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I've been recently trying to to solve the following system of linear 1st order PDE's:

$f:\Omega^d\xrightarrow{}\mathbb{R},\quad A^{(k)}\in\mathbb{R}^{N\times N},\quad B^{(k)}\in\mathbb{R}^N$

$\dfrac{\partial{y(\vec{x})}}{\partial{x_k}} = A^{(k)} y(\vec{x}) + B^{(k)} f(\vec{x}) \quad y(\vec{x}_0)=\vec{0} \quad k\in\{1,...,d\}$

Where $N>1$, $\Omega$ is a compact simply connected subset of $\mathbb{R}^d$. This seems quite standard but haven't managed to find conditions for uniqueness & existence of the solution and/or an explicit form which is it what I'm looking or eventually.

Is there a common book/reference or an explicit solution for these type of PDEs?

Thanks you for your help!

Note: I came across 1 which states a condition for uniqueness & existence in the case of locally integrable coefficients - that the matrices $A^{(k)}$ commute, in that case the homogeneous solution is easy to achieve but it seems that condition is too restricting in my case.

1 - ON SYSTEMS OF FIRST ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH $L^p$ COEFFICIENTS - S. MARDARE

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    $\begingroup$ You absolutely need some integrability conditions. Given a solution, taking second partial derivatives you find $\partial_j\partial_k y = \partial_k\partial_j y$, which requires $$ [A^{(k)}, A^{(j)}] y + A^{(k)}B^{(j)} f - A^{(j)}B^{(k)} f + B^{(k)}\partial_j f - B^{(j)}\partial_k f = 0 $$ Observe that these equations are algebraic for $y$. This puts strong restrictions on what solutions may look like. $\endgroup$ May 19, 2022 at 17:07
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    $\begingroup$ In general equations of the type you consider may well be over-determined. You certainly cannot count on there being simple formulae for explicit solutions in the general case. For more about existence and uniqueness, you may want to look up link.springer.com/book/10.1007/978-1-4613-9714-4 (Exterior Differential Systems by Bryant et al.) $\endgroup$ May 19, 2022 at 17:11
  • $\begingroup$ What is $h$ : should that be $y$? $\endgroup$
    – username
    May 19, 2022 at 18:02
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    $\begingroup$ Once the integrability conditions are taken into account, as explained by Willie, you can apply the Frobenius theorem. You may need to convert the equations to vector fields on the total space (composed of both $x$ and $y$ variables). $\endgroup$ May 20, 2022 at 17:57
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    $\begingroup$ @IgorKhavkine is correct. My answer below is in fact a proof of the Frobenius theorem for this special case. $\endgroup$
    – Deane Yang
    May 20, 2022 at 21:06

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ADDDED: Since equation (1) below is a first order linear ODE,it has an explicit solution (with an integral). It follows that the solution to the entire system of equations can be written explicitly as an iterated integral.

Following up on @WillieWong's comments: If $$ [A^{(k)},A^{(j)}]=0,\ A^{(k)}B^{(j)}f - A^{(j)}B^{(k)}f= 0,\ B^{(k)}\partial_jf - B^{(j)}\partial_kf = 0,$$ then the system is said to be in involution. In that case, given any value for $y(x_0)$, there is a unique solution to the system. This is not hard to prove using the following inductive step:

For simplicity, assume $x_0 = 0$. For each $k \ge 0$, assume you have a solution $y$ along the set $x^{k+1}=\cdots x^{d}=0$. Then for each $(x^1, \dots, x^k) \in \mathbb{R}^k$, the equation \begin{equation}\tag{1} \dfrac{\partial{y(\vec{x})}}{\partial{x_{k+1}}} = A^{(k+1)} y(\vec{x}) + B^{(k+1)} f(\vec{x}) \end{equation} is a constant coefficient linear ODE with respect to $x^{k+1}$ along the curve obtained to restricting to the line where $(x^1, \dots, x^k)$ are held constant and $x^{k+2}=\cdots=x^d=0$. Given $y=0$ when $x^{k+1}=0$, there is a unique solution. If you do this for each $(x^1,\dots,x^k) \in \mathbb{R}^k$, then you get a function $y$ on the set $x^{k+2}=\cdots=x^d=0$. By induction, you get a function on all of $\mathbb{R}^d$.

This is not obviously a solution since each equation is solved only along a subset. But what you do now is the consider the equation, for each $j > k$, $$ \partial_j\left(\partial_ky - A^{(k)}y - B^{(k)}f\right)=0 $$ By the construction of $y$ above, this equation is known to hold when $i$. You now use the assumptions on $A^{(k)}$, $B^{(k)}$, $f$ to show that this system is equivalent to $$ \partial_k\left(\partial_jy - A^{(j)}y - B^{(j)}f\right)=0. $$ An inductive argument now shows that the entire system holds on $\mathbb{R}^n$.

If the system is not in involution, I'm not sure what happens, but I suspect that no solution exists. This theory is usually presented using Cartan's formulation using exterior differential systems or the more modern approach using Spencer cohomology. As far as I know, the only explicit description of this theory applied to a constant coefficient linear first order system of PDEs is my monograph Involutive Hyperbolic Differential Systems. Unfortunately, when I took a look a few years ago, I found it unreadable.

ADDED: The description above of a linear first order involutive system of PDEs was first found by Guillemin and is known as Guillemin normal form. See https://rdcu.be/cN0pn

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    $\begingroup$ rofl at the final paragraph. // (Is it not possible, when the system is not in involution, that for certain special value of $y(x_0)$ that a solution can exist?) $\endgroup$ May 20, 2022 at 19:22
  • $\begingroup$ @WillieWong, I agree. I was thinking of the case when $A^{(k)}=0$, where no solution would exist. $\endgroup$
    – Deane Yang
    May 20, 2022 at 19:58
  • $\begingroup$ Thanks this really helped me! I'll take a look at your monograph as well $\endgroup$
    – IdoAmos
    May 23, 2022 at 13:49

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