# I feel really lost in my PDE class. Can somebody explain some things to me?

1. Mar 1, 2009

### Weilin Meng

I'm not going to blame anyone except for the fact that I'm probably a slow learner. Can somebody explain some of the things I'm learning in layman terms? That way I can have some context when I'm reading about them. Right now, the things I'm reading have no meaning, so it's really hard to understand it.

I would also really appreciate it if someone gave me a really loose proof too. You don't need to go into the math, I'm thinking something along the lines of "Well we take what is called a "blank" equation, and manipulate it until we get an inequality..this is useful because..etc"

Here is what I do understand so far:
Fourier series/transforms/integrals and why that is useful.
Solving first order and quasi-linear PDE's via separation of variables and characteristic method.

Ok here goes:

1. What is the "Energy Method" and what does it do?
2. What are "shocks"? and why is that a problem?
3. What does it mean what they say "Parabolic, Hyperbolic, Elliptical problem?"
4. What do they mean by "well-posedness and uniqueness?"

2. Mar 1, 2009

### yyat

Which book are you reading? Personally, I found "Partial Differential Equations" by L.C. Evans to be great for the beginner and it explains all these things very well.

1. Energy method: This involves some sort of "energy functional", for example

$$\int |\nabla f|^2dx$$

which represents the total energy of f. It is employed, for instance, in uniqueness proofs and variational methods.

2. Shocks are discontinuities in the solution, which can appear even if the initial conditions are continuous. One of the problems is determining which types of shocks are physically correct, i.e. finding the right concept of a weak solution.

3. See http://en.wikipedia.org/wiki/Partial_differential_equation#Classification".

4. Well posedness means that at least one solution exists, uniqueness that not more then one solution exists. Sometimes well posedness also means both of these things and possibly also that the solution depends continuously on the given data.

Last edited by a moderator: Apr 24, 2017
3. Mar 1, 2009

### Weilin Meng

I'm using Beginning Partial Differential Equations by Peter V. O'neil. I found your book on amazon but unfortunately it's too expensive for me. This is supposed to be an undergraduate course assuming only knowledge of multi-calc and linear algebra, yet I feel that the professor teaches it like a graduate course.

Anyway:

1. So the energy method is used only to determine uniqueness?
2. What are weak solutions?
3. Got it.
4. Got it.

4. Mar 1, 2009

### yyat

No, it can be used, for example in proving Dirichlet's principle or proving finite propagation speed for the wave equation. Energy functionals also play an important role in the calculus of variations.

Roughly speaking, these are solutions which are not differentiable but http://en.wikipedia.org/wiki/Weak_derivative". They have to satisfy certain integral equations, which are obtained from the PDE by partial integration.

Last edited by a moderator: May 4, 2017
5. Mar 2, 2009

### maze

One of the main motivating factors behind weak solutions is to extend the idea of a "derivative" to things that classically don't have a derivative, but really "ought to".

For example, f(x)=|x| has a derivative everywhere except at 0, but because of the one problem at zero, technically the whole function is "not differentiable". Therefore if we are looking for a classical solution to a DE, we have to automatically throw out f as a possible solution. This is throwing the baby out with the bathwater - after all, who really cares exactly what happens at a single point? If your PDE is modeling a physical phenomenon like the pressure field of flowing air, consider that no measuring device could actually measure the exact value at a single point. The best real measurement apparatus can do is measure some sort of "local average" near where you are trying to investigate. This is where the idea of weak derivatives and weak solutions comes in.

If you multiply the whole PDE by an arbitrary smooth test function (with compact support) and integrate by parts, this allows you to take derivatives off of the solution function and put them onto the smooth test function. This is the weak formulation. If the weak solution just so happens to be differentiable, then it has to agree with the classical solution since you could just integrate by parts back to where you started. On the other hand, you could have other solutions that are only slightly less "nice", eg |x|. Note that multiplying by a test function and integrating is a sort of "weighted average" of the solution near a point, where the test function is the weighting function. If the solution satisfies the PDE in the weak sense, then it has to hold for any possible test function. I think of this as being "morally equivalent" to having the solution hold for any possible measurement you make on it.

Last edited: Mar 2, 2009
6. Mar 2, 2009

### Weilin Meng

This is all pretty helpful. Does anyone know of a tutorial of using the energy method to determine uniqueness? It is not in my textbook.

