Matlab code for boundary condition

2019 6 27
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更新时间：2019 6 30
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What is the matlab code for below the boundary conditon?
(i) x=0, y=0
(ii) x=0, dy/dx +dz/dx =0
(iii) t=0, x=infinity, y=1, z=c

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Torsten
Torsten 2019-6-27
Depends on the numerical program for which you try to implement these conditions.
y''(x,t)-y^2*y'(x,t)+a*z(x,t)-c*y(x,t)-y'(t)=0
z''(x,t)-y^2*z'(x,t)-a*z(x,t)+c*y(x,t)-z'(t)=0
Boundary condition
(i) x=0, y=0
(ii) x=0, dy/dx +dz/dx =0
(iii) t=0, x=infinity, y=1, z=b
What is the matlab code for this equation?
Torsten
Torsten 2019-6-27
What is the difference between y'(x,t) and y'(t) ?
Same for z.
According to which independent variable do you differntiate when you write y', y'', z', z'' ?
Torsten
Torsten 2019-6-27
编辑：Torsten 2019-6-27
Salai Mathiselvi Salai Mathiselvi's comment moved here:
d^2y/dx^2-y^2*dy/dx+a*z-c*y-dy/dt=0,
d^2z/dx^2-y^2*dz/dx+a*z-c*y-dz/dt=0
Boundary condition
(i) x=0, y=0
(ii) x=0, dy/dx +dz/dx =0
(iii) t=0, x=infinity, y=1, z=b
What is the matlab code for this equation?

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回答（1 个）

Torsten
Torsten 2019-6-27

0 个投票

So you mean
at x=0, you have y = 0 and dy/dx + dz/dx
at x=infinity, you have y=1 and z = b
and at t = 0, you have
y = 1, z = b
for all x in [0;infinity] ?
If you have numerical values for a, b and c, read the documentation of "pdepe".
It will show you how to write the MATLAB code for your equations.

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Matlab Program
function pdex4
m = 2;
x = linspace(0,5);
t = linspace(0,1000);
sol = pdepe(m,@pdex4pde,@pdex4ic,@pdex4bc,x, t);
u1 = sol(:, :, 1);
u2 = sol(:, :, 2);
% A solution profile can also be illuminating.
figure
plot(x,u1(end,:))
% A solution profile can also be illuminating.
figure
plot(x,u2(end,:))
%--------------------------------------------------------------------
function [c, f, s] = pdex4pde(x,t,u,DuDx)
c = [1;1];
f = [1;1].*DuDx;
k = 0.1;
k1 =0.5;
F=k*u(2)-k1*u(1);
F1=-k*u(2)+k1*u(1);
s = [F;F1];
%---------------------------------------------------------------------
function u0 = pdex4ic(x)
u0 =[1;1];
%---------------------------------------------------------------------
function [pl, ql, pr,qr] = pdex4bc(xl,ul,xr,ur,t)
pl = [0;ul(1)];
ql = [1;1];
pr = [ur(1)-1;ur(2)];
qr = [1;1];
Is it correct? if not how should i correct this for different values of 't'?
Torsten
Torsten 2019-6-27
编辑：Torsten 2019-6-27
  1. m = 2 is wrong.
  2. You didn't include the terms -y^2*dy/dx and -y^2*dz/dx in the equations.
  3. Your boundary conditions settings are all wrong.
To include the boundary condition dy/dx + dz/dx = 0 at x=0 for "pdepe", you will have to rewrite your system of equations in terms of y and u:=y+z. This leads to
d^2y/dx^2-y^2*dy/dx+a*(u-y)-c*y-dy/dt=0,
d^2u/dx^2-y^2*du/dx-du/dt = 0
Best wishes
Torsten.
Thank you very much sir. Your anwer was very helpful my work.

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