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)
]
Antworten (1)
Torsten
am 31 Jan. 2023
0 Stimmen
Read this carefully, and you will know how to set up your problem:
1 Kommentar
Komal Goyal
am 1 Feb. 2023
Kategorien
Mehr zu Boundary Value Problems finden Sie in Hilfe-Center und File Exchange
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