# confirm which one is correct?

Let $f(z)=-(x^2+y^2)^{1/2}$ and $\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. Help to confirm which one is correct for $\Delta f$; this or $(x^2+y^2)^{3/2}-\frac{2}{(x^2+y^2)^{1/2}}$?

thankyou. EDIT: sorry, I meant above. Just one minus sign more.

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How did you arrive at the expression shown above? – Sammy Black Mar 19 '13 at 22:22
WA is pretty smart. – Kaster Mar 19 '13 at 22:36
I made an error and noticed it and now I see why. instead $(x^2+y^2)^{3/2}-\frac{2}{(x^2+y^2)^{1/2}}$ I finally have $\frac{(x^2+y^2)(x^2-y^2)^{-1}}{(x^2+y^2)^{1/2}}-\frac{2}{(x^2+y^2)^{1/2}} = \frac{1}{(x^2+y^2)^{1/2}}-\frac{2}{(x^2+y^2)^{1/2}} = -\frac{1}{(x^2+y^2)^{1/2}}$ Thanks for feedback. – alvoutila Mar 20 '13 at 21:21

Evaluation gives : $$\frac 2{\sqrt{x^2+y^2}}-\frac {x^2}{(x^2+y^2)^{3/2}}-\frac {y^2}{(x^2+y^2)^{3/2}}$$ that simplifies to : $$\frac 1{\sqrt{x^2+y^2}}$$

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It is easier to apply the operator by moving to polar coordinates. If $g(r,\theta) = f(x,y) = f(r\cos(\theta), r \sin(\theta))$, then $$\Delta_{r,\theta} g = \dfrac1r \dfrac{\partial}{\partial r} \left(r \dfrac{\partial g}{\partial r}\right) + \dfrac1{r^2} \dfrac{\partial^2 g}{\partial \theta^2}$$ In your case, $g(r,\theta) = r$. Hence, $\dfrac{\partial g}{\partial r} = 1$, $\dfrac{\partial g}{\partial \theta} = 0$ and $\dfrac{\partial}{\partial r} \left(r \dfrac{\partial g}{\partial r}\right) = 1$. Hence,$$\Delta_{r,\theta} g = \dfrac1r \implies \Delta_{x,y} f = \dfrac1{\sqrt{x^2+y^2}}$$

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I've carried out the calculations. Number 1 is correct, and the other one is wrong.

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This is an application of the chain rule.

$$\nabla^2\!f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} .$$

If $f(x,y) = \sqrt{x^2+y^2}$ then the first order partial derivatives are:

$$\begin{array}{ccc} \frac{\partial\! f}{\partial x} &=& \frac{x}{\sqrt{x^2+y^2}} \\ \frac{\partial\! f}{\partial y} &=& \frac{y}{\sqrt{x^2+y^2}} \end{array}$$

Differentiating a second time, gives:

$$\begin{array}{ccc} \frac{\partial^2 f}{\partial x^2} &=& \frac{y^2}{(x^2+y^2)^{3/2}} \\ \frac{\partial^2 f}{\partial y^2} &=& \frac{x^2}{(x^2+y^2)^{3/2}} \end{array}$$

It follows that the Laplacian is given by

$$\begin{array}{ccc} \nabla^2f &=& \frac{y^2}{(x^2+y^2)^{3/2}} + \frac{x^2}{(x^2+y^2)^{3/2}} \\ &=& \frac{x^2+y^2}{(x^2+y^2)^{3/2}} \\ &\equiv& \frac{1}{\sqrt{x^2+y^2}} \end{array}$$

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