Solve Laplacian for sqrt(x^-y^2) & ln(r^2)

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The discussion focuses on finding the Laplacian of the functions sqrt(x^2 - y^2) and ln(r^2). The Laplacian in Cartesian coordinates is defined as the sum of the second partial derivatives with respect to x, y, and z. The user initially attempts to compute the Laplacian for sqrt(x^2 - y^2) but expresses uncertainty about their calculations. They also seek clarification on the derivative of ln(r^2), which they equate to ln(x^2 + y^2). Overall, the thread highlights the challenges in computing the Laplacian for these specific functions.
jlmac2001
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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?
 
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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
 
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?
 
ln(r^2)= ln(x^2+ y^2). Does that help?
 
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
 

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