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Let $x \in (0,L)$, $t \in (0,T)$, and let $u_0 = u_0(x) \in \mathbb{R}^3$, $g=g(t) \in \mathbb{R}^3$, $P = P(x,t) \in \mathbb{R}^3$ and $Q = Q(x,t) \in \mathbb{R}^3$ be continuously differentiable functions.

Denote by $\hat{P}, \hat{Q}$ the matrices \begin{align*} \hat{P} = \begin{bmatrix} 0 & -P_3 & P_2 \\ Q_3 & 0 & -P_1 \\ -P_2 & P_1 & 0 \end{bmatrix}, \qquad \hat{Q} = \begin{bmatrix} 0 & -Q_3 & Q_2 \\ Q_3 & 0 & -Q_1 \\ -Q_2 & Q_1 & 0 \end{bmatrix}. \end{align*}

My question is:

Assuming that the following four conditions hold:

Condition 1: \begin{align*} \partial_x P - \partial_t Q = \hat{P}Q \qquad \text{in }(0,L) \times (0,T), \end{align*} Condition 2: \begin{align*} u_0(0) = g(0), \end{align*} Condition 3: \begin{align*} u_0' = - \hat{Q}u_0 \qquad \text{for } x \in (0,L), \end{align*} Condition 4: \begin{align*} g' = - \hat{P}g \qquad \text{for } t \in (0,T), \end{align*} is there a solution $u = u(x,t) \in \mathbb{R}^3$ to \begin{align*} \begin{cases} \partial_t u = -\hat{P}u &\text{in }(0,L)\times(0,T) \\ \partial_x u = - \hat{Q}u &\text{in }(0,L)\times(0,T) \\ u(x,0) = u_0(x) & \text{for } x \in (0,L)\\ u(0,t) = g(t) & \text{for }t \in (0,T) \end{cases} \end{align*} or not ?

In the above, $f'$, $\partial_t f$a and $\partial_x f$ denote the derivative, partial time derivative and partial space derivative respectively; and $P_k$, $Q_k$, $u_k$, $u_{0k}$ and $g_k$ denote the components of $P$, $Q$, $u$, $u_0$ and $g$ respectively.

My question with all terms developed rewrites as follows. Assuming that the following four conditions hold:

Condition 1: \begin{align*} \begin{cases} \partial_x P_1 - \partial_t Q_1 = P_2 Q_3 - P_3 Q_2\\ \partial_x P_2 - \partial_t Q_2 = P_3 Q_1 - P_1 Q_3\\ \partial_x P_3 - \partial_t Q_3 = P_1 Q_2 - P_2 Q_1 \end{cases} \qquad \text{in }(0,L) \times (0,T), \end{align*} Condition 2: \begin{align*} u_0(0) = g(0), \end{align*} Condition 3: \begin{align*} \begin{cases} u_{01}' = \ \ Q_3(\cdot,0) u_{02} - Q_2(\cdot,0) u_{03} \\ u_{02}' = -Q_3(\cdot,0) u_{01} + Q_1(\cdot,0) u_{03} \\ u_{03}' = \ \ Q_2(\cdot,0) u_{01} - Q_1(\cdot,0) u_{02} \end{cases} \qquad \text{for } x \in (0,L), \end{align*} Condition 4: \begin{align*} \begin{cases} g_1' &= \ \ P_3(0,\cdot) g_2 - P_2(0,\cdot) g_3 \\ g_2' &= -P_3(0,\cdot) g_1 + P_1(0,\cdot) g_3 \\ g_3' &= \ \ P_2(0,\cdot) g_1 - P_1(0,\cdot) g_2 \end{cases} \qquad \text{for } t \in (0,T), \end{align*} is there a solution $u = u(x,t) \in \mathbb{R}^3$ to the problem: \begin{align*} \begin{cases} \partial_t u_1 = \ \ P_3 u_2 - P_2 u_3 &\text{in }(0,L)\times(0,T) \\ \partial_t u_2 = -P_3 u_1 + P_1 u_3 &\text{in }(0,L)\times(0,T) \\ \partial_t u_3 = \ \ P_2 u_1 - P_1 u_2 & \text{in }(0,L)\times(0,T) \\ \partial_x u_1 = \ \ Q_3 u_2 - Q_2 u_3 & \text{in }(0,L)\times(0,T) \\ \partial_x u_2 = -Q_3 u_1 + Q_1 u_3 &\text{in }(0,L)\times(0,T) \\ \partial_x u_3 = \ \ Q_2 u_1 - Q_1 u_2 & \text{in }(0,L)\times(0,T) \\ u(x,0) = u_0(x) & \text{for } x \in (0,L)\\ u(0,t) = g(t) & \text{for }t \in (0,T). \end{cases} \end{align*}

