# When do we use the Laplace equation

## Laplace equation

You already know what partial differential equations are and how you can solve them. In this post we want to apply your theoretical knowledge and look at the Laplace equation.

• Solve the Laplace equation using the separation approach
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• Laplace equation example
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• Use boundary conditions
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### Solve the Laplace equation using the separation approach

The equation named after the mathematician Pierre-Simon Laplace, who was born in 1749, is an elliptical partial differential equation of the second order. The Laplace equation is also used in physics. It can be derived, for example, from the heat conduction equation. Here we solve the Laplace equation on the unit square. For this we use the separation approach.

### Laplace equation example

The Laplace equation is the homogeneous Poisson equation: Both are elliptical differential equations. We want to look at the two-dimensional Laplace equation view on the unit square. The great describes the area in which we consider the differential equation. In this case the unit square.

The boundary conditions are as follows:    ### Solve Laplace's equation

As usual, you choose a product approach And you get from it after inserting, Sorting and equating with the constant  two ordinary differential equations.  ### Solve ordinary DGL

Now you have to solve the ordinary differential equations. But which one do you start with? You can save some time by looking at the boundary conditions. stands out. It is not zero like the other boundary conditions, but the same . So the part that depends on y must be a trigonometric function. That is the case, though is less than zero.
For this we define one , put it in the y-DGL  and set up the characteristic polynomial This gives the eigenvalues so that the solutions look like this. ### Use boundary conditions

As expected, it's a trigonometric function. Now we check the boundary conditions. We write this down for a capital Y first.  The first constraint lists The second constraint lists Either is equal to zero, which would be the trivial solution, or on is zero. That is the case, though a multiple of is.  Let us now consider the differential equation for x. Here we can For deploy,  rearrange the equation and set up the characteristic polynomial. The real eigenvalues ​​result The solution is thus made up of two exponential functions. From the first boundary condition follows, that so we have the solution in the hyperbolic sine can rewrite. As a reminder, the hyperbolic sine is . So we bet for minus one and parentheses two out. What remains is exactly the hyperbolic sine. So that is the general solution the sum over n of the product of the hyperbolic sine function and the sine function. With the last constraint results (1)

1: 2: first of all that all coefficients except equals zero are (2). Solve the rest of the equation results Now you can plug in the constant and the final solution looks like this. You have now solved the Laplace equation on the unit square. In the next post we will solve the Poisson equation.