Find the solution of the heat conduction problem ut = uxx + xe^−t , 0 0, (42) u(x, 0) = 0, 0

Question

Find the solution of the heat conduction problem

[latex]u_{t}=u_{x x}+x e^{-t},\quad 0\lt x\lt 1,\quad t\gt 0,[/latex] (41)

[latex]u(0,t)=0,\quad u_{x}(1,t)+u(1,t)=0,\quad t\gt 0,[/latex] (42)

[latex]u(x,0)=0\quad 0\lt x\lt 1.[/latex] (43)

Accepted Answer

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

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

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

The coefficients [latex]b_{n}[/latex] are determined from the differential equation

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

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

[latex]\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},[/latex] (45)

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

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

The initial condition for equation (44) is

[latex]b_{n}(0)=0[/latex] (46)

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

[latex]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}}.[/latex] (47)

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

[latex]u(x,t)=4\sum_{n=1}^{\infty}{\frac{\left(\sin\sqrt{\lambda_{n}} ight)\left(e^{-t}-e^{-\lambda_{n}t} ight)\sin\left(\sqrt{\lambda_{n}} x ight)}{\lambda_{n}(\lambda_{n}-1)\left(1+\cos^{2}\sqrt{\lambda_{n}} ight)}}.[/latex] (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:

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

Recall from Example 1 in Section 11.1 that the eigenvalues [latex]\lambda_{n}[/latex] 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 [latex]\sum_{n=1}^{\infty}\lambda_{n}^{-2}[/latex] or [latex]\sum_{n=1}^{\infty}n^{-4}[/latex]. 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 [latex]e^{-\lambda_{n}t}[/latex], so its convergence is even more rapid for t > 0; all terms after the first are almost surely negligible.

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

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