# Finding solution of a PDE

1. Mar 24, 2009

### somethingstra

1. The problem statement, all variables and given/known data
Assume we are in the open first quadrant in the (x,y) plane
Say we have u(x,y) a C1 function in the closed first quadrant that satisfies the PDE:

$$u_{y}=3u_{x}$$ in the open first quadrant

Boundary Conditions:
u(0,y)=0 for t greater than or equal to 0
u(x,0)= g(x) for x greater than or equal to 0

and g(x) = 0 for $$0\preceq x\preceq1$$
g(x) = (x-1)^5 for x greater than or equal to 1

Is there a solution? And if there is, is the the only solution?

2. Relevant equations

For first order PDE: solution is f(Ay-Bx) where the PDE is AUx + BUy = 0

3. The attempt at a solution

$$u(x,y) = f(-x-3y)$$

so that means

$$u(0,y) = f(-3y)=0$$

$$u(x,0) = f(-x)=g(x)$$

set -x = w

f(w) = g(-x)

therefore f(-x-3y) = g(x+3y)

and u(x,y) = g(x+3y)

This solution works for u(x,0) but I can't find it to work with u(0,y), if y > 1/3, then the solution would be (x-1)^5 with x greater than or equal to 1, which does not satisfy u(0,y) = 0

so I concluded that there is no solution....this doesn't seem right for me. Any help? I feel that I dont actually have to try to find u in order to determine if a solution exists...

Last edited: Mar 25, 2009
2. Mar 25, 2009

### somethingstra

Hello, maybe in a traditional pde view, it would be more helpful to think of y as t for time?

3. Mar 25, 2009

### somethingstra

bump, am I being too confusing here?

4. Mar 25, 2009

### somethingstra

bump....I have kind of a part two question depending on how correct my answer is.

Last edited: Mar 25, 2009
5. Mar 25, 2009

### somethingstra

Oh, nevermind, I got it =]