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Thread: Heat Equation PDE

  1. #1
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    Heat Equation PDE

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

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

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

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

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

    Then the derivative of $\displaystyle f$ is

    $\displaystyle \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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