Mega thread on coupled ODE

[Yekaterina]

İt's easy to solve coupled systems of differential equations in the form 

d/dt (X,y,z) = (X, y, z)*A where A is a 3x3 matrix with constants as elements. 

Now, what if A is not just constants but functions of t? Below is my investigation into this: 

 

The idea is as follows: 

1. Write the dependent variables into a vector, do the same for the derivatives as well as their non constant coefficients, so that we can apply linear algebra.

2. Map the solution vector X into another vector space U in which the matrix relating the vector and it's derivatives becomes easily diagonalisable. Achieve this by a linear map L acting on the original solution vector. X = L U

3. Now we need to find L and decouple the relationship. 

4. Solve the decoupled equations row by row

5. Map back into X space from U space to obtain the desired solutions.

 

[Yekaterina]

We need to find our L matrix and it must satisfy L°AL-L°L' = a diagonal matrix. 

 

L° represents the inverse of L and L' represents derivative of L with respect to time.

This is the difficult part; for L with constant coefficients, the second term will be 0 which would lead to just the diagonalised A, but if A is not constant then there is a second term containing a derivative of L and inverse L. 

Normally L will be the matrix of eigenvalues of A, in our case and my exampled I forced some eigenvalues and eigenvectors out of A which do contain t. İn the particular exame I used, I was able to satisfy this equation with just the eigenvalues matrix, but I am not sure if the result will still be diagonal if my A was different. I would like some help here, for example proving whether the conditions are always satisfiable. İt would also be great if you can find some alternative methods to calculate L.

The example worked perfectly but it was a very simple matrix.

[Yekaterina]

8 hours ago, m7600 said:

Have you ever taken a crack at the Riemann hypothesis? You'll get a million dollars plus bragging rights if you solve it.

No, because I find it not very useful at this stage. I don't care about million dollars or bragging 

Edited August 20, 2022 by Yekaterina

[Yekaterina]

I have tried more examples but square roots will appear in eigenvalues and that will often result in impossible Integrals.

[Yekaterina]

One may manually find the elements of L by solving 6a and 6b but I find it tedious.