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


1 Answer 1


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 $$\eta\cdot\nabla U=\eta \frac{dU}{dx}$$.


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