**Proof plan.** Let $u_1, u_2$ be two solutions sharing the same data and set $w = u_1 - u_2$. Multiply the equation $\Delta w = 0$ by $w$, integrate over $\Omega$, and apply Green's first identity. The [boundary](/page/Boundary) conditions force the boundary [integral](/page/Integral) to vanish (or be non-positive), leaving $\int_\Omega |\nabla w|^2 \, d\mathcal{L}^n \leq 0$, which forces $w$ to be constant. **Step 1: Equation and boundary conditions for the difference.** Let $w := u_1 - u_2$. By linearity, $\Delta w = f - f = 0$ in $\Omega$. The boundary conditions on $w$ are: - Dirichlet: $w = 0$ on $\partial\Omega$. - Neumann: $\partial_\nu w = 0$ on $\partial\Omega$. - Robin: $\partial_\nu w + \alpha w = 0$ on $\partial\Omega$. **Step 2: Multiply by $w$ and integrate.** Since $\Delta w = 0$, we have $\int_\Omega w \Delta w \, d\mathcal{L}^n = 0$. Applying Green's first identity: \begin{align*} 0 = \int_\Omega w \Delta w \, d\mathcal{L}^n = -\int_\Omega |\nabla w|^2 \, d\mathcal{L}^n + \int_{\partial\Omega} w \, \partial_\nu w \, d\mathcal{H}^{n-1}. \end{align*} Therefore: \begin{align*} \int_\Omega |\nabla w|^2 \, d\mathcal{L}^n = \int_{\partial\Omega} w \, \partial_\nu w \, d\mathcal{H}^{n-1}. \tag{$\ast$} \end{align*} **Step 3: Evaluate the boundary integral in each case.** *Dirichlet:* $w = 0$ on $\partial\Omega$, so the boundary integral vanishes. *Neumann:* $\partial_\nu w = 0$ on $\partial\Omega$, so the boundary integral vanishes. *Robin:* $\partial_\nu w = -\alpha w$ on $\partial\Omega$, so \begin{align*} \int_{\partial\Omega} w \, \partial_\nu w \, d\mathcal{H}^{n-1} = -\alpha \int_{\partial\Omega} |w|^2 \, d\mathcal{H}^{n-1} \leq 0. \end{align*} In all three cases, $(\ast)$ gives $\int_\Omega |\nabla w|^2 \, d\mathcal{L}^n \leq 0$. **Step 4: Conclude.** Since $|\nabla w|^2 \geq 0$, the integral can only vanish, so $\nabla w \equiv 0$ in $\Omega$. Therefore $w = u_1 - u_2$ is constant. For Dirichlet conditions, $w = 0$ on $\partial\Omega$ forces the constant to be zero, giving $u_1 = u_2$. For Robin conditions, $\alpha > 0$ and $\int_{\partial\Omega} |w|^2 \, d\mathcal{H}^{n-1} = 0$ forces $w = 0$ on $\partial\Omega$, hence $w \equiv 0$. For Neumann conditions, the constant is not determined — the best we can say is that $u_1$ and $u_2$ differ by a constant.