Solving the Heat Equation — Question 10

Question 10 – Solution

Strategy. Orthogonality turns energy decay and symmetry restrictions into precise statements about which coefficients are present.

Step 1: Prove the sharp basic energy decay. For $t>0$, integration by parts gives $E'=\kappa\int uu_{xx}=-\kappa\int u_x^2$ because $u=0$ at both ends. With $\lambda_1=(\pi/L)^2$, Parseval yields $$E(t)=\frac L4\sum_{n\ge 1}b_n^2e^{-2\kappa n^2\lambda_1t} \le e^{-2\kappa\lambda_1t}E(0).$$ For fixed $t>0$, every term with $n\ge 2$ has strictly smaller decay factor. Equality for nonzero data holds exactly when $f=b_1\sin(\pi x/L)$ with $b_1\ne 0$. This also proves sharpness of the uniform energy rate.

Step 2: Use antisymmetry to improve the rate. Changing variable $x\mapsto L-x$ in the coefficient integral and using $\sin(n\pi(L-x)/L)=(-1)^{n+1}\sin(n\pi x/L)$ gives $b_n=(-1)^n b_n$. Thus odd coefficients vanish. The smallest remaining index is two, giving $$\boxed{E(t)\le e^{-8\kappa\pi^2t/L^2}E(0).}$$ Nonzero equality data are precisely scalar multiples of $\sin(2\pi x/L)$. The symmetry is preserved by the evolved even-mode series.

Step 3: Identify the sign constraint. If $f\ge 0$ almost everywhere and is nonzero in $L^2$, then it is positive on a set of positive measure. The first sine is strictly positive on $(0,L)$; therefore $b_1=(2/L)\int f\sin(\pi x/L)>0$. Similarly a nonpositive nonzero datum has $b_1<0$. Thus any nonzero real datum with $b_1=0$ must have both positive and negative parts of positive measure. Removing the first mode is possible, but not for a nontrivial nonnegative initial temperature relative to the zero boundaries.

Step 4: Prove the normalized asymptotic statement. After removing the first term, Parseval gives $$\left\|e^{\kappa\lambda_1t}u-b_1\sin(\pi x/L)\right\|_2^2 =\frac L2\sum_{n\ge 2}b_n^2e^{-2\kappa(n^2-1)\lambda_1t} \le e^{-6\kappa\lambda_1t}\|f\|_2^2.$$ The remainder norm is at most $e^{-3\kappa\lambda_1t}\|f\|_2$ and tends to zero. If $b_1=0$, the normalized limit is zero and the actual field decays at least at the second-mode norm rate $e^{-4\kappa\lambda_1t}$. The next nonzero coefficient, if present, can give an even faster decay rate.