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 $\displaystyle k$ 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?
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
I know. But you need to know the boundary conditions are homogenous, watch why we need that.
I am assuming that $\displaystyle k,v_0$ are numbers.∂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.