4
$\begingroup$

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.

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).

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.

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?

$\endgroup$
3
  • 1
    $\begingroup$ Well, the solution isn't uniquely determined in the region $t > x^2/2$, since the points in that region aren't connected by any characteristic curves to the $x$-axis (where the data are given). So your growing solution $u = e^{t-x^2/2}$ isn't the only possible one. $\endgroup$ Commented Feb 6 at 18:31
  • $\begingroup$ Ohh you're absolutely right @HansLundmark. I didn't think about that though my playing around in Desmos3D and Mathematica should have tipped me off there as I kept getting mismatched solutions. I guess Mathematica was just taking something akin to an analytic continuation into that region to make a "unique" solution. It never occurred to me that there wouldn't be a "valid" (read: unique) solution for a first order linear PDE in a region. Do you want to make that an answer? $\endgroup$ Commented Feb 6 at 18:36
  • $\begingroup$ Sure, I just did. Great question, by the way! $\endgroup$ Commented Feb 6 at 20:19

1 Answer 1

3
$\begingroup$

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 $t > x^2/2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .