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

  • Thread starter jlmac2001
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In summary, the conversation discusses finding the Laplacian of two functions, sqrt(x^-y^2) and ln(r^2). There is confusion about whether to take the gradient twice and how to solve for ln(r^2). The Laplacian is explained to be a sum of three partial derivatives in Cartesian coordinates. The correct solution for sqrt(x^2-y^2) is given, but there is uncertainty about the derivative of ln(r^2).
  • #1
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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  • #2
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
 
  • #3
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?
 
  • #4
ln(r^2)= ln(x^2+ y^2). Does that help?
 
  • #5
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
 

What is the Laplacian?

The Laplacian is a mathematical operator that is used to calculate the rate of change or variation of a function in multiple dimensions.

How do you solve the Laplacian for sqrt(x^-y^2) & ln(r^2)?

To solve the Laplacian for this function, first rewrite it as sqrt(x^-y^2) = x^-y and ln(r^2) = 2ln(r). Then, use the Laplacian formula, which is given by: ∇^2f = (∂^2f/∂x^2) + (∂^2f/∂y^2) + (∂^2f/∂z^2). In this case, f = x^-y + 2ln(r). Therefore, the Laplacian of sqrt(x^-y^2) & ln(r^2) is: ∇^2f = -y(x^-y+2ln(r))^2 + 2/r^2.

What is the significance of solving the Laplacian for sqrt(x^-y^2) & ln(r^2)?

Solving the Laplacian for any function allows us to understand its behavior and properties in a given space. In this case, solving the Laplacian for sqrt(x^-y^2) & ln(r^2) can help us understand the rate of change of this function in multiple dimensions.

Are there any practical applications of solving the Laplacian for sqrt(x^-y^2) & ln(r^2)?

Yes, there are many practical applications of solving the Laplacian for this function. Some examples include analyzing heat distribution in a 2D or 3D space, modeling diffusion of particles, and understanding the electric potential in a circuit.

Is there a specific method or algorithm to solve the Laplacian for sqrt(x^-y^2) & ln(r^2)?

Yes, there are various methods and algorithms that can be used to solve the Laplacian for any function, including sqrt(x^-y^2) & ln(r^2). Some common methods include the finite difference method, Fourier transform method, and Green's function method.

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