What I started. I started reasoning the following way. For $x \in (0,L)$ fixed, a function of the form \begin{align*} u_1(x,t) &= u_{01}(x) + \int_0^t P_3(x,s)u_2(x,s) - P_2(x,s)u_3(x,s)ds\\ u_2(x,t) &= u_{02}(x) + \int_0^t -P_3(x,s)u_1(x,s) + P_1(x,s)u_3(x,s)ds\\ u_3(x,t) &= u_{03}(x) + \int_0^t P_2(x,s)u_1(x,s) - P_1(x,s)u_2(x,s)ds, \end{align*} satisfies the initial value problem involving time derivatives, with $u(\cdot, 0) = u_0$, while for $t \in (0,T)$ fixed, a solution of the form \begin{align*} \begin{aligned} u_1(x,t) &= g_1(t) + \int_0^x Q_3(\xi,t)u_2(\xi,t) - Q_2(\xi, t)u_3(\xi, t)d\xi \\ u_2(x,t) &= g_2(t) + \int_0^x -Q_3(\xi, t)u_1(\xi, t) + Q_1(\xi, t) u_3(\xi, t)d\xi\\ u_3(x,t) &= g_3(t) + \int_0^x Q_2(\xi, t)u_1(\xi, t) - Q_1(\xi, t)u_2(\xi, t)d\xi. \end{aligned} \end{align*} satisfies the initial value problem involving space derivatives, with $u(0, \cdot) = g$. I wanted to show, using the four conditions, that both expressions for the solution coincide. Using conditions 3 and 4 we can rewrite both expressions as \begin{align*} u_1(x,t) &= u_{01}(0) + \int_0^x Q_3(\xi, 0)u_{02}(\xi) - Q_2(\xi, 0)u_{03}(\xi)d\xi \\ &\quad + \int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s)ds + \int_0^t \int_0^x \partial_x (P_3 u_2 - P_2 u_3 )(\xi, s)d\xi ds\\ u_2(x,t) &= u_{02}(0) + \int_0^x -Q_3(\xi, 0)u_{01} + Q_1(\xi, 0)u_{03}(\xi)d\xi \\ &\quad + \int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s)ds + \int_0^t \int_0^x \partial_x(-P_3 u_1 + P_1 u_3)(\xi, s)ds\\ u_3(x,t) &= u_{03}(0) + \int_0^x Q_2(\xi, 0)u_{01}(\xi) - Q_1(\xi, 0)u_{02}(\xi)\\ &\quad + \int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s)ds + \int_0^t \int_0^x \partial_x (P_2 u_1 - P_1 u_2)(\xi,s)d\xi ds, \end{align*} and \begin{align*} u_1(x,t) &= g_1(0) + \int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s) ds \\ &\quad + \int_0^x Q_3(\xi,0)u_{02}(\xi) - Q_2(\xi, 0)u_{03}(\xi)d\xi + \int_0^x \int_0^t \partial_t ( Q_3 u_2 - Q_2 u_3)(\xi, s)dsd\xi\\ u_2(x,t) &= g_2(0) + \int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s) ds\\ &\quad + \int_0^x -Q_3(\xi, 0)u_{01}(\xi) + Q_1(\xi, 0) u_{03}(\xi)d\xi + \int_0^x \int_0^t \partial_t (-Q_3 u_1 + Q_1 u_3)(\xi,s)dsd\xi\\ u_3(x,t) &= g_3(0) + \int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s) ds \\ &\quad + \int_0^x Q_2(\xi, 0)u_{01}(\xi) - Q_1(\xi, 0)u_{02}(\xi)d\xi + \int_0^x \int_0^t \partial_t ( Q_2 u_1 - Q_1 u_2)(\xi, s)ds d\xi \end{align*} respectively. It seems that both expressions would coincide if the following equalities hold: \begin{align*} \partial_x (P_3 u_2 - P_2 u_3 ) &= \partial_t ( Q_3 u_2 - Q_2 u_3)\\ \partial_x(-P_3 u_1 + P_1 u_3) &= \partial_t (-Q_3 u_1 + Q_1 u_3)\\ \partial_x (P_2 u_1 - P_1 u_2) &= \partial_t ( Q_2 u_1 - Q_1 u_2). \end{align*} To show that these equalities hold, I would like to use condition 1. However, I do not know how to argue from there.

Any suggestion of how to proceed from here, any alternative way of reasoning, or explanation of why there would be no solution, would be welcome. Thank you.

(This question is also on Mathematics Stack Exchange: https://math.stackexchange.com/questions/3160600/6-linear-pde-for-only-3-unknowns)

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