# Laplace equation

Solution of the Laplace equation on a circular ring with the Dirichlet boundary values ​​u (r = 2) = 0 and u (r = 4) = 4sin (5 * θ)

The Laplace equation (after Pierre-Simon Laplace ) is the elliptical partial differential equation of the second order

${\ displaystyle \ Delta \ Phi = 0}$

for a scalar function in a domain , where represents the Laplace operator . This makes it the homogeneous Poisson equation , that is, the right-hand side is zero. The Laplace equation is the prototype of an elliptic partial differential equation . ${\ displaystyle \ Phi}$ ${\ displaystyle \ Omega \ subset \ mathbb {R} ^ {n}}$${\ displaystyle \ Delta}$

## definition

The math problem is to find a scalar, twofold continuously differentiable function , which is the equation ${\ displaystyle \ Phi}$

${\ displaystyle \ Delta \ Phi = 0}$

Fulfills. The solutions to this differential equation are called harmonic functions . ${\ displaystyle \ Phi}$

The Laplace operator is generally defined for a scalar function as: ${\ displaystyle \ Delta}$

${\ displaystyle \ Delta \ Phi = \ operatorname {div} \ left (\ operatorname {grad} \, \ Phi \ right) = \ nabla ^ {2} \ Phi = \ nabla \ cdot \ nabla \ Phi}$

## Coordinate representations

If a special coordinate system is given, the representation of the Laplace equation can be calculated in these coordinates. In the most commonly used coordinate systems, the Laplace equation can be written as:

In Cartesian coordinates
${\ displaystyle \ Delta = \ sum _ {k = 1} ^ {n} {\ frac {\ partial ^ {2}} {\ partial x_ {k} ^ {2}}}}$,

from which in three-dimensional space:

${\ displaystyle {\ frac {\ partial ^ {2} \ Phi} {\ partial x ^ {2}}} + {\ frac {\ partial ^ {2} \ Phi} {\ partial y ^ {2}}} + {\ frac {\ partial ^ {2} \ Phi} {\ partial z ^ {2}}} = 0}$

results.

In polar coordinates ,
${\ displaystyle {\ frac {1} {\ rho}} {\ frac {\ partial} {\ partial \ rho}} \ left (\ rho {\ frac {\ partial \ Phi} {\ partial \ rho}} \ right) + {\ frac {1} {\ rho ^ {2}}} {\ frac {\ partial ^ {2} \ Phi} {\ partial \ varphi ^ {2}}} = 0}$
In cylindrical coordinates ,
${\ displaystyle {\ frac {1} {\ rho}} {\ frac {\ partial} {\ partial \ rho}} \ left (\ rho {\ frac {\ partial \ Phi} {\ partial \ rho}} \ right) + {\ frac {1} {\ rho ^ {2}}} {\ frac {\ partial ^ {2} \ Phi} {\ partial \ varphi ^ {2}}} + {\ frac {\ partial ^ {2} \ Phi} {\ partial z ^ {2}}} = 0}$
In spherical coordinates ,
${\ displaystyle {\ frac {1} {r ^ {2}}} {\ frac {\ partial} {\ partial r}} \ left (r ^ {2} {\ frac {\ partial \ Phi} {\ partial r}} \ right) + {\ frac {1} {r ^ {2} \ sin \ theta}} {\ frac {\ partial} {\ partial \ theta}} \ left (\ sin \ theta {\ partial \ Phi \ over \ partial \ theta} \ right) + {\ frac {1} {r ^ {2} \ sin ^ {2} \ theta}} {\ frac {\ partial ^ {2} \ Phi} {\ partial \ varphi ^ {2}}} = 0}$.

## Importance in physics

The meaning of the Laplace equation or potential equation , as it is often called in physics, encompasses many sub-areas of physics. This can be guessed by the following examples:

Conduction

A temperature gradient that is constant over time can satisfy the Laplace equation.

