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Simple PDE....

 
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Oct17-11, 08:44 AM   #1
 

Simple PDE....


I'm trying to solve the PDE:

[itex]\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{\partial f(x,t)}{\partial t}[/itex] with [itex]x \in [-1,1][/itex] and boundary conditions f(1,t)=f(-1,t)=0.

Thought that [itex]e^{i(kx-\omega t)}[/itex] would work, but that obviously does not fit with the boundary conditions. Has anyone an idea?
 
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Oct17-11, 08:51 AM   #2
 
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Quote by Aidyan View Post
I'm trying to solve the PDE:

[itex]\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{\partial f(x,t)}{\partial t}[/itex] with [itex]x \in [-1,1][/itex] and boundary conditions f(1,t)=f(-1,t)=0.

Thought that [itex]e^{i(kx-\omega t)}[/itex] would work, but that obviously does not fit with the boundary conditions. Has anyone an idea?
Your equation is the 1D heat equation, the solutions of which are very well known and understood. A google search should yield what you need.

P.S. You will also need some kind of initial condition.
 
Oct17-11, 09:18 AM   #3
 
Quote by Hootenanny View Post
Your equation is the 1D heat equation, the solutions of which are very well known and understood. A google search should yield what you need.

P.S. You will also need some kind of initial condition.
Hmm... looks like it isn't just a simple solution, however. It seems I'm lacking the basics ... I thought this is sufficeint data to solve it uniquely, what is the difference between boundary and initial conditions?
 
Oct17-11, 09:27 AM   #4
 
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Simple PDE....


Quote by Aidyan View Post
I thought this is sufficeint data to solve it uniquely,
Afraid not, without knowing the temperature distribution at a specific time you aren't going to obtain a (non-trivial) unique solution.
Quote by Aidyan View Post
what is the difference between boundary and initial conditions?
The former specifies the temperature on the spatial boundaries of the domain (in this case x=-1 and x=1). The latter specifies the temperature distribution at a specific point in time (usually t=0, hence the term initial condition).
 
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