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Sounds like a retardedly basic PDE problem..

by dingop
Tags: pde
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dingop
#1
May12-11, 08:47 PM
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

[tex]\partial[/tex]u/[tex]\partial[/tex]x+[tex]\partial[/tex]u/[tex]\partial[/tex]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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LCKurtz
#2
May12-11, 08:55 PM
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P: 7,663
Quote Quote by dingop View Post
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

[tex]\partial[/tex]u/[tex]\partial[/tex]x+[tex]\partial[/tex]u/[tex]\partial[/tex]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?
dingop
#3
May13-11, 01:50 AM
P: 2
Quote Quote by LCKurtz View Post
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
#4
May13-11, 01:44 PM
HW Helper
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P: 7,663
Sounds like a retardedly basic PDE problem..

Quote Quote by LCKurtz View Post
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 Quote by dingop View Post
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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