have been wrestling with this n-d coupled PDE for awhile and finding it very difficult to come up with an answer, but am wondering if there is some elusive "trick". have looked through various books with standard forms and havent seen it yet. the context is something of a long story but the eqn is simple to state. has anyone encountered it before? does anyone see a solution? looking for n-d solution in general but even a 2d solution would be helpful. (guess it is technically a linear PDE?)

$\mathbf{J(x) Ax = f(x)}$

$\mathbf{J(x)}$ is the Jacobian (of $\mathbf f(x)$), $\mathbf A$ is a given constant matrix, $\mathbf x$ a column vector; looking to solve for $\mathbf {f(x)}$...? presumably there is some analogy with or generalization of the 1d case which is not hard: $f(x)=x^b, b = \frac{1}{1-a}$. (can discuss further here for anyone interested.)

  • $\begingroup$ update, this appears to be exactly the same eqn (with different var/ fn names) descr on wikipedia under matrix differential eqn. wikipedia does not refer to the jacobian on that pg, but in their eqn $\dot{x}(t) = Ax(t)$ it looks like A is the jacobian. $\endgroup$ – vzn Nov 24 '17 at 17:57

Hint it seems to me that if $x(t)$ is a solution of the ODE $$\frac{ d x(t)}{dt} = A x(t)$$ that is to say $x(t) = e^{t A} x(0)$ then $$\frac{d}{dt}(f(x(t))) = J(x(t)) A x(t) = f(x(t))$$ so that $$f(x(t)) = e^t f(x(0))$$

I don't say it solves the problem but it may be a starting point.

  • $\begingroup$ thx for response but dont forget $\bf A$ is a matrix; have not encountered the concept $e^\bf{A}$, (ie "constant raised to matrix power") do you have some defn or ref? $\endgroup$ – vzn Oct 23 '17 at 17:23
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    $\begingroup$ Sure, you can start with matrix exponential. You can also learn a lot about the method of characteristics for hyperbolic equations. $\endgroup$ – Gribouillis Oct 23 '17 at 17:34

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