how is it possible to give res-function in bvp4c only values of the beginning of the interval ?

Hi

I'm trying to use a boundary value problem in a loop to solve the problem in a piecewise manner in some intervals as [ya yb]. to best of my knowledge bvp should be given value of ya for y1 and yb for y2 or vice versa.

However this is is not suit my problem, as it can be seen in the figure.

In each iteration I set ya(2) as the last value of pervious iteration (y2(end) or yy20)). To get resonable results I should set ya(1) as the last value of pervious iteration (y1(end) or yy10)), but this give me compeletly wrong answer.

how can I set simultaneously ya(1) = yy10 and ya(2) = yy20?

Thank you so, so much in advance! I would deeply appreciate the help.

clear all

global a1 yy10 yy20 b1 tauf

tf = 150; dt = 1; a1 = 0.08;b1 = 1;tauf = 8e4;tn = 150;tm = 15;

yy20 = 0; yy10 =0;

for j = tm:tm:tn

dl = (tm)/dt;

Tl = linspace(j-tm,j,dl); %each step samples

solinit = bvpinit(Tl, [1 1]);

options = bvpset('Stats','on','RelTol',5e-1);%,'NMax',1000

%%--------------------

% ODE's for states and costates

sol1 = bvp4c(@BVP_ode, @BVP_bc, solinit, options);

t = sol1.x ;

y = sol1.y;

y1 = -y(1,:);

y2 = -y(2,:);

yy10 = y(1,end);

yy20 = y(2,end);

pb = ((y2) / b1);

pd =Dl(t)';

figure (101);

subplot(2,1,1);plot(t,y1);hold on;ylabel('y1')

subplot(2,1,2);plot(t,pb);hold on;ylabel('y2')

end

%% -------------------------

function dydt = BVP_ode(t,y)

global a1 b1 yy10 tauf ;

ut1 = -(y(2,:)) / (b1);

dydt = [((-y(1,:)-yy10)/tauf) + ((ut1 - Dl(t)))

-((y(2,:)-0)/tauf) - (a1*(y(1,:) - yy10)) ];

end

% -----------------------------------------------

% The boundary conditions:

function res = BVP_bc(ya,yb)

global yy20 yy10

res = [ yb(1,:)- yy10

ya(2,:)- yy20];

end

%-----------------------------------------------

function d = Dl(t)

d = DFTP75(t)'/1000;

end

function d4 = DFTP75(t)
dddd = load ( 'FTP751.mat');
ddd  = dddd.FTP75(:,2);
dd = 1.3*ddd';
tt3 = 0:1:2475-1;
d4 = interp1(tt3,dd,t);
end

[Torsten]

I don't see the reason why you solve the problem in a loop and why you use bvp4c instead of ode45.

[abs mhr]

Thanks a lot for your reply

The reason I solve bvp4c in a loop is that I'm trying to compare solving the problem on an interval with solving it repeatedly on the interval splitted in several sub-intervals.

I tried my problem with ode45 as you suggest but when I tried to input y0(1) = yy10 and y0(2) = yy20 I get a same result as bvp4c with the same res-function as follows:

It seems that I should somehow define res-function so that it accept res-fun on just beginning of each sub-interval (ya(1) and ya(2)).

Do you have any suggestion for me?

thank you

[Torsten]

Could you state the original problem over the complete interval to which you want to compare the solution obtained with your code from above over the subintervals (equations and initial/boundary conditions) ?

[abs mhr]

The original problem is exactly the same problem when tm =150 (instead of 15). That means only one iteration loop and solving bvp4c once on the compelete interval In this case boundaries are : ya(2,:) =0 yb(1,:) = ya(1,:)

[Torsten]

yb(1,:) = ya(1,:)

This is not a valid boundary condition for bvp4c.

[abs mhr]

Yes you are right Sorry I made a mistake yb(1,:)=0 ya(2,:)=0

[Torsten]

If this is the problem, you should solve it over the complete interval. A step-by-step solution over subintervals is possible, but you must adopt your boundary condition ya(1) depending on what you get for yb(1) in the step-by-step solution and recalculate. In principle, this would be the shooting approach to solve boundary value problems.

[abs mhr]

Thanks for your time and suggestion I will try it Best regards