Extension of Pohozaev's identity to the whole domain $\mathbb R^m$

Question

I tried asking this question on Math SE but received no answers, even with a bounty. So I am asking it here.

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TLDR, I am trying to extend Pohozaev's identity on bounded domains to the unbounded domain $\Omega=\mathbb R^m$ but I am having some difficulty. Would appreciate some help.

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I recently proved the classical Pohozaev inequality on bounded domains, that is:

THEOREM: (Pohozaev's identity.) Let $\Omega\subset \mathbb R^m$ be a connected, open, bounded set. Let $f\in C^0(\mathbb R)$ be a continuous function on the real line and let $F(z):=\int_0^z f(s)\mathrm ds$ be its primitive. Now let $\phi\in C^2(\Omega)\cap C^1(\bar{\Omega})$ satisfy the semi-linear boundary value problem $$\begin{cases} -\Delta\phi=f(\phi) & \text{in}~\Omega \\ \phi=0 & \text{on}~\partial\Omega \end{cases}$$ Then, the following equality holds: $$\left(1-\frac{m}{2}\right)\int_{\Omega}|\nabla \phi|^2\mathrm d\mu^m+m\int_{\Omega} F(\phi)\mathrm d\mu^m=\frac{1}{2}\int_{\partial\Omega}|\nabla \phi(\boldsymbol x)|^2~\boldsymbol x\cdot \boldsymbol n(\boldsymbol x)~\mathrm d^{m-1}\boldsymbol x$$ Where $\boldsymbol n:\partial\Omega\to\mathbb R^m$ is such that $\boldsymbol n(\boldsymbol x)$ is the unit outward normal to $\partial\Omega$ at the point $\boldsymbol x\in\partial\Omega$.

However, when I look up the theorem on Wikipedia, they instead state the theorem for $\Omega=\mathbb R^m$, with the omission of the boundary term, that is, they state the identity as

THEOREM: Pohozaev's identity on the whole space. If $\phi\in C^2(\mathbb R^m)$ satisfies $$-\Delta\phi=f(\phi) \\ \text{in}~\mathbb R^m$$ As well as the growth/integrability conditions $$ \phi\in L^\infty_{\text{loc}}(\mathbb R^m)~,~\nabla\phi\in L^2(\mathbb R^m)~~,~~(F\circ \phi)\in L^1(\mathbb R^m)$$ Then, the following holds: $$(m-2)\int_{\mathbb R^m}|\nabla\phi|^2\mathrm d\mu^m=2m\int_{\Bbb R^m}F(\phi)\mathrm d\mu^m$$

Since I couldn't access the reference that they linked, I tried to prove the result by first generalizing the classical Pohozaev inequality to arbitrary initial data, applying this generalized version to the balls $\mathbb B(0,R)$, and then taking $ R\to\infty$. However, in order for this to work, it is necessary that $F(\phi(\boldsymbol x))$ and $\nabla\phi(\boldsymbol x)$ both decay to $0$ faster than $|\boldsymbol x|^{-m}$, however, this is not implied by the integrability conditions above. Nor is it even implied by the extra condition $\nabla\phi(\boldsymbol x),\phi(\boldsymbol x)\to 0$ as $|\boldsymbol x|\to\infty$. For instance, given the function

$$g:\mathbb R\to\mathbb R \\ g(x)=\sum_{k\in\mathbb N}g_k(x)+g_{-k}(x) \\ g_k(x):=\chi\left(\left[k-\frac{1}{2k^2},k+\frac{1}{2k^2}\right]~,x\right)\cdot(1-2k^2|x-k|)$$ Where $\chi(S,\cdot)$ is the indicator function of a set $S$, then we can see that $g\in L^1(\mathbb R)$, that is $\int_{\mathbb R}|g|~\mathrm d\mu^1<\infty$ (in particular it is equal to $\zeta(2)$). It is also in $L^2(\mathbb R)$, since $|g|\leq 1$, and thus $|g|^2\leq |g|$.

