[proofplan] We first show that the Neumann heat flow preserves the spatial mean, because constants lie in the kernel of the Neumann Laplacian. Subtracting this conserved mean reduces the problem to a zero-mean solution $w$ of the same [heat equation](/page/Heat%20Equation). The energy identity for $w$, together with the variational characterization of the first positive Neumann eigenvalue, gives a differential inequality for $\|w(\cdot,t)\|_{L^2(U)}^2$. Multiplying by the integrating factor $e^{2\lambda_{1,N} t}$ and taking square roots gives the claimed exponential decay. [/proofplan]

[step:Show that the Neumann heat flow preserves the spatial mean]Let $H^1(U)$ denote the first-order [Sobolev space](/page/Sobolev%20Space) of $L^2(U)$ functions with weak first derivatives in $L^2(U)$. Let $\Delta_N: D(\Delta_N)\subset L^2(U)\to L^2(U)$ denote the Neumann realization of the Laplacian, with sign convention $\Delta_N v=\Delta v$ for $v\in D(\Delta_N)$. Let $\mathbb{1}_U: U \to \mathbb{R}$ be the constant function $x \mapsto 1$ on $U$. Define the spatial mass function $M: [0,\infty) \to \mathbb{R}$ by \begin{align*} M(t)=\int_U u(x,t) \, d\mathcal{L}^n(x). \end{align*} For $t > 0$, analyticity of the Neumann heat semigroup, equivalently the spectral theorem for the self-adjoint Neumann Laplacian with discrete spectrum, gives $u(t) \in D(\Delta_N)$ and $\partial_t u(t)=\Delta_N u(t)=\Delta u(t)$ in $L^2(U)$. Therefore differentiating the $L^2$ pairing with $\mathbb{1}_U$ gives \begin{align*} M'(t)=\int_U \partial_t u(x,t) \, d\mathcal{L}^n(x). \end{align*} Using the equation, \begin{align*} M'(t)=\int_U \Delta u(x,t) \, d\mathcal{L}^n(x). \end{align*} The weak Neumann Green identity for the Neumann Laplacian gives \begin{align*} \int_U \Delta u(x,t) \, d\mathcal{L}^n(x) = -\int_U \nabla u(x,t)\cdot \nabla \mathbb{1}_U(x) \, d\mathcal{L}^n(x). \end{align*} Since $\nabla \mathbb{1}_U=0$ in $U$, \begin{align*} M'(t)=0. \end{align*} Thus $M(t)=M(0)$ for all $t \geq 0$ by continuity of $u$ in $L^2(U)$. Therefore \begin{align*} \int_U u(x,t) \, d\mathcal{L}^n(x)=\int_U u_0(x) \, d\mathcal{L}^n(x)=\mathcal{L}^n(U)\bar{u}_0. \end{align*}[/step]

[guided]We first isolate the quantity that should not change under Neumann heat flow: the total spatial mass. Let $\Delta_N: D(\Delta_N)\subset L^2(U)\to L^2(U)$ denote the Neumann realization of the Laplacian, with sign convention $\Delta_N v=\Delta v$ for $v\in D(\Delta_N)$. The solution is the semigroup solution fixed in the theorem statement: $u(t)=e^{t\Delta_N}u_0$ for $t\geq 0$. Let $\mathbb{1}_U: U \to \mathbb{R}$ be the constant function $x \mapsto 1$ on $U$, and define $M: [0,\infty) \to \mathbb{R}$ by \begin{align*} M(t)=\int_U u(x,t) \, d\mathcal{L}^n(x). \end{align*} For $t>0$, analyticity of the Neumann heat semigroup, equivalently the spectral theorem for the self-adjoint Neumann Laplacian with discrete spectrum, gives $u(t)\in D(\Delta_N)$ and $\partial_t u(t)=\Delta_N u(t)=\Delta u(t)$ in $L^2(U)$. Hence differentiating the $L^2$ pairing with $\mathbb{1}_U$ is justified, and \begin{align*} M'(t)=\int_U \partial_t u(x,t) \, d\mathcal{L}^n(x). \end{align*} Substituting the heat equation gives \begin{align*} M'(t)=\int_U \Delta u(x,t) \, d\mathcal{L}^n(x). \end{align*} The weak Neumann Green identity for the Neumann Laplacian applies because $u(t)\in D(\Delta_N)$ and $\mathbb{1}_U\in H^1(U)$. It gives \begin{align*} \int_U \Delta u(x,t) \, d\mathcal{L}^n(x) = -\int_U \nabla u(x,t)\cdot \nabla \mathbb{1}_U(x) \, d\mathcal{L}^n(x). \end{align*} Since $\nabla \mathbb{1}_U=0$ in $U$, the right-hand side vanishes, so \begin{align*} M'(t)=0. \end{align*} Therefore $M(t)=M(0)$ for all $t\geq 0$, using continuity of $u$ as a map $[0,\infty)\to L^2(U)$ to include $t=0$. By the definition of $\bar{u}_0$, \begin{align*} \int_U u(x,t) \, d\mathcal{L}^n(x)=\int_U u_0(x) \, d\mathcal{L}^n(x)=\mathcal{L}^n(U)\bar{u}_0. \end{align*} This proves conservation of the spatial mean.[/guided]

