nonlinear differential equation system

dx1／dt＝x1－x3*x1^2/(1+x1)；
dx2／dt＝x2－x3*x2^2/(1+x2)；
x1+x2=2；

x1(0)=0.5;x2(0)=1.5;
x3(0)=1

three equations with three unknowns，how to solve it by matlab

[Roger Stafford]

Lujia, your three initial conditions are incompatible with your three equations. Add the first two equations, and making use of the third equation which requires that d(x1+x2)/dt = 0, this will give:

0 = 2 - x3(0)*( x1(0)^2/(1+x1(0)) + x2(0)^2/(1+x2(0)) )
  = 2 - 1*(0.5^2/1.5+1.5^2/2.5) = 2 - 1*(1/6+9/10) = 14/15

which is obviously false. Your initial value for x3 would have to be

x3(0) = 1.875

to be compatible.

[LUJIA YAO]

yes，you are right，thanks

Answer 1

[Roger Stafford]

Your three equations can be reduced to one equation. By using the second two equations it can be deduced that

   x2 = 2 - x1
   x3 = (1+x1)*(3-x1)/2

and from this a single equation in x1 can be obtained:

   dx1/dt = x1-3/2*x1^2+1/2*x1^3

From this by partial fractions it follows that:

   log(abs(x1*(x1-2)/(x1-1)^2)) = t-t0

for an arbitrary constant t0, and from this

   x1*(x1-2)/(x1-1)^2 = ±exp(t-t0)

This is a quadratic equation that can be solved for x1 and thereby obtain explicit expressions for x1(t) in terms of exp(t-t0). For that reason you would not have to use the numerical ‘ode’ functions.

[Roger Stafford]

Added note: For your particular initial condition, x1(0) = 0.5 and x2(0) = 1.5, and assuming you correct x3(0) to 1.875, you would have as a solution

   x1(t) = 1 - (1 + exp(t-t0))^(-1/2)
   x2(t) = 1 + (1 + exp(t-t0))^(-1/2)
   x3(t) = (1+x1(t))*(1+x2(t))/2 = 2 - 1/2*(1 + exp(t-t0))^(-1)

where t0 = -log(3).