# Quick PDE question

1. Oct 2, 2004

### bullet_ballet

"Verify that, for any C¹ function f(x), u(x, t) = f(x - ct) is a solution of the PDE u_t + c u_x = 0, where c is a constant and u_t and u_x are partial derivatives."

I managed to get the solution for this and a similar problem by showing that the new variable (x - ct in this case) satisfies the PDE, but why does doing that work? Is it because if x-ct is a solution so is any function of it? That just doesn't sound right since you'd need at least two linearly independent solutions to make that kind of generalization. Or maybe I'm just an idiot. :)

Mucho thanks if anyone can point out what basic thing I'm missing.

Last edited: Oct 2, 2004
2. Oct 2, 2004

### Clausius2

f(x-ct) represents a function that is moving along the x axis with velocity c. The variable x-ct is widely used in wave propagation phenomena, because making this traslation you are travelling just over the wave edge, so that some variables remain constant viewed at that point. It is true that f(x-ct) is the solution of your PDE (named a characteristic equation), but f has to be figured out by means of your boundary conditions.

I'm not sure of having answered you completely.

3. Oct 2, 2004

### bullet_ballet

I don't think I was specific enough. I was asking why I could use x-ct to satisfy the PDE and show that f(x-ct) is therefore a solution.

Having done this for a couple of different problems, it seemed like I could draw the conclusion that if the PDE held for some a(x,t), then the solution to the PDE could be stated as u(x,t) = f(a(x,t)).

4. Oct 2, 2004

### Clausius2

x-ct is NOT a solution of your PDE. It's only a "special variable" which enhances a simpler solution. Why, because viewing your PDE one can establish:

$$\frac{dx}{dt}=c$$ ----> along the rect $$x-ct=x_o$$ u(x,t)=f(xo) is a constant.

f is a solution of your PDE but x-ct is not a solution.

5. Oct 2, 2004

### arildno

Sure, Clausius2!
u(x,t)=x-ct is certainly A solution of that differential equation.
(f is here the identity function)

6. Oct 3, 2004

### Clausius2

You are not allowed to substitute any value for f if you do not know (as me) the boundary conditions. So that, for the problem he posted, f=1 is not a general solution. Perhaps some boundary conditions enhances your reply, but this is not the case.

7. Oct 3, 2004

### arildno

Without boundary conditions, a differential equation has a multitude of solutions.
If you put on boundary conditions and the problem is well-posed, you'll have a unique f satisfying BOTH the differential equation and the boundary conditions.
I never said f=x-ct (or f=1) are a GENERAL solution, I said it was A solution.

8. Oct 3, 2004

### Clausius2

Ok Norwegian (well written?, sorry if not) friend, If you want we could come up with a settlement:

u=x-ct is only a solution for a concrete boundary conditions.
u=f(x-ct) is the most general solution (not general in the sense you wrote, only in the sense of generallity) for that problem.

Is it Ok?

9. Oct 3, 2004

### arildno

u=f(x-ct) is certainly, indisputably, factually and necessarily the most general solution (of the differential equation).

There exist a well-posed problem in which u(x,t)=x-ct is the unique solution.
Is that OK?

(I don't think we had much disagreement in the first place..)

10. Oct 3, 2004

### Clausius2

Ha Ha Ha! :rofl:

OK! We have reached an agreement!

Where are the beers?

(What would say Bullet Ballet of all this stuff? )

11. Oct 3, 2004