The Laplace equation itself can also be obtained from the heat conduction equation . In the stationary case, i.e. in the state of equilibrium, the time derivative in the heat conduction equation is zero. This equation is the Poisson equation. If there are still no sources or sinks, so there is no further heat exchange - for example with the environment - than the one under consideration, then the heat conduction equation becomes the Laplace equation.

An example of this is a metal rod under which there is a candle at one end and the other end of which is cooled by ice water. After some time, a constant temperature gradient will develop on the rod, which fulfills the Laplace equation (temperature exchange with the environment is neglected). The same example, somewhat more practical, can be found in the insulation of houses. The heating inside is the candle and the cold outside air is the ice water.

Electrostatics

In electrostatics, the electrical potential in the charge-free space of the Laplace equation is sufficient . This is a special case of the Poisson equation for electrostatics .

If, for example, a conductive ball is brought into an external electric field , the electrons rearrange themselves on the surface. The result of this rearrangement is that the potential on the surface of the sphere is constant. According to the minimum-maximum principle (see below ) the potential within the sphere is constant.

This is the working principle of the Faraday cage . Since the electrical voltage is defined as the potential difference and the potential is constant, as just said, you are safe from electric shocks inside .

Fluid dynamics

A stationary , two-dimensional, incompressible , eddy-free flow can also be described by means of a potential equation instead of the full Navier-Stokes equations . With the help of such a potential function, simple flows such as B. laminar flows in tubes can be calculated analytically without complex computer programs.

## Boundary value problems

There are three types of boundary value problems . The Dirichlet problem , the Neumann problem and the mixed problem. These differ in the type of additional boundary conditions.

This is generally a restricted area and the edge of . ${\ displaystyle \ Omega \ subset \ mathbb {R} ^ {n}}$${\ displaystyle \ partial \ Omega}$${\ displaystyle \ Omega}$

### Dirichlet problem

In the Dirichlet problem, the continuous mapping is given on the edge . In other words, the values ​​are specified which the solution of the Laplace equation should assume on the boundary. ${\ displaystyle \ varphi (y)}$${\ displaystyle \ partial \ Omega}$

The Dirichlet problem can be formulated in the following way:

${\ displaystyle \ left \ {{\ begin {array} {rcll} \ Delta \ Phi & = & 0 & {\ text {in}} \ \ Omega \\\ Phi & = & \ varphi & {\ text {on}} \ \ partial \ Omega \\\ end {array}} \ right.}$

The solution to the Dirichlet problem is clear.

### Neumann problem

With the Neumann problem, the normal derivative is given on the boundary , which the solution of the Laplace equation should assume. ${\ displaystyle \ partial \ Omega}$

The Neumann problem can be formulated in the following way:

${\ displaystyle \ left \ {{\ begin {array} {rcll} \ Delta \ Phi & = & 0 & {\ text {in}} \ \ Omega \\ {\ frac {\ partial \ Phi} {\ partial n}} & = & h & {\ text {auf}} \ \ partial \ Omega \\\ end {array}} \ right.}$

where denotes the normal derivative of , i.e. the normal component of the gradient of on the surface of . ${\ displaystyle {\ tfrac {\ partial \ Phi} {\ partial n}} = \ nabla \ Phi \ cdot {\ vec {n}}}$${\ displaystyle \ Phi}$${\ displaystyle \ Phi}$${\ displaystyle \ partial \ Omega}$

The solution to the Neumann problem is unambiguous except for one additive constant.

