Question 11.3.2: Find the solution of the heat conduction problem ut = uxx + ......

Again, we use the eigenvalues \lambda_{n} and the normalized eigenfunctions \phi_{n} of the problem (17) and assume that u is given by equation (28)

u(x,t)=\sum_{n=1}^{\infty}b_{n}(t)\,\phi_{n}(x)\qquad\qquad\qquad\qquad\qquad\qquad(28)

u(x,t)=\sum_{n=1}^{\infty}b_{n}(t)\phi_{n}(x).

The coefficients b_{n} are determined from the differential equation

b_{n}^{\prime}+\lambda_{n}b_{n}=\gamma_{n}(t), (44)

where \gamma_{n}(t) is the coefficient of the n^{th} term in the eigenfunction expansion of the nonhomogeneous term x e^{-t}. Thus we have

\gamma_{n}(t)=\int_{0}^{1}x e^{-t}\phi_{n}(x)d x=e^{-t}\int_{0}^{1}x\phi_{n}(x)d x=c_{n}e^{-t}, (45)

where c_{n}=\int_{0}^{1}x\phi_{n}(x)d x is given by equation (21).

c_{n}={\frac{2\sqrt{2}\sin\sqrt{\lambda_{n}}}{\lambda_{n}(1+\cos^{2}\sqrt{\lambda_{n}})^{1/2}}}. (21)

The initial condition for equation (44) is

b_{n}(0)=0 (46)

since the initial temperature distribution (43) is zero everywhere. The solution of the initial value problem (44), (46) is

b_{n}(t)=e^{-\lambda_{n}t}\int_{0}^{t}e^{\lambda_{n}s}c_{n}e^{-s}d s=c_{n}e^{-\lambda_{n}t}\ {\frac{e^{(\lambda_{n}-1)t}-1}{\lambda_{n}-1}}\\=c_{n}{\frac{e^{-t}-e^{-\lambda_{n}t}}{\lambda_{n}-1}}. (47)

Thus the solution of the heat conduction problem (41) to (43) is given by

u(x,t)=4\sum_{n=1}^{\infty}{\frac{\left(\sin\sqrt{\lambda_{n}}\right)\left(e^{-t}-e^{-\lambda_{n}t}\right)\sin\left(\sqrt{\lambda_{n}} x\right)}{\lambda_{n}(\lambda_{n}-1)\left(1+\cos^{2}\sqrt{\lambda_{n}}\right)}}. (48)

The solution given by equation (48) is exact but complicated. To judge whether a satisfactory approximation to the solution can be obtained by using only a few terms in this series, we must estimate its speed of convergence. First we split the right-hand side of equation (48) into two parts:

u(x,t)=4e^{-t}\sum_{n=1}^{\infty}\frac{\sin\sqrt{\lambda_{n}}\,\sin\left(\sqrt{\lambda_{n}} x\right)}{\lambda_{n}(\lambda_{n}-1)\left(1+\cos^{2}\sqrt{\lambda_{n}}\right)}-4\sum_{n=1}^{\infty}\frac{e^{-\lambda_{n}t}\sin\sqrt{\lambda_{n}}\,\sin\left(\sqrt{\lambda_{n}} x\right)}{\lambda_{n}(\lambda_{n}-1)\left(1+\cos^{2}\sqrt{\lambda_{n}}\right)}. (49)

Recall from Example 1 in Section 11.1 that the eigenvalues \lambda_{n} are very nearly proportional to n². In the first series on the right-hand side of equation (49), the trigonometric factors are all bounded as n → ∞; thus this series converges similarly to the series \sum_{n=1}^{\infty}\lambda_{n}^{-2} or \sum_{n=1}^{\infty}n^{-4}. Hence at most two or three terms are required for us to obtain an excellent approximation to this part of the solution. The second series contains the additional factor e^{-\lambda_{n}t}, so its convergence is even more rapid for t > 0; all terms after the first are almost surely negligible.