Solving the Heat Equation — Question 8

Question 8 – Solution

Strategy. The source has one spatial eigenfunction, so the full problem reduces to a forced scalar decay equation.

Step 1: Solve the amplitude equation including the exceptional rate. Substitution gives $q_\alpha'+q_\alpha=e^{-\alpha t}$, $q_\alpha(0)=0$. Multiplying by $e^t$ and integrating yields $$\boxed{q_\alpha(t)=\begin{cases} (e^{-\alpha t}-e^{-t})/(1-\alpha),&\alpha\ne 1,\\ te^{-t},&\alpha=1. \end{cases}}$$ The exceptional formula follows from the same integrating-factor integral; it is not obtained by substituting into a zero denominator.

Step 2: Verify existence, sign and uniqueness. The scalar ODE verifies the PDE, while the sine factor and $q_\alpha(0)=0$ verify endpoints and initial data. The integral representation $$q_\alpha(t)=\int_0^t e^{-(t-s)}e^{-\alpha s}\,ds$$ is strictly positive for $t>0$. The difference of two solutions with the same source solves the homogeneous zero-data problem. Its squared $L^2$ norm is nonincreasing by integration by parts; starting at a positive time and using the zero initial norm proves uniqueness in the stated class.

Step 3: Locate the unique maximum. For $\alpha\ne 1$, $q_\alpha'=(e^{-t}-\alpha e^{-\alpha t})/(1-\alpha)$. The equation $q_\alpha'=0$ is $e^{(\alpha-1)t}=\alpha$, giving $$\boxed{t_\alpha=\frac{\log\alpha}{\alpha-1}>0.}$$ There is exactly one root; $q_\alpha'(0)=1>0$ and the derivative is negative for sufficiently large time, proving it is the unique maximum. As $\alpha\to 1$, $t_\alpha\to 1$. Directly $(te^{-t})'=e^{-t}(1-t)$, with maximum $1/e$ at $t=1$.

Step 4: Interpret the forced heat balance. The total heat is $H=2q_\alpha$. Since $[u_x]_0^\pi=-2q_\alpha$ and the source integral is $2e^{-\alpha t}$, $$H'=-2q_\alpha+2e^{-\alpha t},$$ which agrees with the scalar ODE. At $\alpha=1$, matching the source decay rate to the homogeneous decay produces $te^{-t}$. This field is bounded and tends to zero; its time factor does not imply unbounded oscillatory resonance. For every fixed $\alpha>0$, the source and the solution ultimately decay.