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Math Help - PDEs

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by taypez View Post
    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?
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by taypez View Post
    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.

    ∂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.
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