Well-posedness of the weak conductivity Dirichlet-to-Neumann map

For $w\in H_0^1(\mathrm{\Omega })$, [F3] gives $\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}^2=\mathrm{\parallel }w{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2^2\le 2\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2^2$. By [F1], the form $a_{\gamma }(w,z):={\int }_{\mathrm{\Omega }}\gamma \mathrm{\nabla }w\cdot \overline{\mathrm{\nabla }z}dx$ is bounded by $M\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}$ and satisfies $a_{\gamma }(w,w)\ge (m\mathrm{/}2)\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}^2$. Thus it is bounded and coercive on $H_0^1(\mathrm{\Omega })$.

Choose any representative $U\in f$. The functional $z\mapsto -a_{\gamma }(U,z)$ is bounded and conjugate-linear on $H_0^1$, with norm at most $M\mathrm{\parallel }U{\mathrm{\parallel }}_{H^1}$. Lax–Milgram [F4] applied to step 1.1 gives a unique $w\in H_0^1$ satisfying $a_{\gamma }(w,z)=-a_{\gamma }(U,z)$ for every $z\in H_0^1$. Then $u:=U+w\in f$ satisfies the required weak equation.

If $u_1,u_2\in f$ both solve the weak equation, then $v:=u_1-u_2\in H_0^1$ by the quotient definition in [F2], and $a_{\gamma }(v,z)=0$ for every $z\in H_0^1$. Taking $z=v$ gives $m\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2^2\le a_{\gamma }(v,v)=0$; [F3] then gives $\mathrm{\parallel }v{\mathrm{\parallel }}_2=0$, so $v=0$. This proves uniqueness.

If $V_g,V_g^{\mathrm{\prime }}\in g$, then $V_g-V_g^{\mathrm{\prime }}\in H_0^1$ by [F2], so the weak equation for $u_{\gamma ,f}$ makes $a_{\gamma }(u_{\gamma ,f},V_g)=a_{\gamma }(u_{\gamma ,f},V_g^{\mathrm{\prime }})$. Thus the defining functional is independent of the representative.

For a representative $U\in f$, the solution estimate in [F4] and step 1.1 give $\mathrm{\parallel }{u}_{\gamma ,f}{\mathrm{\parallel }}_{H^1}\le (1+2M\mathrm{/}m)\mathrm{\parallel }U{\mathrm{\parallel }}_{H^1}$. Taking the infimum over representatives gives $\mathrm{\parallel }{u}_{\gamma ,f}{\mathrm{\parallel }}_{H^1}\le (1+2M\mathrm{/}m)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathcal{D}}$. Applying the same representative bound to $g$ yields $\mathrm{\mid }{a}_{\gamma }(u_{\gamma ,f},V_g)\mathrm{\mid }\le M(1+2M\mathrm{/}m{)}^2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathcal{D}}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathcal{D}},$ so ${\mathrm{\Lambda }}_{\gamma }$ is bounded into ${\mathcal{D}}^{\ast }$.

For $f_1,f_2\in \mathcal{D}(\mathrm{\Omega })$ and $c_1,c_2\in \mathbb{C}$, the function $c_1{u}_{\gamma ,f_1}+c_2{u}_{\gamma ,f_2}$ belongs to the trace class $c_1{f}_1+c_2{f}_2$. The weak equation is linear in its solution because the energy form is linear in its first argument, so this combination solves the weak problem for that class. Uniqueness from step 3.1 gives $u_{\gamma ,c_1{f}_1+c_2{f}_2}=c_1{u}_{\gamma ,f_1}+c_2{u}_{\gamma ,f_2}$. The energy definition is also linear in this first argument, hence ${\mathrm{\Lambda }}_{\gamma }$ is linear.

For a constant input $c$, the constant function $u=c$ solves the weak equation and uniqueness gives $u_{\gamma ,c}=c$, whose gradient is zero. For a constant test class $c$, its constant representative has zero gradient. The energy formula therefore vanishes on constant inputs and constant tests, so it induces the stated map on the quotient by constants.