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I'm interested in the way of solving the following recurrence relation:

$$a_{n+1}=a_n'+a_1 a_n-b_1 b_n \\ b_{n+1}=b_n'+b_1 a_n+a_1 b_n$$

Where all $a_n,b_n$ are functions of $x$, and $'$ means derivative w.r.t. $x$. The initial functions $a_1$ and $b_1$ are known. Also, assume we have closed forms the derivatives: $a_1^{(n)}$ and $b_1^{(n)}$ in terms of $n$.

I expect there to be a closed form solution for this case, however, I have trouble deriving it.

One idea I had was to use operator notation and matrix notation, then we have:

$$\begin{bmatrix} a_{n+1} \\ b_{n+1} \end{bmatrix}= \begin{bmatrix} D+a_1 & -b_1 \\ D+b_1 & a_1 \end{bmatrix} \begin{bmatrix} a_n \\ b_n \end{bmatrix}$$

Which obviously can be written as:

$$\begin{bmatrix} a_{n+1} \\ b_{n+1} \end{bmatrix}= \begin{bmatrix} D+a_1 & -b_1 \\ D+b_1 & a_1 \end{bmatrix}^n \begin{bmatrix} a_1 \\ b_1 \end{bmatrix}$$

However, the $n$th power of a matrix is found through its eigenvalues, and I'm not sure how to find eigenvalues for the operator matrix.

We can also rewrite this in a more clear way:

$$\begin{bmatrix} a_{n+1} \\ b_{n+1} \end{bmatrix}= \left(\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \frac{d}{dx} +\begin{bmatrix} a_1 & -b_1 \\ b_1 & a_1 \end{bmatrix} \right)^n \begin{bmatrix} a_1 \\ b_1 \end{bmatrix}$$

And use binomial sum? Not sure. How to correctly write binomials sum with a differential operator inside?

Or maybe there's a more simple way to obtain the closed form?

I would prefer to use strictly real methods if possible, thought I kind of doubt it could work without complex numbers.

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By letting $z_n = a_n + i b_n$ the given equations can be coupled as

$$z_{n+1} = z_n' + z_1 z_n = (D+z_1)z_n = (D+z_1)^n z_1$$

$$z_2=z_1'+z_1^2$$ $$z_3=z_1''+3 z_1 z_1'+z_1^3$$ $$z_4=z_1'''+3 z_1'^2 + 4 z_1 z_1''+6 z_1' z_1^2 + z_1^4$$ and the general form for $z_n$ is given by Faà di Bruno's formula.

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    $\begingroup$ I should have mentioned that I don't want to use the complex numbers. But considering the matrix I have got, it's all just complex numbers in disguise, so I don't believe any real methods are possible here $\endgroup$ – Yuriy S Nov 7 '18 at 7:57

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