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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!

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

2 Answers 2

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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.

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  • $\begingroup$ Personally I would have stuck with the $h_u$ and $h_x:=\omega_x{\circ}\phi$ notation.$$\begin{align}h_u&=2 u~h_x+2v~h_y\\h_{uu}& = 2~h_x+2u~(h_x)_u+2v~(h_y)_u\end{align}$$ $\endgroup$ Mar 16, 2022 at 16:27
  • $\begingroup$ This was very clear, thank you! $\endgroup$
    – Ilay
    Mar 16, 2022 at 16:53
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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 $\mathrm{tr}(\mathbf{H}_x)=0$.

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