# sounds like a retardedly basic PDE problem..

by dingop
Tags: pde
 P: 2 1. The problem statement, all variables and given/known data This is a simple pde I need to solve in order to determine a straightforward expansion for a given overall equation. 2. Relevant equations $$\partial$$u/$$\partial$$x+$$\partial$$u/$$\partial$$y=0 with initial condition: u(x,0)=epsilon*phi(x) 3. The attempt at a solution I used the method of associated equations and got the answer of u=f(x-y); indeed u(x,y)=x-y does satisfy the equation. However, there is no way to input a constant and solve for the constant to satisfy the initial condition. Any help will be appreciated! thanks! 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
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PF Gold
P: 7,005
 Quote by dingop 1. The problem statement, all variables and given/known data This is a simple pde I need to solve in order to determine a straightforward expansion for a given overall equation. 2. Relevant equations $$\partial$$u/$$\partial$$x+$$\partial$$u/$$\partial$$y=0 with initial condition: u(x,0)=epsilon*phi(x) 3. The attempt at a solution I used the method of associated equations and got the answer of u=f(x-y); indeed u(x,y)=x-y does satisfy the equation. However, there is no way to input a constant and solve for the constant to satisfy the initial condition. Any help will be appreciated! thanks! 1. The problem statement, all variables and given/known data
But you have u(x,y) = f(x-y) for any function f. You don't have to just take f(x-y) = x-y. So you have better than a constant of integration; you have an arbitrary function of integration to play with. Your condition becomes:

u(x,0) = f(x-0) = εφ(x)

so what does f need to be?
P: 2
 Quote by LCKurtz But you have u(x,y) = f(x-y) for any function f. You don't have to just take f(x-y) = x-y. So you have better than a constant of integration; you have an arbitrary function of integration to play with. Your condition becomes: u(x,0) = f(x-0) = εφ(x) so what does f need to be?

Right now, I actually don't know what f needs to be-
I just have the PDE equation and the initial condition to satisfy- and my previous sample problems involving associated equations had no initial conditions, so I am pretty much lost..

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Thanks
PF Gold
P: 7,005

## sounds like a retardedly basic PDE problem..

 Quote by LCKurtz But you have u(x,y) = f(x-y) for any function f. You don't have to just take f(x-y) = x-y. So you have better than a constant of integration; you have an arbitrary function of integration to play with. Your condition becomes: u(x,0) = f(x-0) = εφ(x) so what does f need to be?
 Quote by dingop thanks for the reply.! Right now, I actually don't know what f needs to be- I just have the PDE equation and the initial condition to satisfy- and my previous sample problems involving associated equations had no initial conditions, so I am pretty much lost..
It isn't that hard. The answer to my question is in the line above it. You are given φ and you want to choose f to make that equation work. Then u = f(x-y) will be your solution.

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