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Math Help - PDE-Check

  1. #1
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    PDE-Check

    Find all separable exponential solutions of:

    a) u_tt= u_xx - 4 u_x + 2u

    I have
    u(x,t) = X(x)T(t)
    X(x) = Ce^ax

    X'(x) = aCe^ax
    X''(x) = a^2Ce^ax

    X''T-4X'+2X=0
    = T(X''-4X'+2X)=0
    = Ce^ax(a^2-4a+2)

    b) u_xx + u_yy=0

    u(x.y) = X(x)Y(y)

    X''/X = -Y''/Y

    X(x) = Ce^ax

    not sure from here.

    Thanks
    Last edited by taypez; February 18th 2008 at 07:12 PM. Reason: typo
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  2. #2
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    Quote Originally Posted by taypez View Post
    Find all separable exponential solutions of:

    a) u_tt= u_xx - 4 u_x + 2u

    I have
    u(x,t) = X(x)T(t)
    X(x) = Ce^ax

    X'(x) = aCe^ax
    X''(x) = a^2Ce^ax

    X''T-4X'+2X=0
    = T(X''-4X'+2X)=0
    = Ce^ax(a^2-4a+2)

    [snip]
    Let u = X(x) T(t).

    Then X \frac{d^2 T}{dt^2} = T\left( \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + 2X \right)


    \Rightarrow \frac{1}{T} \frac{d^2 T}{dt^2} = \frac{1}{X} \left( \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + 2X \right)


    Therefore

    \frac{1}{T} \frac{d^2 T}{dt^2} = k^2 \, and \, \frac{1}{X} \left( \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + 2X \right) = k^2

    where k^2 is the seperation constant and is necessarily positive for exponential solutions.

    Therefore:

    \frac{d^2 T}{dt^2} - k^2 T = 0\, and \, \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + (2 - k^2) X = 0.

    Now assume exponential solutions of the form T = A e^{\lambda t} \, and \, X =  B e^{\mu x} \, .......
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  3. #3
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    Quote Originally Posted by taypez View Post
    Find all separable exponential solutions of:

    [snip]
    b) u_xx + u_yy=0

    u(x.y) = X(x)Y(y)

    X''/X = -Y''/Y Mr F says: Therefore either:

    1. X''/X = k^2 and -Y''/Y = k^2 => X'' - k^2 X = 0 and Y'' + k^2 Y = 0 and the exponential solution will be in X = X(x), or

    2. X''/X = -k^2 and -Y''/Y = -k^2 => X'' + k^2 X = 0 and Y'' - k^2 Y = 0 and the exponential solution will be in the Y = Y(y).


    X(x) = Ce^ax

    not sure from here.

    Thanks
    ..
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