Heat Equation PDE

**leads** · February 14th 2011, 04:04 PM

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?

**TheEmptySet** · February 14th 2011, 04:13 PM

Originally Posted by leads

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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