Laplace Equation
1 Laplace Equation
The Laplace equation states that the Laplacian of a twice-differentiable function \(u\) is zero
\[ \Delta u = 0, \quad \Delta = \nabla^2 = \sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}. \]
In 2D, the Laplace equation is
\[ u_{xx} + u_{yy} = 0. \]
It is a linear, homogenous, second-order PDE and is the prototypical elliptic PDE equation. Solutions of the Laplace equation are called harmonic functions. The inhomogenous version \(\Delta u=f\) is called the Poisson equation.
1.1 Where does it come from?
- The Laplace equation describes the steady state (no time dependence) of diffusion and wave phenomena. Setting \(u_t=0\) in the heat equation or the wave equation gives \(\Delta u = 0\).
- It also governs electrostatic potential in charge-free regions, gravitational potential in vacuum, and the velocity potential of incompressible, irrotational fluid flow.
2 Properties of Harmonic Functions
2.1 Mean Value Property
If \(u\) is harmonic on a ball \(B_r(x)\), then
\[ u(x) = \frac{1}{|\partial B_r|}\int_{\partial B_r(x)} u(y)\,dS(y) =\frac{1}{|B_r|}\int_{B_r(x)} u\,dy. \]
where \(|\partial B_r|\) is the surface area of the boundary of the ball and \(dS\) is the surface measure. In other words, the value at the center equals the average over the sphere or the ball.
2.2 Maximum Principle
On a bounded domain \(\Omega\) with \(u\in C^2(\Omega)\subset C(\overline{\Omega})\) harmonic then
\[ \max _{\overline{\Omega}} u = \max_{\partial \Omega} u, \quad \]
so there are no local maxima or minima in the interior of the domain.
2.3 Smoothness
Harmonic functions are automatically \(C^\infty\) (infinitely differentiable), and are in fact real-analytic.
2.4 Liouville’s Theorem
A bounded harmonic function on all of \(\mathbb{R}^n\) is constant.
3 Boundary Value Problems
3.1 Dirichlet problem
\(\Delta u = 0\) in \(\Omega\), \(u = g\) on \(\partial\Omega\).
The solution is unique.
3.2 Neumann Problem
\(\Delta u = 0\) in \(\Omega\), \(\partial_\nu u = h\) on \(\partial\Omega\). The solution is unique only up to an additive constant, and a solution exists only if \(\int_{\partial\Omega} h\, dS = 0\).
4 Fundamental solution
The radially symmetric solutions away from the origin are
\[ \Phi(x) = \begin{cases} -\dfrac{1}{2\pi}\ln|x|, & n=2,\\[2mm] \dfrac{1}{n(n-2)\alpha(n)}\,|x|^{2-n}, & n\ge 3, \end{cases} \]
where \(\alpha(n)\) is the volume of the unit ball in \(\mathbb{R}^n\). It satisfies \(-\Delta\Phi = \delta_0\), and convolving with it gives solutions of Poisson’s equation: \(u = \Phi * f\) solves \(-\Delta u = f\).
In polar coordinates the equation becomes
\[ u_{rr} + \frac1r u_r + \frac{1}{r^2}u_{\theta\theta} = 0. \]
Separation of variables with \(u = R(r)\Theta(\theta)\) and \(2\pi\)-periodicity gives the bounded solutions
\[u(r,\theta) = \frac{a_0}{2} + \sum_{k=1}^\infty r^k\left(a_k\cos k\theta + b_k \sin k\theta\right),\]
where \(a_k, b_k\) are the Fourier coefficients of the boundary data \(g(\theta)\). Summing the series yields the Poisson integral formula:
\[u(r,\theta) = \frac{1}{2\pi}\int_0^{2\pi} \frac{1-r^2}{1-2r\cos(\theta-\phi)+r^2}\, g(\phi)\, d\phi.\]
For instance, with \(g(\theta) = \cos\theta\) the solution is \(u = r\cos\theta = x\), which is clearly harmonic.
5 Connection to complex analysis
In 2D, the real and imaginary parts of any holomorphic function \(f = u + iv\) are harmonic. This follows from the Cauchy–Riemann equations \(u_x = v_y\), \(u_y = -v_x\), which give \(u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0\). This is why conformal mapping is such a powerful tool for 2D Laplace problems.
6 References
- Wikipedia: Laplace’s equation
- Wikipedia: Harmonic function
- Paul’s Online Math Notes: Laplace’s Equation
- Evans, Partial Differential Equations, Section 2.2 (fundamental solution, mean value property, and properties of harmonic functions)
- Claude