# How I can derive the Neuman boundary condition of this system of hyperbolic equations in 1D?

I would like to research the Neuman boundary that can verify the following problem

\begin{aligned} &\text { (} P \text { )}\left\{\begin{array}{l} \frac{\partial U}{\partial t}(x, t)+A \frac{\partial U}{\partial x}(x, t)=0, \quad x \in[a, b], t>0 \\ U(x, 0)=U_{0}(x), \\ \frac{\partial U}{\partial \eta}=? \end{array}\right.\\ &U(u, t)=\left(\begin{array}{l} u(x, t) \\ v(x, t) \end{array}\right), \quad U_{0}(x)=\left(\begin{array}{l} cos(x) \\ sin(x) \end{array}\right)\\ &\text { and }\\ &A=\left(\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right) \end{aligned}

I've already find the exact solution using the eigenvalues and the eigenvectors of $$A$$:

$$U=\left(\begin{array}{l} cos(x-t)+sin(x-t)-sin(x+t)\\ \qquad sin(x+t) \end{array}\right)\\\\$$

However, I straggle in calculating $$\frac{\partial U}{\partial \eta}$$

First of all, I know that $$\frac{\partial U}{\partial \eta}=\nabla U.\eta \quad$$ but the operator $$\nabla$$ is define on the space of function $$f: \mathbf{R}^{n} \rightarrow \mathbf{R}, \text { so that } \nabla f: \mathbf{R}^{n} \rightarrow \mathbf{R}^{n}$$, so I conclude that the notation of $$\frac{\partial U}{\partial \eta}$$ is incorrect. For this reason, I presume that I have to research for $$\frac{\partial u}{\partial \eta}$$ and $$\frac{\partial v}{\partial \eta}$$ separately, the problem that I still face is that I'm not sure if $$\eta$$ it is $$(1,0)$$ or $$(0,1)$$ ? to me $$\eta$$ is the normal derivative it need to be normal on the abscise axe, so it should be $$(0,1)$$ but I'm not sure of that. I don't know where I have a problem because I need to approach the exact solution whith the finite difference method using Matlab , but none of the case works for me, so I need to be sure that the error doesn't come from my calculations

The notation $$\frac{\partial U}{\partial \eta}$$ means usually $$\eta \cdot \nabla U$$. This is correct even if the domain is the interval $$[a,b]$$. The normal vector on the interval $$[a,b]$$ @a is $$\eta=-1$$ and @b $$\eta= 1$$ both pointing outwards of the domain. Hence in 1D $$\frac{\partial U}{\partial \eta}$$ means $$\eta\cdot\nabla U=\eta \frac{dU}{dx}$$.