PDEs

• Feb 4th 2008, 03:49 PM
taypez
PDEs
Need some help.

What ordinary differential equations are implied by the method of separation of variables?

∂u/∂t = k ∂^2u/∂x^2 – v_0 ∂u/∂x

Thanks
• Feb 4th 2008, 07:40 PM
ThePerfectHacker
Quote:

Originally Posted by taypez
Need some help.

What ordinary differential equations are implied by the method of separation of variables?

∂u/∂t = k ∂^2u/∂x^2 – v_0 ∂u/∂x

Thanks

I assume that $k$ and $v_0$ are given numbers, right? Also, what are the boundary conditions, seperation of variables not going to work if this is not a specific type of equation. Meaning we need to be solving for $t\geq 0$ and $0\leq x\leq L$ which satisfies $u(0,t) = u(L,t) = 0$ for $t\geq 0$. Is this true?
• Feb 5th 2008, 05:48 AM
taypez
PDEs
This is just asking for the ODEs and not to solve the entire problem. For example,

∂^2u/∂x^2 + ∂^2u/∂y^2 = 0 implies u(x,y) = φ (x)h(y) yields h d^2 φ /dx^2 + φ d^2h/dy^2 = 0.

Then dividing by φ (h) yields 1/ φ d^2φ/dx^2 = - 1/h d^2h/dy^2 = -λ
Or d^2φ/dx^2= -λ φ and d^2h/dy^2 = λh
• Feb 5th 2008, 08:51 AM
ThePerfectHacker
Quote:

Originally Posted by taypez
This is just asking for the ODEs and not to solve the entire problem. For example,

I know. But you need to know the boundary conditions are homogenous, watch why we need that.

Quote:

∂u/∂t = k ∂^2u/∂x^2 – v_0 ∂u/∂x
I am assuming that $k,v_0$ are numbers.

Let $u(x,t) = X(x)T(t)$ then,
$XT' = kX''T- v_0 X'T$.
Divide by $XT$ and we get,
$\frac{T'}{T} = k\frac{X''}{X} - v_0 \frac{X'}{X}$
Now the LHS only has $t$ while RHS only has $x$ thus,
$\frac{T'}{T} = c \mbox{ and }k\frac{X''}{X} - v_0 \frac{X'}{X} = c$ for some real number $c$.
This means that,
$T' - Tc = 0 \mbox{ and }kX'' -v_0X' - cX = 0$.

As far as seperation of variables goes that is the seperated equation.
But if we want to continue we need to determine what $c$ is. That is we need to determine those values of $c$ such that those two equations have non-trivial solutions. This is done by using boundary conditions, thus if they are homogeneous then $X(0) = X(L) = 0$. But it seems you do not have to go that far. So I stop here.