How to give a final value to solve an ODE's, instead of the standard initial values?

I'm solving a first order differential equation with variable constants and a mass matrix. Therefore, I use ODE113. As shown in the code below, I call 2 functions for defining the ODE and the mass matrix. In order to give in the final value as 'initial condition', in the first call of ode113, the independent variable x is transformed to x_end - x. For comparison, in the second call of ode113, the original x is used with the initial value given from the previous solution. The functions called in the second time are identical as in the first time, except that now x is immediately defined by the function input, and has no transformation. Both solutions were plotted but do not coincide. Why is this, is this the correct way of transforming the pre-programmed initial value programs to final value problems?

Thanks in advance! regards, Kenny

% Limit of the differential eq.:
x_end=2.5;
% 
% Parameters:
x0=[0.32;0.15;45]; 
% 
% Final condition is known:
options =odeset('Mass',@(xb) Mass_fromback(x0(1),x0(2),xb,x_end),'MStateDependence','none','RelTol',1e-3);
yfb=ode113(@(xb,y)diffEqn_fromback(x0(1),x0(2),x0(3),xb,y,x_end),[0 x_end], 0,options);
% 
% Initial condition from previous solution:
y0=deval(yfb,x_end);
% Solve by giving in initial condition:
options =odeset('Mass',@(xb) Mass(x0(1),x0(2),xb),'MStateDependence','none','RelTol',1e-3);
y=ode113(@(xb,y)diffEqn(x0(1),x0(2),x0(3),xb,y),[0 x_end], y0,options);

with functions:

function M=Mass_fromback(E,nu,xb,x_end)
% Define variable from back to the front, as such, the initial condition
% will be the deformation at the back of the chamber.
x=x_end-xb;
% 
hx=1.2-x*tan(3*pi/180);
bx=0.9-x*(2*tan(3*pi/180));
A=-E.*(1-nu);
% 
M=A.*hx.*bx;

and

function Mdydx=diffEqn_fromback(E,nu,mu,xb,y,x_end)
% 
% Define variable from back to the front:
x=x_end-xb;
% 
B=E*nu*(mu+(mu+tan(3*pi/180))./(1-mu*3*pi/180));
D= E*(mu +(mu+8*pi/180)/(1-mu*3*pi/180));
% 
Mdydx= B.*x.*y +(1-nu).*D.*x.*log(x+0.1);

For comparison, the two solutions are plotted but they do not coincide:

figure;
plot(x_end-yfb.x,yfb.y,y.x,y.y)
xlabel('x'),ylabel('y'), legend('solved with xb = x\_end - x','solved with x')

[SATISH luke]

what if one variable known at the start and other at the end of time

Answer 1

[Jan]

When you transform x -> c - x, the derivative d/dx needs a change in the sign also.

The easiest way to integrate from the end to the initial time is to revert tspan:

tspan = [x_end, 0];

Then Matlab's ODE integrators handle the sign properly automagically.

[Kenny]

Thank you!

Answer 2

[Juan Camilo Medina]

There are different methods to solve a boundary value problem, which is effectively what you have. The easiest one is the shooting method:

http://en.wikipedia.org/wiki/Shooting_method

Answer 3

[Sana Ben Hmouda]

Hello, i have the same problem but i can't get the suggested solutions. Would any one provide me plz with a solution to my problem? In fact i'm solving dxdt=(A+BK)x by means of ode45, the solution x(t) converges to zero, now i would like to set the final value of the solution according to a desired value of xd, someone suggected that i replace dxdt=(A+Bk)x+xd, but the result still converges to zero. Thanks