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Math Help - Heat Equation PDE

  1. #1
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    Heat Equation PDE

    Show that if u(x,t) is a solution of u_{t}=u_{xx}, then so is
    v(x,t)=t^{\frac{-1}{2}} e^{\frac{-x^{2}}{4t}}u(\frac{x}{t},-\frac{1}{t})

    I'm assuming that I should simply show v_{t}=v_{xx} however I'm not sure how I should differentiate u(\frac{x}{t},-\frac{1}{t}) with respect to t.

    ??? (u(\frac{x}{t},-\frac{1}{t}))'=t^{-4}u(\frac{x}{t},-\frac{1}{t})
    Is this correct?
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  2. #2
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    Quote Originally Posted by leads View Post
    Show that if u(x,t) is a solution of u_{t}=u_{xx}, then so is
    v(x,t)=t^{\frac{-1}{2}} e^{\frac{-x^{2}}{4t}}u(\frac{x}{t},-\frac{1}{t})

    I'm assuming that I should simply show v_{t}=v_{xx} however I'm not sure how I should differentiate u(\frac{x}{t},-\frac{1}{t}) with respect to t.

    ??? (u(\frac{x}{t},-\frac{1}{t}))'=t^{-4}u(\frac{x}{t},-\frac{1}{t})
    Is this correct?
    You need to use the chain rule consider

    If f(\alpha,\beta) and \displaystyle \alpha=\frac{x}{t},\beta=\frac{-1}{t}

    Then the derivative of f is

    \displaystyle \frac{\partial f}{\partial t}=\frac{\partial f}{\partial \alpha}\frac{\partial \alpha }{\partial t}+\frac{\partial f}{\partial \beta} \frac{\partial \beta}{\partial t}
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