I am currently experiencing the following problem:
I am solving a system of nonlinear differential equations with 6 differential equations and 7 boundary conditions (y1=y2=y3=y4=0 at x=x1; y3=1.1,y5=y6=0 at x=x2). Where the boundary condition (y3=1 at x=x2) affects the range of the solution interval and the final numerical results. When I used the bvp4c function that comes with matlab, it was clear that I could only impose 6 boundary conditions (I imposed y1=y2=y3=y4=0 at x=x1; y5=y6=0 at x=x2). Because the solution interval (x1=0, but x2 is unknown) is uncertain, I start with a range given x2, traverse within that range, solve the system of differential equations, and then determine whether y3 is equal to 1 at x=x2; but since the system of differential equations is missing a constraint, there are many cases where: at x=x2, it does satisfy y3=1,y5=0,y6=0, but not the result I want. result. Then I tried (imposing x=x1, y1=y2=y3=0; x=x2, y3=1,y5=y6=0),x2 traversing a range to find the case where x=x1 is exactly satisfied by y4=0, but that case is still not optimal. So I am thinking that the problem should fall under the category of solving edge-valued problems with indeterminate intervals, how do I add more than one of this constraint to the system of differential equations in this case? Is there an optimization algorithm involved? How to deal with it? I hope my friends can give me some advice.
The differential equations are as follows:(
are constants) and the boundry conditions are:(
is an unknown variable) I tried to solve it using the margin problem approach, using matlab's bvp4c function, but ran into the problem I described above.Please help me see if there is a better mathematical way to deal with this problem.My code is as follows and it runs without a problem, just the results are not what I want. For example, inside the solution result, actually y3>0.
clear
clc
tic;
bar = waitbar(0,'读取数据中...');
x2=linspace(8,20,13000);
len = length(x2);
v0=[];
l1=[];
k=0;
for i=1:len
x1=0; delta0=1.1;
xmesh = linspace(x1,x2(i),500);
solinit = bvpinit(xmesh, @guess);
sol = bvp4c(@bvpfcn, @bcfcn, solinit);
str=['计算中...',num2str(100*i/len),'%'];
waitbar(i/len,bar,str)
u0_judge=abs(sol.y(2,1)-0);
while(u0_judge<1)
k=k+1;
v0(k,:)=sol.y(3,:);
x(k,:)=xmesh;
plot(x(k,:),v0(k,:))
hold on
l1(k)=x2(i);
break
end
end
close(bar)
toc;
function res = bcfcn(ya,yb)
res = [ya(1); ya(2); ya(3); yb(3)-1.1; yb(5); yb(6)];
end
function dydx = bvpfcn(x,y,ML,NL)
D11=1142.9; A=1; sigma0=0.1; Iy=1/12;
ML=6; h=1; d0=0.2*h; b=1;
NL=ML*(exp(-h/d0)-1)/(-d0+h/2+(2*d0+h)/2*exp(-h/d0));
dydx = [(-y(5)*sin(y(4))+NL)/D11/A+1;((-y(5)*sin(y(4))+NL)/D11/A+1)*cos(y(4))-1;
((-y(5)*sin(y(4))+NL)/D11/A+1)*sin(y(4));(y(6)+ML)/D11/Iy;
-((-y(5)*sin(y(4))+NL)/D11/A+1)*b*sigma0;((-y(5)*sin(y(4))+NL)/D11/A+1)*y(5)*cos(y(4))];
end
function g = guess(x)
g = [sin(x);cos(x);x;cos(x);cos(x);cos(x)];
end
