1. Apr 3, 2016

kezzstar

Φ=3y12+2y22-3y1y32

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

2. Apr 3, 2016

drvrm

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

divergence also has a meaning in calculus.

3. Apr 3, 2016

SteamKing

Staff Emeritus
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: Apr 3, 2016
4. Apr 3, 2016

Isaac0427

Just to add, I believe you should get 10-6y1.

I think you forgot a negative sign there.

5. Apr 5, 2016

Khashishi

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.