Solving PDE with Initial Conditions for ϕ(x,t)

[ballzac]

Hi,

I have a problem that I am working through, and I am at a point where I'm not sure what to do.

I am solving a PDE using the method of characteristics. This has given me the solution

ϕ(x,t)=F(x+(3ϕ^2-1)t) "for any function " F

I have the initial conditions

ϕ(x,0)=1, x<0
and
ϕ(x,0)=0, x>0

Normally I have no trouble using initial conditions, but this has me stumped. I set t=0 and phi=1 (or 0), and I get F(x)=0, 1

I have no idea what to do with this. Normally I would end up with something like F(x)=sin(x), and then I could say

ϕ(x,t)=sin(x+(3ϕ^2-1)t)

but that isn't the case her. Any help would be much appreciated.

 

[mighty2000]

Hello,

It seems like you are on the right track with using the method of characteristics to solve your PDE. However, the issue here is that the initial conditions given do not provide enough information to determine the function F. This is because the initial conditions are only given for x<0 and x>0, but not for x=0.

To solve this problem, you will need to use the method of characteristics again for x=0. This will give you a new set of characteristics and corresponding initial conditions for x=0. Then, you can use these new initial conditions to determine the function F and complete your solution.

I hope this helps and good luck with your problem!