Sounds like a retardedly basic PDE problem

[dingop]

Homework Statement

This is a simple pde I need to solve in order to determine a straightforward expansion for a given overall equation.

Homework Equations

$$\partial$$u/$$\partial$$x+$$\partial$$u/$$\partial$$y=0
with initial condition:
u(x,0)=epsilon*phi(x)

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!

 

[LCKurtz]

Homework Statement

This is a simple pde I need to solve in order to determine a straightforward expansion for a given overall equation.

Homework Equations

$$\partial$$u/$$\partial$$x+$$\partial$$u/$$\partial$$y=0
with initial condition:
u(x,0)=epsilon*phi(x)

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!

Homework Statement

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?

 

[dingop]

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?

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..

 

[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?

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.