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Regular grid - 1D problem.

Let’s consider elliptic problem (*) on the interval [0, 1], with Laplace operator

Laplace operator 1D

On the ends of the interval we set the boundary condition u(0) = u(1) = 0.

Let’s subdivide our interval on n equal parts and apply finite differences schema with the step h = 1/n.

As a result, we receive the following 3-diagonal matrix

Laplace matrix 1D

The vector of unknowns’ u = (uj)j=0,…,n contains the values of unknown function u from our equation (*) in the nodes xj = j ∙ h, j = 0,…, n, of our 1-dimensional grid.

The right hand vector of the system of equations is the vector (fj)j=0,…,n . Its coordinates are the values of function f = f(x) from the right-hand side of equation (*) in the nodes of the grid xj, j = 0,…, n.

Quantities σ = (σj)j=0,…,n , on the diagonal of this matrix, are the values of function σ = σ(x) from equation (*) in the nodes of the grid. If σ is a constant (in particular if σ = 0), then all numbers σj, j = 0,…, n, are equal to this constant.

Thus, in one-dimensional case the system of equations is quite simple. And it is quite easy to solve this system of equations as soon as the matrix is so simple. It especially simple, when σ is constant (for instance, zero).

As soon as the situation is so simple, there is no need to use our Marlin Solver here.

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