Compute $\vec\nabla^2 \frac{1}{\mid \vec x \mid}$

Question

Compute $\vec\nabla^2 \frac{1}{\| \vec x \|}$ where $\vec x= x\hat i + y \hat j +z\hat k$

I have that $$\frac{1}{\| \vec x \|} = (x^2 + y^2 + z^2)^{-\frac{1}{2}}$$

So it follows that $\vec\nabla\frac{1}{\| \vec x \|}= -\frac{1}{2}\left(\| \vec x \|^2\right)^{-\frac{3}{2}}(2x\hat i + 2y\hat j + 2z \hat k)$

At this point I do not know if this is right or how to continue.

What would $\vec\nabla^2\frac{1}{\| \vec x \|} $ be?

Answer 1

Let $u=x^2+y^2+z^2$, so we have:

$$ f(x,y,z)=u^{-\frac{1}{2}} $$ and: $$ \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}=-\frac{1}{2}u^{-\frac{3}{2}}\cdot 2x=-xu^{-\frac{3}{2}} $$ $$ \frac{\partial^2 f}{\partial x^2}=\frac{\partial }{\partial x}\frac{\partial f}{\partial x}=-u^{-\frac{3}{2}}+x\cdot \frac{3}{2}u^{-\frac{5}{2}}\cdot 2x=-u^{-\frac{3}{2}}+3x^2u^{-\frac{5}{2}} $$ By symmetry we have, in the same way: $$ \frac{\partial^2 f}{\partial y^2}=-u^{-\frac{3}{2}}+3y^2u^{-\frac{5}{2}} $$ and $$ \frac{\partial^2 f}{\partial z^2}=-u^{-\frac{3}{2}}+3z^2u^{-\frac{5}{2}} $$

so: $$ \nabla^2f= \frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2}=-3u^{-\frac{3}{2}}+3u^{-\frac{5}{2}}(x^2+y^2+z^2)=0 $$

Note that this a case of $\mbox{div (grad(} f))=0$

Answer 2

You have to compute a 'symbolic' dot product between $\nabla$ and $\nabla f$, that is: $$\nabla \cdot \nabla f = \frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}$$ In your case

$$\nabla \cdot\nabla\frac{1}{\|\boldsymbol x\|}= \frac{\partial}{\partial x}\left(-x(x^2+y^2+z^2)^{-\frac{3}{2}}\right) +\frac{\partial}{\partial y}\left(-y(x^2+y^2+z^2)^{-\frac{3}{2}}\right)+\frac{\partial}{\partial z}\left(-z(x^2+y^2+z^2)^{-\frac{3}{2}}\right) = -\left(-\frac{3}{2}2x^2(x^2+y^2+z^2)^{-\frac{5}{2}}+(x^2+y^2+z^2)^{-\frac{3}{2}}\right)-\left(-\frac{3}{2}2y^2(x^2+y^2+z^2)^{-\frac{5}{2}}+(x^2+y^2+z^2)^{-\frac{3}{2}}\right)-\left(-\frac{3}{2}2z^2(x^2+y^2+z^2)^{-\frac{5}{2}}+(x^2+y^2+z^2)^{-\frac{3}{2}}\right) = 3(x^2+y^2+z^2)(x^2+y^2+z^2)^{-\frac{5}{2}}-3(x^2+y^2+z^2)^{-\frac{3}{2}}=3(x^2+y^2+z^2)^{-\frac{3}{2}}-3(x^2+y^2+z^2)^{-\frac{3}{2}}=3\frac{1}{\|\boldsymbol x\|^3}-3\frac{1}{\|\boldsymbol x\|^3} = 0$$

Since the function satisfies the Laplace equation $\nabla^2f =0$, it is a harmonic function.

Answer 3

Using index notation the proof is more succinct:

$$\frac{\partial}{\partial x_{j}}\frac{\partial}{\partial x_{j}}(x_{i}x_{i})^{-\frac{1}{2}}=\frac{\partial}{\partial x_{j}}\left[-\frac{1}{2}(x_{k}x_{k})^{-\frac{3}{2}}2x_{i}\frac{\partial x_{i}}{\partial x_{j}}\right]=\frac{\partial}{\partial x_{j}}\left[(x_{k}x_{k})^{-\frac{3}{2}}x_{i}\delta_{ij}\right]=\frac{\partial}{\partial x_{j}}\left[-(x_{k}x_{k})^{-\frac{3}{2}}x_{j}\right]=\frac{3}{2}(x_{k}x_{k})^{-\frac{5}{2}}2x_{p}\frac{\partial x_{p}}{\partial x_{j}}x_{j}-(x_{k}x_{k})^{-\frac{3}{2}}\frac{\partial x_{j}}{\partial x_{j}}=3(x_{k}x_{k})^{-\frac{5}{2}}(x_{j}x_{j})-3(x_{k}x_{k})^{-\frac{3}{2}}=0$$