Solving the Heat Equation — Question 4

Question 4 – Solution

Strategy. Interpret the initial condition in its stated norm; incompatible corners and jumps do not invalidate a positive-time series solution.

Step 1: Compute all modal amplitudes. Integration over the nonzero half of the interval gives $$\boxed{b_n=\frac{2}{\pi n}\bigl(1-\cos(n\pi/2)\bigr),\qquad u(x,t)=\sum_{n\ge 1}b_ne^{-n^2t}\sin(nx).}$$ The point value assigned at $\pi/2$ does not affect the integral. Modes divisible by four vanish; the other coefficients must not all be replaced by the odd-only coefficients for a constant temperature on the entire interval.

Step 2: Verify positive-time smoothness and the norm trace. For $t\ge\tau>0$, $|b_n|\le 4/(\pi n)$ and the exponential makes all finitely differentiated series uniformly convergent. Thus the PDE and endpoints hold. Parseval and the sine expansion of $f$ give $$\|u(\cdot,t)-f\|_2^2=\frac\pi 2\sum_{n\ge 1}b_n^2(1-e^{-n^2t})^2\longrightarrow 0,$$ by summable domination, since $\sum b_n^2<\infty$. This is precisely the specified initial condition.

Step 3: State the pointwise limits correctly. The limits are $1$ on $(0,\pi/2)$, $0$ on $(\pi/2,\pi)$ and $1/2$ at the jump. To justify Gaussian damping of a pointwise convergent Fourier series, apply summation by parts: the decreasing weights $e^{-n^2t}$ average its partial sums, have total weight tending to one, and move past each fixed index as $t\downarrow 0$. Thus they preserve the Fourier limit. At both endpoints $u$ is identically zero for $t>0$, so the limit is zero. A uniform limit of continuous functions on $[0,\pi]$ would be continuous; the jump and the left corner mismatch preclude uniform convergence to these data.

Step 4: Prove uniqueness in the stated class. Let $w$ be the difference of two such solutions. For $t\ge\delta>0$, $$\frac d{dt}\|w\|_2^2=2\int_0^\pi ww_{xx}\,dx=-2\int_0^\pi w_x^2\,dx\le 0.$$ The boundary term vanishes. Hence $\|w(\cdot,t)\|_2\le\|w(\cdot,\delta)\|_2$. Both solutions have the same $L^2$ initial trace, so the right side tends to zero as $\delta\downarrow 0$. Positive-time continuity then makes $w=0$ pointwise. No smoothness at the discontinuous initial trace was assumed.