Poisson Equation

Author

John Robin Inston

Published

October 15, 2026

1 Poisson Equation

References: Evans, Partial Differential Equations, Sections 2.2 and 2.2.4 (Green’s functions).

The Poisson equation is the inhomogeneous version of the Laplace Equation. It is the prototypical elliptic PDE and describes the equilibrium of a system subject to a distributed source.

Let \(\Omega\subset\mathbb{R}^n\) be open and \(f:\Omega\to\mathbb{R}\). The Poisson equation is \[ -\Delta u = f \quad \text{in } \Omega, \] where \(\Delta\) is the Laplace Operator. When \(f\equiv0\) it reduces to the Laplace equation, whose solutions are the harmonic functions.

The minus sign is a convention which makes \(-\Delta\) a positive operator (see below). Many texts write \(\Delta u=f\) instead, so always check the sign.

2 Where does it come from?

Steady state of diffusion. Setting \(u_t=0\) in the non-homogeneous Heat Equation \(u_t-k\Delta u=f\) gives \(-k\Delta u=f\), so \(u\) is the equilibrium temperature (or concentration) under a constant source \(f\).
Electrostatics. The electric potential \(\varphi\) with charge density \(\rho\) satisfies \(-\Delta\varphi=\rho/\varepsilon_0\).
Gravity. The gravitational potential \(\Phi\) with mass density \(\rho\) satisfies \(\Delta\Phi=4\pi G\rho\).
Probability. If \(B_t\) is a Brownian Motion started at \(x\) and \(\tau\) is its exit time from \(\Omega\), then \(u(x)=\mathbb{E}_x[\tau]\) solves \(-\tfrac12\Delta u=1\) with \(u=0\) on \(\partial\Omega\).

Gauss’ law \(\nabla\cdot\vec E=\rho/\varepsilon_0\) together with \(\vec E=-\nabla\varphi\) (see Gradient) gives \[ \nabla\cdot(-\nabla\varphi)=\frac{\rho}{\varepsilon_0} \implies -\Delta\varphi=\frac{\rho}{\varepsilon_0}. \]

3 Boundary Value Problems

As the equation is elliptic it is posed on a bounded domain with one condition on all of \(\partial\Omega\).

Dirichlet: \(u=g\) on \(\partial\Omega\).
Neumann: \(\partial_\nu u=h\) on \(\partial\Omega\), where \(\nu\) is the outward unit normal.
Robin: \(\partial_\nu u+\alpha u=h\) on \(\partial\Omega\).

The Dirichlet problem \(-\Delta u=f\) in \(\Omega\), \(u=g\) on \(\partial\Omega\) has at most one solution \(u\in C^2(\Omega)\cap C(\overline\Omega)\). The Neumann problem has a solution only if the compatibility condition \[ \int_\Omega f~dx=-\int_{\partial\Omega}h~dS \] holds, and then it is unique only up to an additive constant.

Proof. If \(u_1,u_2\) are two solutions of the Dirichlet problem then \(w=u_1-u_2\) satisfies \(-\Delta w=0\) and \(w=0\) on \(\partial\Omega\). Multiplying by \(w\) and integrating by parts (Green’s first identity), \[ 0=-\int_\Omega w\Delta w~dx=\int_\Omega|\nabla w|^2~dx -\int_{\partial\Omega}w\,\partial_\nu w~dS=\int_\Omega|\nabla w|^2~dx. \] So \(\nabla w\equiv0\), \(w\) is constant, and since \(w=0\) on the boundary, \(w\equiv0\).

For the Neumann compatibility, integrate the equation and apply the divergence theorem: \(\int_\Omega f~dx=-\int_\Omega\Delta u~dx=-\int_{\partial\Omega}\partial_\nu u~dS\).

Note: The compatibility condition is a conservation law. In the steady-state heat problem it says that the total heat generated inside must equal the total heat flowing out through the boundary.

4 Properties

Linearity: if \(-\Delta u_i=f_i\) then \(-\Delta(c_1u_1+c_2u_2)=c_1f_1+c_2f_2\). Hence the difference of two solutions with the same \(f\) is harmonic.
Weak maximum principle: if \(-\Delta u\le0\) (\(u\) is subharmonic) then \(\max_{\overline\Omega}u=\max_{\partial\Omega}u\). If \(-\Delta u\ge0\) (\(u\) is superharmonic) the minimum is attained on the boundary.
Regularity: if \(f\in C^\infty\) then \(u\in C^\infty\) in the interior, even if the boundary data is rough (elliptic regularity).
Positivity of \(-\Delta\): with Dirichlet conditions, \(\int_\Omega u(-\Delta u)~dx=\int_\Omega|\nabla u|^2~dx\ge0\).

