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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- February 4th 2008, 03:49 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 - February 4th 2008, 07:40 PMThePerfectHacker
- February 5th 2008, 05:48 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 - February 5th 2008, 08:51 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 then,

.

Divide by and we get,

Now the LHS only has while RHS only has thus,

for some real number .

This means that,

.

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

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