Question

Hi guys,

Answer 1

$\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\cos\theta-y\sin\theta\\x\sin\theta+y\cos\theta\end{bmatrix}$

so that

$\begin{cases}u(x,y)=x\cos\theta-y\sin\theta\\v(x,y)=x\sin\theta+y\cos\theta\end{cases}$

For a function $f(u,v)=f(u(x,y),v(x,y))$, we have by the chain rule,

$\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial x}$

and

$\begin{cases}\frac{\partial u}{\partial x}=\cos\theta\\\\\frac{\partial v}{\partial x}=\sin\theta\end{cases}$

$\implies\dfrac{\partial f}{\partial x}=\cos\theta\dfrac{\partial f}{\partial u}+\sin\theta\dfrac{\partial f}{\partial v}$

Let $g(u,v)=\frac{\partial f}{\partial u}$ and $h(u,v)=\frac{\partial f}{\partial v}$. This substitution is made just to make the application of the chain rule clearer.

$\dfrac{\partial f}{\partial x}=\cos\theta\,g+\sin\theta\,h$

Differentiating again wrt $x$ gives

$\dfrac{\partial^2f}{\partial x^2}=\cos\theta\dfrac{\partial g}{\partial x}+\sin\theta\dfrac{\partial h}{\partial x}$

By the chain rule,

$\dfrac{\partial g}{\partial x}=\dfrac{\partial g}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial g}{\partial v}\dfrac{\partial v}{\partial x}$

and our substitution shows that, for instance,

$\dfrac{\partial g}{\partial u}=\dfrac{\partial}{\partial u}\dfrac{\partial f}{\partial u}=\dfrac{\partial^2f}{\partial u^2}$

and so

$\dfrac{\partial g}{\partial x}=\cos\theta\dfrac{\partial^2f}{\partial u^2}+\sin\theta\dfrac{\partial^2f}{\partial v\partial u}$

Similarly, we find

$\dfrac{\partial h}{\partial x}=\cos\theta\dfrac{\partial^2f}{\partial u\partial v}+\sin\theta\dfrac{\partial^2f}{\partial v^2}$

Putting everything together, we have

$\dfrac{\partial^2f}{\partial x^2}=\cos^2\theta\dfrac{\partial^2f}{\partial u^2}+\cos\theta\sin\theta\dfrac{\partial^2f}{\partial v\partial u}+\sin\theta\cos\theta\dfrac{\partial^2f}{\partial u\partial v}+\sin^2\theta\dfrac{\partial^2f}{\partial v^2}$

and we can similarly find that

$\dfrac{\partial^2f}{\partial y^2}=\sin^2\theta\dfrac{\partial^2f}{\partial u^2}-\cos\theta\sin\theta\dfrac{\partial^2f}{\partial v\partial u}-\sin\theta\cos\theta\dfrac{\partial^2f}{\partial u\partial v}+\cos^2\theta\dfrac{\partial^2f}{\partial v^2}$

Adding together these derivatives, we see the mixed partials cancel, and recalling that $\cos^2\theta+\sin^2\theta=1$, we end up with

$\dfrac{\partial^2f}{\partial x^2}+\dfrac{\partial^2f}{\partial y^2}=\dfrac{\partial^2f}{\partial u^2}+\dfrac{\partial^2f}{\partial v^2}$

as required.