### Mixed problem

The mixed boundary value problem represents a combination of the Dirichlet and the Neumann problem,

${\ displaystyle \ left \ {{\ begin {array} {rcll} \ Delta \ Phi & = & 0 & {\ text {in}} \ \ Omega \\ {\ frac {\ partial \ Phi} {\ partial n}} + c_ {0} \ Phi & = & h & {\ text {auf}} \ \ partial \ Omega \\\ end {array}} \ right.}$

with a constant , whereby further conditions, such as initial values, are necessary to solve this problem . ${\ displaystyle c_ {0}}$

The mixed problem is without known additional conditions such as B. initial values, not clearly solvable. The uniqueness of this problem requires the unique solvability of the differential equation of the values ​​on the boundary:

${\ displaystyle {\ frac {\ partial \ Phi} {\ partial n}} + c_ {0} \ Phi = h \ {\ text {on}} \ \ partial \ Omega}$.

However, if this differential equation can be solved uniquely on the basis of further information, the mixed problem can be converted into a Dirichlet problem, which has a unique solution.

## Gaussian mean theorem

If the area is harmonious, then its function value at this point is equal to the mean value of on the surface of each sphere around with radius , provided that the sphere lies in and the function values ​​of are continuous on the surface, ${\ displaystyle \ Phi}$${\ displaystyle \ Omega}$${\ displaystyle \ Phi (x_ {0})}$${\ displaystyle x_ {0} \ in \ Omega}$${\ displaystyle \ Phi (y)}$${\ displaystyle B (x_ {0}, r)}$${\ displaystyle x_ {0}}$${\ displaystyle r}$${\ displaystyle \ Omega}$${\ displaystyle \ Phi (y)}$

${\ displaystyle \ Phi (x_ {0}) = {\ frac {1} {| S (x_ {0}, r) |}} \ oint _ {S (x_ {0}, r)} \! \ Phi (y) \, \ mathrm {d} S = {\ frac {1} {| B (x_ {0}, r) |}} \ int _ {B (x_ {0}, r)} \! \ Phi (y) \, \ mathrm {d} y \ ,.}$

Here is the spherical surface of the sphere with its center and radius${\ displaystyle S (x_ {0}, r)}$${\ displaystyle B (x_ {0}, r) \ subset \ Omega}$${\ displaystyle x_ {0} \ in \ Omega}$${\ displaystyle r \ ,,}$

${\ displaystyle | S (x_ {0}, r) | = \ omega _ {n} \, r ^ {n-1}}$
${\ displaystyle | B (x_ {0}, r) | = {\ frac {\ omega _ {n} \, r ^ {n}} {n}}}$

with the area of the surface of the -dimensional unit sphere${\ displaystyle \ omega _ {n}}$${\ displaystyle n}$

${\ displaystyle \ omega _ {n} = {\ frac {n \ pi ^ {n / 2}} {(n / 2)!}} = {\ frac {2 \ pi ^ {n / 2}} {\ Gamma ({\ frac {n} {2}})}}}$

Here is the gamma function , the analytical extension of the faculty to non-natural numbers, as they occur for every non-even number . ${\ displaystyle \ Gamma}$${\ displaystyle n}$

## Minimum-maximum principle

From the mean value theorem of Gauss it follows that the solution of the Laplace equation in a restricted area does not have its minimum or its maximum, as long as the values on the boundary are continuous and not constant. This means: ${\ displaystyle \ Phi (x)}$${\ displaystyle \ Omega}$${\ displaystyle \ Phi (y)}$${\ displaystyle \ partial \ Omega}$

${\ displaystyle \ max \ {\ Phi (\ mathbf {x}), \ mathbf {x} \ in \ Omega \} <\ max \ {\ Phi (\ mathbf {x}), \ mathbf {x} \ in {\ overline {\ Omega}} \} = \ max \ {\ Phi (\ mathbf {y}), \ mathbf {y} \ in \ partial \ Omega \} \,}$
${\ displaystyle \ min \ {\ Phi (\ mathbf {y}), \ mathbf {y} \ in \ partial \ Omega \} = \ min \ {\ Phi (\ mathbf {x}), \ mathbf {x} \ in {\ overline {\ Omega}} \} <\ min \ {\ Phi (\ mathbf {x}), \ mathbf {x} \ in \ Omega \} \ ,.}$

