MATLAB Mathematics - SERC - Index of

Partial Differential Equations
The solution u
satisfies the initial conditions
u 1 ( x,
0) ≡ 1
u 2 ( x,
0) ≡ 0
and boundary conditions
∂u
--------- 1
( 0,
t) ≡ 0
∂x
u 2 ( 0,
t) ≡ 0
u 1 ( 1,
t) ≡ 1
∂u
--------- 2
( 1,
t) ≡ 0
∂x
Note The demo pdex4 contains the complete code for this example. The demo
uses subfunctions to place all required functions in a single M-file. To run this
example type pdex4 at the command line. See “PDE Solver Basic Syntax” on
page 5-92 and “Solving PDE Problems” on page 5-94 for more information.
1 Rewrite the PDE. In the form expected by pdepe, the equations are
1
1
.∗
∂
---- u 1
∂t u 2
=
∂
----- 0.024( ∂u 1 ⁄ ∂x)
∂x 0.170( ∂u 2 ⁄ ∂x)
+
– Fu ( 1 – u 2 )
Fu ( 1 – u 2 )
The boundary conditions on the partial derivatives of u have to be written
in terms of the flux. In the form expected by pdepe, the left boundary
condition is
0 1
+
u 2 0
.∗
0.024( ∂u 1 ⁄ ∂x)
0.170( ∂u 2 ⁄ ∂x)
=
0
0
5-101