Therefore, the function $$\tilde{g}(x):=\frac{1}{1+\sqrt{|x|}}~g(x)$$ Is in $L^1(\mathbb R)\cap L^2(\mathbb R)$, and also satisfies $\tilde{g}(x)\to 0$ as $|x|\to\infty$, however, it does NOT decay faster than $|x|^{-1}$, I.e, $\nexists A>0, \alpha>0$ such that $\tilde{g}(x)\leq \frac{A}{|x|^{1+\alpha}}$ for all $x\in\mathbb R$. I suppose the point is that such pathological functions cannot possibly be the solutions to an elliptic semi-linear PDE like $-\Delta\phi=f(\phi)$, but I am not experienced enough in this field to rigorously say why that is.

Can someone help me out with extending Pohozaev's identity to the whole space $\mathbb R^m$ please? I will post my attempt below.

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ATTEMPT:

We can actually extend Pohozaev's identity to arbitrary boundary data, and, in doing so, extend it to the case $\Omega=\mathbb R^m$ as well.

Theorem: (Generalized Pohozaev identity). Let $\Omega\subset \mathbb R^m$ be a connected, open, bounded set. Let $f\in C^0(\mathbb R)$ be a continuous function on the real line and let $F(z):=\int_0^z f(s)\mathrm ds$ be its primitive. Now let $\phi\in C^2(\Omega)\cap C^1(\bar{\Omega})$ satisfy the semi-linear boundary value problem $$\begin{cases} -\Delta\phi=f(\phi) & \text{in}~\Omega \\ \phi=h & \text{on}~\partial\Omega \end{cases}$$ With, obviously, $h\in C^1(\partial\Omega)$. Then, the following equality holds: $$\int_{\Omega}\bigg[mF(\phi)+\left(1-\frac{m}{2}\right)|\nabla \phi|^2\bigg]~\mathrm d\mu ^m=\int_{\partial\Omega}\bigg[\big(\nabla\phi(\boldsymbol x)\cdot \boldsymbol x\big)\big(\nabla\phi(\boldsymbol x)\cdot \boldsymbol n(\boldsymbol x)\big)+\left(F(h(\boldsymbol x))-\frac{1}{2}|\nabla\phi(\boldsymbol x)|^2\right)(\boldsymbol x\cdot \boldsymbol n(\boldsymbol x))\bigg]~~\mathrm d^{m-1}\boldsymbol x$$ Where $\boldsymbol n:\partial\Omega\to\mathbb R^m$ is such that $\boldsymbol n(\boldsymbol x)$ is the unit outward normal to $\partial\Omega$ at the point $\boldsymbol x\in\partial\Omega$.

Proof (sketch): The proof follows easily from our prior proof of the conventional Pohozaev identity. All we need to do is find exactly where we used the assumption that $\phi=0$ on $\partial\Omega$. We in fact used it in two places. The first was right at the end, when we said that $$\int_{\partial\Omega}F(\phi(\boldsymbol x))(\boldsymbol x\cdot \boldsymbol n(\boldsymbol x))~\mathrm d^{m-1}\boldsymbol x=0$$ Hence why this term is included in the above identity, with the substitution $\phi\to h$ as per the BCs. We also used the assumption of $\phi$ vanishing on the boundary when we showed that $$(\nabla\phi(\boldsymbol x)\cdot \boldsymbol x)(\nabla \phi(\boldsymbol x)\cdot \boldsymbol n(\boldsymbol x))=|\nabla\phi(\boldsymbol x)|^2~\boldsymbol x\cdot \boldsymbol n(\boldsymbol x)$$ for $\boldsymbol x\in\partial\Omega$, which follows from the fact that $\nabla\phi=|\nabla\phi|\boldsymbol n$ on $\partial\Omega$, which follows from the fact that $\partial \Omega$ is a level set of $\phi$. In the case of arbitrary Dirichlet conditions, this no longer holds, hence why this term appears unaltered in the above generalized identity. It is easy to check that the above generalized identity reduces to the standard Pohozaev identity if these conditions are met.

