# (RESOLVED) Given $z = f (x, y)$ and $x = r \cos \theta$, $y = r \sin \theta$ prove the following

Question:

Let $$z = f (x, y)$$ and $$x = r \cos \theta$$, $$y = r \sin \theta$$

Show that

$$\frac{\partial^2z}{\partial x^2} + \frac{\partial^2z}{\partial y^2} = \frac{\partial^2z}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2z}{\partial \theta^2} + \frac{1}{r} \frac{\partial z}{\partial r}$$

My attempt is to show that LS=RS, but I am stuck on how to eliminate the $$\frac{1}{r} \frac{\partial z}{\partial r}$$ term . See below

$$LS: \frac{\partial^2z}{\partial x^2} + \frac{\partial^2z}{\partial y^2}$$

• For the RS, first find its given partial derivatives/expressions

First term: $$===>\frac{\partial^2z}{\partial r^2} = \frac{\partial^2z}{\partial x^2}cos^2\theta + \frac{\partial^2z}{\partial y^2}sin^2\theta + 2(\frac{\partial^2z}{\partial x\partial y}sin\theta cos\theta)$$

• This is the part where I get stuck

$$\frac{1}{r^2} \frac{\partial^2z}{\partial \theta^2}=\frac{1}{r^2}[\frac{\partial}{\partial \theta}(-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta)]$$

• Now treat $$(-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta)$$ as a function of $$z$$ hence $$(-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta)$$ can be seen as $$\frac{\partial z}{\partial \theta}$$

$$= \frac{1}{r^2}[\frac{\partial}{\partial x}(-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta)\frac{\partial x}{\partial \theta} + \frac{\partial}{\partial y}(-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta)\frac{\partial y}{\partial \theta}]$$

We know $$\frac{\partial x}{\partial \theta} = -rsin\theta$$ and $$\frac{\partial y}{\partial \theta}=rcos\theta$$, so substitute them in

$$= \frac{1}{r^2}[\frac{\partial}{\partial x}(-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta)(-rsin\theta) + \frac{\partial}{\partial y}(-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta)(rcos\theta)]$$

$$= \frac{1}{r^2}[(\frac{\partial ^2z}{\partial x^2}r^2sin^2\theta-\frac{\partial ^2z}{\partial x\partial y}r^2cos\theta sin\theta) + (-\frac{\partial ^2z}{\partial x\partial y}r^2cos\theta sin\theta+\frac{\partial ^2z}{\partial y^2}r^2cos^2\theta)]$$

Second term:$$===>\frac{1}{r^2} \frac{\partial^2z}{\partial \theta^2} = \frac{\partial ^2z}{\partial x^2}sin^2\theta-2(\frac{\partial ^2z}{\partial x\partial y}cos\theta sin\theta) +\frac{\partial ^2z}{\partial y^2}cos^2\theta$$

• Now for the last term on the right side,

Third term: $$===>\frac{1}{r} \frac{\partial z}{\partial r}= \frac{1}{r}[\frac{\partial z}{\partial x}cos\theta + \frac{\partial z}{\partial y}sin\theta]$$

• Combining all terms of the right side, we have the following $$\frac{\partial^2z}{\partial x^2}cos^2\theta + \frac{\partial^2z}{\partial y^2}sin^2\theta + \frac{\partial^2z}{\partial x^2}sin^2\theta + \frac{\partial^2z}{\partial y^2}cos^2\theta + \frac{1}{r}[\frac{\partial z}{\partial x}cos\theta + \frac{\partial z}{\partial y}sin\theta]$$

$$===>RS: \frac{\partial^2z}{\partial x^2} + \frac{\partial^2z}{\partial y^2} + \frac{1}{r}[\frac{\partial z}{\partial x}cos\theta + \frac{\partial z}{\partial y}sin\theta]$$

which, $$≠LS=\frac{\partial^2z}{\partial x^2} + \frac{\partial^2z}{\partial y^2}$$

My QUESTION:

There is supposed to be an extra term for the second term of the right side that cancels the third term, but I can't see it. Anyways, what did I do wrong? I believe my error is evaluating the second term of the right side, but it seems correct to me.

• I am not sure if I am supposed to multiply the differentials (d/dx, etc) when taking the second derivative. That is what I did to get the terms of the right side Jul 2 '20 at 0:45
• I've finally found the problem. Turns out that I did the chain rule incorrectly for the second term, since it contains the $\theta$ term. Jul 2 '20 at 7:05

By the Chain Rule, we have that $$z_r = f_x x_r + f_y y_r$$ and $$z_\theta = f_x x_\theta + f_y y_\theta$$ so that $$z_{rr} = f_x x_{rr} + f_{xx} x_r^2 + f_y y_{rr} + f_{yy} y_r^2 \text{ and } z_{\theta \theta} = f_x x_{\theta \theta} + f_{xx} x_\theta^2 + f_y y_{\theta \theta} + f_{yy} y_\theta^2$$ by the Product Rule. Observe that $$x_r = \cos \theta,$$ $$y_r = \cos \theta,$$ $$x_\theta = -r \sin \theta,$$ $$y_\theta = r \cos \theta,$$ $$x_{rr} = y_{rr} = 0,$$ $$x_{\theta \theta} = -r \cos \theta,$$ and $$y_{\theta \theta} = -r \sin \theta,$$ from which it follows that \begin{align*} z_{rr} &= \cos^2 \theta f_{xx} + \sin^2 \theta f_{yy} \text{ and} \\ \\ z_{\theta \theta} &= -r \cos \theta f_x + r^2 \sin^2 \theta f_{xx} - r \sin \theta f_y + r^2 \sin^2 \theta f_{yy} \\ \\ &= r^2 \sin^2 \theta(f_{xx} + f_{yy}) - r(\cos \theta f_x + \sin \theta f_y). \end{align*} We have therefore that \begin{align*} z_{rr} + \frac{z_{\theta \theta}}{r^2} + \frac{z_r}{r} &= \underbrace{\cos^2 \theta f_{xx} + \sin^2 \theta f_{yy}}_{z_{rr}} + \underbrace{\sin^2 \theta (f_{xx} + f_{yy}) - \frac{\cos \theta f_x + \sin \theta f_y} r}_{\frac{z_{\theta \theta}}{r^2}} \\ \\ &\phantom{\cos^2 \theta f_{xx} + \sin^2 \theta f_{yy}+..} + \underbrace{\frac{\cos \theta f_x + \sin \theta f_y}{r}}_{\frac{z_r} r} \end{align*}. Of course, the last two terms cancel, and the first four terms simplify to $$f_{xx} + f_{yy} = z_{xx} + z_{yy}.$$