1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.