Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  laovultai Mar 20 '13 at 21:21
add comment

4 Answers

up vote 2 down vote accepted

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 : $$\frac 1{\sqrt{x^2+y^2}}$$

share|improve this answer
add comment

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,$$\Delta_{r,\theta} g = \dfrac1r \implies \Delta_{x,y} f = \dfrac1{\sqrt{x^2+y^2}}$$

share|improve this answer
add comment

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

share|improve this answer
add comment

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.