[proofplan]
We reduce the affine boundary condition to a homogeneous one by writing the unknown as $u = G + w$, where $w \in H^1_0(U)$. This converts the desired variational equation into a coercive variational problem on the Hilbert space $H^1_0(U)$. The new right-hand side is bounded because $F \in H^{-1}(U)$ and $B$ is bounded on $H^1(U) \times H^1_0(U)$. The boundedness of $U$, Poincare's inequality, and the trace-lifting property are part of the boundary-value setting; the reduction itself uses the fixed lifting $G$ and direct coercivity in the $H^1(U)$ norm. The [Lax-Milgram Theorem](/theorems/91) then gives a unique $w$, and translating back by $G$ gives the unique solution in the affine space $G + H^1_0(U)$.
[/proofplan]
[step:Translate the affine problem into a homogeneous unknown]
Using the [Sobolev space](/page/Sobolev%20Space) $H^1_0(U)$, define the affine space
\begin{align*}
G + H^1_0(U) := \{G + w : w \in H^1_0(U)\}.
\end{align*}
For a candidate solution $u \in G + H^1_0(U)$, there is a unique $w \in H^1_0(U)$ such that
\begin{align*}
u = G + w.
\end{align*}
Substituting this expression into the desired identity gives, for every $v \in H^1_0(U)$,
\begin{align*}
B[G+w,v] = F(v).
\end{align*}
By bilinearity of $B$, this is equivalent to
\begin{align*}
B[w,v] = F(v) - B[G,v].
\end{align*}
Thus it is enough to solve a homogeneous variational problem for $w \in H^1_0(U)$.
[guided]
The boundary condition is encoded by the affine space $G + H^1_0(U)$. This means that every admissible solution differs from the fixed lifting $G$ by an element with homogeneous boundary data. We therefore introduce the unknown
\begin{align*}
w := u - G.
\end{align*}
Since $u \in G + H^1_0(U)$, this definition gives $w \in H^1_0(U)$, and conversely every $w \in H^1_0(U)$ produces an admissible function $u = G + w$.
Now impose the variational equation on $u = G + w$. For every [test function](/page/Test%20Function) $v \in H^1_0(U)$, the desired identity is
\begin{align*}
B[G+w,v] = F(v).
\end{align*}
Because $B$ is bilinear on $H^1(U) \times H^1_0(U)$, and because both $G \in H^1(U)$ and $w \in H^1_0(U) \subset H^1(U)$, we may split the first argument:
\begin{align*}
B[G+w,v] = B[G,v] + B[w,v].
\end{align*}
Therefore the equation for $u$ is equivalent to
\begin{align*}
B[w,v] = F(v) - B[G,v]
\end{align*}
for every $v \in H^1_0(U)$. The advantage is that the unknown now lies in the linear Hilbert space $H^1_0(U)$ rather than in the affine space $G + H^1_0(U)$.
[/guided]
[/step]
[step:Build the bounded functional for the homogeneous problem]
Let
\begin{align*}
a: H^1_0(U) \times H^1_0(U) \to \mathbb{R}
\end{align*}
be the restricted [bilinear form](/page/Bilinear%20Form) defined by
\begin{align*}
a[w,v] := B[w,v].
\end{align*}
Define
\begin{align*}
\Lambda: H^1_0(U) \to \mathbb{R}
\end{align*}
by
\begin{align*}
\Lambda(v) := F(v) - B[G,v].
\end{align*}
Linearity of $\Lambda$ follows from linearity of $F$ and bilinearity of $B$ in its second argument.
Since $B$ is bounded, there exists $M_B > 0$ such that
\begin{align*}
|B[z,v]| \leq M_B \|z\|_{H^1(U)} \|v\|_{H^1(U)}
\end{align*}
for every $z \in H^1(U)$ and every $v \in H^1_0(U)$. Since $F \in H^{-1}(U) = (H^1_0(U))^*$, where $H^{-1}(U)$ is the [topological dual](/page/Topological%20Dual) of $H^1_0(U)$, there exists $\|F\|_{H^{-1}(U)} \geq 0$ such that
\begin{align*}
|F(v)| \leq \|F\|_{H^{-1}(U)} \|v\|_{H^1(U)}
\end{align*}
for every $v \in H^1_0(U)$. Hence
\begin{align*}
|\Lambda(v)| \leq \left(\|F\|_{H^{-1}(U)} + M_B \|G\|_{H^1(U)}\right)\|v\|_{H^1(U)}.
\end{align*}
Thus $\Lambda \in (H^1_0(U))^* = H^{-1}(U)$.
[/step]
[step:Apply Lax-Milgram to solve for the homogeneous correction]
The space $H^1_0(U)$, equipped with the $H^1(U)$ [inner product](/page/Inner%20Product), is a [Hilbert space](/page/Hilbert%20Space). The bilinear form $a$ is bounded because it is the restriction of the bounded bilinear form $B$. By the coercivity hypothesis, there exists a constant $\alpha > 0$ such that, for every $w \in H^1_0(U)$,
\begin{align*}
a[w,w] = B[w,w] \geq \alpha \|w\|_{H^1(U)}^2.
\end{align*}
Therefore, by the real Hilbert-space form of the [Lax-Milgram Theorem](/theorems/91), applied to the Hilbert space $H^1_0(U)$, the bounded coercive bilinear form $a$, and the bounded linear functional $\Lambda$, there exists a unique $w \in H^1_0(U)$. This version requires boundedness and coercivity of the bilinear form, but does not require symmetry. The element $w$ satisfies
\begin{align*}
a[w,v] = \Lambda(v)
\end{align*}
for every $v \in H^1_0(U)$. By the definitions of $a$ and $\Lambda$, this means
\begin{align*}
B[w,v] = F(v) - B[G,v]
\end{align*}
for every $v \in H^1_0(U)$.
