# Proving the uniqueness of a PDE's solution

Let $\Omega$ be a bounded open set in $\mathbb{R}^2$. I have

\begin{equation} -\Delta u (X) + \alpha \partial_x u(X) + u(X) = f(X), \ \forall X \in \Omega, \ \alpha \in \mathbb{R} \end{equation}

with \begin{equation} \partial_n u = g \ on \ \partial \Omega \end{equation}

Let $f_1, f_2, g_1, g_2$ be regular functions. Let $u_1$ (resp. $u_2$) be solutions of the system with $f = f_1$ and $g = g_1$ ($\ f = f_2$ and $g = g_2$ resp.). Also, there exists a constant $c_2 > 0$ such that for all function $u \in \mathcal{C}^1(\overline{\Omega})$,

\begin{equation} \int_{\partial \Omega} |u|^2 d\sigma \leq c_2\Big(\int_{\Omega}|u(X)|^2dX + \int_{\Omega}|\nabla u(X)|^2 dX \Big) \end{equation}

By rewriting the intial equation (i.e. replacing $u$ by $u_1 - u_2$ and such) and multiplying by $(u_1 - u_2)$ on both sides (so that I may apply Green's formula), I obtain

\begin{equation} \int_{\Omega} |\nabla u_1(X) - \nabla u_2(X)|^2 dX - \int_{\partial \Omega} (g_1-g_2)(u_1-u_2)d\sigma + \frac{\alpha}{2} \int_{\partial \Omega} |u_1-u_2|^2 n_x d\sigma + \int_{\Omega} |u_1(X) - u_2(X)|^2 dX = \int_{\Omega} (f_1(X) - f_2(X))(u_1(X) - u_2(X))dX. \end{equation}

Furthermore, by rearranging and then using the absolute value integral inequality, as well as the aforementioned inequality, I obtain

\begin{equation} \int_{\Omega} |\nabla u_1(X) - \nabla u_2(X)|^2 dX + \int_{\Omega} |u_1(X) - u_2(X)|^2 dX \end{equation} \begin{equation} \leq \int_{\Omega} |f_1(X) - f_2(X)||u_1(X) - u_2(X)|dX + \int_{\partial \Omega} |g_1-g_2||u_1-u_2|d\sigma + c_2 \frac{\alpha}{2} \Big(\int_{\Omega} |\nabla u_1(X) - \nabla u_2(X)|^2 dX + \int_{\Omega} |u_1(X) - u_2(X)|dX \Big) . \end{equation}

Now suppose $\alpha < 1$. How do I prove that there exists $c_3$ such that

\begin{equation} \int_{\Omega} |\nabla u_1(X) - \nabla u_2(X)|^2 dX + \int_{\Omega} |u_1(X) - u_2(X)|^2 dX \end{equation}

\begin{equation} \leq c_3 \Big(\int_{\Omega} |f_1(X) - f_2(X)|^2 dX + \int_{ \partial \Omega} |g_1 - g_2|^2d\sigma \Big)? \end{equation}

Thank you.

## 1 Answer

If you are proving uniqueness, you may as well take $f_1=f_2$ and $g_1=g_2$. Then $u:=u_1-u_2$ solves

$-\Delta u + \alpha \partial_x u + u = 0$

with $\partial_n u = 0$ on the boundary. Multiplying by $u$ on both sides and integrating over $\Omega$ we have

$\int_\Omega |Du|^2 + u^2 \, dx = -\alpha \int_\Omega \partial_x u \cdot u \, dx$,

where I applied integration by parts on the first term and the fact that $\partial_n u = 0$. Now use Cauchy's inequality $ab \leq \frac{1}{2}a^2 + \frac{1}{2}b^2$ on the right hand side to get

$\int_\Omega |Du|^2 + u^2 \, dx \leq \frac{|\alpha|}{2} \int_\Omega |\partial_x u|^2 + u^2 \, dx \leq \frac{|\alpha|}{2}\int_\Omega |Du|^2 + u^2 \, dx.$

If $|\alpha| < 2$ we can absorb the term on the right into the left hand side to obtain

$(1-\frac{|\alpha|}{2})\int_\Omega |Du|^2 + u^2 \, dx \leq 0.$

Therefore $u=0$, and so $u_1=u_2$. If $|\alpha|>2$ I'm not sure if energy methods will work, you might try the maximum principle instead.