I have this BVP: \begin{cases} -\Delta_{p}(u)=\lambda_1 |x|^{\theta}|u|^{q-2} u+f(x,u)-h ~~\text{in}~\Omega,\\ u\in W^{1,p}_0(\Omega). \end{cases}

where $\Delta_p$ is denoting the $p-$Laplacian operator, i.e., $\Delta_p u= \text{div}~ (|\nabla u|^{p-2}\nabla u),$

What is the Euler functional associated to this problem please.

The first step is:

$-\int_{\Omega} \Delta_p(u) v dx-\lambda_1 \int_{\Omega} |x|^{\theta} |u|^{q-2} u v dx-\int_{\Omega} f(x,u) v dx+\int_{\Omega} h v dx=0$

The second step is to do apply the Green formula,then we obtain:

$$(J'(u),v)=\int_{\Omega}|\nabla u|^{p-2}\nabla u \nabla v dx-\lambda_1 \int_{\Omega} |u|^{q-2} uv dx -\int_{\Omega} f(x,u)v dx+\int_{\Omega} hv dx$$

My question is what is $J(u)$ ?

For a problem like this:

\begin{cases} -\Delta_{p}(u)=\lambda_1 |u|^{q-2} u+f(x,u)-h ~~\text{in}~\Omega,\\ u\in W^{1,p}_0(\Omega). \end{cases}

$J(u)=\frac1p \int_{\Omega} |\nabla u|^p dx -\frac{\lambda_1}{q} \int_{\Omega} |u|^q dx +\int_{\Omega} F(x,u) dx+\int_{\Omega} h u dx$

But what happen when there is $|x|^{\theta}$ in the problem ?

thank you.


1 Answer 1


The derivative of $$ J_0(u)= \frac1{q}\int_\Omega |x|^\theta |u|^{q} dx $$ with respect to $u$ is $$ J_0'(u)v = \int_\Omega |x|^\theta |u|^{q-2}uv dx. $$ When deriving the Euler-Lagrange equation, you do not need to differentiate with respect to the coordinate $x$.

  • $\begingroup$ but we have dx in the integral ? $\endgroup$
    – Vrouvrou
    Commented Mar 25, 2015 at 9:34
  • $\begingroup$ As the energy is minimized with respect to $u$, the term $|x|^\theta$ acts like a constant in this setting. $\endgroup$
    – daw
    Commented Mar 25, 2015 at 9:47
  • $\begingroup$ Please an other question if i have $J(u)=\frac1q \int_{\Omega}(|x|^{\theta} |u|)^q dx$ then $J'(u)v=\int |x|^{q\theta} |u|^{q-2}uv dx$ right ? $\endgroup$
    – Vrouvrou
    Commented Mar 25, 2015 at 11:16

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