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We consider the Poisson equation $$-\Delta u=f \quad \text{in} \quad\mathbb{R}^{n},$$

where $f\in L^{2}(\mathbb{R}^{n})$. We say $ u\in W_{loc}^{1,2}(\mathbb{R}^{n}) $ is a weak solution of the equation, if $$\int_{\mathbb{R}^{n}}\nabla u\cdot \nabla v dx=\int_{\mathbb{R}^{n}}fv dx, \forall v \in C_{0}^{\infty}(\mathbb{R}^{n}).$$

I want to prove

  • There is always a weak solution of the equation.

More generally, we can consider the $p$-Laplace equation $$-\nabla\cdot(|\nabla u|^{p-2}\nabla u)=f \quad \text{in} \quad\mathbb{R}^{n},$$

where $1<p<\infty.$ and $f\in L^{q}(\mathbb{R}^{n}),q=\frac{p}{p-1}.$ We say $ u\in W_{loc}^{1,p}(\mathbb{R}^{n}) $ is a weak solution of the equation, if $$\int_{\mathbb{R}^{n}}|\nabla u|^{p-2}\nabla u\cdot \nabla v dx=\int_{\mathbb{R}^{n}}fv dx, \forall v \in C_{0}^{\infty}(\mathbb{R}^{n}).$$

I know that we can't have $W^{1,p}(\mathbb{R}^{n})$ solutions in general, neither the uniqueness of the solution, so I want to know the existence of $W_{loc}^{1,p}(\mathbb{R}^{n})$ solution. In fact, I am able to use the variational method to prove that

  • If $\mathbb{R}^{n}$ is replaced by a bounded domain $\Omega$, then there is a unique weak solution in $W_{0}^{1,2}(\Omega)$ of the $p$-Laplace equation.

However, we do not have the Poincare inequality and compactness results in $\mathbb{R}^{n}$, so the variational method fails. I have also tried to approximate by solutions on bounded domains, but I can't get nice apriori estimates.

Holomodric
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1 Answers1

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I will do the case $p=2$. The trick is to consider the perturbed problem $$-\Delta u+\frac{1}{k}u=f$$ For this one, you have existence and you can find a solution $u_k$ such that $$\int_{\mathbb{R}^n}\frac{1}{k^2}u^2_k+\frac{1}{k}|\nabla u_k|^2+|\nabla^2u_k|^2\le C\int_{\mathbb{R}^n}f^2$$ Then you consider $v_k=u_k-\text{affine function}$ where the affine function is chosen so that both $v_k$ and $\nabla v_k$ have average zero in the unit ball. Using Poincaré inequality you show that $v_k$ is bounded in $H^2(B(0,R))$ for each $R>1$. So you can use a diagonal limit to find a function $u$ in $H^2_{\text{loc}}(\mathbb{R}^n)$ and a subsequence not relabeled such that $v_n$ converges weakly to $u$ in $H^2$ of bounded sets and $\nabla^2 v_k=\nabla^2 u_k$ goes weakly to $\nabla^2 u$ in $L^2(\mathbb{R}^n)$. The details are in Chapter 13 of fractional book or Mueller lecture notes I am not sure for $p\ne 2$ since the proof uses Nirenberg difference quotient regularity theory which only works for $p=2$.

If you take $n=1$ so that $u^{\prime\prime}=f$ you can see that the only estimate in the entire space that you can hope for is $\|u’’\|_{L^2}\le \|f\|_{L^2}$ but by direct integration neither $u$ nor $u’$ will be bounded in $L^2$ of the entire space. This is why all the constants will depend on $R$ if you try to bound the norm $H^1(B(0, R)$.

Gio67
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