[guided]
We now solve the homogeneous correction equation in the Hilbert space $H^1_0(U)$. The form used by Lax-Milgram must be a bounded coercive bilinear form, so we verify each condition for
\begin{align*}
a: H^1_0(U) \times H^1_0(U) \to \mathbb{R}, \qquad a[w,v] := B[w,v].
\end{align*}
Because $B$ is bilinear by hypothesis, its restriction $a$ is bilinear. Because $B$ is bounded on $H^1(U) \times H^1_0(U)$, there is $M_B > 0$ such that
\begin{align*}
|a[w,v]| = |B[w,v]| \leq M_B \|w\|_{H^1(U)}\|v\|_{H^1(U)}
\end{align*}
for all $w,v \in H^1_0(U)$. Because the restriction of $B$ to $H^1_0(U) \times H^1_0(U)$ is coercive in the $H^1(U)$ norm, there is $\alpha > 0$ such that
\begin{align*}
a[w,w] = B[w,w] \geq \alpha \|w\|_{H^1(U)}^2
\end{align*}
for every $w \in H^1_0(U)$.
The functional $\Lambda$ was constructed as a bounded [linear map](/page/Linear%20Map) from $H^1_0(U)$ to $\mathbb{R}$. Therefore all hypotheses of the real Hilbert-space form of the [Lax-Milgram Theorem](/theorems/91) are satisfied: the underlying space is Hilbert, the form is bounded and coercive, and the right-hand side is bounded and linear. Hence there exists a unique $w \in H^1_0(U)$ such that
\begin{align*}
a[w,v] = \Lambda(v)
\end{align*}
for every $v \in H^1_0(U)$. Substituting the definitions of $a$ and $\Lambda$ gives
\begin{align*}
B[w,v] = F(v) - B[G,v]
\end{align*}
for every $v \in H^1_0(U)$.
[/guided]
[/step]
[step:Translate the homogeneous solution back to the affine solution]
Define
\begin{align*}
u := G + w.
\end{align*}
Since $w \in H^1_0(U)$, we have $u \in G + H^1_0(U)$. For every $v \in H^1_0(U)$, bilinearity of $B$ gives
\begin{align*}
B[u,v] = B[G+w,v] = B[G,v] + B[w,v].
\end{align*}
Using the equation satisfied by $w$, we obtain
\begin{align*}
B[u,v] = B[G,v] + F(v) - B[G,v] = F(v).
\end{align*}
Thus $u$ is a solution of the inhomogeneous Dirichlet problem.
[/step]
[step:Prove uniqueness in the affine space]
Let $u_1,u_2 \in G + H^1_0(U)$ be two solutions. Define $w_1,w_2 \in H^1_0(U)$ by
\begin{align*}
w_1 := u_1 - G
\end{align*}
and
\begin{align*}
w_2 := u_2 - G.
\end{align*}
For every $v \in H^1_0(U)$, the solution identities for $u_1$ and $u_2$ give
\begin{align*}
B[u_1,v] = F(v)
\end{align*}
and
\begin{align*}
B[u_2,v] = F(v).
\end{align*}
Subtracting these equalities and using bilinearity yields
\begin{align*}
B[u_1-u_2,v] = 0.
\end{align*}
Since $u_1-u_2 = w_1-w_2 \in H^1_0(U)$, this says
\begin{align*}
a[w_1-w_2,v] = 0
\end{align*}
for every $v \in H^1_0(U)$. Taking $v = w_1-w_2$ and using coercivity gives
\begin{align*}
\alpha \|w_1-w_2\|_{H^1(U)}^2 \leq a[w_1-w_2,w_1-w_2] = 0.
\end{align*}
Because $\alpha > 0$, we have $\|w_1-w_2\|_{H^1(U)} = 0$, hence $w_1 = w_2$ in $H^1_0(U)$. Therefore $u_1 = G+w_1 = G+w_2 = u_2$. This proves uniqueness in $G + H^1_0(U)$ and completes the proof.
[guided]
To prove uniqueness, suppose $u_1,u_2 \in G + H^1_0(U)$ both solve the variational problem. Define the corresponding homogeneous corrections by
\begin{align*}
w_1 := u_1 - G
\end{align*}
and
\begin{align*}
w_2 := u_2 - G.
\end{align*}
Then $w_1,w_2 \in H^1_0(U)$, and $u_1-u_2 = w_1-w_2$ also belongs to $H^1_0(U)$ because $H^1_0(U)$ is a linear subspace.
For every $v \in H^1_0(U)$, the two solution identities give
\begin{align*}
B[u_1,v] = F(v)
\end{align*}
and
\begin{align*}
B[u_2,v] = F(v).
\end{align*}
Subtracting these identities and using bilinearity of $B$ in the first argument gives
\begin{align*}
B[u_1-u_2,v] = 0.
\end{align*}
Since $u_1-u_2 = w_1-w_2$, this is
\begin{align*}
a[w_1-w_2,v] = 0
\end{align*}
for every $v \in H^1_0(U)$. We choose the particular test function $v = w_1-w_2$, which is admissible because $w_1-w_2 \in H^1_0(U)$. Coercivity gives
\begin{align*}
\alpha \|w_1-w_2\|_{H^1(U)}^2 \leq a[w_1-w_2,w_1-w_2] = 0.
\end{align*}
Because $\alpha > 0$, this forces $\|w_1-w_2\|_{H^1(U)} = 0$, so $w_1=w_2$ as elements of $H^1_0(U)$. Translating back to the affine variables gives $u_1=G+w_1=G+w_2=u_2$. Thus the solution in $G+H^1_0(U)$ is unique.
[/guided]
[/step]
