Example 3

This example is derived from example 2. We notice that in the three equations of example 2 the second degree terms of $x,y$ are for all functions $x^2+y^2$. Thus by subtracting the first function to the second and third we get a linear system in $x,y$. This system is solved and the value of $x,y$ are substituted in the first function. We get thus a system of one equation in the unknown $\theta$ (see section 15.1.2). The roots of this equation are 0,-0.8067834380. The test program is Test_Solve_Gradient_General2. The IntervalGradient function is written as:

 
INTERVAL_MATRIX IntervalGradient (int l1,int l2,INTERVAL_VECTOR & in)
{
INTERVAL_MATRIX Grad(1,1);
Grad(1,1)=-3788.0+(2692.0+(7576.0-4800.0*Cos(in(1)))*Cos(in(1)))*Cos(in(1))+(
25912.0+(-39320.0+14520.0*Cos(in(1)))*Cos(in(1)))*Sin(in(1));
return Grad;
}

With epsilonf=0 and epsilon=0.001 we get the solution by using 8 boxes:

$$\theta=[-0.0005316396,0]~~~~~ \theta=[-0.8071015,-0.8063247]$$

for whatever order. The solution intervals contain the roots of the equation. If we use epsilon=0 and epsilonf=0.1 we get by using 8 boxes:

$$\theta=[-0.8067839277]~~~~ \theta=[-5.4136579e-16]$$

Here we get a unique solution and a range solution. But we notice that the solution 0 is exactly the middle point of the test domain: Moore test will fail as 0 will always be an end-point of the range. If we break the symmetry of the test domain we will get exactly both solutions.