1
$\begingroup$

Subjected to the following boundary conditions:

$\left.\frac{\partial \theta}{\partial x}\right|_{x=0,t}=c_1\theta\bigg\vert_{x=0,t}$

$-\left.\frac{\partial \theta}{\partial x}\right|_{x=1,t}=c_2\theta\bigg\vert_{x=1,t}$

$\theta\bigg\vert_{x,t=0}=0$

Where $c_1$ and $c_2$ are nonzero constants.

I know I can solve it with fourier sin and cos expansion, but can I have analytical solution?

I accept "as far I know, it doesn't" or just tell me the method I should study.

(Physical meaning: It is a dimensionless transient heat conduction for 1D with uniform source and both boundary conditions with convection)

$\endgroup$
0

1 Answer 1

1
$\begingroup$

$\theta(x,t)=t$ is an analytic solution.

EDIT: With the introduction of the boundary conditions, I would say: No, I don't know any other method not involving Fourier transforms or separation of variables.

$\endgroup$
6
  • $\begingroup$ I'm sorry, now it has boundary conditions. $\endgroup$ Jul 6, 2016 at 23:43
  • 2
    $\begingroup$ Can you give some additional informations to the boundary conditions? At least, I cannot understand the right hand side, since it involves some $\theta$ which is also the name of your solution...? And what are $c_{1}$ and $c_{2}$? Constants? functions? $\endgroup$
    – Alex
    Jul 6, 2016 at 23:46
  • $\begingroup$ @Alex , question updated! $\endgroup$ Jul 7, 2016 at 0:21
  • $\begingroup$ As it seems to me, a stationary solution should do the trick: Try $\theta(t,x)=-\frac{x^{2}}{2}+bx+c$ with some constants $b$ and $c$ yet to be determined. If you plug in the boundary condition, you get at least one solution... $\endgroup$
    – Alex
    Jul 7, 2016 at 0:47
  • $\begingroup$ @Vitor: Now, you have again changed the problem. What is wrong with proposing the full problem right from the beginning? $\endgroup$
    – Alex
    Jul 7, 2016 at 17:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .