Consider $p>2$ and the Sobolev space $W^{1,p}(\mathbb{R}\times [0,1],\mathbb{R}^{2})$. Since $p>2$ just define this as being the sobolev space $W^{1,p}(\mathbb{R}\times (0,1),\mathbb{R}^{2})$ where we extend by continuity the functions to the boundary.
Now consider an $S\in L^p(\mathbb{R}\times (0,1), \mathbb{R}^{2\times 2})$ and the differential operator $L_S=\partial_s-J_0\partial_t-S(s,t)$. Suppose that $Y$ is a solution of $L_SY=0$ on some interval $\mathbb{R}\times [0,\epsilon[$. I was wondering if we assume that $Y(s,0)\subset 0\times \mathbb{R}$, then we can extend $Y$ to a function $Y'$ defined on $\mathbb{R}\times (-\epsilon,\epsilon)$ of sobolev class $W^{1,p}$, and also extend $S$ to $\tilde S$, so that $L_{\tilde S}\tilde Y=0$.
I have tried defining $\tilde Y$ for the negative values of $t$ as $(-Y_1(s,-t),Y_2(s,-t))$ and analogous for $\tilde S$. One needs to check that $\tilde Y$ is of the desired sobolev class. I tought I was able to see this however the condition on $Y$ seemed superfluous, I did not use it.
So I would like to know if this is possible independent of the boundary condition on $Y$ or if we in fact do need this boundary condition and we're it's used.
Any enlightment is appreciated, thanks in advance.