Equation of Laplace

In vectorial Analysis, the equation of Laplace is a partial derivative equation of the second order, whose name is a homage to the physicist mathematician Pierre-Simon Laplace.

Introduced for the needs for the Newtonian Mechanical , the equation of Laplace appears in many other branches of the theoretical physics: electrostatic Astronomy, , Mechanical of the fluids, propagation of heat, diffusion, Brownian Movement, Mechanical quantum.

Any function solution of the equation of Laplace is known as harmonic.

Equation of Laplace with three dimensions

Of Cartesian coordinates in an Euclidean space of Dimension 3, the problem consists in finding all the functions with three real variables $\ varphi \left(X, there, Z\right)$ which check the partial derivative equation of the second order:

$\ frac \left\{\ partial^2 \ varphi\right\} \left\{\ partial x^2\right\} \ + \ \ frac \left\{\ partial^2 \ varphi\right\} \left\{\ partial y^2\right\} \ + \ \ frac \left\{\ partial^2 \ varphi\right\} \left\{\ partial z^2\right\} \ = \ 0$

To simplify the writing, one introduces a differential Opérateur noted $\ Delta$ and called operator of Laplace , or simply Laplacian , such as the partial derivative equation preceding is written in a compact way:

$\ Delta \ varphi \ = \ 0$

Equation of Laplace with two dimensions

Of Cartesian coordinates in an Euclidean space of Dimension 2, the problem consists in finding all the functions with two real variables $V \left(X, there\right)$ which check:

$\ frac \left\{\ partial^2 V\right\} \left\{\ partial x^2\right\} \ + \ \ frac \left\{\ partial^2 V\right\} \left\{\ partial y^2\right\} \ = \ 0$

It is shown that all holomorphic Fonction gives solutions of the equation of Laplace to two dimensions by their real part and their imaginary part; moreover, these solutions are orthogonal in any point.

Recalls on the holomorphic functions

All polynomial function with Coefficient S complexes is holomorphic on $\ mathbb C$; to also are it the goniometrical functions and the exponential function. (The goniometrical functions are in fact relatively close to the exponential function since they can be defined from this one by using the formulas of Euler).

• the function Logarithme is holomorphic on the whole of the complex numbers private of the half-line of negative realities (one can choose an unspecified half-line resulting from 0).

• the function square root can be defined by

$\ sqrt \left\{Z\right\} = e^$
and is thus holomorphic everywhere where the function logarithm is.
• the function reverses $z \ mapsto 1/z$ is holomorphic on $\ mathbb C^*$.

• the reciprocal goniometrical functions have same manners of the seams and are holomorphic safe everywhere with the seams.

Results on the equation of Laplace and the functions holomorphic

Demonstration

The complex variable is introduced: $z=x+iy$ where $i^2 = -1$, and one defines the holomorphic function $F \left(Z\right)$. By derivation, one obtains that:

$\ frac \left\{\ partial F\right\} \left\{\ partial X\right\} \ = \ \ frac \left\{\ mathrm D F\right\} \left\{\ mathrm dz\right\} \ \ frac \left\{\ partial Z\right\} \left\{\ partial X\right\} \ = \ F\text{'} \left(Z\right)$

whereas:

$\ frac \left\{\ partial F\right\} \left\{\ partial there\right\} \ = \ \ frac \left\{\ mathrm dF\right\} \left\{\ mathrm dz\right\} \ \ frac \left\{\ partial Z\right\} \left\{\ partial there\right\} \ = \ I \ F\text{'} \left(Z\right)$.

By deriving one second time, one obtains in a similar way:

$\ frac \left\{\ partial^2 F\right\} \left\{\ partial x^2\right\} \ = \ F$ (Z)

whereas:

$\ frac \left\{\ partial^2 F\right\} \left\{\ partial y^2\right\} \ = \ - \ F$ (Z)

The sum is null, therefore the holomorphic function F is well a solution of the equation of Laplace:

$\ frac \left\{\ partial^2 F\right\} \left\{\ partial x^2\right\} \ + \ \ frac \left\{\ partial^2 F\right\} \left\{\ partial y^2\right\} \ = \ 0$

Note: the holomorphic function always admits a partly real decomposition and imaginary part:

$F \left(Z\right) \ = \ V \left(X, there\right) \ + \ I \ \ phi \left(X, there\right)$

By cancelling the real part and the imaginary part separately, one obtains two equations of independent Laplace:

$\ frac \left\{\ partial^2 V\right\} \left\{\ partial x^2\right\} + \ frac \left\{\ partial^2 V\right\} \left\{\ partial y^2\right\} =0 \ qquad \ mathrm \left\{and:\right\} \ qquad \ frac \left\{\ partial^2 \ partial phi\right\} \left\{\ x^2\right\} + \ frac \left\{\ partial^2 \ partial phi\right\} \left\{\ y^2\right\} =0$

Demonstration

One can write:

$\ frac \left\{\ partial F\right\} \left\{\ partial X\right\} \ = \ F\text{'} \left(x+iy\right) \ = \ \ frac \left\{\ partial V \left(X, there\right)\right\}\left\{\ partial X\right\} \ + \ I \ \ frac \left\{\ partial \ Phi \left(X, there\right)\right\}\left\{\ partial X\right\}$

and:

$\ frac \left\{\ partial F\right\} \left\{\ partial there\right\} \ = \ I F\text{'} \left(x+iy\right) \ = \ \ frac \left\{\ partial V \left(X, there\right)\right\}\left\{\ partial there\right\} \ + \ I \ \ frac \left\{\ partial \ Phi \left(X, there\right)\right\}\left\{\ partial there\right\}$

One from of deduced:

$\ frac \left\{\ partial V \left(X, there\right)\right\}\left\{\ partial X\right\} \ = \ \ frac \left\{\ partial \ Phi \left(X, there\right)\right\}\left\{\ partial there\right\} \ qquad \ mathrm \left\{and:\right\} \ qquad \ frac \left\{\ partial \ Phi \left(X, there\right)\right\}\left\{\ partial X\right\} \ = \ - \ \ frac \left\{\ partial V \left(X, there\right)\right\}\left\{\ partial there\right\}$

that is to say finally:

$\ frac \left\{\ partial V \left(X, there\right)\right\}\left\{\ partial X\right\} \ cdot \ frac \left\{\ partial \ Phi \left(X, there\right)\right\}\left\{\ partial X\right\} \ + \ \ frac \left\{\ partial V \left(X, there\right)\right\}\left\{\ partial there\right\} \ cdot \ frac \left\{\ partial \ Phi \left(X, there\right)\right\}\left\{\ partial there\right\} \ = \ 0$
One recognizes the scalar Produit there of the two vectors:

$\ overrightarrow \ operatorname \left\{grad\right\} \ \left(V\right) \ cdot \ overrightarrow \ operatorname \left\{grad\right\} \ \left(\ Phi\right) \ = \ 0$

One from of deduced that the curves with “= constant” and “= constant” are perpendicular (transformation in conformity). With the result that if “= constant” represents the curves in the same way potential, then “= constant” the lines of electrostatic Electric field in represents

Poisson's equation

If the member of right-hand side is a given function $f \left(X, there, Z\right)$, one obtains the Poisson's equation:

$\ Delta \ varphi = f$

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