A]sin 0 (2m + 1)πx dx [18] 2L e) What is the answer to part d if f(x) = u(x,0) = U, a constant? (This question was not included on the exam!) In this case equation [18] for Cm = becomes 2 . Cm = L L L L ( (2m + 1)πx dx [19] ( 2m + 1)πx 2(U − A) 2m + 1)πx 2B dx = dx − ∫ [U − Bx − A]sin ∫ sin ∫ x sin 0 2L L 0 2L L 0 2L The first integral above, with the (U – A) factor, is similar to the one found in equation [11]. The result for this integral is given below. First midterm solutions Page 4 (2m + 1)πx dx = 4(U − A) 2(U − A) sin ∫ (2m + 1)π L 2L 0 L [20] The second integral in equation [19] is found from integral tables. 1 x= L 2 L ( (2m + 1)πx  ( 2B 2m + 1)πx 2B  2 Lx 2m + 1)πx  2 L  −  sin +  cos x sin dx = 2L 2L 2L L L  (2m + 1)π   (2m + 1)π  0  x =0 2 2   (2m + 1)πL +  2 L  sin (2m + 1)πL − 0 − sin 0 2B 2L − [21] . cos  (2m + 1)π  2L 2L L  (2m + 1)π      2 m (2m + 1)π = 8BL(− 1) 2B  2L    sin = −0 + 2 L  (2m + 1)π  (2m + 1)2 π 2 ∫ Combining equations [19], [20], and [21] gives L L ( ( 2(U − A) 2m + 1)πx 2B 2m + 1)πx 4(U − A) 8BL(− 1)m sin Cm = dx − x sin dx = − [22] (2m + 1)π (2m + 1)2 π 2 2L 2L L L ∫ . ∫ 0 0 WE can replace m by n in this equation and then substitute the result for Cn into equation [17] to get the problem solution.  4(U − A) 8BL(− 1)n  e − u ( x, t ) = ∑   (2n + 1)2 π 2  n = 0  (2n + 1)π − ( 2 n +1)2 π 2 αt ∞ 4 L2 sin (2n + 1)πx (2n + 1)π 2L + Bx + A [23] We can divide this equation by U to obtain the following result. e 4 u ( x, t ) ∞  A BL 8(− 1)m  − = 1 −   2 2 U  U  (2m + 1)π U (2m + 1) π  n =0  ∑ − (2 n+1)2 π2 αt 4 L2 sin (2n + 1)πx (2n + 1)π 2L + BL x A + U L U [23] As before the solution for u(x,t)/U depends on x/L and ατ/L2, but this solution also depends on the ratios BL/U and A/U. 1 x 1 ∫ x sin axdx = − a cos ax + a 2 sin ax + C First midterm solutions Page 5 2. Solve the following problem which is a model of heat transfer in an electronic component. ∂ 2T ∂T ∂x x =0 =0 ∂T ∂x ∂x 2 x= L + ∂ 2T ∂y 2 = 0 0 ≤ x ≤ L, 0 ≤ y ≤ H h + (Tx = L − T∞ ) = 0 k ∂T ∂y y =0 = GS ∂T ∂y ( y=H ) h + T y = H − T∞ = 0 k In this problem, T∞, k, h, and GS are constants. We can solve this problem by defining a new variable, Θ = (T – T∞). With this substitution we have the following problem. ∂ 2Θ ∂Θ ∂x x =0 =0 ∂ 2Θ = 0 0 ≤ x ≤ L, 0 ≤ y ≤ H ∂x 2 ∂y 2 ∂Θ ∂Θ ∂Θ h + Θ x= L = 0 = GS ∂x x = L k ∂y y =0 ∂y + y=H h + Θ y=H = 0 k [2] This problem has homogenous boundary conditions at all boundaries except the boundary at y = 0. The general separation of variables solution for the Laplace equation with an eigenfunction expansion in the x direction to fit the boundary condition at y = 0 is Θ( x, y ) = [A sin(λx) + B cos(λx)][C sinh(λy ) + D cosh(λy )] [3] Taking the first derivative of this solution gives. ∂Θ = λ [A cos(λx) − B sin(λx)][C sinh(λy ) + D cosh(λy )] ∂x [4] We can satisfy the boundary condition this derivative is zero at x = 0 by setting A = 0. The mixed boundary condition at x = L requires the following relation (after setting A = 0). ∂Θ h + Θ x = L = −λB sin(λL)[C sinh(λy ) + D cosh(λy )] + ∂x x = L k h B cos(λL)[C sinh(λy ) + D cosh(λy )] = 0 k [5] We can satisfy this boundary condition by the following equation that can be solved for the eigenvalues, α = λL. The eigenvalues will depend on the problem parameter ratio hL/k. h − λ sin(λL) + cos(λL) = 0 ⇒ k k (λL ) = cot (λL ) hL [6] We now have to match the boundary conditions at y = 0 and y = H. Taking the y derivative of the solution in equation [3] (after setting A = 0 as noted above) gives [1] First midterm solutions Page 6 ∂Θ = λB cos(λx)[C cosh(λy ) + D sinh(λy )] ∂y [7] The boundary condition at y = H requires that the following equation