Solve $(A- \operatorname{diag}(x) ) \,\nabla_x f(x) - c f(x)=0, \, f(0)=1$

Question

Let $f : \mathbb{R}^n \to \mathbb{R}$. How to solve the following differential equation

$$ (A- \operatorname{diag}(x) ) \nabla_x f(x) - c f(x)=0, \qquad f(0)=1. $$

where $\operatorname{diag}(x)$ is a diagonal matrix with vector $x$ on the main diagonal, $A$ is some $n \times n$ matrix and $ c \in \mathbb{R}^n$?

In the scalar case, this is easy to solve since it is a first-order linear ordinary differential equation (ODE) whose solution is given by

$$f(x) = a^{c} (a - x)^{-c}.$$

I don't have have much experience solving matrix differential equations and would appreciate some references on this topic.

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Edit: Using the approach of NN2 the above PDE can be reformulated as: \begin{align} (A- \operatorname{diag}(x) ) \nabla_x g(x)=c, \qquad g(0)=0. \end{align}

Answer 1

My first answer isn't good. I edit the answer by taking account the comment of @loupblanc. I haven't had yet the solution. I think we can simplify the problem by using this transformation.

From the equation, we have $$ (A- \operatorname{diag}(x) ) \frac{\nabla_x f(x)}{f(x)} - c=0 $$ $$ \iff (A- \operatorname{diag}(x) ) \nabla_x \ln{f(x)} - c=0 $$ Put $g(x) = \ln{f(x)}$ $$ \iff (A- \operatorname{diag}(x) ) \nabla_x g(x) - c=0 $$

Answer 2

We consider the ODE in the form obtained by NN2:

$(*)$ $\nabla_xg(x)=(A-diag(x))^{-1}c=[v_1,\cdots v_n]^T$.

EDIT 1.

In general $(*)$ has no solutions except if $curl(V)=0$, that is if, for every $i<j$,

$\dfrac{\partial v_i}{\partial x_j}=\dfrac{\partial v_j}{\partial x_i}$. This NS condition reflects the fact that $\dfrac{\partial^2 g}{\partial x_jx_i}=\dfrac{\partial^2 g}{\partial x_ix_j}$.

For example, for $n=2$, the conditions on $A=[a_{i,j}],c=[c_i]$ are

$a_{1,2}c_2=a_{2,1}c_1=0.$

For $n=3$, the conditions are

$a_{2,1}c_1=a_{3,1}c_1=a_{1,2}c_2=a_{3,2}c_2=a_{1,3}c_3=a_{2,3}c_3=0$.

In particular, if, for every $i$, $c_i\not= 0$, then $A$ is diagonal (at least when $n=2,3$) and the integration of the ODE is easy.

EDIT 2. To the OP. Since I have pity on you, I'll give you a basic example that may convince you that there is no solution except for exceptional choices of $A$ and $c$.

We choose $A=\begin{pmatrix}0&1\\1&0\end{pmatrix},c=[1,1]^T$.

The considered equation is $\Delta_x g(x)=(A-diag(x))^{-1}c$, that is

$[\dfrac{\partial g}{\partial x_1},\dfrac{\partial g}{\partial x_2}]^T=[\dfrac{-x_2-1}{x_1x_2-1},\dfrac{-x_1-1}{x_1x_2-1}]^T$. Then necessarily

$\dfrac{\partial^2 g}{\partial x_1x_2}=\dfrac{1+x_1}{(x_1x_2-1)^2}$ and $\dfrac{\partial^2 g}{\partial x_2x_1}=\dfrac{1+x_2}{(x_1x_2-1)^2}$.

The last two quantities are not equal; therefore, there are no solutions $g$ that are $C^2$.