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Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorzskoda
• CommentTimeJul 22nd 2011
• (edited Jul 22nd 2011)

I want to solve the nonhomogeneous system of $n^2$ ODEs with constant coefficients for $n^2$ functions $P_{i j} = P_{i j}(\mu)$:

$\frac{d P_{i j}}{d\mu} - t \sum_{s=1}^n k_{i s}P_{j s} = k_{i j}$

where $i,j = 1,\ldots,n$ for arbitrary $n$ where $t$ is a parameter and the initial condition is

$P_{i j}(\mu = 0)= q_{i j}$

In matrix form one writes $d P/d\mu = k - t k P^T$, and $P(0) = q$, where $P^T$ is the transpose of $P$ and $k P^T$ assumes the matrix product.

For small $n$ one can try to solve by brute force (solving the homogeneous first and then, say, variation of constants method), but I am not sure I can get a formula for general $n$. Help from more skillful ones appreciated. Of course, if the parameter $t = 0$ the solution is $P_{i j} = \mu k_{i j} + q_{i j}$ but I want for nonzero $t$.

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeJul 22nd 2011
• (edited Jul 22nd 2011)

No! I am wrong. The differential equation for my purpose should be (if I am not mistaken again) without the transpose (cf.coproduct+for+Ugln+dual (zoranskoda)), i.e.

$\frac{d P_{i j}}{d\mu} - t \sum_{s=1}^n k_{i s}P_{s j} = k_{i j}$ $P_{i j}(\mu = 0)= q_{i j}$

and in a matrix form $d P/d\mu = k - t k P$, $P(0)=q$.

Thus we can use, for $t\neq 0$, the matrix substitution $R = t P-I$ to get simply the homogeneous matrix equation $d R/d\mu = -t k R$ (here $t$ is a scalar, $k$ and $R$ are $n\times n$-matrices and $k R$ denotes the matrix product) with the initial condition $R(0) = t q -I$ i.e. $R(0)_{i j} = t q_{i j} - \delta_{i j}$. This is now much easier than the one with the transpose I had before.

To relate it to known examples, one sees that the equation in this case, without the transpose, do not mix different $j$-s, so one has $n$ independent systems, one for each $j$ (what is not the case in 1 where we had a transpose).

So for a fixed $j$, call (for emphasis) $R_{i j} =: \tilde{R}_i$ and we have simply

$d \tilde{R}_i/d\mu = -t \sum_s k_{i s}\tilde{R}_s$

what is in a rather standard form for homogeneous systems of first order ODEs with standard exponential matrix solution.

Thus the formal solution is

$R = e^{-\mu t k} R(0)$

i.e. $R_{i j} = \sum_s exp(-\mu t k)_{i s} R(0)_{s j}$. where $exp(-\mu t k)$ is the matrix exponential.

Now this, with $R(0)_{s j} = t q_{s j}-\delta_{s j}$ gives, in matrix form,

$t P - I = e^{-\mu t k} (t q- I)$

hence

$P(\mu) = e^{-\mu t k} q + \frac{I - e^{-\mu t k}}{t}$

Note that $q$ and $k$ are constant matrices and $t$,$\mu$ are scalars. For $\mu = 0$ we have $P(0) = q$ and for small $t$ we have $P = \mu k + q + (\mu^2 k^2/2 - \mu k q) t + O(t^2)$, agreeing with (i.e. with continuous limit to) $t = 0$ result $P = \mu k + q$.

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeJul 22nd 2011

Now one needs (for my original purpose) also to compute $P(K^{-1}(k),q)$ where $K^{-1}$ is inverse of the function $k\mapsto P(k,0)$, and $P(k,q)$ denotes the function $P(\mu = 1)$ calculated above with $k$ and $q$ as above. In our case, we need to invert

$k \mapsto \frac{I - e^{-t k}}{t}$

what gives $k \mapsto -\frac{1}{t} ln(I-t k)$ where we symbolically use the matrix logarithm. Thus we obtain

$P(K^{-1}(k),q) = k + q - t k q$

what gives a simple quadratic coproduct.