Thus, the function values ​​are always between the minimum and the maximum of the values ​​on the edge: ${\ displaystyle \ Omega}$

${\ displaystyle \ min \ {\ Phi (\ mathbf {y}), \ mathbf {y} \ in \ partial \ Omega \} <\ Phi (\ mathbf {x}) <\ max \ {\ Phi (\ mathbf {y}), \ mathbf {y} \ in \ partial \ Omega \} \,}$for everyone .${\ displaystyle \ mathbf {x} \ in \ Omega}$

The exception to the above-mentioned principle is the trivial case that the boundary values ​​are constant, because in this case the overall solution is constant.

## Solution of Laplace's equation

### Fundamental solution

To find the fundamental solution of the Laplace equation, it is advisable to use the rotational invariance of the Laplace operator. It puts this on, with the Euclidean norm of designated. With the help of the chain rule , Laplace's equation for is transformed into an ordinary second-order differential equation of . The following dimension- dependent formula is obtained for the function that is only dependent on : ${\ displaystyle \ gamma \ colon \ mathbb {R} ^ {n} \ to \ mathbb {R}}$${\ displaystyle \ gamma (x) = g (| x |)}$${\ displaystyle | x |}$${\ displaystyle x}$${\ displaystyle \ gamma}$${\ displaystyle g}$${\ displaystyle | x |}$${\ displaystyle \ gamma}$

${\ displaystyle \ gamma (x): = \ left \ {{\ begin {array} {rl} - {\ frac {1} {2 \ pi}} \ ln {| x |} \, & n = 2 \\ {\ frac {1} {(n-2) \, \ omega _ {n}}} {\ frac {1} {| x | ^ {n-2}}} \, & n> 2 \\\ end { array}} \ right.}$

with the area of the surface of the -dimensional unit sphere${\ displaystyle \ omega _ {n}}$${\ displaystyle n}$

${\ displaystyle \ omega _ {n} = {\ frac {n \ pi ^ {n / 2}} {\ Gamma ({\ frac {n} {2}} + 1)}} = {\ frac {2 \ pi ^ {n / 2}} {\ Gamma ({\ frac {n} {2}})}}}$.

Here is the gamma function , the analytical extension of the faculty to non-natural numbers, as they occur for every non-even number . ${\ displaystyle \ Gamma (\ cdot)}$${\ displaystyle n}$

It should be noted here that the fundamental solution is not an actual solution of the Laplace equation if the origin is in , since it has a singularity at this point. ${\ displaystyle \ gamma}$${\ displaystyle \ Omega}$

The solution to the Dirichlet problem is discussed below. It should be noted that the Neumann problem and the mixed problem can be converted into a Dirichlet problem by solving the differential equation of the boundary values.

### Solution using the Greenscher function

The core problem is the construction of Green's function , which does not have to exist in every case. Finding these is generally difficult, especially since Green's function depends on the area in which Laplace's equation is satisfied. However, if Green's function is known, the Dirichlet problem can be clearly solved with its help. ${\ displaystyle \ Omega}$

The fundamental solution of the Laplace equation is the basis for determining the Green function . ${\ displaystyle \ gamma}$

In addition, an auxiliary function must be constructed, which in is continuously differentiable twice and steadily on with meeting the following conditions: ${\ displaystyle h (x, y)}$${\ displaystyle \ Omega}$${\ displaystyle \ partial \ Omega}$${\ displaystyle x \ in \ Omega}$

${\ displaystyle \ left \ {{\ begin {array} {rcll} \ Delta _ {y} h (x, y) & = & 0 \, & x \ in {\ overline {\ Omega}}, \, y \ in \ Omega \\ h (x, y) & = & - \ gamma (x, y) \, & x \ in \ Omega, \, y \ in \ partial \ Omega \\\ end {array}} \ right.}$

Finding this auxiliary function is the central step in determining Green's function.