Solve \(-u''=1\) on \((0,1)\) with \(u(0)=u(1)=0\). Integrating twice, \(u=-x^2/2+ax+b\) and the boundary conditions give \(b=0\), \(a=1/2\), so \[ u(x)=\frac{x(1-x)}{2}. \] This is the equilibrium temperature of a rod with uniform heating and both ends held at zero. By the maximum principle for superharmonic functions \(u\ge0\), and the maximum \(u(1/2)=1/8\) is in the interior, which is allowed because \(-u''\ge0\) makes \(u\) superharmonic, not subharmonic.

5 Solutions

5.1 Fundamental Solution

The fundamental solution of the Laplacian on \(\mathbb{R}^n\) is \[ \Phi(x)= \begin{cases} -\dfrac{1}{2\pi}\ln|x|, & n=2,\\[2mm] \dfrac{1}{n(n-2)\alpha(n)}|x|^{2-n}, & n\ge3, \end{cases} \] where \(\alpha(n)\) is the volume of the unit ball in \(\mathbb{R}^n\). It satisfies \(-\Delta\Phi=\delta_0\).

For \(n=3\) this is \(\Phi(x)=1/(4\pi|x|)\), the Coulomb potential of a unit point charge.

If \(f\in C_c^2(\mathbb{R}^n)\) then \[ u(x)=(\Phi*f)(x)=\int_{\mathbb{R}^n}\Phi(x-y)f(y)~dy \] is \(C^2\) and solves \(-\Delta u=f\) on \(\mathbb{R}^n\).

Note: This is the elliptic analogue of the heat kernel solution. Each point source \(f(y)\delta_y\) produces the potential \(f(y)\Phi(x-y)\), and by linearity the solution is the superposition of all of these (see Convolution). For the charge distribution \(\rho\) this is exactly Coulomb’s law.

5.2 Green’s Function

On a bounded domain the point source must also satisfy the boundary condition. The Green’s function is \(G(x,y)=\Phi(x-y)-h^y(x)\) where the corrector \(h^y\) is harmonic with \(h^y=\Phi(\cdot-y)\) on \(\partial\Omega\), so that \[ -\Delta_xG(x,y)=\delta_y(x)\ \text{in }\Omega,\qquad G(x,y)=0\ \text{for }x\in\partial\Omega. \] The solution of the Dirichlet problem is then \[ u(x)=\int_\Omega G(x,y)f(y)~dy-\int_{\partial\Omega}g(y)\,\partial_{\nu_y}G(x,y)~dS(y). \] \(G\) is symmetric, \(G(x,y)=G(y,x)\), and is explicitly known for the half-space and ball (by the method of images), where it reproduces the Poisson integral formula on the disk from the Laplace Equation note.

5.3 Separation of Variables and Eigenfunctions

Let \(\{\varphi_k\}\) be the Dirichlet eigenfunctions \(-\Delta\varphi_k=\lambda_k\varphi_k\), \(\varphi_k=0\) on \(\partial\Omega\), with \(0<\lambda_1\le\lambda_2\le\cdots\). Expanding \(f=\sum f_k\varphi_k\) and \(u=\sum u_k\varphi_k\) gives \(\lambda_ku_k=f_k\), so \[ u=\sum_{k=1}^\infty\frac{f_k}{\lambda_k}\varphi_k,\qquad f_k=\langle f,\varphi_k\rangle. \] The Green’s function is \(G(x,y)=\sum_k\varphi_k(x)\varphi_k(y)/\lambda_k\).

On \((0,1)\) the eigenfunctions are \(\sin(k\pi x)\) with \(\lambda_k=(k\pi)^2\). For \(f=1\) the sine coefficients (normalised, \(\sqrt2\sin k\pi x\)) are \(f_k=\sqrt2\,(1-(-1)^k)/(k\pi)\), which vanish for even \(k\). Hence \[ u(x)=\sum_{k\text{ odd}}\frac{4}{(k\pi)^3}\sin(k\pi x), \] which converges to \(x(1-x)/2\) from the previous example.

5.4 Numerical Solution

On a uniform grid with spacing \(h\) in two dimensions the five-point stencil approximates the Laplacian, \[ \Delta u(x,y)\approx\frac{u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1}-4u_{i,j}}{h^2}, \] which leads to a sparse linear system \(A\boldsymbol u=\boldsymbol f\) with \(A\) symmetric positive definite, matching the positivity of \(-\Delta\). Each interior value is a discrete mean value property: \(u_{i,j}\) is the average of its four neighbours plus \(h^2f_{i,j}/4\). See Finite-Difference Methods.