# Poisson's equation

The Poisson equation , named after the French mathematician and physicist Siméon Denis Poisson , is an elliptical partial differential equation of the second order that is used as part of boundary value problems in large parts of physics.

## Mathematical formulation

The Poisson equation is general

${\ displaystyle - \ Delta u = f}$

Here designated

• ${\ displaystyle \ Delta}$the Laplace operator
• ${\ displaystyle u}$ the solution you are looking for
• ${\ displaystyle f}$a function. If the equation becomes Laplace's equation .${\ displaystyle f \ equiv 0}$

To solve Poisson's equation, additional information must be given, e.g. B. in the form of a Dirichlet boundary condition :

${\ displaystyle {\ begin {cases} - \ Delta u & = f & {\ text {in}} & \ Omega \\\ quad u & = g & {\ text {on}} & \ partial \ Omega \ end {cases}} }$

with open and restricted. ${\ displaystyle \ Omega \ subset \ mathbb {R} ^ {n}}$

In this case, we construct a solution using the fundamental solution of Laplace's equation: ${\ displaystyle \ Phi}$

${\ displaystyle \ Phi (x): = {\ begin {cases} - {\ dfrac {1} {2 \ pi}} \ ln | x | & n = 2 \\ {\ dfrac {1} {(n-2 ) \ omega _ {n}}} \ cdot {\ dfrac {1} {| x | ^ {n-2}}} & n \ geq 3 \ end {cases}}}$

It denotes the area of ​​the unit sphere in the -dimensional Euclidean space . ${\ displaystyle \ omega _ {n} = {\ tfrac {2 \ pi ^ {\ frac {n} {2}}} {\ Gamma ({\ frac {n} {2}})}}}$${\ displaystyle n}$

The convolution gives a solution to Poisson's equation. ${\ displaystyle (\ Phi * f)}$

In order to meet the boundary condition can be the Green's function use

${\ displaystyle G (x, y): = \ Phi (yx) - \ phi ^ {x} (y)}$

${\ displaystyle \ phi ^ {x}}$ is a correction function that

${\ displaystyle {\ begin {cases} \ Delta \ phi ^ {x} = 0 & {\ text {in}} \ \ Omega \\\ phi ^ {x} = \ Phi (yx) & {\ text {on} } \ \ partial \ Omega \ end {cases}}}$

Fulfills. It is generally dependent on and is easy to find only for simple areas. ${\ displaystyle \ Omega}$

If one knows , then a solution of the boundary value problem from above is given by ${\ displaystyle G (x, y)}$

${\ displaystyle u (x) = - \ int _ {\ partial \ Omega} g (y) {\ frac {\ partial G (x, y)} {\ partial n}} \ mathrm {d} \ sigma (y ) + \ int _ {\ Omega} f (y) G (x, y) \ mathrm {d} y}$

where the surface dimension denotes. ${\ displaystyle \ sigma}$${\ displaystyle \ partial \ Omega}$

The solution can also be found with the help of the platform method or a variation approach .

## Applications in physics

The Poisson equation, for example, is satisfied by the electrostatic potential and the gravitational potential , each with symbols . The function is proportional to the electrical charge density or to the mass density${\ displaystyle \ Delta \ Phi (\ mathbf {r}) = f (\ mathbf {r})}$${\ displaystyle \ Phi}$${\ displaystyle f}$ ${\ displaystyle \ rho (\ mathbf {r}).}$

If it is known everywhere, the general solution of Poisson's equation, which approaches zero for large distances, is the integral ${\ displaystyle f (\ mathbf {r})}$

${\ displaystyle \ Phi (\ mathbf {r}) = - {\ frac {1} {4 \, \ pi}} \ int \ mathrm {d} ^ {3} r '\, {\ frac {f (\ mathbf {r} ')} {| \ mathbf {r} - \ mathbf {r}' |}}}$.

In words: every charge at the location in the small area of ​​size contributes additively to the potential at the location with its electrostatic or gravitational potential: ${\ displaystyle \ mathrm {d} ^ {3} r '\, f (\ mathbf {r}')}$${\ displaystyle \ mathbf {r} '}$${\ displaystyle \ mathrm {d} ^ {3} r '}$${\ displaystyle \ Phi}$${\ displaystyle \ mathbf {r}}$

${\ displaystyle - {\ frac {\ mathrm {d} ^ {3} r '\, f (\ mathbf {r}')} {4 \, \ pi \, | \ mathbf {r} - \ mathbf {r } '|}}}$

### Electrostatics

Since the electrostatic field is a conservative field , it can be expressed in terms of the gradient of a potential : ${\ displaystyle \ nabla \ Phi}$ ${\ displaystyle \ Phi (\ mathbf {r})}$

