Laplacian Identity

Question

Given

$\Delta f :=$ $d^2 f \over d x^2 $ + $d^2 f \over dy^2$,

I have to prove

$\Delta (fg) $=$ f$ * $\Delta$ $g + 2(\nabla f)(\nabla g)+ g$ $\Delta$ $f $.

This was part of one of our exams. The question might be similar to this one

Laplacian identity.,

but we never introduced the general product rule for the Laplacian. Is there another (short) way to show this equation?

Edit

I receive

$f_x$ $=$ $df \over dx^2$ $*$ $g + f$ $*$ $dg \over dx^2$

$f_{xx}$ $=$ ($d^2 f \over dx^2$ $*$ $g$ $+$ $df \over dx^2$ $*$ $dg \over dx^2$) + ($df \over dx^2$ $*$ $dg \over dx^2$ $+$ $f$ $*$ $d^2g \over dx^2$)

$f_y$ $=$ $df \over dy^2$ $*$ $g + f$ $*$ $dg \over dy^2$

$f_{yy}$ $=$ ($d^2 f \over dy^2$ $*$ $g$ $+$ $df \over dy^2$ $*$ $dg \over dy^2$) + ($df \over dy^2$ $*$ $dg \over dy^2$ $+$ $f$ $*$ $d^2g \over dy^2$)

Edit 2

Now, I found all the necessary components I need for the equality, but how do I express the gradient in the middle? I would need canonical vectors to put this all toghether. Am I allowed to assume that they exist in this case? What's left is:

$df \over dx^2$ $*$ $dg \over dx^2$ + $df \over dx^2$ $*$ $dg \over dx^2$ $+$ $df \over dy^2$ $*$ $dg \over dy^2$ $+$ $df \over dy^2$ $*$ $dg \over dy^2$

score 2 · Accepted Answer · answered Jul 04 '16 at 12:53

2

By product rule, begin with $\partial_x fg$.

Then derivate once more : $\partial_{xx}fg$.

The same works with $\partial_y$ and $\partial_{yy}$.

Finally, put everything together.

answered Jul 04 '16 at 12:53

anonymus

1,408

I gave it a try - the derivatives first. Would you mind taking a look at it? :-) – Julian Jul 04 '16 at 14:29
Of course, show me :) – anonymus Jul 04 '16 at 19:35
I editet it in the main post. – Julian Jul 05 '16 at 08:18
You made a mistake : $\partial_x f$ is ok but $\partial_{xx}f$ is not. Try again (with the product formula again). – anonymus Jul 05 '16 at 09:04
Tried it again, edited in the main post. – Julian Jul 05 '16 at 11:26
The $f_{xx}, f_{yy}$ in your 1st edit are correct. Now $\nabla f.\nabla g =.. ?$ – anonymus Jul 05 '16 at 12:22
Yes, that's why I asked in Edit 2. I guess I'm able to build them by multiplying the remaining terms with the canonical vectors (1, 0) and (0,1). – Julian Jul 05 '16 at 12:25
I don't know why you are talking about canonical vectors. Just write $\nabla f$ as a vector, do the same for $\nabla g$ and then do the scalar product :) – anonymus Jul 05 '16 at 12:27
Ohhh, so the remaining terms are the result from the scalar product of $\nabla f$ and $\nabla g$, that's why they equal $2 \nabla f \nabla g$. Got it! :D – Julian Jul 05 '16 at 12:28

Laplacian Identity

1 Answers1