# Energy argument to prove that the solution to a PDE is unique

Background Information:

From Walter A Strauss Chapter 2.

Imagine an infinite string with constants $\rho$ and $Y$. Then $\rho u_{tt} = T u _{xx}$ for $-\infty < x < \infty$. From physics we know that the kinetic energy is $\frac{1}{2}m v^2$ which in our case takes the form $KE = \frac{1}{2}\rho \int u_t^{2} dx$. This integral, and the following ones, are evaluated from $-\infty$ to $\infty$. To be sure that the integral converges, we assume that $\phi(x)$ and $\psi(x)$ vanish outside an interval $\{|x|\leq R\}$. As mentioned above, $u(x,t)$ [and therefore $u_t(x,t)$] vanish for $|x| > R + ct$. Differentiating the kinetic energy, we can pass the derivatives under the integral sign to get $$\frac{d KE}{dt} = \rho \int u_t u_{tt} dx$$ Then we substitute the PDE $\rho u_{tt} = T u_{xx}$ and integrate by parts to get $$\frac{d KE}{dt} = T\int u_t u_{xx} dx = T u_t u_x - T \int u_{tx} u_x dx$$ The term, $T u_t u_x$ is evaluated at $x = \pm \infty$ and so it vanishes. But the final term is pure derivative since $u_{tx} u_x (\frac{1}{2} u_x^{2})_{t}$. Therefore, $$\frac{d KE}{dt} = -\frac{d}{dt}\int \frac{1}{2} T u_{x}^{2} dx$$ Let $PE = \frac{1}{2}T\int u_{x}^{2}dx$ and let $E = KE + PE$. Then $d KE/dt = - d PE/dt$ or $dE/dt = 0$. Thus $$E = \frac{1}{2}\int_{-\infty}^{\infty}(\rho u_t^{2} + T u_{x}^{2})dx$$ is a constant independent of $t$. This is the law of conservation of energy.

Question:

Let $D$ be the domain in $\mathbb{R}^3$. Use an energy argument to prove that any solution to $$\begin{cases} \Delta u(x) = f(x) \ \ \ &\text{for} \ x\in D\\ \frac{\partial u}{\partial n} + au(x) = h(x)\ \ \ &\text{for} \ x\in \partial D \end{cases}$$ is unique, where $f$ is a function defined on $D$, $h$ is a function defined on $\partial D$, and $a$ is a positive constant.

I normally don't post a question without somewhat of an attempt of an solution but I am not sure how to proceed with this. Any suggestions are greatly appreciated.

Based on the comments above I will now attempt a proof.

Attempted proof - Assume that we have two solutions $u_1(x)$ and $u_2(x)$. Let $v(x) = u_1(x) - u_2(x)$. Then we have $$\begin{cases}\Delta v(x) = 0 \ \ &\text{for} \ x\in D\\ \frac{\partial v}{\partial n}(x) + a v(x) = 0 \ \ &\text{for} \ x\in \partial D \end{cases}$$ So,

\begin{align*} 0 = \int_D v \Delta v dx &= \int_{D} v\cdot div(grad (v))dx\\ &= \int_{\partial D} (v \cdot grad(v))\cdot n dS - \int_{D} grad(v) \cdot grad(v)dx\\ &= \int_{\partial D}v \frac{\partial v}{\partial n}dS - \int_{D} |\nabla v|^2 dx\\ &= -\int_{\partial D}v\alpha v dS - \int_{D} |\nabla v|^2 dx\\ &= -\int_{\partial D}\alpha v^2 dS - \int_{D} |\nabla v|^2 dx\\ &= -\int_{\partial D}\alpha v^2 dS + \int_{\partial D}\alpha v^2 dS\\ &= 0 \end{align*} So we ended up with $$0 = - \int_{\partial D}a v^2 - \int_{D} |\nabla v|^2 dx$$ From this we get that $$\int_{D}|\nabla|^2dx = = - a\int_{\partial D}v^2 dS$$ Now the left-hand side is non-negative and the right-hand side is non-positive, which implies that we must have $\Delta v(x) = 0$ so then $u_1(x) - u_2(x) = 0 \Rightarrow u_1(x) = u_2(x)$.

Not sure if this is right or where to go from here though to conclude that we have a unique solution, one of the commentators below stated to think of positive. Any suggestions on where to go from here are greatly appreciated.

• For the reader, it would be helpful to know what precisely is an "energy argument". I tried googling it, but turned up nothing relevant. Commented Jul 13, 2017 at 14:58
• @AlvinLepik I added a background information section to the question if this does not help at all please let me know Commented Jul 13, 2017 at 15:28
• Hint: look at the homogeneous equations; for each equation multiply by $u$ and integrate (over $D$ and $\partial D$ respectively) Commented Jul 13, 2017 at 15:35
• Well, you are almost there, aren't you? Let's rewrite your equation: $\int_{D} \lvert \nabla v \rvert^2 \, \text{d}x = - a \int_{\partial D} v^2 \, \text{d}S$. If we assume that $v$ is real-valued, then you should get from this equality that $v$ is constant in $D$ (recall that $a$ is positive). You also get from the above equality that $v$ vanishes almost everywhere on $\partial D$. Combining both observations, you find that $v = 0$ in $D$.
– Mike
Commented Jul 13, 2017 at 17:45
• You tested your equation with $v$ and using integration by parts you ended up with $0 = - \int_{\partial D} a v^2 \, \text{d}S - \int_{D} \lvert \nabla v\rvert^2 \, \text{d}x$. From this equation, you get that $\int_{D} \lvert \nabla v\rvert^2 \, \text{d}x = - a \int_{\partial D} v^2 \, \text{d}S$. Now, the left-hand side is non-negative and the right-hand side is non-positive, which implies... The rest should be easy for you.
– Mike
Commented Jul 14, 2017 at 7:42