${\ displaystyle \ mathbf {E} (\ mathbf {r}) = - \ nabla \ Phi (\ mathbf {r}).}$

Applying the divergence results in

${\ displaystyle \ nabla \ cdot \ mathbf {E} (\ mathbf {r}) = - \ Delta \ Phi (\ mathbf {r})}$

with the Laplace operator . ${\ displaystyle \ Delta}$

According to the first Maxwell equation , however, also applies

${\ displaystyle \ nabla \ cdot \ mathbf {E} (\ mathbf {r}) = {\ frac {\ rho (\ mathbf {r})} {\ varepsilon}}}$

With

• the charge density ${\ displaystyle \ rho (\ mathbf {r})}$
• the permittivity .${\ displaystyle \ varepsilon = \ varepsilon _ {\ mathrm {r}} \ cdot \ varepsilon _ {0}}$

It follows for the Poisson equation of the electric field

${\ displaystyle \ Delta \ Phi (\ mathbf {r}) = - {\ frac {\ rho (\ mathbf {r})} {\ varepsilon}}}$

The special case for each location in the area under consideration is called the Laplace equation of electrostatics . ${\ displaystyle \ rho (\ mathbf {r}) = 0}$

### Electrodynamics of stationary currents

The emitter of a silicon solar cell , which can be described as purely two-dimensional to a good approximation, is considered here as an example . The emitter is located in the xy-plane, the z-axis points into the base. The lateral surface current density in the emitter depends on the z-component of the (volume) current density of the base occurring at the emitter , which is indicated by the continuity equation in the form ${\ displaystyle \ mathbf {j} (x, y)}$${\ displaystyle J_ {z} (x, y, z = 0)}$

${\ displaystyle \ nabla _ {2} \ cdot \ mathbf {j} (x, y) = - J_ {z} (x, y, z = 0)}$

can be described (with the two-dimensional Nabla operator ). The surface current density depends on the local Ohm's law with the lateral electric field in the emitter together: ; here is the specific sheet resistance of the emitter assumed to be homogeneous . If you write (as discussed in the section on electrostatics) the electric field as a gradient of the electric potential, you get a Poisson equation for the potential distribution in the emitter in the form ${\ displaystyle \ nabla _ {2}}$${\ displaystyle \ mathbf {j} (x, y) = R _ {\ Box} ^ {- 1} \ mathbf {E} (x, y)}$${\ displaystyle R _ {\ Box}}$${\ displaystyle \ mathbf {E} (x, y) = - \ nabla _ {2} \ Phi (x, y)}$

${\ displaystyle \ Delta _ {2} \ Phi (x, y) = R _ {\ Box} J_ {z} (x, y, z = 0).}$

### Gravity

Just like the electrostatic field

${\ displaystyle \ mathbf {E} (\ mathbf {r}) = {\ frac {1} {4 \ pi \ varepsilon}} \ int \ rho _ {\ mathrm {el}} \ left (\ mathbf {r} '\ right) {\ frac {\ mathbf {r} - \ mathbf {r}'} {\ left | \ mathbf {r} - \ mathbf {r} '\ right | ^ {3}}} \; dx' \, dy '\, dz' = - \ nabla \ Phi _ {\ mathrm {el}} (\ mathbf {r})}$,

the gravitational field  g is also a conservative field:

${\ displaystyle \ mathbf {g} (\ mathbf {r}) = - G \ int \ rho _ {\ mathrm {m}} \ left (\ mathbf {r} '\ right) {\ frac {\ mathbf {r } - \ mathbf {r} '} {\ left | \ mathbf {r} - \ mathbf {r}' \ right | ^ {3}}} \; dx '\, dy' \, dz '= - \ nabla \ Phi _ {\ mathrm {m}} (\ mathbf {r})}$.

It is

• G is the gravitational constant
• ${\ displaystyle \ rho _ {\ mathrm {m}} (\ mathbf {r} ')}$ the mass density.

Since only the charges are replaced by masses and by , analogously to the first Maxwell equation applies ${\ displaystyle {\ frac {1} {4 \ pi \ varepsilon}}}$${\ displaystyle -G}$

${\ displaystyle \ nabla \ cdot \ mathbf {g} = -4 \ pi G \ rho _ {\ mathrm {m}} (\ mathbf {r})}$.

This results in the Poisson equation for gravity

${\ displaystyle \ nabla \ mathbf {g} = - \ nabla \ cdot (\ nabla \ Phi _ {\ mathrm {m}} (\ mathbf {r})) = - 4 \ pi \ cdot G \ cdot \ rho _ {\ mathrm {m}} (\ mathbf {r}) \ Leftrightarrow}$
${\ displaystyle \ Delta \ Phi _ {\ mathrm {m}} (\ mathbf {r}) = 4 \ pi \ cdot G \ cdot \ rho _ {\ mathrm {m}} (\ mathbf {r})}$.

## Individual evidence

1. Wolfgang Nolting: Basic course in theoretical physics . [Online excl. the] 8th [dr.]. 3. Electrodynamics. Springer, Berlin, ISBN 978-3-540-71252-7 .