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Math Help - Change of Variable in a second-order PDE

  1. #1
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    Change of Variable in a second-order PDE

    Hello!

    I am working on a question involving a P.D.E. in a adv. calc. course.

    Consider the general homogenous second-order partial differential equation
    a\frac{\delta^2u}{\delta x^2} + 2b\frac{\delta^2 u}{\delta x\delta y} + c\frac{\delta^2 u}{\delta y^2} = 0

    with constant coefficients (a,b,c).

    If ac - b^2 = 0, show that the substitution s = bx - ay, \ t = y reduces the above equation to \frac{\delta^2 u}{\delta t^2} = 0


    So we must consider the change of variable from u(x,y) to u(s,t), which is u(bx-ay, y) and take partial derivatives? I can't seem to derive the desired result, however, and I have double-checked my messy algebra...

    Any help appreciated. Thanks!
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  2. #2
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    I recommend solving for x = x(s,t) and y = y(s,t). Compute their partial derivatives with respect to s and t. Then simply compute

    u_{t}=u_{x}x_{t}+u_{y}y_{t}, and the subsequent u_{tt}. Using both facts that are given to you (that is, the original pde, and the condition on a,b,c), you can show the desired result.

    What do you get?
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  3. #3
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    I see! Let me try to fill in the blanks... you wrote x = x(s,t) = \frac{s + at}{b} and y = y(s,t) = t

    Then by the chain rule,
    u_t = u_xx_t + u_yy_t

    but we can see that x_t = \frac{a}{b} and y_t = 1, so
    u_t = u_x\frac{a}{b} + u_y

    then another chain rule yields

    u_{tt} = (u_{xx}\frac{a}{b})x_t + u_{yy}y_t

    If that is correct, then...
    u_{tt} = (u_{xx}\frac{a}{b})\frac{a}{b} + u_{yy}
    or u_{tt} = (u_{xx}\frac{a^2}{b^2}) + u_{yy}
    am I close enough here? I don't see any algebraic manipulation I could do with the condition and PDE to derive the result...

    I appreciate your post, thank you!
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  4. #4
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    When you compute your second derivative, you need to do this:

    u_{tt}=\dfrac{\partial}{\partial x}\left(u_{t}\right)\,\dfrac{\partial x}{\partial t}+\dfrac{\partial}{\partial y}\left(u_{t}\right)\,\dfrac{\partial y}{\partial t}=\dfrac{\partial}{\partial x}\left(\dfrac{a}{b}\,u_{x}+u_{y}\right)\,\dfrac{\  partial x}{\partial t}+\dfrac{\partial}{\partial y}\left(\dfrac{a}{b}\,u_{x}+u_{y}\right)\,\dfrac{\  partial y}{\partial t}\dots

    Once you've brought that correction through, why not solve the pde for a u_{xx} and plug into your expression for u_{tt}? Also, don't forget to use your b^{2}=ac relation.
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  5. #5
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    OK! Two chain rules "compounds" those partials.... so
    u_{tt} = (\frac{a}{b}u_{xx} + u_{yx})(\frac{a}{b}) + (\frac{a}{b}u_{xy} + u_{yy})

    Then,
    u_{tt} = \frac{a^2}{b^2}u_{xx} + 2\frac{a}{b}u_{xy} + u_{yy}

    But b^2 = ac

    u_{tt} = \frac{a}{c}u_{xx} + 2\frac{a}{b}u_{xy} + u_{yy}

    Also, a = \frac{b^2}{c}

    u_{tt} = \frac{a}{c}u_{xx} + 2\frac{b}{c}u_{xy} + u_{yy}

    Multiply through by c?

    cu_{tt} = au_{xx} + 2bu_{xy} + cu_{yy}

    RHS = 0 from assumption

    cu_{tt} = 0

    c \neq 0 so

    u_{tt} = 0

    Woo-hoo! Thank you so much!!! I was really fudging the chain rule. I think everything I have done here is OK?
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  6. #6
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    Yep, that looks fine. My route to the end was a bit different, but definitely mathematically equivalent. Good job.
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