Using the Energy Method to show there exists at most one solution

Question

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Question:

Consider the following heat system with $U\subset\mathbb{R}^n $ open and bounded $$\left\{\begin{array}{l} u_{t}-\Delta u=f(x,t), x \in U, t>0, \\ u+a\frac{\partial u}{\partial \nu}=0 \space\text{on}\space\partial U,\\ u(x, 0)=g(x), x \in U. \end{array}\right.$$

where $ν$ is the outward unit normal to the boundary. Suppose that $a(x)\ge0$ for any $x\in\partial U$ $(\text{recall that } \frac{\partial u}{\partial ν} := \nabla u\cdot v)$. Use the energy method to show that there is at most one solution $u\in C^{2}_{1}(U\times(0,\infty)) \cup C(U \times[0, \infty))$ to the above system.

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My attempt

$$\text{Suppose } \alpha,\beta\text{ are solutions. Let }\gamma=\alpha-\beta.$$

$$\left\{\begin{array}{l} \gamma_{t}-\Delta \gamma=f(x,t), x \in U, t>0, \\ \gamma+a\frac{\partial \gamma}{\partial v}=0 \space\text{on}\space\partial U,\\ \gamma(x, 0)=g(x), x \in U. \end{array}\right. $$

$$\text{Let: } E(t)=\int_U \gamma^2(t)dx\ge0;\space E(0)=\int_U (g(x))^2dx$$

$$\Rightarrow E'(t)=2\int_U \gamma\cdot\gamma_tdx=2\int_U\gamma\cdot\Delta\gamma dx =-2\int|\nabla\gamma|^2dx+2\int_{\partial U}\gamma\cdot\frac{\partial\gamma}{\partial v}dS(x)$$ $$=-2\int|\nabla\gamma|^2dx-2\int_{\partial U}\frac{\gamma ^2}{a}dS(x)\le0.\quad \Rightarrow 0\le E(t)\le 0 \space\therefore\text{ by Sandwich Principle } E(t)=0$$

$$\Rightarrow \gamma=\alpha-\beta=0 \Rightarrow \alpha=\beta \space\therefore\text{ there must be one solution.}$$

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Help!

Feel like this is pure waffle. Any alternative solutions or help would be wonderful :)

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Answer 1

Suppose $\alpha, \beta$ be solutions to the PDE problem in $u$. Let $\gamma=\alpha-\beta$. Then $\gamma$ solves the system

$$\left\{\begin{array}{l} \begin{align} \gamma_{t} - \Delta \gamma &= f(x,t)-f(x,t) = 0 & x \in U, t>0 \\ \gamma + a \frac{\partial \gamma}{\partial v} &=0 & \text{on } \partial U \\ \gamma(x, 0) &= g(x)-g(x) = 0 & x \in U \end{align} \end{array}\right. $$

Let

\begin{align} E(t) &= \int_U \gamma^2(x,t)dx \ge 0 \tag 1 \\ \implies E(0) &= \int_U \gamma^{2}(x,0)dx = 0 \tag 2 \end{align}

Computing the derivative of $E$

\begin{align} E'(t) &= 2 \int_U \gamma \cdot \gamma_t dx \\ &= 2 \int_U \gamma \cdot \Delta \gamma dx \\ &= -2 \int \lvert \nabla \gamma \rvert^2 dx + 2 \int_{\partial U} \gamma \cdot \frac{\partial \gamma}{\partial v} dS(x) \\ &= -2\int \lvert \nabla \gamma\rvert^2 dx - 2 \int_{\partial U} a \left(\frac{\partial \gamma}{\partial v} \right)^2 dS(x) \\ &\le0 \end{align}

where the last integral follows from $\gamma \cdot \frac{\partial \gamma}{\partial v}=-a\left(\frac{\partial\gamma}{\partial v}\right)^2$.

As $E'$ is non-positive then $E$ is decreasing or constant in time, meaning that $E$ is bounded above by $E(0) = 0$. Therefore, by $(1)$ and $(2)$, we have $0 \le E(t) \le 0$ and so by the Sandwich Principle $E$ is identically zero. As the integral of a non-negative function is non-negative, this implies $\gamma = \alpha - \beta = 0$ and hence $\alpha = \beta$, implying that the PDE system in $u$ has a unique solution.