[step:Subtract the conserved mean to obtain a zero-mean heat solution]Define the $L^2(U)$-valued map \begin{align*} w: [0,\infty) \to L^2(U), \quad t \mapsto u(t)-\bar{u}_0\mathbb{1}_U. \end{align*} Here $\mathbb{1}_U: U\to\mathbb{R}$ denotes the constant function $x\mapsto 1$, and all identities involving $w(\cdot,t)$ are understood as identities of $L^2(U)$ equivalence classes. Since $\bar{u}_0\mathbb{1}_U$ is constant in space and time, $w \in C([0,\infty);L^2(U))$, $w(t) \in D(\Delta_N)$ for $t>0$, and \begin{align*} \partial_t w(\cdot,t)-\Delta w(\cdot,t)=0 \quad \text{in } L^2(U). \end{align*} Moreover the Neumann boundary condition is preserved because the normal derivative of a constant is zero. From conservation of the spatial mean, \begin{align*} \int_U w(x,t) \, d\mathcal{L}^n(x)=\int_U u(x,t) \, d\mathcal{L}^n(x)-\mathcal{L}^n(U)\bar{u}_0=0 \end{align*} for every $t \geq 0$.[/step]

[guided]We now remove the conserved part of the solution. Define the $L^2(U)$-valued map \begin{align*} w: [0,\infty) \to L^2(U), \quad t \mapsto u(t)-\bar{u}_0\mathbb{1}_U. \end{align*} Here $\mathbb{1}_U: U\to\mathbb{R}$ is the constant function $x\mapsto 1$, and all identities involving $w(\cdot,t)$ are interpreted in $L^2(U)$, hence up to equality almost everywhere. The number $\bar{u}_0$ is constant in both $x$ and $t$, so subtracting it does not change the heat equation or the Neumann boundary condition. More precisely, $w \in C([0,\infty);L^2(U))$, $w(t)\in D(\Delta_N)$ for $t>0$, and \begin{align*} \partial_t w(\cdot,t)-\Delta w(\cdot,t)=0 \quad \text{in } L^2(U). \end{align*} The Neumann condition is preserved because the normal derivative of a constant is zero. The point of this subtraction is that $w$ has zero spatial mean. Using the conservation of mass just proved and the definition of $\bar{u}_0$, we get \begin{align*} \int_U w(x,t) \, d\mathcal{L}^n(x)=\int_U u(x,t) \, d\mathcal{L}^n(x)-\mathcal{L}^n(U)\bar{u}_0=0 \end{align*} for every $t\geq 0$. Thus the entire decay problem has been reduced to proving decay for the zero-mean Neumann heat solution $w$.[/guided]

[step:Derive the energy differential inequality for the zero-mean part]Define the energy function $E: [0,\infty) \to [0,\infty)$ by \begin{align*} E(t)=\|w(\cdot,t)\|_{L^2(U)}^2. \end{align*} For $t>0$, analyticity of the Neumann heat semigroup gives $w(t)\in D(\Delta_N)$ and $\partial_t w=\Delta_N w=\Delta w$ in $L^2(U)$. Differentiating the $L^2$ norm gives \begin{align*} E'(t)=2\int_U w(x,t)\partial_t w(x,t) \, d\mathcal{L}^n(x). \end{align*} Hence \begin{align*} E'(t)=2\int_U w(x,t)\Delta w(x,t) \, d\mathcal{L}^n(x). \end{align*} The weak Neumann Green identity applied to $w(t)\in D(\Delta_N)$ gives \begin{align*} \int_U w(x,t)\Delta w(x,t) \, d\mathcal{L}^n(x)=-\int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} E'(t)=-2\int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x). \end{align*} Since $U$ is connected, the zero eigenspace of the Neumann Laplacian consists exactly of the constant functions. Hence the first positive Neumann eigenvalue $\lambda_{1,N}$ controls the orthogonal complement of the constants. Because $w(t)\in H^1(U)$ has zero spatial mean, the variational characterization of $\lambda_{1,N}$, equivalently the Neumann Poincare inequality on zero-mean functions assumed in the theorem statement, yields \begin{align*} \int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x) \geq \lambda_{1,N} \int_U |w(x,t)|^2 \, d\mathcal{L}^n(x). \end{align*} Thus \begin{align*} E'(t)\leq -2\lambda_{1,N} E(t). \end{align*}[/step]

