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

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- Feb 4th 2008, 02:49 PMtaypezPDEs
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, 06:40 PMThePerfectHacker
I assume that $\displaystyle k$ and $\displaystyle v_0$ 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 $\displaystyle t\geq 0$ and $\displaystyle 0\leq x\leq L$ which satisfies $\displaystyle u(0,t) = u(L,t) = 0$ for $\displaystyle t\geq 0$. Is this true?

- Feb 5th 2008, 04:48 AMtaypezPDEs
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, 07:51 AMThePerfectHacker
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

Let $\displaystyle u(x,t) = X(x)T(t)$ then,

$\displaystyle XT' = kX''T- v_0 X'T$.

Divide by $\displaystyle XT$ and we get,

$\displaystyle \frac{T'}{T} = k\frac{X''}{X} - v_0 \frac{X'}{X}$

Now the LHS only has $\displaystyle t$ while RHS only has $\displaystyle x$ thus,

$\displaystyle \frac{T'}{T} = c \mbox{ and }k\frac{X''}{X} - v_0 \frac{X'}{X} = c$ for some real number $\displaystyle c$.

This means that,

$\displaystyle T' - Tc = 0 \mbox{ and }kX'' -v_0X' - cX = 0$.

As far as seperation of variables goes that is the seperated equation.

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