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We can now use this to adapt the identity to the whole space ($\Omega=\mathbb R^m$). That is,

THEOREM: Pohozaev's identity on the whole space. If $\phi\in C^2(\mathbb R^m)$ satisfies $$-\Delta\phi=f(\phi) \\ \text{in}~\mathbb R^m$$ As well as the growth/integrability conditions $$\phi(\boldsymbol x),\nabla\phi(\boldsymbol x)\to 0~~~\text{as}~|\boldsymbol x|\to\infty \\ \phi\in L^\infty_{\text{loc}}(\mathbb R^m)~,~\nabla\phi\in L^2(\mathbb R^m)~~,~~(F\circ \phi)\in L^1(\mathbb R^m)$$ Then, the following holds: $$(m-2)\int_{\mathbb R^m}|\nabla\phi|^2\mathrm d\mu^m=2m\int_{\Bbb R^m}F(\phi)\mathrm d\mu^m$$

Proof: This easily follows from applying the generalized Pohozaev identity on the domains $\mathbb B(0,R)$ and taking $R\to\infty$. One of the neat side effects of this approach is that we can write $\boldsymbol n(\boldsymbol x)=\frac{\boldsymbol x}{|\boldsymbol x|}$, which allows us to write $$\big(\nabla\phi(\boldsymbol x)\cdot \boldsymbol x\big)\big(\nabla\phi(\boldsymbol x)\cdot \boldsymbol n(\boldsymbol x)\big)=\frac{1}{|\boldsymbol x|}\big|\nabla\phi(\boldsymbol x)\cdot \boldsymbol x\big|^2 \\ \leq|\boldsymbol x|~|\nabla\phi(\boldsymbol x)|^2$$ By Cauchy-Schwartz, and similarly $\boldsymbol x\cdot \boldsymbol n(\boldsymbol x)=|\boldsymbol x|$. Therefore, we get the following inequality: $$\int\limits_{\mathbb B(0,R)}\bigg[mF(\phi)+\left(1-\frac{m}{2}\right)|\nabla \phi|^2\bigg]~\mathrm d\mu ^m\leq \int\limits_{\partial\mathbb B(0,R)}\bigg[\frac{1}{2}|\boldsymbol x|~|\nabla\phi(\boldsymbol x)|^2+F(\phi(\boldsymbol x))|\boldsymbol x|\bigg]~~\mathrm d^{m-1}\boldsymbol x$$ And since $|\boldsymbol x|=R$ on $\partial\mathbb B(0,R)$, this simplifies further to $$\int\limits_{\mathbb B(0,R)}\bigg[mF(\phi)+\left(1-\frac{m}{2}\right)|\nabla \phi|^2\bigg]~\mathrm d\mu ^m\leq \frac{R}{2}\int\limits_{\partial\mathbb B(0,R)}|\nabla\phi|^2~\mathrm d\mu^{m-1}+R\int\limits_{\partial\mathbb B(0,R)}F(\phi)~\mathrm d\mu^{m-1}$$

At this point, the analysis gets a bit complicated. We essentially have to show that $\nabla\phi$ and $F(\phi)$ decay faster than $|\boldsymbol x|^{-m}$....

(It is at this point that I get stuck!)

Answer 1

The key point is that since $$\int_{\mathbb{R}^m}|\nabla u|^2 dx=\int_{0}^{+\infty} \int_{\partial B_{r}} |\nabla u|^2 d\sigma dr, \int_{\mathbb{R}^m}F(u) dx=\int_{0}^{+\infty} \int_{\partial B_{r}} F(u) d\sigma dr,$$ then it must exist a sequence $r_n\to+\infty$ such that $$r_n\int_{\partial B_{r_n}} |\nabla u|^2 d\sigma,~ r_n\int_{\partial B_{r_n}}F(u) d\sigma \to 0.$$ Otherwise, if $\liminf\limits_{r\to+\infty}r\int_{\partial B_{r}} |\nabla u|^2 d\sigma>0$， then $\int_{0}^{+\infty} \int_{\partial B_{r}} |\nabla u|^2 d\sigma dr=+\infty$.

Now you can use the local Pohozaev identity in each $B_{r_n}$, and by letting $n\to +\infty$, you will find that all the boundary terms must go to zero.