These are actually 4 question about a proof given in this paper. Any hint to solutions for any of these questions would be much appreciated!

Lemma 1. Assume $v$ is a solution to the equation \begin{equation} ic \partial_1 v + \Delta v + v(1-\vert v \vert^2) =0 \tag{GP} \end{equation} in $L^1_{loc}(\mathbb{R}^3)$ of finite energy, then $v$ is regular, bounded and its gradient belongs to all the spaces $W^{k,p}(\mathbb{R}^3)$ for $k \in \mathbb{3}$ and $p \in [2,\infty]$.

Here "finite energy" means, that \begin{equation} \frac{1}{2} \int_{\mathbb{R}^N}\vert \nabla v \vert^2+\frac{1}{4}\int_{\mathbb{R}^N}(1-\vert v \vert^2)^2 < \infty. \end{equation}

Proof. The authors consider a point $z_0 \in \mathbb{R}^3$ and denote by $\Omega$ the unit ball with center $z_0$. Then the consider the solutions $v_1$ and $v_2$ of \begin{equation} \begin{cases} \Delta v_1=0 & \mbox{on } \Omega \\ v_1 = v & \mbox{on } \partial \Omega\end{cases} \tag{1} \end{equation} and \begin{equation} \begin{cases} \Delta v_2=g(v):=v(1-\vert v\vert^2) + ic \partial_1 v & \mbox{on } \Omega \\ v_2 = 0 & \mbox{on } \partial \Omega\end{cases} \tag{2} \end{equation}

Question 1. It seams odd to consider $v \in L^1_{loc}(\mathbb{R}^3)$. A function in this space can not even be a weak solution to (GP), can it? Also equation (1) does not need have a solution if $v$ is only $L^1$, does it?

The authors go on and show that $g(v)$ is uniformly bounded in $L^{\frac{4}{3}}(\Omega)$, i.e. $\Vert g(v) \Vert_{L^{\frac{4}{3}}(\Omega)}$ is bounded by a constant not depending on $z_0$. The infer that $v_1$ is unformly bounded in $L^4(\Omega)$ and $v_2$ in $W^{2,\frac{4}{3}}(\Omega)$. I do already understand this part (some elliptic arguments and Sobolev embeddings).

But now they denote by $\omega$ the ball with center $z_0$ and radius $\frac{1}{2}$ and tell me to use Cacioppoli inequalities to show that $v_1$ is uniformly bounded on $W^{2,\frac{4}{3}}(\omega)$ and $W^{3,\frac{12}{11}}(\omega)$. This would show that $v$ is uniformly bounded in $W^{2,\frac{4}{3}}(\Omega)$.

Question 2. How do you do that? By Caccioppoli inequalities for the Laplace equation I can show that $\Vert Dv_1 \Vert_{L^2(\omega)} \Vert \leq C \Vert v_1 \Vert_{L^2(\omega)}$ but nothing else. How do they do this? Why do they even need the second statement (i.e. $W^{3,\frac{12}{11}}$)?

They go on and compute \begin{equation} \nabla g(v) = \nabla v(1-\vert v\vert^2) - 2 (v.\nabla v)v + ic \partial_1 \nabla v \end{equation} which is easy. But from this they infer that $g(v)$ is uniformly bounded in $L^\frac{12}{11}(\omega)$ and finally $v$ is uniformly bounded in $C^{0,\frac{1}{12}}(\omega)$.

Question 3. I don't understand any of these steps. For the first one I try to estimate $\int \vert \nabla v - \nabla v \vert v \vert^2 \vert^\frac{12}{11} \leq \int \vert \nabla v \vert^\frac{12}{11} + \int \vert \nabla v \vert^\frac{12}{11} \vert v \vert^\frac{24}{11}$ which I can't manage. For the second one I don't understand why they pick the $\frac{1}{12}$-Hölder space of all Hölder spaces.

Last but not least they set \begin{equation} h(w)=w(1-\vert v \vert^2) - 2 (v.w)v + \left(\frac{c^2}{2}+2\right)w \end{equation} where $w=\nabla v$ and state that $h(w)$ belongs to $L^2(\mathbb{R}^N)$ and $w$ to $H^2(\mathbb{R}^N)$.

Question 4. Why is $\nabla v$ even differentiable enough to be plugged in $h$? How do they infer their two results? I have absolutely no idea.

Thank you so much in advance!


1 Answer 1


Question 1. The function is "in $L^1_{\rm loc}(\mathbb R^3)$ of finite energy". The part "in $L^1_{\rm loc}(\mathbb R^3)$" is actually redundant here, since the finiteness of energy gives $L^4_{\rm loc}$ directly (and $L^6_{\rm loc}$ via Sobolev embedding of $W^{1,2}$ in three dimensions). Also, since $v\in W^{1,2}$, the Dirichlet problem (1) for $v_1$ makes sense: the boundary condition $v_1=v$ is understood as $v_1-v\in W^{1,2}_0(\Omega)$. Observe that since $v_1$ minimizes Dirichlet energy subject to this boundary condition, we have $$\int_{\Omega}|\nabla v_1|^2\le \int_{\Omega}|\nabla v|^2\le 2E(v) \tag{A}$$

The weak form of (GP) is $$\int \left\{ - ic v\phi_1 - \nabla v\cdot \nabla \phi + v(1-\vert v \vert^2)\phi\right\} =0\tag{WGP}$$ for all smooth compactly supported functions $\phi$. This makes sense as long as $\nabla v$ and $v^3$ are integrable, which again follows from the finiteness of energy.

