Second variation corresponding to the functional

Question

I am facing difficulty to calculate the second variation to the following functional.

Define $J: W_{0}^{1,p}(\Omega)\to\mathbb{R}$ by $J(u)=\frac{1}{p}\int_{\Omega}|\nabla u|^p\,dx$ where $p>1$.

I am able to calculate the first variation as follows: $J'(u)\phi=\int_{\Omega}\,|\nabla u|^{p-2}\nabla u\cdot\nabla\phi\,dx$ which I have got by using the functional $E:\mathbb{R}\to\mathbb{R}$ defined by $E(t)=J(u+t\phi)$.

But I am unable to calculate the second variation.

Any type of help is very much appreciated.

Thanks.

Answer 1

Calculating $\delta^2J$ by using the standard definition is a bit tedious and tricky.
Notation: I assume that $\Omega\in\mathbb{R}^n$, $n\ge 2$. Also, to denote the scalar product of two vectors $\boldsymbol{a},\boldsymbol{b}\in\mathbb{R}^n$ I will use both the notations $\boldsymbol{a}\cdot\boldsymbol{b}$ and $\langle\boldsymbol{a},\boldsymbol{b}\rangle$ since the former is quickly understandable while this last one is easier to handle when its arguments are sums of vectors. Then $$ |\boldsymbol{a}|=(\boldsymbol{a}\cdot\boldsymbol{a})^{1\over 2}=\langle\boldsymbol{a}\cdot\boldsymbol{a}\rangle^{1\over 2}=\sqrt[2]{\sum_{i=1}^n a_i^2}\qquad\boldsymbol{a}=(a_1,\dots,a_n) $$ By using the standard definition of second variation of a functional,we have $$ \begin{split} \delta^2 J(u,\phi)&=\frac{\mathrm{d}^2}{\mathrm{d}t^2} J(u+t\phi)|_{t=0}\\ &=\frac{\mathrm{d}}{\mathrm{d}t}\left[{1\over p}\frac{\mathrm{d}}{\mathrm{d}t}\int\limits_\Omega |\nabla u+t\nabla\phi|^p\mathrm{d}x\right]_{\,t=0}\\ &=\frac{\mathrm{d}}{\mathrm{d}t}\left[{1\over 2}\int\limits_\Omega |\nabla u+t\nabla\phi|^{p-2}\frac{\mathrm{d}}{\mathrm{d}t}\langle\nabla u+t\nabla\phi,\nabla u+t\nabla\phi\rangle\mathrm{d}x\right]_{\,t=0} \end{split}\label{1}\tag{1} $$ Now we have
$$ \begin{split} \frac{\mathrm{d}}{\mathrm{d}t}\langle\nabla u+t\nabla\phi,\nabla u+t\nabla\phi\rangle&=\frac{\mathrm{d}}{\mathrm{d}t}\sum_{i=1}^n(\partial_{x_i} u+t\partial_{x_i}\phi)^2\\ &=\frac{\mathrm{d}}{\mathrm{d}t}\sum_{i=1}^n\big[(\partial_{x_i} u)^2+2t(\partial_{x_i} u\,\partial_{x_i}\phi)+t^2(\partial_{x_i}\phi)^2\big]\\ &=2\sum_{i=1}^n\big[(\partial_{x_i} u\,\partial_{x_i}\phi)+t(\partial_{x_i}\phi)^2\big]=2\big(\nabla u\cdot\nabla\phi+t|\nabla\phi|^2\big) \end{split}\label{2}\tag{2} $$ and by using \eqref{2} in \eqref{1} jointly with Leibnitz's rule we obtain $$ \begin{split} \delta^2 J(u,\phi)=&\frac{\mathrm{d}^2}{\mathrm{d}t^2} J(u+t\phi)|_{t=0}\\ =&\frac{\mathrm{d}}{\mathrm{d}t}\left[\int\limits_\Omega |\nabla u+t\nabla\phi|^{p-2}\big(\nabla u\cdot\nabla\phi+t|\nabla\phi|^2\big)\mathrm{d}x\right]_{\,t=0}\\ =&\left[(p-2)\int\limits_\Omega |\nabla u+t\nabla\phi|^{p-4}\big(\nabla u\cdot\nabla\phi+t|\nabla\phi|^2\big)^2\mathrm{d}x\right.\\ &+\left.\int\limits_\Omega |\nabla u+t\nabla\phi|^{p-2}|\nabla\phi|^2\mathrm{d}x\right]_{\,t=0}\\ =&(p-2)\int\limits_\Omega |\nabla u|^{p-4}(\nabla u\cdot\nabla\phi)^2\mathrm{d}x+ \int\limits_\Omega |\nabla u|^{p-2}|\nabla\phi|^2\mathrm{d}x \end{split}\label{3}\tag{3} $$ Notes

• The deduction above is only formal since, without further hypotheses, if $0< p<2$ we cannot be sure that $|\nabla u|\neq 0$ on a set of zero Lebesgue measure non we can assume the integrability of its inverse powers $|\nabla u|^{p-2}$ and $|\nabla u|^{p-4}$.

• Perhaps in this case, instead of the standard definition of $\delta^2J$, the use of the classical definition of the second derivative of a functional (as can be seen in [1], §1.3, pp. 23-26 and also §1.4, pp. 26-29) could have eased the task: according to this definition, which dates back at least to the work of Vito Volterra, $$ \delta^2J(u,\phi,\psi)=\frac{{\partial}^2}{\partial t\partial s} J(u+t\phi+s\psi)|_{t,s=0} $$ This definition has the advantage that it allows the calculation of the second variation directly from the first, without having to take care of terms that will vanish at the end of calculations. Another advantage of this definition is that it works also when $u,\phi$ and $\psi$ are more abstract objects, that for which a concept of multiplication is not univocally defined (generalized functions and the likes). However, I preferred to follow the standard route because the OP asked so and because it is interactive.

[1] A. Ambrosetti and G. Prodi (1995), A primer of nonlinear analysis (English) Cambridge Studies in Advanced Mathematics. 34. Cambridge: Cambridge University Press, pp. 180, ISBN: 0-521-48573-8, MR1336591, Zbl 0818.47059.

Answer 2

I propose a quick shortcut, based on dimensional analysis. This method does not yield the complete result, however.

The second derivative $J''(u)\phi$ must be $p-2$-homogeneous in $\nabla u$ and quadratic in $\nabla \phi$, thus it must be that $$ J''(u)\phi = C_1\int_{\Omega} |\nabla u|^{p-2}|\nabla\phi|^2 + C_2\int_{\Omega} |\nabla u|^{p-4}(\nabla u \cdot \nabla\phi)^2,$$ for some constants $C_1$ and $C_2$ that cannot be determined by homogeneity alone. However, for $\phi=u$ it must be that $$ \begin{split} J''(u)u =&=\left.\frac{d^2}{d\epsilon^2}\right|_{\epsilon=0}\frac1p\int_{\Omega}|\nabla(u+\epsilon u)|^p\\ &= \left.\frac{d^2}{d\epsilon^2}\frac{(1+\epsilon)^p}{p}\right|_{\epsilon=0}\int_\Omega|\nabla u|^p=(C_1+C_2)\int_\Omega|\nabla u|^p, \end{split}$$ so $C_1+C_2=p-1$.

I don't know how to find another equation, thus determining $C_1$ and $C_2$, so this answer is incomplete. But I do think that this method may be useful. It is always good to have these shortcuts available, when performing long computations with a high chance of mistake (or computations with high entropy, in the words of the book Street fighting mathematics).