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Apply the divergence theorem to $\mathbf{F}=\psi \nabla \varphi$, for functions smooth enough on the closure of the bounded piecewise smooth volume. The product rule givesThus Green's first theorem, also called Green's first identity, isThe normal is outward. This identifies the boundary term and its sign directly from the divergence theorem.

Use the Wirtinger derivative $q_z=(q_x-i{q}_y)/2$ and write $z=x+iy$. Since $q$ solves the Laplace equation, $q_z$ is a holomorphic function. The transforms in the question select solutions with sufficient decay at infinity. In this class the solution, when it exists, is unique; hence reflection of the symmetric data about $y=\ell /2$ givesFor $\gamma >0$, this uniqueness follows directly from Green's first identity: the homogeneous problem has integral $\int |\nabla q{|}^2+\gamma {\int }_{y=0,\ell }{q}^2=0$. For arbitrary real $\gamma$, the decaying-class qualifications are explained in part (c). Without a condition at infinity, symmetry of the data alone would not force symmetry of every solution.

Put $c=q(0,0)$, $f(x)=q(x,0)$, and introduce a known transformThe bottom Robin boundary condition gives $q_y=\gamma f$, while the top one gives $q_y=-\gamma f$. The bottom side is traversed from infinity to zero. Integration by parts therefore givesThe top side is traversed from $i\ell$ towards infinity, and its exponential supplies $e^{k\ell }$, givingOn the vertical side, $dz=idy$ and the prescribed Neumann boundary condition is $q_x=g$. ThusBoth the side orientations and the factor $1/2$ from the Wirtinger derivative are essential to these signs.

Work with real-valued smooth functions on the bounded region enclosed by $S$, with the oriented surface element pointing outward. Set $\psi =\varphi -{\varphi }_1$. It is a harmonic function in $V$ and has zero Dirichlet boundary conditions on $S$. Applying Green's first identity to $\psi$ with itself givesBoth terms on the right vanish by the boundary condition and harmonicity. The integrand is nonnegative and continuous, so $\nabla \psi =0$ throughout $V$. Consequently $\psi$ is constant on each connected component; every bounded component meets the prescribed boundary, where that constant is zero. ThereforeThis is the energy proof of Uniqueness of the Dirichlet problem. Connectedness of $V$ is not essential if the boundary values are prescribed on every component. Boundedness, or suitable decay and integrability at infinity, is needed for the boundary/energy argument; the finite enclosed-volume interpretation is used here.

Put $h=u-\varphi -C$. Then $h=0$ on $S$, and the constant does not change a gradient, so $\nabla u=\nabla \varphi +\nabla h$. Expand the unnormalized Dirichlet energy:The cross term is zero by Green's first identity:Thus the useful stronger identity isEquality holds exactly when $h$ has zero gradient. It is then constant on every connected component and zero on that component’s boundary, hence $h=0$. Therefore equality holds precisely when $u=\varphi +C$ throughout $V$. This is the Dirichlet principle; an additive boundary constant does not alter the minimizing energy. The energy here omits the optional factor $1/2$, which does not affect minimizers or the inequality.

Let $E(t)={\int }_V|\nabla w{|}^{2}dV$, using real-valued smooth $w$ and a fixed bounded region. Differentiation under the integral sign and Green's first identity yieldThe boundary term vanishes because $w_t=0$ on $S$, and the heat equation gives $w_t=\Delta w$ in $V$. ConsequentlyThe integrand is continuous and nonnegative. Thus at any fixed time, equality is equivalent to $\Delta w=0$ everywhere in $V$, also giving $w_t=0$ there at that time. This proves Dirichlet energy dissipation for the heat equation.

The boundary condition fixes the boundary temperature in time; it is not a zero normal derivative condition. No sign condition on ${\partial }_{n}w$ is used. The stated smoothness licenses the time derivative, spatial integration by parts and boundary trace; without that regularity an appropriate weak energy argument would be required.

For a scalar $\varphi$, the displayed basis identities giveTaking the divergence and differentiating the basis vectors yields the Laplacian in polar coordinatesConsequently,

The regular harmonic mode with angular dependence $\cos 2\theta$ is $C{r}^2\cos 2\theta$. Its radial derivative at $r=a$ is $2Ca\cos 2\theta$, so $C=1/(2a)$. A constant is invisible to the Neumann boundary condition, givingIf two solutions existed, their difference $u$ would satisfy ${\nabla }^{2}u=0$ and ${\partial }_{n}u=0$. Green's first identity gives ${\int }_D|\nabla u{|}^2={\int }_{\partial D}u{\partial }_{n}u-{\int }_{D}u{\nabla }^{2}u=0$, so $\nabla u=0$ and $u$ is constant. Thus these are all the solutions.

Green's first identity and the boundary condition giveThus $\nabla u=0$, so $u$ is constant on each connected component of $V$; its zero boundary value forces $\boxed{u=0}$.

Since $w-v=0$ on $S$ and ${\nabla }^{2}v=0$, another application of Green's first identity givesConsequentlyExpanding $\nabla w=\nabla v+\nabla (w-v)$ and using the vanishing cross term yields the Dirichlet principle

Suppose $u_1,u_2$ are solutions and put $w=u_1-u_2$. Then $\Delta w=0$ and $w$ satisfies the homogeneous Robin boundary conditionFor $0<\lambda <1$, Green's first identity givesThe left side is nonnegative, so both sides vanish and $w=0$. For $\lambda =0$, the Dirichlet boundary condition $w=0$ on $S$ and the same identity again imply $w=0$. The solution is therefore unique for every $0\le \lambda <1$.

For a regular spherically symmetric solution, the Laplacian equation isRegularity at the origin gives $u^{\prime }=2r$, hence $u=r^2+C$. Applying the boundary condition at $r=b$ yieldsAt $\lambda =1$ the Neumann boundary condition would require $u^{\prime }(b)=2b=0$, which is impossible for $b>0$. Equivalently, its compatibility condition would require ${\int }_{V}6dV=0$.

Let ${\varphi }_1,{\varphi }_2$ be two solutions with the same Neumann boundary condition, and set $u={\varphi }_1-{\varphi }_2$. ThenMultiplying by $u$ and applying Green's first identity givesThe integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the modified Helmholtz equation is unique.

If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the Laplace equation with prescribed normal derivative are unique only up to an additive constant, so

Multiply the Poisson equation by $v\in V$ and apply Green's first identity. The Dirichlet boundary condition makes the trace of $v$ vanish on ${\Gamma }_1$, while the Neumann boundary condition makes the boundary flux vanish on ${\Gamma }_2$. Thus the weak formulation is: find $u\in V$ such thatwhere

First choose a test function $v\in C_c^{\infty }(U)$. The weak formulation and integration by parts giveThe fundamental lemma of the calculus of variations implies $-\Delta u=f$ pointwise because $u\in C^2(\overline{U})$ and $f$ is continuous. The Dirichlet boundary condition on ${\Gamma }_1$ already follows from $u\in V$ and continuity of $u$.

For arbitrary $v\in V$, Green's first identity and the interior equation now reduce the weak identity toThe traces of smooth members of $V$ can be chosen freely on compact subsets of ${\Gamma }_2$. Another application of the fundamental lemma of the calculus of variations, now on the boundary, gives ${\partial }_{\nu }u=0$ pointwise on ${\Gamma }_2$. Hence $u$ is a classical solution of the complete mixed boundary value problem.