Solving a Simple PDE: Need Assistance!

  • Thread starter Thread starter Aidyan
  • Start date Start date
  • Tags Tags
    Assistance Pde
Aidyan
Messages
182
Reaction score
14
Simple PDE...

I'm trying to solve the PDE:

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

Thought that e^{i(kx-\omega t)} would work, but that obviously does not fit with the boundary conditions. Has anyone an idea?
 
Physics news on Phys.org


Aidyan said:
I'm trying to solve the PDE:

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

Thought that e^{i(kx-\omega t)} 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.
 


Hootenanny said:
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 ... :confused: I thought this is sufficeint data to solve it uniquely, what is the difference between boundary and initial conditions?
 


Aidyan said:
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.
Aidyan said:
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).
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Replies
36
Views
4K
Replies
3
Views
3K
Replies
5
Views
3K
Replies
3
Views
4K
Replies
7
Views
3K
Replies
3
Views
2K
Back
Top