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Let a homogeneous plate $0<x<L$ begin at uniform temperature $T_0$, cool with zero-temperature Dirichlet boundary conditions until time $t_b$, then switch to insulated Neumann boundary conditions. For thermal diffusivity $\kappa >0$, the Fourier sine series during cooling is $T=(4T_0/\pi ){\sum }_{n\ge 1,n\text{ odd}}\sin (n\pi x/L)e^{-\kappa (n\pi /L{)}^{2}t}/n$. Integrating the insulated heat equation shows that its integral average is conserved. All nonconstant Fourier cosine series modes then decay, proving that the final uniform temperature is $T_{\infty }=(8T_0/{\pi }^2){\sum }_{n\ge 1,n\text{ odd}}{e}^{-\kappa (n\pi /L{)}^2{t}_b}/n^2$. Its limits are $T_0$ at zero bath time and zero at infinite bath time. Instantaneous changes of boundary condition can create initial-boundary corner incompatibilities without invalidating the positive-time solution.

On a bar of length $L$, let the initial temperature increase linearly from zero at either endpoint to height $H$ at the midpoint. With thermal diffusivity $D>0$ and insulated Neumann boundary conditions, a Fourier cosine series gives the displayed solution. Its constant term is the initial integral average $H/2$. Integration by parts on the two straight segments gives zero coefficients for odd modes and modes divisible by four; the remaining coefficients are $-4H/[{\pi }^2(2j+1{)}^2]$. Nonconstant modes decay, leaving the uniform temperature $H/2$, consistent with conservation of the total heat. This contrasts with the zero-temperature boundary solution in heat flow from a triangular initial temperature, where heat can leave the bar and the limit is zero. The initial endpoint slopes do not satisfy the insulated condition, so derivatives need not converge to their initial endpoint values as $t\downarrow 0$; the series satisfies the Neumann condition for every $t>0$ and recovers the continuous initial profile.

The integral average of an integrable function over a measurable set $E$ of finite positive measure is $⟨v{⟩}_E=|E{|}^{-1}{\int }_{E}v$. Normalization removes the volume factor when comparing estimates on balls of different radii.

The Weak Harnack inequality is an estimate for a nonnegative weak supersolution. Write $F=(f^1,\dots ,f^n)$, let $B_{2R}(x_0)\subset \subset \Omega$, and assume $u\ge 0$ almost everywhere on this ball. There are $p>0$ and $C<\infty$, depending only on the dimension, the uniform ellipticity bounds, $q$, and the scaled norms of the lower-order coefficients, such thatHere $⟨v{⟩}_E=|E{|}^{-1}{\int }_{E}v$ is the integral average. The weak supersolution inequality uses the sign convention $Lu\le \mathrm{div}F+g$. For instance, the coefficient dependence can be expressed usingThus one may use the same $p,C$ on all sufficiently small balls when the global coefficient norms and uniform ellipticity bounds are fixed. The nonnegativity condition is essential to this formulation; a signed weak supersolution may first be shifted, with the resulting change in its forcing included. As usual, supremum and infimum statements for functions in a Sobolev space mean essential suprema and essential infima. The standard multidimensional statement uses $n\ge 2$, so $q/2>1$. In dimension one the analogous formulation requires $q\ge 2$; the printed restriction $q>n$ alone would allow $q/2<1$, which does not suffice. To see the obstruction, smooth the nonnegative capped function $\min \{1,|x|/\epsilon \}$ by a mollifier of width $\tau$. Its positive second derivative is a bump of mass $2/\epsilon$ and has $L^{q/2}$ size of order ${\epsilon }^{-1}{\tau }^{2/q-1}$. For fixed $\epsilon$ and $1<q<2$, this size and the minimum both tend to zero as $\tau \downarrow 0$, whereas the average on a fixed larger interval stays positive. Thus the Weak Harnack inequality cannot have the displayed uniform forcing bound in that range. Shrinking $\epsilon$ and choosing $\tau$ still smaller likewise defeats a uniform Hölder estimate based on that norm alone.

