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.