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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.

Use the printed coordinate derivative $u_x(0,t)$, not an outward normal derivative; at the left endpoint these have opposite signs. Assume the usual bounded or admissibly growing solution at infinity, with an initial trace $b_0=u_0(0)$ and forcing possessing a Laplace transform. Set $a=\alpha /2$, so $\beta =a^2$, and use $z=\sqrt{p}$ on the principal square root branch. DefineThe half-line Dirichlet Green function for $p-{\partial }_x^2$ givesWrite the transformed solution as $U=U_D+e^{-zx}B(p)$. The Laplace transform of a derivative in the dynamic boundary condition is $pB-b_0$, soThe special choice of $\beta$ gives the perfect squareIn particular, the initial boundary trace cannot be omitted. Withan integral representation isChoose the Bromwich contour to the right of the forcing's growth abscissa and every pole; $\gamma >\max (0,a^2)$ and sufficiently large for the data is a safe choice for real $a$. No sign of $\alpha$ is explicitly imposed in this question.

The inverse transform can also be performed explicitly. For the repeated-root dynamic-boundary heat kernel, setFirst integrate the Heat Poisson kernel against $e^{ar}$, or differentiate the result with respect to $a$:Since differentiation of $(\sqrt{p}-a{)}^{-1}$ with respect to $a$ produces $(\sqrt{p}-a{)}^{-2}$, and ${\partial }_a{E}_a=(-x+2at)E_a+2\sqrt{t/\pi }{e}^{-x^2/(4t)}$, the desired kernel isHere $\mathrm{erfc}$ is the complementary error function. A direct integral characterization, also proving the sign and the transform, isThe convolution theorem for Laplace transforms now gives the fully real time-domain representationAll quantities here are known from the prescribed data. For $x>0$ it tends to $u_0(x)$ as $t\downarrow 0$. At $x=0$, $J_a(0,t)\to 1$ and the spatial correction tends to zero, recovering $b_0$. A solution classical through the initial corner additionally needs $u_0^{\prime \prime }(0)+\alpha u_0^{\prime }(0)+a^2{b}_0=f(0)$; weaker corner regularity does not invalidate the formula for positive times.

The repeated-root unstable heat boundary mode gives a useful sign check on the dynamic boundary condition for the heat equation. If $a>0$, $(\sqrt{p}-a{)}^2$ has a genuine double pole at $p=a^2$ on the physical branch, and the exact homogeneous mode $e^{-ax+a^{2}t}$ satisfies both the heat equation and the printed boundary condition. Generic data can also excite a $t{e}^{a^{2}t}$ contribution. A contour deformation must retain this repeated-pole contribution. If $a<0$, the putative root $\sqrt{p}=a$ is outside the chosen branch and is not a physical pole. At $a=0$, $J_0=\mathrm{erfc}(x/(2\sqrt{t}))$, as expected when the boundary trace satisfies $b^{\prime }(t)=f(t)$.

For the dynamic boundary condition for the heat equation with $\alpha =2a$, $\beta =a^2$, the boundary resolvent denominator is $(\sqrt{p}-a{)}^2$. With the principal square root, its inverse kernel isEquivalently, $J_a={\int }_0^{\infty }r{e}^{ar}P(x+r,t)dr$ using the Heat Poisson kernel. This representation fixes the repeated-root sign and remains valid for either sign of $a$.