# Why are the tangential derivatives in this diffraction problem zero? [closed]

I'm considering the diffraction problem described in section 3.16 of "Linear and quasilinear elliptic equations" of Ladyzhenskaya and Uraltseva (1968). Let $\Omega$ be an open bounded subset in $\mathbb{R}^n$, and suppose the surface $\Gamma$ partition $\Omega$ into sub-regions $\Omega_1$ and $\Omega_2$ where $\Omega = \Omega_1 \cup \Omega_2$. For simplicity, I assume $\Gamma$ to be the surface $\{x_n=0\}$. Now suppose $u(x)$ satisfies the following condition in $\Omega_1$ and $\Omega_2$ $$Lu \equiv \partial_{x_i} (a^{ij}(x) \partial_{x_j}u) + b^i(x) \partial_{x_i}u + c(x) u = f(x) \quad (16.1)$$ and the conditions $$u|_{\partial\Omega} = 0, \quad [u]|_\Gamma = 0, \quad \left.\left[ \frac{\partial u}{\partial N} \right]\right|_\Gamma = 0 \quad (16.3)$$ on $\partial\Omega$ and $\Gamma$.

Here the symbol $[v]|_\Gamma$ denotes the jump in the function $v$ as it crosses $\Gamma$. Also, $\displaystyle \left.\frac{\partial u}{\partial N} \right|_\Gamma = a^{ij} (\partial_{x_j}u) \cos(n, x_i)$, where $n$ is the normal to $\Gamma$.

$u \in H^1_0(\Omega)$ is a weak solution of the problem (16.1) and (16.3) if $u$ satisfies $$\int_\Omega \left[ a^{ij}(\partial_{x_j}u) (\partial_{x_i}v) -b^i (\partial_{x_i}u) v - cuv \right] \,dx = - \int_\Omega fv \,dx \quad (16.5)$$ for all $v \in H^1_0(\Omega)$. Assume boundedness of the coefficients $a^{ij}, \partial_{x_k}a^{ij}, b^i, c$, the uniformly elliptic condition, and $f \in L^{q/2}(\Omega)$ for $q > n$.

Suppose we already showed $u \in H^2(\Omega_i)$. In (16.5), take $v(x) = \partial_{x_s} \eta$ for $s \le n-1$, where $\eta$ is a sufficiently smooth function with compact support in $K_\rho$, where $K_\rho$ is a sphere inside $\Omega$ with center on $\Gamma$. Let $\Gamma_1 := K_\rho \cap \Gamma$.

Substituting into (16.5) and integrating by parts we get $$\int_\Omega \left[ a^{ij} (\partial_{x_s}\partial_{x_j}u) (\partial_{x_i}\eta) + \partial_{x_s} a^{ij} (\partial_{x_j}u) (\partial_{x_i}\eta) + b^i (\partial_{x_i}u) (\partial_{x_s}\eta) + cu(\partial_{x_s}\eta) \right] \,dx = \int_\Omega f (\partial_{x_s} \eta) \,dx. \quad (16.13)$$ The author stated "On the basis of the conditions (16.13), we have $\left.[\partial_{x_s}u] \right|_{\Gamma_1} = 0$ for $s \le n-1$.

I don't understand why it is the case. I'm new to this type and really want to understand it.

Thank you.

EDIT: Following John Jiang's comment, I did the computation out. Substituting $v = \partial_{x_s} \eta$ into (16.5) and integrating by parts the first term, we get (justifying derivatives of $u$ makes sense) $$\int_{K_\rho} \left[ -\partial_{x_s}(a^{ij}\partial_{x_j}u)(\partial_{x_i}\eta) - b^i(\partial_{x_i}u)(\partial_{x_s}\eta) - cu(\partial_{x_s}\eta) \right] \,dx + \int_{\Gamma_1} \big[a^{ij}(\partial_{x_j}u)(\partial_{x_i}\eta)\big]\Big|_\Gamma \,dS = -\int_{K_\rho} f(\partial_{x_s}\eta) dx.$$ The integral over $\Gamma_1$ vanishes by the hypothesis that $\displaystyle \left.\left[ \frac{\partial u}{\partial N} \right]\right|_\Gamma = 0$. However, I haven't seen why the tangential derivatives of $u$ across $\Gamma_1$, $[\partial_{x_s}u]|_{\Gamma_1}$ must be 0.

## closed as off-topic by Michael Renardy, Stefan Kohl, Wolfgang, Myshkin, Willie WongMar 21 '16 at 15:58

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You can exchange the partial derivatives on $\nu = \partial_{x_s} \eta$. Taking one partial of $u$ is allowed in the whole $\Omega$ since $u \in H^1_0(\Omega) \subset W^1_0(\Omega)$. So $\partial_{x_s} u$ makes sense there. Mixed second partials of $u$ are justified by equation 16.10, which you didn't mention; it's established by assuming $u$ to be a solution of the original equation 16.1 satisfying condition 16.3. I believe 16.10 alone tells you there is no gap of $u$ across $\Gamma$. This is of course in the almost sure sense since $u$ is an equivalence class of functions. Any gap of positive measure in $\Gamma$ would show up as a boundary term upon integration by parts. For the simplest situation refer to this. I encourage you to write down the details carefully.
• I added the details into my question, but I haven't seen why the tangential derivatives of $u$ across $\Gamma$ must be 0. Can you please elaborate more? Also, there is no gap of $u$ across $\Gamma$ by the hypothesis, isn't it? – dh16 Mar 20 '16 at 16:48
• If you exchange partial derivative with respect to s and j, if there is a gap in $\partial_s u$ across $\Gamma$, how do you make sense of its derivative with respect to $n$? – John Jiang Mar 20 '16 at 22:02
• By gap of u I meant gap of $\partial_s u$. – John Jiang Mar 20 '16 at 22:03