Laplacian, partial derivatives

In summary, the Laplacian of F = sin(k_x x)sin(k_y y)sin(k_z z) is equal to -3k^2(sin). When taking the partial derivatives, the k terms are pulled out using the chain rule and the sine function is replaced with its derivative, cosine. The sum of the three second derivatives with respect to x, y, and z results in -3k^2(sin).
  • #1
SpY]
65
0

Homework Statement


Find the Laplacian of F = [tex]sin(k_x x)sin(k_y y)sin(k_z z)[/tex]


Homework Equations



[tex]\nabla^2 f = \left( \frac{\partial}{\partial x} +\frac{\partial}{\partial y} + \frac{\partial}{\partial z} \right) \cdot \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} \right) \cdot F[/tex]
Where F is a scalar function

The Attempt at a Solution



Biggest problem is with partial derivatives. I don't know how to approach taking a partial derivative of such a big multivariate product :( Just got the derivative of sin(x) is cos(x) and the second derivative -sin(x)

Just want to makes sure, but [tex]\frac{\partial}{\partial x}y = 0 [/tex] but [tex]\frac{\partial}{\partial x}xy = y [/tex] ?
 
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  • #2
Yes to everything you've written. The dot of the two vector operators is the sum of the second derivatives of F. There is no dot product with F. F is a scalar. What do you get if you add the three second derivatives of F with respect to x, y and z?
 
  • #3
It will just be the sum that is the Laplacian.. as you just said :O

Problem is how to do one partial derivative, for starters, on
[tex]
sin(k_x x)sin(k_y y)sin(k_z z)
[/tex]
 
  • #4
What is the derivative, with respect to x, of [itex]AB sin(k_x x)[/itex] with A and B constants?

What is the derivative, with respect to y, of [itex]AB sin(k_y y)[/itex] with A and B constants?

What is the derivative, with respect to z, of [itex]AB sin(k_z z)[/itex] with A and B constants?
 
  • #5
I think when you're taking the partial derivative of the first term, the k term gets pulled out by the chain rule, then just change it to cos. Taking the second derivative brings the k out again by the chain rule, leaving k^. So it will be -k^2(sin-sin-sin)
 

Related to Laplacian, partial derivatives

1. What is a Laplacian?

A Laplacian, also known as the Laplace operator, is a mathematical operator used in vector calculus to measure the curvature and change of a function in a multi-dimensional space. It is denoted by the symbol ∇².

2. How is a Laplacian calculated?

The Laplacian of a function is calculated by taking the sum of the second partial derivatives of the function with respect to each variable. In other words, it is the divergence of the gradient of the function.

3. What is the significance of the Laplacian in physics?

In physics, the Laplacian is used to represent the rate of change of a physical quantity in space. It is commonly used in fields such as fluid dynamics, electromagnetism, and quantum mechanics to describe the behavior of physical systems.

4. How is the Laplacian related to partial derivatives?

The Laplacian is related to partial derivatives because it is essentially the sum of all second order partial derivatives of a function. It is a way to combine multiple partial derivatives into a single operator, making it useful for solving equations with multiple variables.

5. What are some real-world applications of the Laplacian?

The Laplacian has numerous real-world applications, including image and signal processing, computer vision, and machine learning. It is also used in the study of heat transfer, fluid flow, and electrical circuits. Additionally, the Laplacian plays a crucial role in the study of differential equations and their solutions.

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