The Heat Equation — Question 6

Question 6 – Solution

Strategy. Differentiate the amplitude and width together; local warming can coexist with a falling global peak and conserved heat.

Step 1: Verify the differential equation. Put $v=u-T_a$ and $\sigma=t+\tau>0$. Logarithmic differentiation gives $$v_t=v\left(-\frac 1{2\sigma}+\frac{x^2}{4\alpha\sigma^2}\right),\qquad v_{xx}=v\left(-\frac 1{2\alpha\sigma}+\frac{x^2}{4\alpha^2\sigma^2}\right).$$ Thus $v_t=\alpha v_{xx}$. Omitting the derivative of $\sigma^{-1/2}$ would remove the first term and invalidate the equation.

Step 2: Integrate the excess heat. With $z=x/\sqrt{4\alpha\sigma}$, $$\boxed{\mathcal H=C\int_{\mathbb R}v\,dx=2CB\sqrt{\pi\alpha}.}$$ The factor from the increasing width cancels the decreasing prefactor. The Gaussian flux tends to zero at both infinities, consistent with no net heat loss from the infinite rod.

Step 3: Measure spreading. Symmetry gives center zero. The provided Gaussian integrals yield $$\boxed{\frac{\int_{\mathbb R}x^2v\,dx}{\int_{\mathbb R}v\,dx} =2\alpha\sigma,\qquad \max_x v=\frac{B}{\sqrt{\sigma}}.}$$ Thus the root-mean-square width grows as $\sqrt{2\alpha(t+\tau)}$, while the peak decreases as $(t+\tau)^{-1/2}$. The inflection points occur at $x=\pm\sqrt{2\alpha\sigma}$.

Step 4: Locate local warming. Since $v>0$, $u_t>0$ precisely when $x^2>2\alpha(t+\tau)$. A fixed point reaches its maximum on $t\geq 0$ at $$\boxed{t_*=\max\!\left(0,\frac{x^2}{2\alpha}-\tau\right).}$$ Outside the initial inflection points it first warms and then cools. Inside them it cools from the start; equality gives an initially zero derivative. Therefore diffusion does not force temperature to decrease at every location. The graph uses excess temperature $v$, so the ambient offset is removed.