Partial Differential Equations

Printable View

July 9th 2007, 06:57 AM
mathgrad

Partial Differential Equations

Hi I am stuck on a PDE boundary question, I think I have shown the equation seperates and understand what the boundarys mean but I am not sure how to show the general solution and then solve the boundary value problems. Please help:confused::confused:

Assume solution $u(x,t) = X(x)T(t)$ to the modified diffusion equation $u_t - Du_{xx} - au = 0$ .

First it says show the equation seperates, I think the follwing is correct?:

substituting the solution into the equation:
$u_t - au = Du_{xx}$

$ XT' - aXT = DX''T $

so $\frac {T'}{T} - a = D \frac {X''}{X}$

Then it wants the general solution for X(x) and T(t)

and then assuming $D>0, a>=0, L>0$ solve the boundary value problem:
$u_t - Du_{xx} - au = 0$ for all $0=<x=<L , t>=0$

$u(0,t) = u(L,t) = 0$ for all $t>=0$

$ u(x,0) = sin(\pi x/L) + sin(2\pi x/L) $

and finally characterise the differenc between solutions for $a>D\pi^2 / L^2$ and $a<D\pi^2 / L^2$

I have found a similiar problem in Boyce's elementary Diff equations book but just cant work it out!

Thanks
July 9th 2007, 09:38 AM
CaptainBlack

Quote:

Originally Posted by mathgrad

Hi I am stuck on a PDE boundary question, I think I have shown the equation seperates and understand what the boundarys mean but I am not sure how to show the general solution and then solve the boundary value problems. Please help:confused::confused:

Assume solution $u(x,t) = X(x)T(t)$ to the modified diffusion equation $u_t - Du_{xx} - au = 0$ .

First it says show the equation seperates, I think the follwing is correct?:

substituting the solution into the equation:
$u_t - au = Du_{xx}$

$ XT' - aXT = DX''T $

so $\frac {T'}{T} - a = D \frac {X''}{X}$

Assume that this is correct (your job to check), than the right hand side is a function of $x$ only and the left hand side is a function of $y$ only, so as they are equal they are also equal to a constant, which leaves you with the two ordinary differential equations:

${T'}= (K+a) T$

and:

$X''=\frac{K}{D}X$

for some constant $K$

RonL
July 11th 2007, 12:38 PM
mathgrad

Thanks, ok I understand that but still dont know how to find the general solution and solve the boundary problem? Sorry if I'm missing the point here.:confused: