# Laplacian identity.

Consider $f$ and $g$ smooth functions. How to prove the following identity:

$$\Delta\left(\frac{f}{g}\right)=\frac{1}{g}\Delta f-\frac{2}{g}\nabla\left(\frac{f}{g}\right).\nabla g-\frac{f}{g^2}\Delta g, \ \ \ \ \ g\neq0$$

Thank you very much.

• I read that this is a general identity! $f$ and $g$ would be $C^\infty$ functions. Jan 27, 2014 at 2:13
• (A) what is the Laplacian of $1/g,$ and (B) what is the Laplacian of $f \cdot h?$ Jan 27, 2014 at 2:26
• We have a general product rule for the Laplacian: $$\Delta(f.g)=f\Delta g+2\nabla f.\nabla g+g\Delta f.$$ Jan 27, 2014 at 2:39
• So all you need now is $\Delta (1/g),$ and combine Jan 27, 2014 at 2:54

Apply the product rule $$\Delta(uv)=u\Delta v+2\nabla u\cdot\nabla v+v\Delta u \tag{1}$$ with $u=f/g$ and $v=g$. The result is $$\Delta f=\frac{f}{g}\Delta g+2\nabla \left(\frac{f}{g}\right)\cdot \nabla g+g\,\Delta \left(\frac{f}{g}\right)$$ which can be rearranged as $$\Delta \left(\frac{f}{g}\right) =\frac{1}{g}\Delta f-\frac{2}{g}\nabla \left(\frac{f}{g}\right)\cdot \nabla g-\frac{f}{g^2}\Delta g$$

• Nice one, very nice indeed, sir. And a nice "+1"! Jan 27, 2014 at 7:45
• Elegant solution. Thank you very much! Jan 27, 2014 at 14:29

I'm going to use $\nabla^2$ for the Laplacian, so herein, "$\nabla^2$" = "$\Delta$".

$\nabla^2(uv) = v\nabla^2u + 2\nabla u \cdot \nabla v + u \nabla^2 v, \tag{0}$

a proof of which may be found in my answer to this question: Show that $\nabla^2(fg)= f\nabla^2g+g\nabla^2f+2\nabla f\cdot\nabla g$. Then, setting $f = u$ and $g^{-1} = v$, we have

$\nabla^2(fg^{-1}) = g^{-1}\nabla^2 f + 2\nabla f \cdot \nabla (g^{-1}) + f\nabla^2(g^{-1}); \tag{1}$

we proceed by first calculating $\nabla (g^{-1})$ and then $\nabla^2(g^{-1})$. $\nabla (g^{-1})$ looks like the easier one; indeed, for any differentiable function $G(g)$,

$\nabla (G(g)) = G'(g)\nabla g, \tag{2}$

which easily follows from the chain rule; for example, in Cartesian coordinates $x_i$ on $\Bbb R^n$, with

$\nabla w = (\dfrac{\partial w}{\partial x_1}, \dfrac{\partial w}{\partial x_2}, . . . , \dfrac{\partial w}{\partial x_n})^T; \tag{3}$

taking $w = G(g)$, we have

$\dfrac{\partial G(g)}{\partial x_i} = \dfrac{d G}{d g}\dfrac{\partial g}{\partial x_i} = G'(g)\dfrac{\partial g}{\partial x_i}, \tag{4}$

from which (2) readily follows. Equation (2) may also be established via the standard, coordinate-independent definition of $\nabla w$ as the unique vector field satisfying

$\langle \nabla w, X \rangle = X[w] \tag{5}$

for all vector fields $X$, where $\langle \cdot, \cdot \rangle$ is the inner product; for then

$\langle \nabla G(g), X \rangle = X[G(g)] = G'(g)X[g] = G'(g)\langle \nabla g, X \rangle = \langle G'(g) \nabla g , X \rangle, \tag{6}$

demonstrating (2) by a somewhat different argument. Taking $G(g) = g^{-1}$ then yields

$\nabla (g^{-1}) = - g^{-2}\nabla g. \tag{7}$

We move on to finding $\nabla^2 (g^{-1})$:

$\nabla^2 (g^{-1}) = \nabla \cdot \nabla (g^{-1}) = \nabla \cdot (-g^{-2} \nabla g), \tag{8}$

using (7); we evaluate $\nabla \cdot (-g^{-2} \nabla g)$ by exploiting the formula

$\nabla \cdot (wX) = \nabla w \cdot X + w \nabla \cdot X \tag{9}$

which may also be found in the linked question cited above, along with some explanatory remarks. (8) and (9) yield, with the aid of (2),

$\nabla^2 (g^{-1}) = \nabla \cdot (-g^{-2} \nabla g) = \nabla (-g^{-2}) \cdot \nabla g -g^{-2} \nabla^2g = 2g^{-3}\nabla g \cdot \nabla g -g^{-2} \nabla^2g, \tag{10}$

and when (7) and (10) are inserted into (1) we find that

$\nabla^2(fg^{-1}) = g^{-1}\nabla^2 f + 2\nabla f \cdot (- g^{-2}\nabla g) + f(2g^{-3}\nabla g \cdot \nabla g -g^{-2} \nabla^2g)$ $= g^{-1}\nabla^2 f - fg^{-2} \nabla^2g - 2g^{-1}(\nabla f \cdot g \nabla g - fg^{-2}\nabla g \cdot \nabla g)$ $= g^{-1}\nabla^2 f - fg^{-2} \nabla^2g - 2g^{-1}(g \nabla f - fg^{-2}\nabla g)\cdot \nabla g. \tag{11}$

We finally evaluate the expression $g \nabla f - fg^{-2}\nabla g$ which occurs inside the parentheses in the last line of (11), using the formula

$\nabla (uv) = u\nabla v + v \nabla u, \tag{12}$

a variant of the ordinary Leibniz rule for the derivative of a product, a discussion of which is also available in the linked citing. From (12) and yet again (2),

$\nabla (fg^{-1}) = g^{-1}\nabla f + f\nabla g^{-1} = g^{-1}\nabla f - fg^{-2}\nabla g. \tag{13}$

Using (13) in (11) and performing some minor re-arrangements we finally arrive at our destination, or perhaps I should say, like Darth Vader in Star Wars, our Destiny:

$\nabla^2(fg^{-1}) = g^{-1}\nabla^2 f - 2g^{-1}\nabla (fg^{-1}) \cdot \nabla g - fg^{-2}\nabla^2 g. \tag{14}$

(14) is, of course, the requisite result.

Hope this helps. May The Force be with us, one and all,

and of course,

Fiat Lux!!!

• Thank you very much for help me. Jan 27, 2014 at 14:42