MATLAB Mathematics - SERC - Index of

Partial Differential Equations
At x = 0 and x = 1 , the solution satisfies the boundary conditions
u( 0,
t) = 0
π e – t + ∂u ------ ( 1,
t)
= 0
∂x
Note The demo pdex1 contains the complete code for this example. The demo
uses subfunctions to place all functions it requires in a single M-file. To run
the demo type pdex1 at the command line. See “PDE Solver Basic Syntax” on
page 5-92 for more information.
1 Rewrite the PDE. Write the PDE in the form
c⎛x, t, u,
∂u ------ ⎞ ∂u
------ = x – m ------ ∂ x m f⎛x, t, u,
∂u ------ ⎞
⎝ ∂x⎠
∂t ∂x ⎝
⎛ ⎝ ∂x⎠⎠
⎞ + s ⎛ x , t , u ,
∂u ------ ⎞
⎝ ∂x⎠
This is the form shown in Equation 5-3 and expected by pdepe. See
“Introduction to PDE Problems” on page 5-90 for more information. For this
example, the resulting equation is
π 2 ∂u ------ = x 0 ------ ∂ ⎛x 0 ∂u ------ ⎞ + 0
∂t ∂x ⎝ ∂x⎠
with parameter
m = 0
and the terms
c⎛x, t, u,
∂u ------ ⎞ = π 2
⎝ ∂x⎠
f x, t, u,
∂u ------
⎝
⎛ ∂x⎠
⎞ =
∂u ------
∂x
s⎛x, t, u,
∂u ------ ⎞ = 0
⎝ ∂x⎠
5-95