Two-Point Boundary Value Problem

Hi there, I am currently trying to solve a two point boundary value problem for a system of 2 ordinary linear differential equation.

dx/dt=A(x/t)+By dy/dt=C(x/t^2)+D(y/t)

B.C y at 1=10 and y at 2=0

I have terrible matlab experience and knowledge.

This is my code so far (It doesnt work)

if true
  % %Mechanical Properties of Material
E=200e9; 
nu=0.3;
P=100E6;
%Constants A,B,C,D in the Equations
a11= (1/E);
a12= (-nu/E);
a33= (1/E);
A= (a12)/(a11+a12);
B= ((a33)-((2*a12^2)/(a11+a12)));
C= (a11)/(a11^2-a12^2);
D= (2*a12+a11)/(a11+a12);
%Defining the System of 2ODES
syms x(t) y(t)
t=1;
eqns = [diff(x,t)==A*(x/t)+B*y, diff(y,t)==-C*(x/t^2)-D*y];
cond = [y(0) == 0, y(1.2)==-P];
withSimplifications = dsolve(eqns, cond)
withoutSimplifications = dsolve(eqns, cond, 'IgnoreAnalyticConstraints', false)
[xSol(t), ySol(t)] = dsolve(eqns, cond)
end

Would anyone be able to give me any solutions to this problem?

[Torsten]

We already told you several times that your equations don't permit an analytical solution (so "dsolve" is not the appropriate tool) and that you should use a numerical integrator (bvp4c or bvp5c) instead.

Why don't you follow the advice ?

Best wishes

Torsten.

[Ketav Majumdar]

I have tried BVP4C and 5C it doesnt work. There is a analytical solution for it, This is a cauchy euler equation. you can make x or y the subject of one equation find the derivative and substitute into the second equation. I have the hand calculations. Im not sure why matlab cant solve this

[Torsten]

Then I suggest you try "dsolve" for the resulting Cauchy-Euler equation for x (or y).

I think MATLAB (like me) does not recognize that your system leads to this equation form.

Best wishes

Torsten.

[Torsten]

Maybe

http://www.wolframalpha.com/input/?i=1%2FB*t%5E2*y''(t)%2B(-A%2FB-D%2FB)*t*y'(t)%2B(A%2FB%2BD*A%2FB-C)*y(t)%3D0

helps.

Best wishes

Torsten.

[Bill Greene]

Exactly what problem did you have solving these equations with bvp4c? Can you post your code using bvp4c? (I would have thought it could easily solve this system.) Beyond that, the description of your boundary conditions in the original post don't appear to be consistent with the cond variable you pass to dsolve. Can you clarify precisely the problem you are trying to solve?

[Torsten]

http://www.wolframalpha.com/input/?i=t%5E2*y''(t)%2BU*t*y'(t)%2BV*y(t)%3D0,y(1)%3D10,y(2)%3D0

with

U=2-D-A and V=-D+A*D-B*C

gives the solution for y.

Then

x = 1/C*t^2*dy/dt-D/C*t*y

gives the solution for x.

Best wishes

Torsten.

Answer 1

[Tamir Suliman]

You will have to differneitate then solve Since

   x'=A(x/t)+By  --- differnetiate A(x/t)+By   for y relative to t 
  Y' = A*(x *-1/t^2 +1/t*x' ) + BY'

at t= 1 Y =10 at t =2 y = 0

(1-B)*C(x/t^2) + D (y/t) = A*(x *-1/t^2 +1/t*x' )
sub y =10 then solve  for t =1 again sub y = 0 then solve  for t =2
then use diff and dsvolve with  t = 1 and t =2