# Multivariable chain rule problem with second partial derivatives

I've recently learned the chain rule for multivariable calculus, and received the following question on our latest assignment:

$$w:\mathbb{R}^2\rightarrow\mathbb{R}$$ is a twice differentiable function such that $$\triangle w=w_{xx}+w_{yy}=0$$

Define: $$\phi:\mathbb{R}^2\rightarrow\mathbb{R^2},\quad \phi(u,v)=(u^2-v^2,2uv)$$

Let $$h$$ be the composition of $$w$$ and $$\phi$$: $$h:\mathbb{R}^2\rightarrow\mathbb{R},\quad h=w\circ\phi$$ Prove $$\triangle h=h_{uu}+h_{vv}=0$$

This is what I have done so far:

Define: $$x(u,v)=u^2-v^2,\quad y(u,v)=2uv$$ So, using the chain rule, we will find the first partial derivative of $$h$$ with respect to $$u$$: $$\frac{\partial h}{\partial u}=\frac{\partial h}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial h}{\partial y}\frac{\partial y}{\partial u}=2u\cdot\frac{\partial h}{\partial x}+2v\cdot\frac{\partial h}{\partial y}$$

And the second partial derivative with respect to $$u$$ will be: $$\frac{\partial^2 h}{\partial^2 u}=\frac{\partial}{\partial u}(2u\cdot\frac{\partial h}{\partial x}+2v\cdot\frac{\partial h}{\partial y})$$

I do the same thing with the partial derivatives of $$h$$ with respect to $$v$$, and I get: $$\frac{\partial^2 h}{\partial^2 v}=\frac{\partial}{\partial v}(-2v\cdot\frac{\partial h}{\partial x}+2u\cdot\frac{\partial h}{\partial y})$$

Here I am stuck. I am not sure what I can do with the partial derivatives $$\frac{\partial h}{\partial y}$$ and $$\frac{\partial h}{\partial x}$$.

Any help will be appreciated, Thanks!

• What is $w \circ \phi$? Mar 16, 2022 at 8:15
• It's a function composition, I'll edit the post for clarity
– Ilay
Mar 16, 2022 at 8:17
• Okay, I asked because you originally wrote that the range of $\phi$ is $\mathbb R$. Mar 16, 2022 at 8:29
• Yep, I've edited that too, thanks :)
– Ilay
Mar 16, 2022 at 8:29

I am not sure what I can do with the partial derivatives $$\tfrac{∂h}{∂y}$$ and $$\tfrac{∂h}{∂x}$$.

Leave them, as they are cleaner to write.

They are actually the compositions: $$\tfrac{\partial h}{\partial x}=\tfrac{\partial \omega}{\partial x}\circ\phi$$ and $$\tfrac{\partial h}{\partial y}=\tfrac{\partial \omega}{\partial y}\circ\phi$$ .

Therefore $$\tfrac{\partial^2 h}{\partial x~^2}+\tfrac{\partial^2 h}{\partial y~^2}=0$$, because $$\left[\tfrac{\partial^2 \omega}{\partial x~^2}+\tfrac{\partial^2 \omega}{\partial y~^2}\right]\circ\phi=0$$.

So your aim is just to show that when evaluating $$\tfrac{\partial^2 h}{\partial u~^2}+\tfrac{\partial^2 h}{\partial v~^2}$$, you have a product of some factor and the sum $$\tfrac{\partial^2 h}{\partial x~^2}+\tfrac{\partial^2 h}{\partial y~^2}$$, and all other terms vanish.

Up next is the product rule:\begin{align}\dfrac{\partial^2 h}{\partial u^2}&=\dfrac{\partial~~}{\partial u}\left(2u\dfrac{\partial h}{\partial x}+2v\dfrac{\partial h}{\partial y}\right)\\&=2\dfrac{\partial h}{\partial x}+2u\left(\dfrac{\partial ~~}{\partial u}\dfrac{\partial h}{\partial x}\right)+0+2v\left(\dfrac{\partial ~~}{\partial u}\dfrac{\partial h}{\partial y}\right)\end{align}

... and then the chain rule again. Likewise for the double v derivative. Finally, add and see what cancels.

• Personally I would have stuck with the $h_u$ and $h_x:=\omega_x{\circ}\phi$ notation.\begin{align}h_u&=2 u~h_x+2v~h_y\\h_{uu}& = 2~h_x+2u~(h_x)_u+2v~(h_y)_u\end{align} Mar 16, 2022 at 16:27
• This was very clear, thank you!
– Ilay
Mar 16, 2022 at 16:53

Denote by $$\mathbf{J}=2\begin{pmatrix} u & -v \\ v & u \end{pmatrix}$$ the Jacobian matrix so that $$d\mathbf{x}=\mathbf{J}\cdot d\mathbf{u}$$.

It is easy to see that gradients in both coordinates are linked by the relation $$\mathbf{g}_u = \mathbf{J}^T \mathbf{g}_x$$.

Differentiating this expression, yields $$\begin{eqnarray} d\mathbf{g}_u &=& (d\mathbf{J})^T \mathbf{g}_x + \mathbf{J}^T d\mathbf{g}_x = \left[ \mathbf{A} + \mathbf{J}^T \mathbf{H}_x \mathbf{J} \right] d\mathbf{u} \\ \end{eqnarray}$$ Note: I leave to you to compute the $$\mathbf{A}$$ matrix.

The bracket term is the Hessian in $$(u,v)$$ coordinates. Doing computations, you will see that the trace of the bracket term is indeed null if $$\mathrm{tr}(\mathbf{H}_x)=0$$.