Need help with PDE problem

**Mierin** · February 10th 2010, 01:10 AM

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.

**kjchauhan** · February 10th 2010, 03:50 AM

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..

**Danny** · February 10th 2010, 04:31 AM

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 $u = a(x^2+2t) + bx + v$ , then you can choose $a$ and $b$ such that you BC's becomes $v_x = 0$ at both boundaries.

**HallsofIvy** · February 10th 2010, 06:12 AM

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

If you let $v(x,t)= u(x,t)- \frac{x^2}{2l}+ x$ , then $\frac{\partial^2 v}{\partial x^2}= \frac{\partial^2 u}{\partial u^2} - \frac{1}{l}$ $= \frac{1}{\kappa^2}\frac{\partial u}{\partial t}= \frac{1}{\kappa^2}\frac{\partial v}{\partial t}$ .

That is, you now have the differential equation
$\frac{1}{\kappa^2}\frac{\partial v}{\partial t}= \frac{\partial^2 v}{\partial x^2}- \frac{1}{l}$

with boundary conditions
$\frac{\partial v}{\partial x}(0)= \frac{\partial u}{\partial x}(0)- \frac{(0}{l}+ 1= -1+ 1= 0$

and initial condition $v(x, 0)= u(x, 0)- \frac{x^2}{2l}+ x= -\frac{x^2}{2l}+ x$

With those boundary conditions, you can write v as a Fourier cosine series:

$v(x,t)= \sum_{n=0}^\infty C_n(t)sin(\frac{2\pi}{l}x)$ .
$\frac{\partial v}{\partial x}(l)= \frac{\partial u}{\partial x}(l)- \frac{(0}{l}+ 1= 0 -1+ 1= 0$

**Mierin** · February 10th 2010, 07:40 PM

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

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

**Danny** · February 11th 2010, 05:49 AM

Originally Posted by Mierin

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

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

It's a typo - I think he meant

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

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

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