bvp4c for coupled interface boundary conditions
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Here I am trying to solve a system of ODE with coupled boundary conditions, but got stucked. I dont know how to define coupled boundary condition in Matlab. Can anybody please help me. If any other information is required please let me know. Moreover, one PDF file is also attached there with name bvp4c with the complete details of the ODE and respective boundary conditions.
function pp(solver)
infinity =1;
maxinfinity=infinity;
pp1=infinity;
pp2=maxinfinity;
%CONSTANTS
%--------------------------------------------------------------------------
p=0.01;
e=0.783;
H=1;
M=1;
a=1.92;
%--------------------------------------------------------------------------
solver = 'bvp4c';
bvpsolver = fcnchk(solver);
infinity = pp1;
maxinfinity = pp2;
Mva=[0.5 1 1.5 2];
colors = ['k','m','r','b'];
for j = 1:4
Le=Mva(j);
for d=Mva(j)
solinit = bvpinit(linspace(0,infinity,45),[1 1 1 0 1 0 1 0 0 1 1 1 1 1 1 1]);
sol = bvpsolver(@fsode,@fsbc,solinit);
eta = sol.x;
f = sol.y;
solinit = bvpinit(linspace(0,0.5,100),[0.001 0 0 0]);
sol = bvp4c(@pp2_ode,@pp2_bc,solinit,options);
x = sol.x;
y1 = sol.y(1,:);
y2 = sol.y(2,:);
y3 = sol.y(3,:);
y4 = sol.y(4,:);
figure(5)
plot(real(y(1,:)+p*exp(1i*t)*y(3,:)),eta,colors(j),'LineWidth',2)
plot(real(y(5,:)+p*exp(1i*t)*y(7,:)),eta,colors(j),'LineWidth',2)
hold on
%--------------------------------------------------------------------------
function dydx = pp2_ode(x,y)
dydx = [y(2)
(((M^2)*y(1))-((H)^2))
y(4)
(((M^2)+(i*((H)^2)))*y(3))-((H)^2)
y(6)
-a*((H)^2)
y(8)
((i*a*(H)^2)*y(7))/e-(a*((H)^2))];
end
%--------------------------------------------------------------------------
function res = pp2_bc(ya,yb)
res = [ ya(1)
]
Answers (1)
Torsten
on 31 Jan 2023
0 votes
Read this carefully, and you will know how to set up your problem:
1 Comment
Komal Goyal
on 1 Feb 2023
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