The Green function results from: ${\ displaystyle G (x, y)}$

${\ displaystyle G (x, y) = h (x, y) + \ gamma (xy)}$,

hence the solution of the Dirichlet problem in can be calculated: ${\ displaystyle \ Phi (x)}$${\ displaystyle \ Omega}$

${\ displaystyle \ Phi (x) = - \ int _ {\ partial \ Omega} \! \ Phi (y) {\ frac {\ partial G (x, y)} {\ partial n (y)}} \, \ mathrm {d} S}$

### Solution in two dimensions

The basis of this solution is the Fourier method. The Dirichlet problem is considered in polar coordinates

${\ displaystyle \ left \ {{\ begin {array} {rcll} \ Delta \ Phi (r, \ phi) & = & 0 & {\ text {in}} \ \ Omega \\\ Phi & = & \ Phi (r_ {0}, \ phi _ {0}) & {\ text {auf}} \ \ partial \ Omega \\\ end {array}} \ right.}$

and the function you are looking for is split into two independent functions by separating the variables . The chosen approach is thus: ${\ displaystyle \ Phi (r, \ phi)}$

${\ displaystyle \ Phi (r, \ phi) = v (r) \, w (\ phi).}$

Substituting this approach into Laplace's equation and using a separation approach leads the problem back to two ordinary differential equations.

The solutions to these ordinary differential equations are:

${\ displaystyle \ left \ {{\ begin {array} {rcl} w (\ phi) & = & c_ {1, n} \ cos (n \ phi) + c_ {2, n} \ sin (n \ phi) \, \\ v (r) & = & d_ {1} \, r ^ {n} + d_ {2} \, r ^ {- n} \. \\\ end {array}} \ right.}$

Here, , , , constants, and wherein - the constant from the separation approach - is positive and real, whereby (in obtaining the solutions) which is satisfied periodicity of angle. This periodicity can also be interpreted as the continuity of the values ​​of on the edge . ${\ displaystyle c_ {1, n} \,}$${\ displaystyle c_ {2, n} \,}$${\ displaystyle d_ {1} \,}$${\ displaystyle d_ {2} \,}$${\ displaystyle n = {\ sqrt {\ lambda}}}$${\ displaystyle \ lambda}$${\ displaystyle 2 \ pi}$${\ displaystyle \ Phi (r, \ phi)}$${\ displaystyle \ partial \ Omega}$

If there were a singularity , which in turn contradicts the condition of continuity in . So is . ${\ displaystyle d_ {2} \ neq 0}$${\ displaystyle r = 0}$${\ displaystyle \ Omega}$${\ displaystyle d_ {2} = 0}$

If these solutions are used in the separation approach chosen above and added up over all possible solutions according to the principle of superposition , the result is the solution of the Laplace equation:

${\ displaystyle \ Phi (r, \ phi) = \ sum _ {n = 0} ^ {\ infty} \ Phi _ {n} (r, \ phi) = {\ frac {1} {2}} a_ { 0} + \ sum _ {n = 1} ^ {\ infty} r ^ {n} (a_ {n} \ cos (n \ phi) + b_ {n} \ sin (n \ phi)).}$

where , and are the Fourier coefficients of the values ​​of . ${\ displaystyle a_ {0} \,}$${\ displaystyle a_ {n} \,}$${\ displaystyle b_ {n} \,}$${\ displaystyle \ Phi (r_ {0}, \ phi _ {0})}$

## literature

• Klemens Burg, Herbert Haf, Friedrich Wille : Partial differential equations. Higher mathematics for engineers, scientists and mathematicians. 1st edition. = 3rd revised and expanded edition. Teubner, Stuttgart et al. 2004, ISBN 3-519-22965-X .
• Lawrence C. Evans : Partial Differential Equations. American Mathematical Society, Providence RI 1998, ISBN 0-8218-0772-2 ( Graduate studies in mathematics 19).