Write $E(t)={\int }_{B_t(x_0)}|\nabla u{|}^2$. For $n\ge 2$, choose $c$ in the annular Caccioppoli inequality to be the integral average of $u$ over $A_R$. The supplied Poincare inequality on an annulus givesMoving the $E(R)$ term to the left is the hole-filling argument:Consequently $E(2^{-k}{r}_0)\le {\theta }^{k}E(r_0)$. Put $\beta =-{\log }_2\theta >0$ and choose $\mu =\min \{\beta /2,1/2\}\in (0,1)$. For $2^{-k-1}{r}_0<r\le 2^{-k}{r}_0$, monotonicity givesDyadic endpoints can be assigned to either adjacent interval. Since $E(r_0)\le {\int }_B|\nabla u{|}^2$, the requested dyadic energy decay isBoth constants depend only on the dimension and uniform ellipticity bounds, not on $u,x_0,r,r_0$.

There is a dimensional detail in the printed hint: an annulus is disconnected in dimension one, so that Poincaré inequality with a single average is false there. The conclusion still holds. In one dimension the weak equation gives $a(x)u^{\prime }(x)=J$ almost everywhere for a constant flux for a one-dimensional divergence-form equation $J$. Since $\lambda \le a\le \Lambda$,This implies the required estimate with, for example, $\mu =1/2$ and $K=(\Lambda /\lambda {)}^2$. If $J=0$, the estimate is immediate. Thus the proof also covers dimension one without using the inapplicable hint.

The square wave alternates between a low interval of length five and a high interval of length five; its integral average is three. In the real Fourier series conventionintegration over one period givesandThus only odd harmonics occur:The Dirichlet-Jordan convergence theorem makes the sum equal to the function at its continuity points and to the mean of its one-sided limits at a jump. Consequently the sum is 3 at each of $x=-5,0,5$, regardless of any separately assigned values of the original function at those endpoints.

By Fourier's law, insulation means zero outward heat flux. The Neumann boundary conditions are ${\theta }_x(0,t)={\theta }_x(L,t)=0$. The constant mode equals the initial integral average $L/4$. For $n\ge 1$, the cosine coefficients areOdd modes and modes divisible by four vanish; write the remaining indices as $n=2(2j+1)$. The heat flow from a triangular initial temperature with insulated endpoints givesThe nonconstant modes decay, so $\boxed{{\theta }_N\to L/4}$ uniformly as $t\to \infty$. Integrating the heat equation gives $d({\int }_0^L\theta dx)/dt=\nu [{\theta }_x{]}_0^L=0$, explaining the conserved mean. The cooled case tends instead to zero because heat leaves through its ends.

The initial triangle has nonzero endpoint slopes, so it is not compatible with the insulated condition at the corner $t=0$. The series nevertheless recovers the continuous initial temperature and satisfies the insulated condition for every $t>0$; one should not assert continuity of its spatial derivative at that initial corner.

After bath cooling followed by insulated equilibration, the remaining mean temperature is conserved: integrating the insulated heat equation givesFor constant density, specific heat capacity and cross-section, this is conservation of energy. Its Fourier cosine series has a constant mode equal to the initial integral average; every nonconstant mode decays. Therefore the eventual uniform temperature isFor each mode with an odd integer index in part (c), $L^{-1}{\int }_0^L\sin ((2j+1)\pi x/L)dx=2/[(2j+1)\pi ]$. Integrating the bath-cooled series yieldsFor $t_f>0$, uniform convergence justifies termwise integration; the $t_f=0$ result also follows by dominated convergence of the summable positive coefficients. As a check, ${\sum }_{j\ge 0}(2j+1{)}^{-2}={\pi }^2/8$ gives $T_f=T_0$ for zero bath time, and $T_f\to 0$ as $t_f\to \infty$. More bath time strictly lowers the retained average temperature when $T_0>0$ and $\kappa >0$.

A volume average is the normalized integral average of a field over a three-dimensional region of finite positive volume. For a fixed region, differentiating in time commutes with this average under the usual regularity conditions. The divergence theorem converts the average of a divergence into its net boundary flux divided by volume.