Solving the Heat Equation — Question 2

Question 2 – Solution

Strategy. Exact coefficients control the initial convergence, while Gaussian mode decay justifies later differentiation.

Step 1: Compute the sine coefficients. Put $k=n\pi/L$. Twice integrating by parts, using $f(0)=f(L)=0$ and $f''=-2$, gives $\int_0^L f\sin(kx)\,dx=2(1-(-1)^n)/k^3$. Thus $b_n=4L^2(1-(-1)^n)/(n^3\pi^3)$, and $$\begin{gathered} \boxed{u(x,t)=\frac{8L^2}{\pi^3} \sum_{\substack{n\ge 1\\n\text{ odd}}}\frac{\sin(n\pi x/L)}{n^3} e^{-\kappa(n\pi/L)^2t}.} \end{gathered}$$

Step 2: Verify the full initial-value solution. The coefficients are absolutely summable, so the series at $t=0$ converges uniformly. Fourier convergence identifies its sum as $f$ on $[0,L]$, including the zero endpoints. Dominated convergence of the coefficient sum then gives $\|u(\cdot,t)-f\|_\infty\to 0$. For $t\ge\tau>0$, any fixed number of spatial or time derivatives only adds powers of $n$; the factor $e^{-\kappa(n\pi/L)^2\tau}$ makes these series uniformly summable. Hence $u_t=\kappa u_{xx}$ termwise for positive time. At an initial corner, $f''=-2$ conflicts with the zero boundary time derivative; continuity of all second spatial derivatives through those corners is not claimed.

Step 3: Use symmetry and integrate the solution. Since $f(L-x)=f(x)$ and $\sin(n\pi(L-x)/L)=(-1)^{n+1}\sin(n\pi x/L)$, even coefficients vanish. Integrating the uniformly convergent series gives $$H(t)=\frac{16L^3}{\pi^4}\sum_{n\text{ odd}}\frac{e^{-\kappa(n\pi/L)^2t}}{n^4}, \qquad H(0)=\int_0^L x(L-x)\,dx=\frac{L^3}{6}.$$

Step 4: Check heat loss and the mode content. For $t>0$, differentiation gives $$H'(t)=-\frac{16\kappa L}{\pi^2}\sum_{n\text{ odd}} \frac{e^{-\kappa(n\pi/L)^2t}}{n^2}=\kappa[u_x]_0^L.$$ The equality on the right follows by differentiating the spatial sine series and using the odd-mode cosine endpoint values. Every summand is positive, so $H'<0$. Infinitely many nonzero coefficients decay at distinct rates; in particular the ratio of the third coefficient to the first changes with time. A single product would have fixed ratios wherever its time factor is nonzero.