Question 2. I don't know what form of Caccioppoli's inequality the author had in mind, but one does not need Caccioppoli to estimate the derivatives of a harmonic function. Indeed, $|\nabla v_1|^2$ is a subharmonic function, which means its value at any point is bounded by the average over a ball centered at that point. If $x\in \frac{3}{4}\Omega$ (ball of radius $3/4$), then $B(x,1/4)\subset \Omega$, hence $|\nabla v_1(x)|^2\le \frac{1}{\pi/16}\int_{\Omega}|\nabla v_1|^2 $, where the integral is controlled by (A). This gives a uniform upper bound on the values of $\nabla v_1$ in $\frac{3}{4}\Omega$. Since $\nabla v_1$ is harmonic, the interior regularity for harmonic equations (Theorem 2.10 in Gilbarg-Trudinger) gives you control over all derivatives of $v_1$ in $\omega$ in terms of the supremum of $|\nabla v_1|$ on the boundary of the ball $\frac{3}{4}\Omega$. You have uniform boundedness in any Sobolev space you want.

The part "so, $v$ is uniformly bounded in $W^{2;4/3}(\omega)$" follows by recalling that $v=v_1+v_2$ and $v_2$ is uniformly bounded in $W^{2;4/3}(\omega)$.

Question 3. Recall that $v\in L^6(\omega)$ with a uniform bound coming from the Sobolev embedding. The formula for $\nabla g(v)$ has things like $|\nabla v| v^2$, which is the product of $L^2$ function with $L^3$ function. By Hölder's inequality, such a product is in $L^p$ with $\frac{1}{p}=\frac{1}{2}+\frac{1}{3}$, hence $p=6/5$. There's also $\partial_1 \nabla v$ which is in $L^{4/3}$ because $v\in W^{2,4/3}$. All in all, $\nabla g\in L^{6/5}$. This places $v_2$ into $W^{3,6/5}$ from where it lands (via Morrey-Sobolev) into $C^{0,1/2}$. The numerology is $$\frac{1}{\text{integrability you have}} - \frac{\text{derivatives you give up}}{\text{dimension}} = \begin{cases} 1/q \\ -\alpha/\text{dimension} \end{cases} $$ where $q$ is the new integrability exponent (if the result is positive) and $\alpha$ is the Hölder exponent (if the result is negative).

I don't know why the author has $12/11$ and $1/12$ for the Sobolev and Hölder exponents, they are worse than what I got above.

Question 4. Since $v$ is in $W^{3,6/5}$, the equation (GP) not only holds in the strong sense, but leaves the room for differentiating it once. The expression $h(w)$ is misleadingly named since it obviously depends on $v$ as well. Anyway, $w\in L^2$ because $u$ has finite energy and multiplication by something involving $v$ is harmless ($v$ is bounded). This is why $h(w)$ is in $L^2$.

The reason why the author added $(c^2/2+2)w$ to both sides is to have a uniformly elliptic operator $L$ on the left side of $Lw=h(w)$. The symbol of $L$ is $\xi_1^2+\xi_2^2+\xi_3^2+c\xi_1+(c^2/2+2)$ which is appropriately positive in $\mathbb R^3$. Ellipticity and $Lw\in L^2$ give $w\in H^2$.

  • $\begingroup$ Thank you so much for all our effort! This is really helpful! $\endgroup$
    – mjb
    Aug 14, 2013 at 8:05
  • $\begingroup$ One more question, if I may: In your answer to question 1 you infer $v \in L^4_{loc}$, which is fine, but then you say $v \in W^{1,2}$ which would imply that $v \in L^2$. Is that correct? Or did you mean $loc$? In fact we know by the finiteness of the Energy that $1-\vert v \vert^2 \in L^2$. But then $\int 1 - 2\vert v \vert^2+\vert v \vert^4 =: \int A-B+C < \infty$. Since $\int A+C = \infty$ cleary $\int B =\infty$ and therefore $v \not\in L^2$, no? $\endgroup$
    – mjb
    Aug 14, 2013 at 12:09
  • 1
    $\begingroup$ @mjb I did not attach any domain to $W^{1,2}$ but I meant $\Omega$, not $\mathbb R^3$. $\endgroup$
    – user90090
    Aug 14, 2013 at 17:37
  • $\begingroup$ Ah of course! Thank you so much! $\endgroup$
    – mjb
    Aug 15, 2013 at 7:46
  • 1
    $\begingroup$ @mjb A function harmonic in any sensible way is classically harmonic, by Weyl's lemma. $\endgroup$
    – user90090
    Aug 15, 2013 at 15:58

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