Div(gradΦ): Solve 3y12+2y22-3y1y32

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Discussion Overview

The discussion focuses on calculating the divergence of the gradient of a scalar function, specifically Φ = 3y1² + 2y2² - 3y1y3². Participants explore the steps involved in finding div(gradΦ) and clarify concepts related to gradients and divergence.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the function Φ and expresses uncertainty about whether their calculations for the gradient are correct.
  • Another participant reiterates the need to find div(gradΦ) and discusses the meaning of the gradient and divergence in calculus.
  • A third participant confirms the partial derivatives calculated and introduces the concept that the divergence of the gradient is known as the Laplacian.
  • One participant suggests a correction regarding a sign in the expression for y3 and proposes a result of 10 - 6y1.
  • Another participant points out a potential misuse of notation in expressing the partial derivative, suggesting clearer alternatives for notation.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the calculations and notation used, indicating that the discussion remains unresolved regarding the accuracy of the steps taken and the final result.

Contextual Notes

There are indications of missing assumptions and potential misunderstandings regarding notation and expressions, which may affect the clarity of the discussion.

kezzstar
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Φ=3y12+2y22-3y1y32

Find: div(gradΦ).

I broke it down to find the partial derivative of all 3 coordinates y1,y2 and y3, so using the continuity equation:

y1=6y1-3y32
y2=4y2
y3=6y3y1

However, I'm not sure if this correct, would the first part give the gradient? Could someone confirm or solve this?

Regards
 
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kezzstar said:
Find: div(gradΦ).

I broke it down to find the partial derivative of all 3 coordinates y1,y2 and y3, so using the continuity equation:

y1=6y1-3y32
y2=4y2
y3=6y3y1

However, I'm not sure if this correct, would the first part give the gradient? Could someone confirm or solve this?

what is the gradient of a function ? its the slope of the curve representing the function/

divergence also has a meaning in calculus.
 
kezzstar said:
Φ=3y12+2y22-3y1y32

Find: div(gradΦ).

I broke it down to find the partial derivative of all 3 coordinates y1,y2 and y3, so using the continuity equation:

y1=6y1-3y32
y2=4y2
y3=6y3y1

However, I'm not sure if this correct, would the first part give the gradient? Could someone confirm or solve this?

Regards
The divergence of the gradient of a function is also known as the Laplacian of that function:

https://en.wikipedia.org/wiki/Laplace_operator

What you have calculated so far is:

$$\frac{∂Φ}{∂y_1} = 6y_1-3y_3^2$$
$$\frac{∂Φ}{∂y_2}=4y_2$$
$$\frac{∂Φ}{∂y_3}=-6y_3y_1$$

or

##∇Φ = (\frac{∂Φ}{∂y_1},\frac{∂Φ}{∂y_2},\frac{∂Φ}{∂y_3}) = (6y_1-3y_3^2,4y_2,-6y_1y_3)##

The final step is

##div(grad Φ) = ∇ ⋅ ∇Φ = \frac{∂^2Φ}{∂y_1^2}+\frac{∂^2Φ}{∂y_2^2}+\frac{∂^2Φ}{∂y_3^2}##

(Note: edited to fix sign error in third term.)
 
Last edited:
Just to add, I believe you should get 10-6y1.

kezzstar said:
y3=6y3y1
I think you forgot a negative sign there.
 
You wrote
##y_1=6y_1-3y_3^2##
when you should have written
##\frac{\partial\Psi}{\partial y_1}=6y_1-3y_3^2##.
I don't know if you made a mistake in understanding or used a lazy abuse of notation, but you really need to avoid misuse of the equality expression. If you want to write it as a semantic pairing of the left and right, but not an equality, you are better off using a colon or comma, or making a table.
 

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