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210 CHAPTER 3. BOUNDARY VALUE PROBLEMS Given the property (3.3.45), the appropriate choice of Fourier series is a sine series. If the boundary conditions are
210 CHAPTER 3. BOUNDARY VALUE PROBLEMS Given the property (3.3.45), the appropriate choice of Fourier series is a sine series. If the boundary conditions are of the type in (3.3.9), then the appropriate choice of the particular solution is a cosine series (3.3.11). 3.3.5 Exercises The following exercises provide some more practice in solving boundary value problems. 1. Construct the solution to the following dierential equations and the given boundary conditions. Note that homogeneous versions of these exercises appear in exercise 1 of Sec. 3.3.4. (a) d2 f df (z) + 5 (z) + 6 f (z) = z , dz 2 dz subject to df df ( L) = 1 , (L) = 1 . dz dz Answer: With L = 1, f (1/2) = 1.171. (b) d2 w dw (y) + 2 (y) + 5 2 w(y) = e dy 2 dy subject to w(0) = 0 , Answer: For = 20, w(1/8) = 2y , dw (1/4) = 0 . dy 0.26. (c) d2 y (x) + k 2 y(x) = sin(3x) , dx2 subject to y(0) = 0 and y(L) = 0. The parameter k > 0. Answer: For k = 3, L = 1 and = 6, y(0.5) = 7.03. 2. Consider what happens when the characteristic equation has just one root. Solve d2 y dy (x) + 4 (x) + 4 y(x) = , 2 dx dx subject to y(0) = y(L) = 0. Answer: For L = 1, y(1/2) = 0.452 . The next exercises consider various applications of boundary value problems. 3. Let's consider some specic examples of the inuence of the ambient temperature on the heated plate. The dierential equation is (3.3.10) and assume that the ends are insulated (3.3.9). Consider the following proles for the ambient temperature. 3.3. FORCING EFFECTS 211 (a) T0 (x) = + sin (b) T0 (x) = ( , 2 4 (2x/L 2 1) , x . L for L/2 < |2x L| < L , for |2x L| < L/2 . Construct the solutions and compare the results by graphing them. Answer: The proles for the two cases are shown below for the choice L = 5. T(x)/ 2 1.5 1 (a) (b) 0.5 0 0 0.2 0.4 0.6 0.8 1 x/L 4. Repeat the previous exercise but now assume the boundary conditions are given by (3.3.21). Answer: The proles for the two cases are shown below for the choice = 5, Ta / = 1 and Tb / = 2. T(x)/ 2 1.5 1 (a) (b) 0.5 0 0 0.2 0.4 0.6 x/L 0.8 1 \f\f
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