# Uniquely Determined? - Partial Differential Equations

1. Mar 1, 2009

### bobcat817

1. The problem statement, all variables and given/known data

(a) Solve the equation $$yu_{x} + xu_{y}$$ = 0 with the condition u(0,y) = $$e^{-y^{2}}$$. Okay. My tex has gone wrong. Those are supposed to be subscripts in the the equation. I'm not sure why they aren't.
(b) In which region of the xy plane is the solution uniquely determined.

2. Relevant equations

3. The attempt at a solution

I have solved part (a) e$$^{x^{2}-y^{2}}$$. What I don't understand is the meaning of uniquely determined. I can't find it in my textbook or online. I'm really not interested in the solution, I only included the problem to give a background for what I was asking. So, if someone can explain to me what uniquely determined means, that would be brilliant. Cheers.

Last edited by a moderator: Mar 2, 2009
2. Mar 1, 2009

### Bacat

Differential equations describe the slopes of functions. Unless there is a boundary condition to uniquely determine the solution, the solution is a family of equations (all with the same slope and same differential equations). This is a consequence of integration being equal up to a constant C, and why you should always include the constant C when you perform indefinite integrals.

So part b seems to be asking in what regions of the cartesian plane you have boundary conditions for.

Any region where you dont have a boundary condition is not uniquely determined. u(0,y) is your boundary condition.

3. Mar 2, 2009

### HallsofIvy

Staff Emeritus
I'm not at all clear what Bacat is saying. It doesn't make sense to me. Solutions to artial differential equations involve unknown functions, not constants. u(0,y) is an "initial condition", not a "boundary condition", and there is no region where "you don't have a boundary condition".

bobcat817, yes, the "general solution" if $$u(x,y)= f(x^2- y^2)$$, where f can be any differentiable function and the soltution to this particular problem is $$u(x,y)= e^{x^2- y^2}$$.

But notice that on the two lines where $$x^2- y^2= 0$$, the "characteristics", we could replace that solution with u(x,y)= constant, using the differentiable function f= constant. That is, we can "patch" together two different solutions to get a new one. The solution is "uniquely determined" as long as we do not cross one of those lines.

By the way, it is better to put whole equations or expresions inside the 'tex' labels, not just individual symbols.

4. Mar 2, 2009

### bobcat817

Thanks for the tex advice.

This does clarify things a bit, but I have a few more questions. To begin with, can you give me an example of "patching" together two different solutions to get a new one? I think I am following but I'm not entirely sure. Also, u(x,y)= constant because a function of 0 is necessarily a constant, correct? So, the two lines (y=x and y=-x) break the xy plane into four regions, but I can think of them as {$${(x,y)|x^2 < y^2}$$} and {$${(x,y)|x^2 > y^2}$$}. The second one doesn't satisfy the initial condition. So, then I would think the answer is {$${(x,y)|x^2 < y^2}$$}, but when I looked at the answer that was given, it is {$${(x,y)|x^2 \leq y^2}$$}. I don't understand why that is the case. Is it okay for it to equal the line, just not "cross" it?

I really want to understand this, so any additional help would be greatly appreciated.

5. Mar 2, 2009

### Bacat

Hello HallsofIvy, you can think of an initial condition as a boundary condition, and in most physics or engineering problems that use PDEs they are referred to as boundary conditions. This is because we use them to describe heat flow or hydraulic conductivity or quantum mechanics where we have definite boundaries.

You're right that the solutions are not always constants. I think I may have connotated that when I said that the family of solutions all have the same slope. It would be more correct to say that the family of solutions all have the same functional forms, up to a constant.

In this case, the only region where there is a boundary condition, or initial condition, is at $$u(0,y)$$. So the uniquely determined solution is one where $$x=0$$. If you take away that condition, there is no unique solution at all since x and y can take any value. That region is called the y-axis.

I'm certainly no math expert, so if I have made an error, please let me know. I'm just trying to help.