Explaining the Laplace Operator and Its Properties for Homework

[Ed Aboud]

Homework Statement

Let x = (x,y,z) .
Recall that the vector x is determined by its direction and length
r = |x| = $\sqrt{x^2 + y^2 + z^2}$

and assume we are given a function f which depends only on the length of x

f = f(r)

Show that

$$\Delta f = f'' + \frac{2}{r} f'$$

where $$f' = \frac{\partial f}{\partial r}$$

Homework Equations

The Attempt at a Solution

$$u = x^2 + y^2 + z^2$$
$$r = \sqrt{u}$$
$$\frac{\partial r}{\partial x} = \frac{\partial \sqrt{u}}{\partial u} \frac{\partial u}{\partial x} = \frac{1}{2}(\frac{1}{\sqrt{r}})(2x) = \frac{x}{\sqrt{r}}$$

$$\frac{\partial ^2 r}{\partial x^2} = \frac{\partial (x)}{\partial x } \frac{1}{\sqrt{u}} + x \frac{\partial \frac{1}{\sqrt{u}}}{\partial u} \frac{\partial u }{\partial x}$$


$$= \frac{1}{\sqrt{u}} - x^2 \frac{1}{\sqrt{u^3}}$$

Since f and u are symmetric in x,y,z

$$\frac{\partial ^2 r}{\partial y^2} = \frac{1}{\sqrt{u}} - y^2 \frac{1}{\sqrt{u^3}}$$

$$\frac{\partial ^2 r}{\partial z^2} = \frac{1}{\sqrt{u}} - z^2 \frac{1}{\sqrt{u^3}}$$

$$x^2 + y^2 + z^2 = u$$

$$\Delta r = (\frac{1}{\sqrt{u}}) - x^2 (\frac{1}{\sqrt{u^3}}) + (\frac{1}{\sqrt{u}}) - y^2 (\frac{1}{\sqrt{u^3}}) + (\frac{1}{\sqrt{u}}) - z^2 (\frac{1}{\sqrt{u^3}})$$

$$= \frac{3}{\sqrt{u}} - (x^2 + y^2 + z^2) \frac{1}{\sqrt{u^3}}$$

$$= \frac{2}{\sqrt{u}}$$

$$= \frac{2}{r}$$


I see that this is a part of the solution but I have no idea what to do to get the rest.

Any help would be greatly appreciated because I'm lost and it has to be in tomorrow morning.

 

[yyat]

Compute

$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)f(r(x,y,z))$$

using the chain rule.

 

[Ed Aboud]

I'm not really sure how to apply the chain rule in this case. Is there any general formula that I can use?

 

[Ed Aboud]

Actually its cool, I got it.

Thanks for the help!