Skip to main content
MathOverflow

Return to Question

added 6 characters in body
Source Link
JumpJump
JumpJump
  • 679
  • 3
  • 13

Given the following PDE $$ \begin{cases} -\Delta u+\alpha=u^q &x\in\Omega\\ u=0 &x\in\partial\Omega \end{cases} $$ where $\Omega\subset\mathbb R^3$ is open bounded with smooth boundary, $1<q<5$, and $\alpha>0$ is a constant.

I am trying to show this PDE has at least a positive weak solution. Here is what I tried so far:

Define the energy functional $$ E[u]=\frac{1}{2}\int_\Omega |\nabla u|^2dx +\alpha \int_\Omega |u|\,dx $$ and the admissible set $$ \mathcal A:=\{u\in H_0^1(\Omega),\,\|u\|_{L^{q+1}(\Omega)}=1 \}$$ Hence by directly method we have a minimizer $u$ over admissible set $\mathcal A$. Also, notice that $E[u]=E[|u|]$ and hence we could assume the minimizer $u$ is non-negative. Therefore, by Lagrange Multiplier we have, for some $k\in\mathbb R$, $$ \int_\Omega\nabla u\nabla v\,dx+\alpha\int_\Omega \frac{u}{u}v\,dx=k\int_\Omega u^qv\,dx$$ for all $v\in H_0^1(\Omega)$.

Now the last step is to show that $u\neq 0$ a.e. and hence the term $$ \alpha\int_\Omega \frac{u}{u}v\,dx$$$$ \alpha\int_\Omega \frac{u}{u}v\,dx\tag 1$$ is well-judged. However, I got stuck here... Any help is really welcome!

PS: I have another way to obtain a positive (classical) solution of this PDE but it takes too many pages but I don't want my paper to contain too many tactical details about this part...

Given the following PDE $$ \begin{cases} -\Delta u+\alpha=u^q &x\in\Omega\\ u=0 &x\in\partial\Omega \end{cases} $$ where $\Omega\subset\mathbb R^3$ is open bounded with smooth boundary, $1<q<5$, and $\alpha>0$ is a constant.

I am trying to show this PDE has at least a positive weak solution. Here is what I tried so far:

Define the energy functional $$ E[u]=\frac{1}{2}\int_\Omega |\nabla u|^2dx +\alpha \int_\Omega |u|\,dx $$ and the admissible set $$ \mathcal A:=\{u\in H_0^1(\Omega),\,\|u\|_{L^{q+1}(\Omega)}=1 \}$$ Hence by directly method we have a minimizer $u$ over admissible set $\mathcal A$. Also, notice that $E[u]=E[|u|]$ and hence we could assume the minimizer $u$ is non-negative. Therefore, by Lagrange Multiplier we have, for some $k\in\mathbb R$, $$ \int_\Omega\nabla u\nabla v\,dx+\alpha\int_\Omega \frac{u}{u}v\,dx=k\int_\Omega u^qv\,dx$$ for all $v\in H_0^1(\Omega)$.

Now the last step is to show that $u\neq 0$ a.e. and hence the term $$ \alpha\int_\Omega \frac{u}{u}v\,dx$$ is well-judged. However, I got stuck here... Any help is really welcome!

PS: I have another way to obtain a positive (classical) solution of this PDE but it takes too many pages but I don't want my paper to contain too many tactical details about this part...

Given the following PDE $$ \begin{cases} -\Delta u+\alpha=u^q &x\in\Omega\\ u=0 &x\in\partial\Omega \end{cases} $$ where $\Omega\subset\mathbb R^3$ is open bounded with smooth boundary, $1<q<5$, and $\alpha>0$ is a constant.

I am trying to show this PDE has at least a positive weak solution. Here is what I tried so far:

Define the energy functional $$ E[u]=\frac{1}{2}\int_\Omega |\nabla u|^2dx +\alpha \int_\Omega |u|\,dx $$ and the admissible set $$ \mathcal A:=\{u\in H_0^1(\Omega),\,\|u\|_{L^{q+1}(\Omega)}=1 \}$$ Hence by directly method we have a minimizer $u$ over admissible set $\mathcal A$. Also, notice that $E[u]=E[|u|]$ and hence we could assume the minimizer $u$ is non-negative. Therefore, by Lagrange Multiplier we have, for some $k\in\mathbb R$, $$ \int_\Omega\nabla u\nabla v\,dx+\alpha\int_\Omega \frac{u}{u}v\,dx=k\int_\Omega u^qv\,dx$$ for all $v\in H_0^1(\Omega)$.

Now the last step is to show that $u\neq 0$ a.e. and hence the term $$ \alpha\int_\Omega \frac{u}{u}v\,dx\tag 1$$ is well-judged. However, I got stuck here... Any help is really welcome!

PS: I have another way to obtain a positive (classical) solution of this PDE but it takes too many pages but I don't want my paper to contain too many tactical details about this part...

Source Link
JumpJump
JumpJump
  • 679
  • 3
  • 13

Showing existence of positive weak solution of a PDE by CoV

Given the following PDE $$ \begin{cases} -\Delta u+\alpha=u^q &x\in\Omega\\ u=0 &x\in\partial\Omega \end{cases} $$ where $\Omega\subset\mathbb R^3$ is open bounded with smooth boundary, $1<q<5$, and $\alpha>0$ is a constant.

I am trying to show this PDE has at least a positive weak solution. Here is what I tried so far:

Define the energy functional $$ E[u]=\frac{1}{2}\int_\Omega |\nabla u|^2dx +\alpha \int_\Omega |u|\,dx $$ and the admissible set $$ \mathcal A:=\{u\in H_0^1(\Omega),\,\|u\|_{L^{q+1}(\Omega)}=1 \}$$ Hence by directly method we have a minimizer $u$ over admissible set $\mathcal A$. Also, notice that $E[u]=E[|u|]$ and hence we could assume the minimizer $u$ is non-negative. Therefore, by Lagrange Multiplier we have, for some $k\in\mathbb R$, $$ \int_\Omega\nabla u\nabla v\,dx+\alpha\int_\Omega \frac{u}{u}v\,dx=k\int_\Omega u^qv\,dx$$ for all $v\in H_0^1(\Omega)$.

Now the last step is to show that $u\neq 0$ a.e. and hence the term $$ \alpha\int_\Omega \frac{u}{u}v\,dx$$ is well-judged. However, I got stuck here... Any help is really welcome!

PS: I have another way to obtain a positive (classical) solution of this PDE but it takes too many pages but I don't want my paper to contain too many tactical details about this part...