# Thread: Need help with PDE problem

1. ## Need help with PDE problem

Could anyone help me with this? The question is under the spoiler tag.

Spoiler:

The PDE equation is separable, but the boundary conditions are non-homogeneous and are in partial derivatives.

I can answer if the equations are all homogeneous or if the boundary conditions are not homogeneous and not derivatives. But I'm not quite sure how to proceed with this one.

I would appreciate any help.

2. Originally Posted by Mierin
Could anyone help me with this? The question is under the spoiler tag.

Spoiler:

The PDE equation is separable, but the boundary conditions are non-homogeneous and are in partial derivatives.

I can answer if the equations are all homogeneous or if the boundary conditions are not homogeneous and not derivatives. But I'm not quite sure how to proceed with this one.

I would appreciate any help.

It looks like a Diffusion Equation..

3. Originally Posted by Mierin
Could anyone help me with this? The question is under the spoiler tag.

Spoiler:

The PDE equation is separable, but the boundary conditions are non-homogeneous and are in partial derivatives.

I can answer if the equations are all homogeneous or if the boundary conditions are not homogeneous and not derivatives. But I'm not quite sure how to proceed with this one.

I would appreciate any help.
If you let $\displaystyle u = a(x^2+2t) + bx + v$, then you can choose $\displaystyle a$ and $\displaystyle b$ such that you BC's becomes $\displaystyle v_x = 0$ at both boundaries.

4. A function that is equal to -1 at x= 0 and 0 at x= l is $\displaystyle \frac{x}{l}- 1$. Integrating that, a function whose derivative is -1 at x= 0 and 0 at x= l is $\displaystyle \frac{x^2}{2l}- x$.

If you let $\displaystyle v(x,t)= u(x,t)- \frac{x^2}{2l}+ x$, then $\displaystyle \frac{\partial^2 v}{\partial x^2}= \frac{\partial^2 u}{\partial u^2} - \frac{1}{l}$$\displaystyle = \frac{1}{\kappa^2}\frac{\partial u}{\partial t}= \frac{1}{\kappa^2}\frac{\partial v}{\partial t}. That is, you now have the differential equation \displaystyle \frac{1}{\kappa^2}\frac{\partial v}{\partial t}= \frac{\partial^2 v}{\partial x^2}- \frac{1}{l} with boundary conditions \displaystyle \frac{\partial v}{\partial x}(0)= \frac{\partial u}{\partial x}(0)- \frac{(0}{l}+ 1= -1+ 1= 0 and initial condition \displaystyle v(x, 0)= u(x, 0)- \frac{x^2}{2l}+ x= -\frac{x^2}{2l}+ x With those boundary conditions, you can write v as a Fourier cosine series: \displaystyle v(x,t)= \sum_{n=0}^\infty C_n(t)sin(\frac{2\pi}{l}x). \displaystyle \frac{\partial v}{\partial x}(l)= \frac{\partial u}{\partial x}(l)- \frac{(0}{l}+ 1= 0 -1+ 1= 0 5. Thanks a lot for your help. But I still can't seem to understand how this part: \displaystyle \frac{\partial^2 v}{\partial x^2}= \frac{\partial^2 u}{\partial u^2} - \frac{1}{l}$$\displaystyle = \frac{1}{\kappa^2}\frac{\partial u}{\partial t}= \frac{1}{\kappa^2}\frac{\partial v}{\partial t}$ came about.

I can see that $\displaystyle \frac{1}{\kappa^2}\frac{\partial u}{\partial t}$ is obtained from the equation given in the question, but I don't understand $\displaystyle \frac{\partial^2 v}{\partial x^2}= \frac{\partial^2 u}{\partial u^2} - \frac{1}{l}$

6. Originally Posted by Mierin
Thanks a lot for your help. But I still can't seem to understand how this part: $\displaystyle \frac{\partial^2 v}{\partial x^2}= \frac{\partial^2 u}{\partial u^2} - \frac{1}{l}$$\displaystyle = \frac{1}{\kappa^2}\frac{\partial u}{\partial t}= \frac{1}{\kappa^2}\frac{\partial v}{\partial t}$ came about.

I can see that $\displaystyle \frac{1}{\kappa^2}\frac{\partial u}{\partial t}$ is obtained from the equation given in the question, but I don't understand $\displaystyle \frac{\partial^2 v}{\partial x^2}= \frac{\partial^2 u}{\partial u^2} - \frac{1}{l}$
It's a typo - I think he meant

$\displaystyle \frac{\partial^2 v}{\partial x^2}= \frac{\partial^2 u}{\partial x^2} - \frac{1}{l}$

although if you follow my idea your PDE will be the same and hence the usual separation of variables will work.