disclaimer

This post contains explicit math. In fact, there’s so much math, your head just might explode. If you don’t like math, you’re better off doing something else.

Like having a tea party with the girlfriends.

Wuss.

This post is going to be a little bit different.

For starters, there aren’t any pictures of cats.

Instead, I’m going to spend a little bit of time writing about what happens when the solution to a partial differential equation has a jump discontinuity.

Before we get started, let’s look at two cases.

Case A: If the above didn’t make any sense whatsoever…

You should read my last post. If you’re still bored, you can drool over my choco-licious lime biscotti. And if that doesn’t do it, you can laugh at how I looked as a kid.

Otherwise, come back tomorrow.

Case B: If the above gets your blood flowing and produces a strong tingle in your special place…

Read on.

Actually, this topic is pretty easy to understand. What’s odd is that most references I’ve managed to dig up on the problem explain it horrendously at best. But as long as you have a clear understanding of how to solve PDEs using the method of characteristics, you should be well on your way to bigger and better things after this lil’ talk.

Let’s begin with a generalized first-order quasi-linear equation,

We’re going to still be interested in finding solutions u(x,y), but now we’d like to relax the condition that solutions have to be differentiable everywhere.

Weak Discontinuity: Let u(x,y) be defined in some domain D. Let C be a curve parametrized by x = x(t), and y = y(t). Assume that u(x,y) is continuous and differentiable everywhere except on C, across which a first-order derivative of the solution may be discontinuous. This is a weak discontinuity.

Now we can look at the limtiing value of the functions as the curve is approached from either side, so from the chain rule,

But what does that mean? Be sure to distinguish the two types of derivatives above. The dot, as in , is the derivative along the curve (with respect to t). This is in contrast to the partial derivatives, which measure the rates of change of u(x,y) in the x and y directions.

But actually, because u is continuous,

Why? Imagine it this way: Cars are travelling along the road C. An observer to the right of the road measures the speed of the car as . But on the other side, his buddy measures the speed of the car as . Assuming that both observers are sufficiently close (continuity), they’ll obviously measure the same speed.

And so, subtracting (1) from (2) we have,

where the square bracket denotes the ‘jump’ in the expression across C.

Remember that u(x, y) should satisfy the quasi-linear PDE except on C, so by subtraction we have,

Notice the c disappears because of the continuity of u, and this also explains why a and b can be evaluated on C without ambiguity from any direction.

(3) and (4) provide us with a system of equations for the size of the jump discontinuities. And of course, since we’re interested in non-zero jump discontinuities, we’ll require the determinant,

But this is the precise equation for the characteristic curves of the PDE!

Wowsers!

Solutions with Weak Discontinuities: If the solution of a quasi-linear PDE has a weak discontinuity across some curve C, then C must be a characteristic projection.

Actually, we can squeeze a little bit more information out of this theory. It turns out the size of the jumps across C can be calculated from a first order ordinary differential equation. For example, in the case of a linear PDE,

where we follow the characteristic curves and ,

And so if the jump is non-zero at some point in space (x,y), it will continue to be non-zero along a characteristic. More importantly, if there is initially a weak discontinuity at t = 0, then this jump will propogate along the characteristic.

Propogation of Weak Discontinuities: Suppose that there is initially a jump in the derivatives of the initial boundary data (Cauchy problem). Then this jump will propogate along its characteristic curve. Furthermore, if there is no jump, then the solution will be well behaved (around the characteristic).

Let’s finish with a bang.

Let’s solve the following PDE

subject to the Cauchy conditions:

using the method of characteristics, it’s quite easy to show that the solution is given by,

Take a look at the graph.

Notice once you’ve pieced everything back together, the solution is continuous for t > 0. The important observation is of course that the solution is smooth everywhere, except along the two special characteristics , where the discontinuity from the Cauchy data propogates along the characteristic.

And that about completes our look into what happens when there’s a jump discontinuity in the solutions of a PDE.

But we’ve only gotten started!

The more interesting and deeper question is what happens when the solution itself is discontinuous. These solutions contain so-called shock waves and are termed weak solutions. Does this happen in real life? Sure. Take a look at this picture.

Notice that a description of the surface waves would be discontinuous in the region where the waves are about to break, or the region where the wave is about to ‘bend’ over itself.

The basic ideas behind studying discontinuous solutions remains the same as for the above problem. Except now we can’t really use expressions like,

Right? Because since u(x,y) is itself discontinuous at certain points, we can’t be so cavalier about taking derivatives! The key to approach these types of problems is to consider equivalent integral formulations, as we know that integration is more robust with respect to discontinuous solutions.

But that’s all for another day.

Phew. I’m going to go lie down now.