How Do You Find a Matrix A Such That Y(t) Spirals Toward a Plane?

[djinteractive]

I am having trouble getting started with this problem.. I guess I'll explain it first.

Find one possible matrix A for which the solution to Y'=AY with initial condition Y(0)=(6,13,9) has the following property. As time progresses, the solution Y(t) spirals toward the plane 2x+3y+4z=0 where it continues to circultate about a radius five circle.

I know I have to find some vectors Vr(real) and Vi(imaginary) that span the plane and are orthogonal to it. The eigen vectors if I remember correctly should come in a pair Vr + iVi and Vr-iVi that I can normalize with PDP-1(inverse) but I am having a problem with how to get these eigenvalues that span the plane.. will someone PLEASE (I'm begging you) help me.. this is driving me insane
PS the 3rd (in z direction should just be a real lambda value I think)

 

[kleinwolf]

A way to attack the problem could be :

Take a vector in the plane : a=(2,0,-1)/Sqrt(5), the normal vector to the plane : n=(2,3,4) and a vector in the plane, orthogonal to the first : b=(3,-10,6)/Sqrt(145).

Then you notice that you're LUCKY : you can easily compute the distance of Y(0) to the direction given by n : it is 5.

Hence the curve is not a 3d spiral but just a screw type one !


Hence you know that in the (a,b,n) basis, your vector expresses as :

x(t)=(5*cos(wt),5*sin(wt),A*exp(-Dt))

hence in the (a,b,n) basis : $$x'(t)=\left(\begin{array}{ccc} 0& -w & 0\\ w & 0 & 0\\0&0&-D\end{array}\right)x(t)$$

then you just have to change the basis to the canonical one and fit the initial condition.

 

[M98Ranger]

First of all, don't panic! Solving this problem may seem daunting at first, but with a clear understanding of the concepts involved, you will be able to find a solution.

To start off, let's review the given information. We have a system of differential equations, Y' = AY, with an initial condition Y(0) = (6,13,9). We are looking for a matrix A that will cause the solution Y(t) to spiral towards the plane 2x+3y+4z=0 and continue to circulate about a radius five circle.

To achieve this, we need to find a matrix A that has eigenvalues that lie on the plane 2x+3y+4z=0. This means that the characteristic polynomial of A should have roots that satisfy this equation. In other words, the eigenvalues of A should be of the form λ = a+bi, where a and b are real numbers, and a+bi satisfies the equation 2x+3y+4z=0.

Now, let's consider the eigenvectors of A. Since A has complex eigenvalues, its eigenvectors will also be complex. However, we need real and imaginary eigenvectors that span the plane 2x+3y+4z=0 and are orthogonal to it. This means that we need to find two complex eigenvectors, Vr+iVi and Vr-iVi, where Vr and Vi are real vectors, that satisfy the following conditions:

1. They lie on the plane 2x+3y+4z=0.
2. They are orthogonal to the plane, i.e. their dot product with any vector on the plane is zero.

To find these eigenvectors, we can use the following steps:

1. Choose a random vector on the plane 2x+3y+4z=0, for example, (1,-2,1). This vector will be orthogonal to the plane.
2. Find the cross product of this vector with the normal vector of the plane, which is (2,3,4). This will give us a vector that lies on the plane and is orthogonal to the vector chosen in the previous step.
3. Normalize this vector to get Vr.
4. Set Vi as the cross product of Vr with the normal vector of the plane.

Now that we have our eigenvectors, we can construct