Laplacian of sqrt(x² - y²) and ln(r²)

[jlmac2001]

Here's the problem:

Find the Laplacian of sqrt(x^-y^2) and ln(r^2).

Will i just take the gradient of each one of these twice?

 

[quantumdude]

Originally posted by jlmac2001
Will i just take the gradient of each one of these twice?

No, the Laplacian is (in Cartesian coordinates):

∂²/∂x²+∂²/∂y²+∂²/∂z²

 

[jlmac2001]

is this right?

For sqrt(x^2-y^2), I got:

((x^2-y^2)^-1/2 + x^2(x^2-y^2)^-3/2)i + ((x^2-y^2)^-1/2 -y^2(x^2-y^2)^3/2)j

I'm not sure how to do the ln(r^2). Can someone help?

 

[HallsofIvy]

ln(r^2)= ln(x^2+ y^2). Does that help?

 

[jlmac2001]

is this right

First off, did I do the Sqrt(x^2-y^2) right? For ln(r^2), I'm not sure if I'm doing the derivative right. i got, -2/r^3 i