be satisfied ∂Θ ∂y h + Θ y = H = λB cos(λx)[C cosh(λH ) + D sinh(λH )] k y=H h + λB cos(λx)[C sinh(λH ) + D cosh(λH )] = 0 k [8] This equation will be satisfied if λ [C cosh(λH ) + D sinh(λH )] + h [C sinh(λH ) + D cosh(λH )] = 0 k [9] Dividing this equation by cosh(λH), using the definition that tanh(λH) = sinh(λH)/cosh(λH) and rearranging gives h h    C λ + tanh(λH ) + D λ tanh(λH ) +  = 0 k k    [10] Thus the constants C and D are related by the following equation. D = −C h k λ + tanh(λH ) h λ tanh(λH ) + k [11] Substituting this result and into equation [3] (with A = 0) gives h   λ n + tanh(λ n H )   k Θ( x, y ) = BC cos(λx) sinh(λy ) − cosh(λy ) h   λ n tanh(λ n H ) +   k [12] This is our desired eigenfunction solution. We can replace the product of constants, BC, by a single constant, Bn for each eigenvalue and write the general solution as the sum of all the eigenfunctions h   λ n + tanh(λ n H )   k cosh(λ n y ) Θ( x, y ) = ∑ Bn cos(λ n x) sinh(λ n y ) − h n =1   λ n tanh(λ n H ) + k   ∞ [13] We have to match the condition that ∂Θ/∂y = GS.at y = 0. Taking the y derivative of equation [13] and setting y = 0 gives the following results. First midterm solutions Page 7 h   λn + tanh(λn H )   ∂Θ ∞ k sinh(λn y ) = ∑ Bn cos(λn x)λn cosh(λn y ) − h ∂y n =1   λn tanh(λn H ) + k   ∂Θ ∂y [14] h   λn + tanh(λn H )   ∞ k = GS = ∑ Bn cos(λn x)λn cosh(λn 0) − sinh(λn 0) = ∑ Bn λn cos(λn x) h n =1   n =1 λn tanh(λn H ) + k   ∞ y =0 Multiplying this equation by cos(λmx), where λm is a different eigenvalue, and integrating from 0 to L gives the following equation that can be solved for the expansion coefficients, Bn. L L ∞ 0 0 n =1 GS ∫ cos(λm x)dx = ∫ ∑ Bn cos(λn x)λn cos(λm x)dx [16] As usual, the orthogonality of the eigenfunctions leaves only one term in the sum, the one for which n = m. L L 0 0 GS ∫ cos(λm x)dx = Bm λm ∫ cos 2 (λm x)dx [17] The integrals in this equation are found as follows. x=L L 1  1 ∫0 cos(λm x )dx =  λm sin (λm x ) = λm sin (λm L ) x =0 [18] x=L L x  L 1 1 ∫0 cos (λm x)dx =  2 + 4λm sin(2λm x) = 2 + 4λm sin(2λm L) x =0 2 [19] Combining equations [17], [18], and [19], and multiplying top and bottom by L to obtain the eigenvalue, αm = λmL, gives the following result for Bm. Bm = GS 1 λm sin(λm L) L  1 sin(2λm L) λm  +  2 4λm  = 4GS sin(λm L) 4G L sin(λm L) = 2 2 S 2λ L + λm sin(2λm L) 2λm L + λm L sin(2λm L) 2 m Substituting this expression for Bm into the general solution in equation [13] gives [20] [15] First midterm solutions Page 8 h   λn + tanh(λn H )  4GS L sin(λm L) cos(λn x)  k Θ( x , y ) = ∑ 2 2 cosh(λn y ) [21] sinh(λn y ) − h n=1 2λm L + λm L sin( 2λm L )   λn tanh(λn H ) + k   ∞ b) Show that the solution to this problem can be expressed in terms of the ratio (T – T∞)/GSL. Show that this ratio is a function of the following variables: x/L, y/H, H/L, and hL/k. We can substitute the definition of Θ = T – T∞ in equation [21] and divide that temperature difference by GSL to get the desired ratio. In addition we can multiply the fraction with the tanh terms top and bottom by L and introduce the roots of the eigenvalue equation [6], αn = λnL throughout equation [21] to obtain the following result.  hL H x   sin(α n ) cos α n   tanh α n  αn + T − T∞ y H  y H k L L     sinh cosh α − α =4     n n 2 H hL H L H L GS L        ( ) 2 sin 2 α + α α n n n n =1 α n tanh α n  +   L k    ∞ ∑ We see that the variables that are explicitly shown in this equation are the ones listed in the problem statement. In addition, the eigenvalues, αn, depend only on the ratio hL/k. (See equation [6].) Thus, equation [22] shows the desired functional behavior: (T – T∞)/(GSL) = f(x/L, y/H, H/L, hK/k). [22]