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


    So we must consider the change of variable from $\displaystyle u(x,y)$ to $\displaystyle u(s,t)$, which is $\displaystyle u(bx-ay, y)$ 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

    $\displaystyle u_{t}=u_{x}x_{t}+u_{y}y_{t},$ and the subsequent $\displaystyle u_{tt}.$ 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 $\displaystyle x = x(s,t) = \frac{s + at}{b}$ and $\displaystyle y = y(s,t) = t$

    Then by the chain rule,
    $\displaystyle u_t = u_xx_t + u_yy_t$

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

    then another chain rule yields

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

    If that is correct, then...
    $\displaystyle u_{tt} = (u_{xx}\frac{a}{b})\frac{a}{b} + u_{yy}$
    or $\displaystyle u_{tt} = (u_{xx}\frac{a^2}{b^2}) + u_{yy}$
    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:

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

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

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

    But $\displaystyle b^2 = ac$

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

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

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

    Multiply through by c?

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

    RHS = 0 from assumption

    $\displaystyle cu_{tt} = 0$

    $\displaystyle c \neq 0$ so

    $\displaystyle u_{tt} = 0$

    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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