# Write the functional derivative of the Dirichlet energy

I have the following functional:

$$J(u) = \int_{\Omega} \frac{|\nabla u|^p}{p}\,d\Omega$$

and I want to compute its functional derivative along the direction of an arbitrary test function $$v\in H_0^1$$.

I tried applying the limit definition

$$J'(u)v = \lim_{\varepsilon \to 0} \frac{J(u+\varepsilon v)-J(u)}{\varepsilon}$$

but I have some problems with the non linear part of the integral.

I know that the expected result should be

$$J'(u)v=\int_{\Omega}|\nabla u|^{p-2}\nabla u \cdot \nabla v \,d\Omega$$

since imposing $$J'(u)v=0$$ is equivalent to solve weakly the homogeneus p-Laplace equation, but I don't know how to proceed computing this derivative.

• I am able to manage the case $p=2$ because I can use the fact $|\nabla u|^2=\nabla u \cdot \nabla u$ and then compute the limit I wrote above. This thing cannot be done so easily for a general p since I cannot split it into a scalar product of this kind. – Dadeslam Jun 26 at 10:13
• For the case of a general $p$, is the limit definition the right approach to follow or is there a more operative definition to get the result? My problem is how to apply the functional $J'(u)$ to $v$ and how to get $\nabla v$ inside the integral. – Dadeslam Jun 26 at 10:15
• Thanks, I've understood the whole procedure except for the part in which you say that the derivative of $|\nabla u|^2$ is $2\nabla u \cdot \nabla v$, isn't it just $2\nabla u$? – Dadeslam Jun 26 at 15:37
• I am a bit imprecise here. I was referring to the map $\epsilon \mapsto |\nabla (u+\epsilon v)|^2$ and by derivative I mean $\frac{\partial}{\partial \epsilon}$ of that function at $\epsilon =0$. – Arctic Char Jun 26 at 15:40
• Ok thanks, now it is clear. – Dadeslam Jun 26 at 15:41

## 1 Answer

In general when $$f: \Bbb{R}^n \to \Bbb{R}$$ is the euclidean norm, defined by $$\begin{equation} f(x) = |x| = \sqrt{\sum_{i=1}^n x_i^2} \end{equation}$$ then away from the origin, $$f$$ is $$C^{\infty}$$ and its derivative is given by $$\begin{equation} f'(x)(\eta) = \left\langle \dfrac{x}{|x|}, \eta \right\rangle \end{equation}$$ (for all $$x \neq 0$$, and for all $$\eta \in \Bbb{R}^n$$). Now, for convenince, define $$g: \Bbb{R} \times \Bbb{R}^n \to \Bbb{R}^n$$, $$\begin{equation} g(\varepsilon,x) = \nabla u(x) + \varepsilon \nabla v(x) \end{equation}$$

The energy functional's derivative you wish to calculate is given by \begin{align} J'(u)(v) &= \dfrac{d}{d \varepsilon} \bigg|_{\varepsilon = 0} J(u+ \varepsilon v) \\ &= \dfrac{d}{d \varepsilon} \bigg|_{\varepsilon = 0} \int_{ x \in\Omega} \dfrac{\left[ (f \circ g)(\varepsilon, x) \right]^p}{p} \\ &= \int_{ x \in\Omega} \dfrac{\partial}{\partial \varepsilon} \bigg|_{\varepsilon = 0} \dfrac{\left[ (f \circ g)(\varepsilon, x) \right]^p}{p} , \end{align} This last equality is by Leibniz's integral rule for differentiation under the integral sign. Now, inside, we just have to make use of the chain rule inside. Doing so yields \begin{align} J'(u)(v) &= \int_{ x \in\Omega} \dfrac{p \cdot \left[ f(g(0,x)) \right]^{p-1}}{p} \cdot f'[g(0,x)] \left( \dfrac{\partial g}{\partial \varepsilon}(0,x) \right) \\ &= \int_{ x \in\Omega} |\nabla u(x)|^{p-1} \cdot \left \langle \dfrac{\nabla u(x)}{|\nabla u(x)|} , \nabla v(x) \right \rangle \\ &= \int_{x \in \Omega} |\nabla u(x)|^{p-2} \cdot \left \langle \nabla u(x), \nabla v(x) \right \rangle \end{align} Modulo some notation, this is precisely what you wanted to prove.

By the way, my entire answer assumes that all the functions are nice and differentiable, and in particular that if $$p< 2$$, then $$\nabla u(x)$$ doesn't vanish anywhere, so that the norm is differentiable. If this isn't the case, then the argument will necessarily have to be more involved... I'm not too sure how one might proceed in this case though.

If $$p\geq 2$$, then we can write $$|\nabla u|^p = \langle \nabla u, \nabla u\rangle^{p/2}$$. In this case, the functions will be differentiable even regardless of whether or not $$\nabla u$$ vanishes; this is because the function $$t \mapsto t^{p/2}$$ is everywhere differentiable, and the inner product $$\langle \cdot , \cdot \rangle$$ is also everywhere differentiable, hence their composition will be as well. Of course, in this case, we will have to slightly modify the above argument, but the final answer is the same
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