The solution to the one-dimensional heat equation $u_{t} = ku_{xx}, -\infty < x < +\infty, t > 0$, with initial condition $u(x,0) = e^{-x}$ is given by $u(x,t) = e^{kt-x}$. According to the maximum principle, on any bounded interval, say, $x\in[0,l]$, the solution should not grow with time, but this particular example however, grows with t. Strauss explains that this is because
the left side of the rod is initially very hot [u(x,0) $\rightarrow +\infty$ as $x \rightarrow -\infty$] and the heat gradually diffuses throughout the rod.
The weak maximum stated in the book says:
If $u(x, t)$ satisfies the diffusion equation in a rectangle (say, $0 \leq x \leq l, 0 \leq t \leq T)$ in space-time, then the maximum value of $u(x, t)$ is assumed either initially $(t=0)$ or on the lateral sides $(x=0$ or $x=l)$.
Although it restricts the solution on the interval $[0,l]$, while our initial condition is on the whole real line, but I think we restrict the initial value on a bounded interval as well. So, why does the maximum principle "fail" here?
Best Answer
The maximum principle does not fail. For any $\ell>0$ and $T>0$, the maximum of $u(x,t) = e^{kt-x}$ in the closed rectangle $[0,\ell] \times [0,T]$ occurs at $(x,t)=(0,T)$. This is on the lateral side $x=0 $ so there is no contradiction.