When do we use the 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 approachin the text
- Laplace equation examplein the text
- in the text
- in the text
- Use boundary conditionsin the text
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.
It leads to the solutions
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
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
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.
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