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How does one solve a "differential equation" for $\sigma$ of the form $$ \sigma(v)w_i(v)={\partial \over \partial v_j}\left[\sigma(v)A_{ij}(v)\right] \quad i=1,\dots,n. $$ where the summation convention applies.

$w,v$ are an $n$-D vectors, $\sigma$ is a scalar function, $A$ is an invertible $n\times n$ matrix?

Perhaps there is a general solution form? References (links) for the treatment of such an equation is also appreciated.

Thank you.

Added:

In light of drak's suggestion, here is a bit more

Some thoughts:

It might be friendlier to change "variables" to $A\sigma$?

Is there a more familiar expression for the index notation ${\partial \over \partial v_j}M_{ij}(v)$ such as one in terms of $\nabla$? It would seem to me that it is taking the divergence of each row of the matrix $M$.

Some more thoughts: since the function $\sigma$ appears on both sides of the equation, it is likely that it is an exponential.

A simplified version: What if we suppose that $A$ is a constant matrix?

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    Since you are a new user, here are a few things about the site you should know: 1. To get the best possible answers, it is helpful if you say where the problem originated, 2. You will get a better response, if you indicate, what you have already tried to answer the question yourself. and finally: Welcome to math.SE!2012-07-03
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    @draks : Thank you, I have added something to my post.2012-07-03
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    Consider the ODE $\frac{d}{dt} p(t)u(t) = q(t)u(t)$ ($u$ is the unknown). How would you solve it? Can you always "change variables" like $v(t)=p(t)u(t)$?2012-07-03
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    @Siminore: hmm, perhaps not... but then what can I do?2012-07-03
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    @Siminore: what if we start with $A$ is a constant matrix?2012-07-03
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    If $A$ is constant, and since $\sigma$ is a real-valued function, the equation should be easier.2012-07-03
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    @Siminore: thank you, is there a general form of solution for such an equation? I suspect it is some exponential function, but i am new to matrix/vector differential equations so i don;t know how it should look like. could you please help me out?2012-07-03
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    Formally it is like the ODE case. But you need to learn the exponentiation of a matrix.2012-07-03
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    If this can help, your equation is equivalent to $$\nabla\sigma(v)=\sigma(v)A^{-1}[w(v)-B(v)]$$ where $B(v):=(\text{div}A_1(v),\ldots,\text{div}A_n(v))$ with $A_i(v)$ the $i$-th row from $A$.2012-07-06

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