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  Parabolic PDEs-Explicit Method

Parabolic PDEs - Explicit Method:

Heat Flow and Diffusion:

In the preceding sections we studied PDE that represent steady-state heat problem. There was no time variable in the equation. In this section we start to study how to solve equations that involve time that is we compute temperature profiles that are changing.

The conduction of heat as well as diffusion of a chemical happen to be modelled by the same differential equation. The cause for this is that they both involve similar processes. Heat conduction take places when hot fast moving molecules bump into slower molecules and transfer some of their energy. In a solid this engages moles of molecules all moving in different, nearly random ways however the net effect is that the energy eventually spreads itself out over a larger region. The diffusion of a chemical in a gas or else liquid similarly involves large numbers of molecules moving in different nearly random ways.

These molecules finally spread out over a larger region.

In three dimensions the equation that oversees both of these processes is the heat/diffusion equation.

ut= cΔu ,

where c is the coefficient of conduction or else diffusion as well as Δu(x, y, z) = uxx+uyy+uzz. The symbol Δ in this background is called the Laplacian. If there is as well a heat/chemical source after that it is incorporated a function g(x, y, z, t) in the equation as:

ut= cΔu + g.

In a few problems the z dimension is irrelevant either because the object in question is very thin or u doesn’t change in the z direction. In this circumstance the equation is:

ut= cΔu = c(uxx+ uyy).

Finally in some cases merely the x direction matters. In this case the equation is simply:

ut= cuxx

or

ut= cuxx+ g(x, t)

In this lecture we will be trained a straight-forward technique for solving pervious equations. It is very alike to the finite difference method we used for nonlinear boundary value problems.

It is value mentioning a related equation:

ut= cΔ(uγ) for γ >1 ,

This is called the porus media equation. This equation models diffusion in a solid however porus, material such as sandstone or else an earthen structure. We will not resolve this equation numerically however the methods introduced here would work. Several equations that involve 1 time derivative and 2 spatial derivatives are parabolic and the methods introduced here will work for most of them.

Explicit Method Finite Differences:

The one dimensional heat/diffusion equation ut = cuxx has two independent variables t and x and therefore we have to discretize both. Ever since we are considering 0 ≤ x ≤ L we subdivide [0,L] into m equal subintervals that is let,

h = L/m

and

(x0, x1, x2, . . . , xm−1, xm) = (0, h, 2h, . . . ,L− h,L).

Likewise if we are interested in solving the equation on an interval of time [0, T] let

k = T/n

and

(t0, t1, t2, . . . , tn−1, tn) = (0, k, 2k, . . . , T − k, T).

We will then indicate the approximate solution at the grid points by:

uij≈ u(xi, tj).

The equation ut= cuxx can be then be replaced by the difference equations:
(ui,j+1 − ui,j)/ k=c/h2(ui−1,j− 2ui,j+ ui+1,j).

Here we have utilized the forward difference for utand the central difference for uxx. This equation can be solved for ui,j+1 to produce:

ui,j+1 = rui−1,j + (1 − 2r)ui,j + rui+1,j

for 1 ≤ i ≤ m − 1, 0 ≤ j ≤ n − 1, where

r = ck/h2

The formula permits us to compute all the values of u at step j +1 using the values at step j. Notice that ui,j+1 depends on ui,j, ui−1,j and ui+1,j. That is u at grid point i depends on its preceding value and the values of its two nearest neighbours at the previous step.

Initial Condition:

To solve the partial differential equation or else we need an initial condition. This represents the state of the system when we begin that is the initial temperature distribution or initial concentration profile. This is represented through

u(x, 0) = f(x).

To execute this in a program we let

ui,0 = f(xi).

1568_initial condition.jpg

The worth at grid point i depends on its previous values and the previous Values of its nearest neighbours.

Boundary Conditions:

To solve the biased differential equation or we as well need boundary conditions. Just as in the preceding section we will have to specify something about the ends of the domain that is at x = 0 and x = L. One probability is fixed boundary conditions which we can implement just as we did for the ODE boundary value problem.

A second probability is called variable boundary conditions. This is symbolized by time dependent functions:

u(0, t) = g1(t) and u(L, t) = g2(t).

In a heat problem g1 as well as g2 would represent heating or cooling applied to the ends. These are effortlessly implemented in a program by letting:

u0,j= g1(tj) and um,j= g2(tj).

Implementation:

The following program as well available on the web page executes the explicit method. It add in variable boundary conditions at both ends. To run it you should define functions fg1 and g2. Notice that the major loop has only one line. The values of u are reserved as a matrix. It is habitually convenient to define a matrix of the right dimension containing all zeros as well as then fill in the calculated values as the program runs.

Run this program by means of L = 2, T = 20, f(x) = .5x, g1(t) = 0, and g2(t) = cos(t). Note that g1(t) must be input as g1 = inline(’0*t’).

function [t x u] = myheat(f,g1,g2,L,T,m,n,c)
% function [t x u] = myheat(f,g1,g2,L,T,m,n,c)
% solve u_t = c u_xx for 0<=x<=L, 0<=t<=T
% BC: u(0, t) = g1(t); u(L,t) = g2(t)
% IC: u(x, 0) = f(x)
% Inputs:
% f -- inline function for IC
% g1,g2 -- inline functions for BC
% L -- length of rod
% T -- length of time interval
% m -- number of subintervals for x
% n -- number of subintervals for t
% c -- rate constant in equation
% Outputs:
% t -- vector of time points
% x -- vector of x points
% u -- matrix of the solution, u(i,j)~=u(x(i),t(j))
h = L/m; k = T/n;
r = c*k/h^2; rr = 1 - 2*r;
x = linspace(0,L,m+1);
t = linspace(0,T,n+1);
%Set up the matrix for u:
u = zeros(m+1,n+1);
% assign initial conditions
u(:,1) = f(x);
% assign boundary conditions
u(1,:) = g1(t); u(m+1,:) = g2(t);
% find solution at remaining time steps
for j = 1:n
u(2:m,j+1) = r*u(1:m-1,j) + rr*u(2:m,j) + r*u(3:m+1,j);
end
% plot the results
mesh(x,t,u’)

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