Transform Laplace's Equation from Cartesian to Parabolic Coordinates "Show that Laplace's equation in Cartesian coordinates, $f_{xx}+f_{yy}=0$, transforms into the same equation in parabolic coordinates: $x=1/2(u^2-v^2)$ and $y=uv$."
($f_{xx}$ means $\partial^2f/\partial x^2 $)
I am able to rewrite $f_{xx}+f_{yy}$ in terms of $f_u,f_{uu},f_v,f_{vv},f_{uv}$ and $u_x,u_y,u_{xx},u_{yy},v_x,v_y,v_{xx},v_{yy}$. I was hoping to simplify this to $f_{uu}+f_{vv}$.
However, I am stuck on finding $u_{xx},v_{xx},u_{yy},v_{yy}$. I found $u_x,v_x,u_y,v_y$ by finding the Jacobian matrix $\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix}$ and inverting it. This gives me expressions in terms of $u$ and $v$ for $u_x,v_x,u_y,v_y$. Do I need to partially differentiate these expressions, which are in terms of $u$ and $v$, w.r.t $x$?
Am I on the right track? Is there a simpler way of approaching this problem?
 A: Applying the chain rule repeated times you will see it's enough to show that
$$\begin{align}
u_x^2+u_y^2 &= 1 \tag{1} \\
u_{xx}+u_{yy} &= 0 \tag{2}\\
u_xv_x+u_yv_y &= 0 \tag{3}\\
v_x^2+v_y^2 &= 1 \tag{4}\\
v_{xx}+v_{yy} &= 0 \tag{5}
\end{align}$$
I think you have already checked that $(1), (3)$ and $(4)$ hold, so you only need to check the other two.
$$\begin{aligned}
u_{xx}+u_{yy} &= 0 \\
v_{xx}+v_{yy} &= 0
\end{aligned}$$
These two equations say that $u$ and $v$ are harmonic functions. To check this, it is sufficient to note that $u$ and $v$ satisfy the Cauchy-Riemann equations
$$\begin{aligned}
u_{x} &= v_{y} \\
v_{x} &= -u_{y}
\end{aligned}$$
which I assume you can note from the expresions you got for $u_x,u_y,v_x$ and $v_y$.
Indeed, for any pair of (twice continuously differentiable) functions $u,v$ which satisfy the Cauchy-Riemann equations, you have
$$\begin{aligned}
u_{xx}
&= v_{yx} \\
&= v_{xy} \\
&= -u_{yy}
\end{aligned}$$
which is $(2)$. You can reason similarly to get $(5)$.
This idea came from complex analysis, where the Cauchy-Riemann equations come from. You can see there is a certain connection to complex analysis, since your change of variables $(u,v)\mapsto((u^2-v^2)/2,uv)$ is nothing but the real expression of the holomorphic function $w\mapsto w^2/2$, where $w=u+iv$. Then by a classical theorem in complex analysis, the local inverses of $w\mapsto w^2/2$ are holomorphic as well (wherever they are defined) so the real and imaginary parts satisfy the C-R equations.
