Finite Difference Method for Elliptic PDEs

Finite Difference Method for Elliptic PDEs:

Examples of Elliptic PDEs:

Elliptic PDE’s are equations with second derivatives in space as well as no time derivative. The most significant examples are Laplace’s equation:

Δu = uxx+ uyy+ uzz= 0

and the Poisson equation:

Δu = f(x, y, z).

These equations are utilized in a large variety of physical situations such as- steady state chemical distributions, steady state heat problems, electrostatic potentials and elastic deformation and steady state fluid flows.

For the sake of clarity we will merely consider the two dimensional problem. A excellent model problem in this dimension is the elastic deflection of a membrane. Presume that a membrane such as a sheet of rubber is stretched across a rectangular frame. If a few of the edges of the frame are bent or if forces are applied to the sheet then it will deflect by an amount u(x, y) at each point (x, y). This u will satisfy the boundary value problem:

uxx+ uyy= f(x, y) for (x, y) in R,

u(x, y) = g(x, y) for (x, y) on ∂R,

Where R is the rectangle, f(x, y) is the force density (pressure) applied at each point, ∂R is the edge of the rectangle and g(x, y) is the deflection at the edge.

The Finite Difference Equations:

Presume the rectangle is described by:

R = {a ≤ x ≤ b, c ≤ y ≤ d}.

We will divide R in sub-rectangles. If we have m subdivisions in the x direction as well as n subdivisions in the y direction then the step size in the x and y directions respectively are

h = (b – a)/m and k= (d – c)/ n

We acquire the finite difference equations for by replacing uxx and uyy by their central differences to acquire:

(ui+1,j− 2uij+ ui−1,j)/ h2 + (ui,j+1− 2uij+ ui,j−1)/ k2= f(xi, yj) = fij

for 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1. The boundary conditions are commence by:

u0,j= g(a, yj), um,j= g(b, yj), ui,0 = g(xi, c), and ui,n= g(xi, d).

Straight Solution of the Equations:

Notice that since the edge values are imposed there are (m − 1) × (n − 1) grid points where we require to determine the solution. Note as well that there are exactly (m − 1) × (n − 1) equations in. Ultimately observe that the equations are all linear. Therefore we could solve the equations precisely using matrix methods. To do this we would first require to express the uij’s as a vector rather than a matrix. To do this there is a standard process let u be the column vector we get by placing one column subsequent to another from the columns of (uij). Therefore we would list u:,1 first then u:,2, and so on.. Next we would require writing the matrix Athat contains the coefficients of the equations and incorporate the boundary conditions in a vector b. After that we could resolve an equation of the form:

Au = b

Setting up as well as solving this equation is called the direct method.

An advantage of the direct method is that solving is able to be done relatively quickly and accurately. The disadvantage of the direct method is that one must set up u, A and b, which is confusing. Additionally the matrix A has dimensions (m − 1)(n − 1) × (m − 1)(n − 1) which are able to be rather large. Although A is large several of its elements are zero. Such a matrix is called sparse as well as there are special methods intended for efficiently working with sparse matrices.

Iterative Solution:

A typically preferred alternative to the direct method described above is to solve the finite difference equations iteratively. To do this primary solve for uij which yields:

uij= 1/(2(h2 + k2)) (k2(ui+1,j+ ui−1,j) + h2(ui,j+1 + ui,j−1) − h2k2fij)

This method is another instance of a relaxation method. Utilizing this formula along with, we can update uij from its neighbours simply as we did in the relaxation method for the nonlinear boundary value problem. If this method converges after that the result is an approximate solution.

The iterative solution is executed in the program my poisson.m. You will observe that maxit is set to 0. Therefore the program will not do any iteration however will plot the initial guess. The preliminary guess in this case consists of the proper boundary values at the edges and zero everywhere in the interior. To observe the solution evolve, steadily increase maxit.

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