7. Mar 2, 2009

### Weilin Meng

Ok I checked my notes and my professor wrote down how to use the energy method to determine uniqueness via. an example. I don't quite understand the example, but I will show it here and ask my questions in between the proof:

Let div(a∇u) = F in Ω
and u=g on ∂Ω

Assume there are two solutions u1 and u2 with same initial conditions
v= u1-u2 : Now we have a linear situation

By Linearity,
div(a∇u) = 0 in Ω
v=0 on ∂Ω
*Question* I don't understand why that is by linearity. And if v=0, then wouldn't that already show u1=u2? Why go through the rest of the proof then?

Technique: multiply the equation by v and integrate over the domain Ω via integration by parts

∫div(a∇u)vdx = 0 Where did the dx come from?

= $$\int_{\partial \Omega }^{}a\partial v/\partial n*vd\sigma - \int_{\Omega }^{}a\bigtriangledown u\cdot \bigtriangledown vdx$$
Ok I get that this is from integration by parts..but how did we get these boundaries? Why was one assigned ∂Ω and the other Ω?

v=0 so then:
$$\int_{\partial \Omega }^{}a\partial v/\partial n*vd\sigma = 0$$

and $$\bigtriangledown u\cdot \bigtriangledown v = |\bigtriangledown v|^{2}$$ Why?

so it all equals to:
$$\int a|\triangledown v|^{2}dx = 0$$

assume a(x) > 0 for all x in Ω (to get one sign only)

So, $$|\triangledown v(x)| =0$$ for all x since the integral is zero.
∇v = 0 in Ω (connected)
Did it matter that we assigned a(x) > 0 if |v(x)| = 0? Wouldn't we get the same thing for any a?

therefore v(x) = constant in Ω
But v=0 on ∂Ω
so v=0 in Ω

therefore u1=u2 and it is unique

8. Mar 2, 2009

### maze

Oh, I can see how this would be confusing. Here they are not multiplying by a test function and integrating for the purposes of forming a weak solution, but rather they are multiplying by the original function and integrating for other purposes.

I will try to answer your questions. The good news is that most of your confusion is because of typographical errors in notation and writing the equations down - switching up u's and v's and the like.

This looks like a typo here, it should go like this:
div(a∇u1) = F in Ω
div(a∇u2) = F in Ω
=> div(a∇u1-u2) = 0 in Ω
=> div(a∇v) = 0 in Ω

It should be ∫div(a∇v)v = 0, or more precisely ∫div(a∇v(x))v(x)dx = 0. v:Ω->R is a function of x, and the integrand is being integrated over Ω.

Again the notation is a little wonky here, they should both be v's. Should read
= $$\int_{\partial \Omega }^{}a\partial v/\partial n \cdot vd\sigma - \int_{\Omega }^{}a\bigtriangledown v\cdot \bigtriangledown vdx$$

Then of course it doesn't matter which way you integrated by parts since both are v. In other situations you might have u's and v's that are different, and then you might integrate by parts either way depending on your goal.

The whole point of the proof is to show that v=0, so for the purposes of proving that, we need to assume a(x) isn't negative or identically zero.

yep looks good.

9. Mar 2, 2009

### maze

As a side note - if you know a little electrostatics, recall that poisson's equation (the PDE div(grad(v))=F describes the electric potential v, where the electric field is the gradient of the potential. E=grad(v). Then also recall that the energy stored in a static electric field is proportional to E^2 = |grad(v)|^2. Thus the term "energy method" is somewhat appropriate.

The same sort of thing applies in other situations described by the same PDE (eg: steady-state heat equation, darcy flow, diffusion, and so on). |grad(v)|^2 will describe an energy. If you have some physics background go ahead and think about some of these situations a bit. It is kind of surprising that it all works out like that for so many widely different physical phenomenon.

10. Mar 2, 2009

### Weilin Meng

wow thanks! that made it a whole lot better. Unfortunately i've only gotten up to intro physics mechanics and E&M, so right now it's hard for me to think of these things in proper context. But your explanation does make the the term "energy method" make sense.

11. Apr 7, 2009

### Weilin Meng

Hello, I'm back with another question.

I realize that I have no idea what the Duhamel principle is and what it is used for. Is there somebody who can clearly explain Duhamel's principle and maybe provide an example too?