[guided]The goal is to turn the heat equation into a scalar inequality for the size of the solution. We measure the size of the zero-mean part $w$ by the squared $L^2$ energy \begin{align*} E(t)=\|w(\cdot,t)\|_{L^2(U)}^2=\int_U |w(x,t)|^2 \, d\mathcal{L}^n(x). \end{align*} For $t>0$, analyticity of the Neumann heat semigroup gives $w(t)\in D(\Delta_N)$ and $\partial_t w(t)=\Delta_N w(t)=\Delta w(t)$ in $L^2(U)$, so differentiating the $L^2$ norm is justified. This gives \begin{align*} E'(t)=2\int_U w(x,t)\partial_t w(x,t) \, d\mathcal{L}^n(x). \end{align*} Substituting the heat equation $\partial_t w=\Delta w$ gives \begin{align*} E'(t)=2\int_U w(x,t)\Delta w(x,t) \, d\mathcal{L}^n(x). \end{align*} Now the Neumann boundary condition is used exactly through the weak Neumann Green identity. For functions in $D(\Delta_N)$, this identity says \begin{align*} \int_U v(x)\Delta v(x) \, d\mathcal{L}^n(x)=-\int_U |\nabla v(x)|^2 \, d\mathcal{L}^n(x). \end{align*} Applying it with $v=w(\cdot,t)$ gives \begin{align*} E'(t)=-2\int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x). \end{align*} This identity says that the heat equation dissipates $L^2$ energy at a rate equal to twice the Dirichlet energy. To convert this dissipation into decay of $E(t)$ itself, we use the zero-mean condition. Since \begin{align*} \int_U w(x,t) \, d\mathcal{L}^n(x)=0, \end{align*} connectedness implies that the zero eigenspace of the Neumann Laplacian is exactly the constants. Since $w(\cdot,t)$ has zero mean, it is orthogonal in $L^2(U)$ to that eigenspace. Therefore the variational characterization of the first positive Neumann eigenvalue applies to $w(\cdot,t)$; equivalently, the Neumann Poincare inequality on zero-mean functions assumed in the theorem statement yields \begin{align*} \int_U |\nabla w(x,t)|^2 \, d\mathcal{L}^n(x) \geq \lambda_{1,N} \int_U |w(x,t)|^2 \, d\mathcal{L}^n(x). \end{align*} Combining the previous two displays gives \begin{align*} E'(t)\leq -2\lambda_{1,N} E(t). \end{align*} This is the point where connectedness and the zero-mean condition enter: they ensure that the first nonzero Neumann eigenvalue controls the entire zero-mean component.[/guided]

[step:Integrate the scalar differential inequality]Define $F: [0,\infty) \to [0,\infty)$ by \begin{align*} F(t)=e^{2\lambda_{1,N} t}E(t). \end{align*} For $t>0$, the product rule and the inequality from the previous step give \begin{align*} F'(t)=e^{2\lambda_{1,N} t}\left(E'(t)+2\lambda_{1,N} E(t)\right)\leq 0. \end{align*} Hence $F(t)\leq F(0)$ for every $t\geq 0$, using continuity at $t=0$. Therefore \begin{align*} E(t)\leq e^{-2\lambda_{1,N} t}E(0). \end{align*} Taking square roots gives \begin{align*} \|w(\cdot,t)\|_{L^2(U)}\leq e^{-\lambda_{1,N} t}\|w(\cdot,0)\|_{L^2(U)}. \end{align*} Since $w(\cdot,t)=u(\cdot,t)-\bar{u}_0$ and $w(\cdot,0)=u_0-\bar{u}_0$, this is exactly \begin{align*} \|u(\cdot,t)-\bar{u}_0\|_{L^2(U)} \leq e^{-\lambda_{1,N} t}\|u_0-\bar{u}_0\|_{L^2(U)}. \end{align*} This proves the claimed exponential decay to the conserved spatial mean.[/step]

[guided]We finish by integrating the scalar differential inequality obtained from the energy estimate. Define $F: [0,\infty) \to [0,\infty)$ by \begin{align*} F(t)=e^{2\lambda_{1,N} t}E(t). \end{align*} For $t>0$, the product rule gives \begin{align*} F'(t)=e^{2\lambda_{1,N} t}\left(E'(t)+2\lambda_{1,N} E(t)\right). \end{align*} The previous step proved $E'(t)\leq -2\lambda_{1,N} E(t)$, so the factor in parentheses is nonpositive. Therefore \begin{align*} F'(t)\leq 0. \end{align*} Thus $F$ is nonincreasing on $(0,\infty)$. Since $w\in C([0,\infty);L^2(U))$, the energy $E$ is continuous at $t=0$, and hence $F(t)\leq F(0)$ for every $t\geq 0$. Unwinding the definition of $F$ gives \begin{align*} E(t)\leq e^{-2\lambda_{1,N} t}E(0). \end{align*} Because $E(t)=\|w(\cdot,t)\|_{L^2(U)}^2$ and both sides are nonnegative, taking square roots gives \begin{align*} \|w(\cdot,t)\|_{L^2(U)}\leq e^{-\lambda_{1,N} t}\|w(\cdot,0)\|_{L^2(U)}. \end{align*} Finally, by the definition of $w$, \begin{align*} w(\cdot,t)=u(\cdot,t)-\bar{u}_0 \end{align*} and \begin{align*} w(\cdot,0)=u_0-\bar{u}_0. \end{align*} Substituting these identities yields \begin{align*} \|u(\cdot,t)-\bar{u}_0\|_{L^2(U)} \leq e^{-\lambda_{1,N} t}\|u_0-\bar{u}_0\|_{L^2(U)}. \end{align*} This is exactly the claimed exponential decay to the conserved spatial mean.[/guided]