Solving the Heat Equation — Question 3

Question 3 – Solution

Strategy. The corner in the initial profile appears as a slope jump in the coefficient calculation, not as a failure of the positive-time heat solution.

Step 1: Compute the contribution of the slope jump. With $k=n\pi/L$, integration by parts once gives $\int f\sin(kx)=k^{-1}\int f'\cos(kx)$ since $f$ vanishes at both endpoints. Using $f'=1/a$ to the left and $-1/(L-a)$ to the right gives $$\int_0^L f\sin(kx)\,dx=\frac{L\sin(ka)}{a(L-a)k^2}.$$ Therefore $$\boxed{b_n=\frac{2L^2\sin(n\pi a/L)}{\pi^2n^2a(L-a)},\qquad u=\sum_{n=1}^\infty b_n e^{-\kappa(n\pi/L)^2t}\sin(n\pi x/L).}$$

Step 2: Separate initial regularity from later regularity. Since $|b_n|\le C/n^2$, the sine series converges absolutely and uniformly, and Fourier convergence identifies its initial sum as the continuous tent. Summable domination gives uniform convergence to $f$ as $t\downarrow 0$. For every $\tau>0$, polynomial factors from any finite number of derivatives are dominated by Gaussian decay for $t\ge\tau$. Thus the series is smooth for positive time, satisfies the PDE termwise and retains zero endpoints. Its initial derivative jump does not persist as a corner in a positive-time profile.

Step 3: Recover the peak location from two modes. Put $\theta=\pi a/L\in(0,\pi)$. Then $b_1>0$ and $$\frac{b_2}{b_1}=\frac{\sin 2\theta}{4\sin\theta} =\tfrac 12\cos\theta,\qquad \boxed{a=\frac L\pi\arccos(2b_2/b_1).}$$ The ratio must lie strictly between $-1/2$ and $1/2$. Cosine is strictly decreasing on $(0,\pi)$, so this reconstruction is unique in the stated family.

Step 4: Specialize and interpret the profiles. For $L=\kappa=1$, $a=1/3$, $b_n=9\sin(n\pi/3)/(\pi^2n^2)$. Exactly the multiples of three vanish, and $b_2/b_1=1/4$ recovers $a=1/3$. The figure uses the exact piecewise initial trace and rapidly convergent positive-time series. The peak rounds as the profile diffuses toward zero.