I'm currently teaching a PDE course and running through a lot of examples with students to build intuition. I came across an example that has me a bit stumped though.
Many first order linear (homogeneous) PDEs can be viewed as transport equations translating the initial data along characteristic curves. The classic example is the one-way wave equation $u_x \pm u_t = 0$ which will translate initial data along $t = \pm x$, respectively. Similar behavior can be seen for other first order linear PDEs, but some others break convention like the PDE
$$ u_x + x u_t = 0. $$
With initial condition $u(x,0) = e^{-x^2/2}$, the solution is $u(x,t) = e^{t-x^2/2}$. This does translate the initial data along characteristic curves, but it also has a growing aspect to it that I can't reason from the PDE directly. See the image below for the behavior over time.
![The solution to the PDE u_x + x u_t = 0 with initial condition e^{-x^2/2} on [-5,5]×[0,2]. The solution grows over time.](https://i.stack.imgur.com/ekLS4.png)
The characteristic curves are of the form $t = \frac{1}{2}x^2 + C$. Below is a plot of the solution with some of the characteristic curves laid on the surface for the solution $u(x,t)$ in blue and the initial data in black. It seems as if the growing behavior is captured in the region where the initial data cannot reach by moving along characteristic curves (above the parabola $t = \frac{1}{2}x^2$ in the $xt$ plane).
This growing behavior is not universal, for instance it does not happen with $1_{[-1,1]}(x^2-1)$ and yet it does for $\cos(x)$ (because $\cosh$ pops out of it due to imaginary solutions) and not for $\cos(x^2)$ (because the appearance of $x^2$ allows this to play nicely with the characteristic curve equations).
How do we physically interpret this behavior? Is there a way to see that the solution should do this directly from the PDE and initial conditions and not by exploration?
The solution is only uniquely determined at those points $(x,t)$ which are connected by a characteristic curve to a point on the $x$-axis (where the data are given) – in other words, at the points in the region $t \le x^2/2$.
Put differently, any classical solution $u \in C^1(\mathbf{R}^2)$ to the PDE must be constant along characteristic curves, so it must take the form $$ u(x,t) = f(t-x^2/2) $$ for some $f \in C^1(\mathbf{R})$. The condition $$ u(x,0) = f(-x^2/2) = \exp(-x^2/2) \qquad (x \in \mathbf{R}) $$ determines that $f(s)=\exp(s)$ for all $s \le 0$, but the values of $f(s)$ for $s>0$ can be assigned arbitrarily (except that $f$ has to be of class $C^1$, of course).
So the solution is highly non-unique in the region $t > x^2/2$.