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Let $u\in H^1(B)$ be a minimising value of $E:H^1(B)\to \mathbb{R}_+$. Then $E'(u)=0$ in the sense that

$$ E'(u)v = \int_B [\nabla u \cdot \nabla v + 4(u^2-1)uv ] dx - \int_{\partial B} Q'(u)v \hspace{.5pc}d\mathcal{H}^2 = 0 $$

for all $v\in H^1(B)$. This implies, in particular, that

$$ \int_B [-u \Delta v + 4(u^2-1)uv] \hspace{.5pc}dx = 0 $$

for $v\in \mathscr{D}(B)$ (this uses the fact that $v$ and all its derivatives vanish on $\partial B$ if it has compact support in $B$). Having this last equation hold for all $v\in \mathscr{D}(B)$ is exactly the statement that

$$ \Delta u = 4(u^2-1)u \hspace{1pc}\mbox{in $B$ } $$

in the sense of distributions. The interior equation in the distributional sense is therefore just what you get from regarding $E'(u)$ itself as a distribution (i.e. by restricting it to $\mathscr{D}(B))$.

For each compact subset $K$ of $B$, take $\psi_K\in \mathscr{D}(B)$ with $\psi_{|K}=1$. Then considering the action of $E'(u)$ on $\{v(1-\psi_K):v\in C^\infty(\bar B),\mbox{$K\subset B$ compact}\}$ should - I think - also give you a weak form of your boundary condition (possibly modulo a minus sign).

I might be wrong, but I think that it's only once you have your interior equation in a (weak or distributional) form which does a priori involve the assumption that $u\in H^2(B)$ that it's time time to start worrying about regularity. In particular, you may be able to establish that $u\in H^2(B)$ using the fact that it is a $H^1$ solution to your PDE.

Let $u\in H^1(B)$ be a minimising value of $E:H^1(B)\to \mathbb{R}_+$. Then $E'(u)=0$ in the sense that

$$ E'(u)v = \int_B [\nabla u \cdot \nabla v + 4(u^2-1)uv ] dx - \int_{\partial B} Q'(u)v \hspace{.5pc}d\mathcal{H}^2 = 0 $$

for all $v\in H^1(B)$. This implies, in particular, that

$$ \int_B [-u \Delta v + 4(u^2-1)uv] \hspace{.5pc}dx = 0 $$

for $v\in \mathscr{D}(B)$ (this uses the fact that $v$ and all its derivatives vanish on $\partial B$ if it has compact support in $B$). Having this last equation hold for all $v\in \mathscr{D}(B)$ is exactly the statement that

$$ \Delta u = 4(u^2-1)u \hspace{1pc}\mbox{in $B$ } $$

in the sense of distributions. The interior equation in the distributional sense is therefore just what you get from regarding $E'(u)$ itself as a distribution (i.e. by restricting it to $\mathscr{D}(B))$.

For each compact subset $K$ of $B$, take $\psi_K\in \mathscr{D}(B)$ with $\psi_{|K}=1$. Then considering the action of $E'(u)$ on $\{v(1-\psi_K):v\in C^\infty(\bar B),\mbox{$K\subset B$ compact}\}$ should - I think - also give you a weak form of your boundary condition.

I might be wrong, but I think that it's only once you have your interior equation in a (weak or distributional) form which does a priori involve the assumption that $u\in H^2(B)$ that it's time time to start worrying about regularity. In particular, you may be able to establish that $u\in H^2(B)$ using the fact that it is a $H^1$ solution to your PDE.

Let $u\in H^1(B)$ be a minimising value of $E:H^1(B)\to \mathbb{R}_+$. Then $E'(u)=0$ in the sense that

$$ E'(u)v = \int_B [\nabla u \cdot \nabla v + 4(u^2-1)uv ] dx - \int_{\partial B} Q'(u)v \hspace{.5pc}d\mathcal{H}^2 = 0 $$

for all $v\in H^1(B)$. This implies, in particular, that

$$ \int_B [-u \Delta v + 4(u^2-1)uv] \hspace{.5pc}dx = 0 $$

for $v\in \mathscr{D}(B)$ (this uses the fact that $v$ and all its derivatives vanish on $\partial B$ if it has compact support in $B$). Having this last equation hold for all $v\in \mathscr{D}(B)$ is exactly the statement that

$$ \Delta u = 4(u^2-1)u \hspace{1pc}\mbox{in $B$ } $$

in the sense of distributions. The interior equation in the distributional sense is therefore just what you get from regarding $E'(u)$ itself as a distribution (i.e. by restricting it to $\mathscr{D}(B))$.

For each compact subset $K$ of $B$, take $\psi_K\in \mathscr{D}(B)$ with $\psi_{|K}=1$. Then considering the action of $E'(u)$ on $\{v(1-\psi_K):v\in C^\infty(\bar B),\mbox{$K\subset B$ compact}\}$ should - I think - also give you a weak form of your boundary condition.

I might be wrong, but I think that it's only once you have your interior equation in a (weak or distributional) form which does a priori involve the assumption that $u\in H^2(B)$ that it's time time to start worrying about regularity. In particular, you may be able to establish that $u\in H^2(B)$ using the fact that it is a $H^1$ solution to your PDE.

