The normal derivative in PDE problems (how is the weak form defined?)

Question

For smooth functions, we know that $$-\int_{\Omega}\Delta v w = \int_\Omega \nabla v \nabla w - \int_{\partial \Omega}\nabla v \cdot \nu w.$$ Thus we can define a generalised normal derivative $\frac{\partial }{\partial \nu}\colon H^{1}(\Omega) \to H^{-\frac{1}{2}}(\Gamma)$ with the formula $$\left\langle \frac{\partial v}{\partial \nu}, w \right \rangle_{H^{-\frac 1 2}(\partial\Omega), H^{\frac 1 2}(\partial\Omega)} := \int_\Omega \nabla v \nabla W+\langle \Delta v, W\rangle_{H^{-1}, H^1}$$ where $W \in H^1(\Omega)$ is an extension of $w \in H^{\frac 1 2}(\Gamma)$.

Questions:

• Is this really how the normal derivative functional is defined? I only saw this in one source.
• since $v \in H^1$, $\Delta v \in L^2$ makes no sense, but the author has taken it to be an element of $H^{-1}$ which is OK, but what on earth is it??
• I guess it doesn't matter which extension one uses of $w$.

Answer 1

You cannot define a normal derivative trace $\frac{\partial }{\partial \nu}$ acting from $H^{1}(\Omega)$ to $H^{-\frac{1}{2}}(\Gamma)$ such that it has reasonable properties:

For $\varphi \in C_0^\infty(\Omega)$, we would expect $\frac{\partial}{\partial \nu} \varphi = 0$, since $\varphi$ has compact support. But now, $C_0^\infty(\Omega)$ is (by definition) dense in $H_0^1(\Omega)$. This means, that every function in $H_0^1(\Omega)$ has not only zero trace but also a zero normal derivative trace. But we do not expect this for an arbitrary function from $H_0^1(\Omega) \cap C^1(\bar\Omega)$.

In order to define a normal derivative trace, you need a stronger norm in the domain, e.g. the norm of the space $$\{u \in H^1(\Omega) : \Delta u \in L^2(\Omega)\}.$$ Then, you can use your formula to define the normal derivative trace.

Answer 2

In fact, your questions are all about presumably ill-defined traces. Indeed, a trace on $\Gamma=\partial\Omega$ cannot be defined for all elements of $L^2(\Omega)$, since an embedding $L^2(\Omega)\hookrightarrow H^{-\frac{1}{2}}(\Gamma)$ does not hold. But traces on certain subspaces, dense in $L^2(\Omega)$, can well be defined. Of course, such self-evident cases as the subspace $C(\overline{\Omega})$ are not of any interest. It must be a subspace elements of which seem to possess no trace. Your example is exactly the case, which could be the news not today but, say, some 50 years ago. Another example is the subspace $$H({\rm div};\Omega)\overset{def}{=}\{v\in L^2(\Omega;\mathbb{R}^n)\,\colon\; {\rm div}\,v\in L^2(\Omega)\}, $$ widely employed in the Mathematical Fluid Dynamics and Modern Maxwell Electrodynamics, where $L^2(\Omega;\mathbb{R}^n)$ denotes the Lebesgue space of vector fields $v\colon\,\Omega \to \mathbb{R}^n$ square-summable over a domain $\Omega\subset\mathbb{R}^n$, $n\geqslant 2$, while the divergence operator is understood in a weak sense. Seen today as classical, the example of subspace $H({\rm div};\Omega)$ was the first successful attempt to define a generalized trace. To see how it works, denote by $\nu$ an outward unit normal to $\Gamma=\partial\Omega$ and let us try to define a generalized trace of the normal component $v\cdot\nu$ on $\Gamma$ in terms of an integral identity evident for sufficiently smooth $v$, but taken for definition when $v\in H({\rm div};\Omega)$. Namely, consider an identity $$\int\limits_{\Omega}\bigl(v\cdot\nabla\varphi+ ({\rm div\,}v)\varphi\bigr)\,dx = \int\limits_{\Gamma}v\cdot\nu\,\varphi\,ds \quad \forall\,\varphi\in H^1(\Omega)$$ with the left-hand side being a bounded linear functional in $\varphi$ on $H^1(\Omega)$. To define a generalized trace of the normal component $v\cdot\nu$ on $\Gamma$ for all $v\in H({\rm div};\Omega)$, consider a linear functional $$\Lambda(\varphi)\overset{def}{=}\int\limits_{\Gamma}v\cdot\nu\,\varphi\,ds=\int\limits_{\Omega}\bigl(v\cdot\nabla\varphi+ ({\rm div\,}v)\varphi\bigr)\,dx \tag{$\ast$}\quad \forall\,\varphi\in H^1(\Omega)$$ Since the functional $\Lambda$ takes one and the same value on all elements of every coset in the quotient $H^1(\Omega)/H^1_0(\Omega)$, values of $\Lambda$ are independent of the way a test function $\varphi\in H^{\frac{1}{2}}(\Gamma)$ is extended from $\Gamma$ to $\Omega$ to become an element of $H^1(\Omega)$. Hence, the functional $(\ast)$ correctly defines a generalized trace of the normal component $v\cdot\nu$ on $\Gamma$ as an element of the space $H^{-\frac{1}{2}}(\Gamma)$ dual to $H^{\frac{1}{2}}(\Gamma)$, which immediately implies the embedding $$ \{u=v\cdot\nu\;\colon\; v\in H({\rm div};\Omega)\} \hookrightarrow H^{-\frac{1}{2}}(\Gamma).$$ The above text is a slightly edited version of what you can find along with further examples on pages 129-130 of the textbook An Introduction to the Navier-Stokes Equation and Oceanography by L. Tartar (Springer,2006). In conclusion, notice that the functional $(\ast)$ retains the property that identifies every correct generalization. Namely, a generalized trace of the normal component $v\cdot\nu$ coincides with the classical trace, i.e. a restriction to $\Gamma$, whenever $v\in C(\overline{\Omega})\cap H({\rm div};\Omega)$, which well explains what-on-earth-it-is.