Show that there is only one function \(u\) which takes given values on the (closed) boundary of a region and satisfies Laplace's equation \(\nabla^{2} u=0\) in the interior of the region. Hints: Suppose \(u_{1}\) and \(u_{2}\) are both solutions with the same boundary conditions so that \(U=u_{1}-u_{2}=0\) on the boundary. In Green's first identity (Chapter 6, Problem 10.16), let \(\phi=\Psi=U\) to show that \(\nabla U \equiv 0 .\) Thus show \(\bar{U} \equiv 0\) everywhere inside the region.

Boundary Value Problems are a class of differential equations with solutions that need to satisfy specific conditions at the boundaries of the domain. Here, we focus on solving these problems for Laplace's equation. This equation describes phenomena like temperature distribution and electrostatic potential inside a region. When solving for a function, the boundary conditions ensure that we get a unique and physically meaningful solution.

To illustrate this process, let's consider having two functions, \(u_1\) and \(u_2\), both satisfying Laplace's equation and the same boundary conditions. By defining a new function \(U = u_1 - u_2\), and noting that \(U\) must be zero on the boundary, we can simplify our problem and focus on the interior and its implications.

Green's First Identity is a powerful mathematical tool when working with solutions to Laplace’s equation. It relates the integrals of functions over a volume and their behavior over the boundary of that volume. Specifically, it states: \[ \int_{V} (\phi \abla^{2} \psi + \abla \phi \cdot \abla \psi ) \, dV = \oint_{S} \phi \abla \psi \cdot \mathbf{n} \, dS \] where \(V\) is the volume, \(S\) is the boundary of \(V\), and \(\mathbf{n}\) is the outward-pointing normal vector on \(S\). By applying this identity with \(\phi = U\) and \(\psi = U\) for the difference function \(U\), we get: \[ \int_{V} \abla U \cdot \abla U \, dV = \oint_{S} U \abla U \cdot \mathbf{n} \, dS \].

Due to the boundary condition \(U = 0\), the surface integral vanishes, leaving us with \[ \int_{V} \|\abla U\|^2 \, dV = 0 \]. This shows that \(abla U\) must be zero everywhere inside the region, implying that \(U\) is a constant. Since \(U = 0\) at the boundary, it must be zero throughout the entire region.

Laplace's Equation \(\abla^{2}u = 0\) is a second-order partial differential equation encountered frequently in physics and engineering. It describes steady-state phenomena, meaning the system does not change over time. Solutions to Laplace's equation are known as harmonic functions, and one of their remarkable properties is the uniqueness under given boundary conditions.

Suppose we have two solutions to Laplace's equation, \(u_1\) and \(u_2\), with the same boundary values. By defining \(U = u_1 - u_2\), we leverage properties from Green's First Identity to show that \(U\) cannot vary within the region. Essentially, the steps we follow demonstrate that \(abla U\) being zero implies \(U\) is a constant, and with boundary conditions forcing \(U = 0\), we see that \(u_1 = u_2\) within the whole domain. This proves the uniqueness of solutions for Laplace's equation with specified boundary conditions.