# Showing that Laplace's equation holds

1. Jan 23, 2013

### Cole A.

1. The problem statement, all variables and given/known data

(From Plonsey, R. and R. C. Barr, "Bioelectricity: A Quantitative Approach")

Show
$$\nabla^2 r = 0$$
given
$$r = \sqrt{x^2 + y^2 + z^2}.$$

2. Relevant equations
$$\nabla = \frac{\partial}{\partial x}{\bf i} + \frac{\partial}{\partial y}{\bf j} + \frac{\partial}{\partial z}{\bf k}$$ (the "abuse of notation")

3. The attempt at a solution

This is my first attempt at a post on this forum, and I'm embarrassed to say I'm really struggling with this question. (Not a homework problem per se, just an in-text side note). It has been a good while since I've studied vector calculus, so if somebody could point out my error, I would appreciate it.

Beginning with
$$r = (x^2 + y^2 + z^2)^{1/2}$$
\begin{aligned} \nabla r &= \frac{\partial r}{\partial x} {\bf i} + \frac{\partial r}{\partial y} {\bf j} + \frac{\partial r}{\partial z} {\bf k} \\ &= \left[ \frac{x}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf i} + \left[ \frac{y}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf j} + \left[ \frac{z}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf k}. \end{aligned}
Now taking the divergence of this gradient $-$ which amounts cumulatively to taking the Laplacian of $r -$ gives
\begin{aligned} \nabla \cdot \nabla r &= \nabla^2 r \\ &= \frac{\partial}{\partial x}\left[ \frac{x}{(x^2 + y^2 + z^2)^{1/2}} \right] + \frac{\partial}{\partial y}\left[ \frac{y}{(x^2 + y^2 + z^2)^{1/2}} \right] + \frac{\partial}{\partial z}\left[ \frac{z}{(x^2 + y^2 + z^2)^{1/2}} \right] \\ &= \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{x^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{y^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] \\ &= \left[ \frac{y^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{x^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{x^2 + y^2}{(x^2 + y^2 + z^2)^{3/2}} \right] \\ &= \frac{2}{(x^2 + y^2 + z^2)^{1/2}} \end{aligned}
which is not zero, although I very much wish it was. (Sorry for the long-winded calculations.)

2. Jan 23, 2013

### Dick

Very good job. You got the right answer. I think the problem is with the original question. The laplacian of r isn't zero. The laplacian of 1/r is zero. Are you sure that's not what it says??

3. Jan 23, 2013

### Cole A.

Ugh, I was starting to suspect I was wasting time because of a typo in the book. I attached the relevant portion of the textbook page as an image, where the above equality is listed along with a couple other vector identities.

Thanks for the extra input (and sanity check).

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Last edited: Jan 24, 2013
4. Jan 23, 2013

### Dick

It's a typo alright. Pretty bad one too. Jeez. I'm aghast. Shocked, really. That's pretty fundamental, you'd think they would have checked that.

Last edited: Jan 24, 2013