1. ## 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?

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

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