Let $u\in H^1(B)$ be a minimising value of $E:H^1(B)\to \mathbb{R}_+$. Then $E'(u)=0$ in the sense that

$$ E'(u)v = \int_B [\nabla u \cdot \nabla v + 4(u^2-1)uv ] dx - \int_{\partial B} Q'(u)v \hspace{.5pc}d\mathcal{H}^2 = 0 $$

for all $v\in H^1(B)$. This implies, in particular, that

$$ \int_B [-u \Delta v + 4(u^2-1)uv] \hspace{.5pc}dx = 0 $$

for $v\in \mathscr{D}(B)$ (this uses the fact that $v$ and all its derivatives vanish on $\partial B$ if it has compact support in $B$). Having this last equation hold for all $v\in \mathscr{D}(B)$ is exactly the statement that

$$ \Delta u = 4(u^2-1)u \hspace{1pc}\mbox{in $B$ } $$

in the sense of distributions. The interior equation in the distributional sense is therefore just what you get from regarding $E'(u)$ itself as a distribution (i.e. by restricting it to $\mathscr{D}(B))$.

For each compact subset $K$ of $B$, take $\psi_K\in \mathscr{D}(B)$ with $\psi_{|K}=1$. Then considering the action of $E'(u)$ on $\{v(1-\psi_K):v\in C^\infty(\bar B),\mbox{$K\subset B$ compact}\}$ should - I think - also give you a weak form of your boundary condition (possibly modulo a minus sign).

I might be wrong, but I think that it's only once you have your interior equation in a (weak or distributional) form which does a priori involve the assumption that $u\in H^2(B)$ that it's time time to start worrying about regularity. In particular, you may be able to establish that $u\in H^2(B)$ using the fact that it is a $H^1$ solution to your PDE.

Let $u\in H^1(B)$ be a minimising value of $E:H^1(B)\to \mathbb{R}_+$. Then $E'(u)=0$ in the sense that

$$ E'(u)v = \int_B [\nabla u \cdot \nabla v + 4(u^2-1)uv ] dx - \int_{\partial B} Q'(u)v \hspace{.5pc}d\mathcal{H}^2 = 0 $$

for all $v\in H^1(B)$. This implies, in particular, that

$$ \int_B [-u \Delta v + 4(u^2-1)uv] \hspace{.5pc}dx = 0 $$

for $v\in \mathscr{D}(B)$ (this uses the fact that $v$ and all its derivatives vanish on $\partial B$ if it has compact support in $B$). Having this last equation hold for all $v\in \mathscr{D}(B)$ is exactly the statement that

$$ \Delta u = 4(u^2-1)u \hspace{1pc}\mbox{in $B$ } $$

in the sense of distributions. The interior equation in the distributional sense is therefore just what you get from regarding $E'(u)$ itself as a distribution (i.e. by restricting it to $\mathscr{D}(B))$.

Considering the action of $E'(u)$ on suitable functions in $\{v(1-\psi): v\in C^\infty(\bar B), \psi\in \mathscr{D}(B)\}$ also - I think - gives you a weak form of your boundary condition.

I might be wrong, but I think that it's only once you have your interior equation in a (weak or distributional) form which does a priori involve the assumption that $u\in H^2(B)$ that it's time time to start worrying about regularity. In particular, you may be able to establish that $u\in H^2(B)$ using the fact that it is a $H^1$ solution to your PDE.

Let $u\in H^1(B)$ be a minimising value of $E:H^1(B)\to \mathbb{R}_+$. Then $E'(u)=0$ in the sense that

$$ E'(u)v = \int_B [\nabla u \cdot \nabla v + 4(u^2-1)uv ] dx - \int_{\partial B} Q'(u)v \hspace{.5pc}d\mathcal{H}^2 = 0 $$

for all $v\in H^1(B)$. This implies, in particular, that

$$ \int_B [-u \Delta v + 4(u^2-1)uv] \hspace{.5pc}dx = 0 $$

for $v\in \mathscr{D}(B)$ (this uses the fact that $v$ and all its derivatives vanish on $\partial B$ if it has compact support in $B$). Having this last equation hold for all $v\in \mathscr{D}(B)$ is exactly the statement that

$$ \Delta u = 4(u^2-1)u \hspace{1pc}\mbox{in $B$ } $$

in the sense of distributions. The interior equation in the distributional sense is therefore just what you get from regarding $E'(u)$ itself as a distribution (i.e. by restricting it to $\mathscr{D}(B))$.

For each compact subset $K$ of $B$, take $\psi_K\in \mathscr{D}(B)$ with $\psi_{|K}=1$. Then considering the action of $E'(u)$ on $\{v(1-\psi_K):v\in C^\infty(\bar B),\mbox{$K\subset B$ compact}\}$ should - I think - also give you a weak form of your boundary condition.

I might be wrong, but I think that it's only once you have your interior equation in a (weak or distributional) form which does a priori involve the assumption that $u\in H^2(B)$ that it's time time to start worrying about regularity. In particular, you may be able to establish that $u\in H^2(B)$ using the fact that it is a $H^1$ solution to your PDE.