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Assume sufficient smoothness and a regular boundary for the integrations below, with $n$ outward. Put $\delta =w-\varphi$, so $\delta =0$ on $S$. Expanding the Dirichlet energy givesBy Green's first identity, the middle integral isThus the Dirichlet principle follows with an explicit remainder:Equality means $w-\varphi$ is constant on each connected component; its zero boundary value forces that constant to be zero.

Now let $w$ solve the heat equation and set $E(t)={\int }_V|\nabla w{|}^{2}dV$. Differentiation followed by Green's first identity givesUnder time-independent Dirichlet boundary conditions, $w_t=0$ on $S$, while $w_t=\Delta w$ in $V$. ThereforeIf equality holds at a particular time, continuity makes $\Delta w=0$ throughout $V$ at that time. Its boundary values are still $f$, so uniqueness of the Dirichlet problem implies $w=\varphi$. Conversely $w=\varphi$ has $\Delta w=0$ and gives equality. This proves the full equality criterion for Dirichlet energy dissipation for the heat equation.

For the dynamic boundary condition for the heat equation, substitute $w_t=-\alpha {\partial }_{n}w$ directly into the boundary term. The interior equation is unchanged, soThis is Dirichlet energy dissipation with a dynamic heat boundary condition. It remains valid when $\alpha$ vanishes, because no division by $\alpha$ is used. The last part preserves energy monotonicity; its stationary states need not have the earlier prescribed Dirichlet data. For example, with strictly positive $\alpha$ on a connected boundary, a stationary harmonic state has zero outward normal derivative and is constant, as follows by applying Green's first identity to that state itself.