Divergence of a function product with functions in $L^\infty (\Omega)$ and $H^1(\Omega)$

Question

I am doing the weak formulation of the equation:

$\nabla ( k \nabla u) = f \ \text{in } \Omega$

I want to find the correct space for k in order to have $\nabla ( k \nabla u)$ well defined. At the moment, I am sure that I have to require $u\in H^2(\Omega)$ during the process to be able to use the Green's formula (or even the Stoke's theorem). Also, I think it is needed that $k\nabla u\in H^1(\Omega)$ in order to have the divergence correctly defined. First of all, I impose $k\in L^\infty(\Omega)$ so $k\nabla u\in L^2(\Omega)$. Does this condition also imply that $k\nabla u\in H^1(\Omega)$?

I was thinking on $k\in W^{1,\infty}(\Omega)$ to have this product well defined, but my professor told me that we can assume that the strong formulation allows us to have well defined the divergence of that product.

I am not sure if I need to impose derivatives of $k$ in $L^\infty (\Omega)$ to have everything well defined, or just think this derivative in the way of distributions and impose $k\in L^\infty(\Omega)$.

Thank you for your time.

Answer 1

I believe you would need $k \nabla u \in H^1$ to have these defined directly. Assuming that $k \in L^\infty$ and $\nabla u \in H^1$ in general isn't enough to make the product in $H^1$ (take $k$ to be very not-differentiable, but still bounded, and take $\nabla u$ to be nonzero where $k$ is badly behaved). However, you could assume the weaker condition that $k \nabla u \in L^2$, and then define $\nabla (k \nabla u)$ weakly by the formula $$ \int_\Omega \phi \nabla(k \nabla u)\, dx = -\int_\Omega \nabla \phi \cdot k \nabla u\, dx $$ for all $\phi$ smooth and compactly supported in $\Omega$ (or all $\phi \in H^1_0(\Omega))$. The condition $k\nabla u \in L^2$, for instance, would follow if $u \in H^1$ and $k \in L^\infty$.

When doing weak formulation of PDE, it is usually this second approach that is used. The unknown $u$ is generally required to be $H^1(\Omega)$ (or $H^1_0(\Omega)$ if we have Dirichlet boundary conditions), and the coefficients $k$ may by $L^\infty$ or whatever we suppose for the problem.

Answer 2

Let me reach a bit further back.

Your notation is a bit sloppy, which may mislead and thus confuse you. The correct expression for the differential operator is $\nabla \cdot (k \nabla u) = \operatorname{div} (k \nabla u)$, i.e. the divergence of $k \nabla u$, while $\nabla (k \nabla u)$ would denote the gradient of the vector field $k \nabla u$, i.e. the derivative in all directions (the Jacobian). This distinction is important because requiring $\operatorname{div}(k \nabla u)$ to be well-defined is a much weaker condition than requiring $\nabla(k \nabla u)$ to be well-defined. In the following I assume you wanted to ask what conditions are necessary for $\operatorname{div}(k \nabla u)$ to be well-defined.

To be nit-picky (or maybe not), you must specify in which sense that expression should be well-defined. I assume you meant that $\operatorname{div}(k \nabla u) \in L^2(\Omega)$ since then, under the additional assumption that $f \in L^2(\Omega)$, one can say that the differential equation $\operatorname{div}(k \nabla u(x)) = f(x)$ holds for $x \in \Omega$ almost everywhere.

If you don't assume from the get-go that $k \nabla u \in (H^1(\Omega))^d$ then the question arises what $\operatorname{div}(k \nabla u) \in L^2(\Omega)$ should even mean. You stated in your comments that you never found a use for distributional derivatives, so this is a good time as any to make use of those concepts (and maybe clear up some more confusion). For any vector field $v \in (L^2(\Omega))^d$, with the special case $v = k \nabla u$ in mind, the distributional divergence of $v$ is the distribution $\operatorname{div} v \in C^\infty_0(\Omega)'$ defined by $$ [\operatorname{div} v](\varphi) := -\int_{\Omega} v \cdot \nabla \varphi \, \mathrm dx \quad \forall \varphi \in C^\infty_0(\Omega). $$

This mirrors the definition of the distributional derivatives which you should already know. Indeed, you can define the distributional version of any linear differential operator in a similar way. It's important to realize that while this may look like "just" partial integration / Green's identities, the basic idea of distributional derivatives is that the right-hand side is well-defined even if only $v \in (L^2(\Omega))^d$ and nothing more. Only if $v$ is more regular, partial integration allows us to a-posteriori identify the distributional derivative with more a classical function. This identification of distributional divergence is done (as for distributional derivatives) by saying that $\operatorname{div} v \in C^\infty_0(\Omega)'$ is represented by a function $h \in L^2(\Omega)$ if there holds $$ \int_{\Omega} h \varphi \,\mathrm dx = v(\varphi) \left( = -\int_{\Omega} v \cdot \nabla \varphi \, \mathrm dx \right)$$ for all $\varphi \in C^\infty_0(\Omega)$. One then, casually one may say, writes $\operatorname{div} v = h$. This is the sense in which $\operatorname{div} v \in L^2(\Omega)$, e.g. $\operatorname{div}(k \nabla u) \in L^2(\Omega)$, must be understood. Of course one could just say that a vector field $v$ has "a divergence in $L^2(\Omega)$" if there is a $h \in L^2(\Omega)$ satisfying this equation without ever mentioning "distribution" or "distributional derivative", but this hides the generality of the ideas behind the construction.

With this preparation, I claim that $k \in L^\infty(\Omega)$ is sufficient that any solution $u \in H^1(\Omega)$ of your weak equation $$ \int_\Omega k \nabla u \cdot \nabla \varphi \,\mathrm dx = \int_\Omega f \varphi \,\mathrm dx $$ for all $\varphi \in H^1_0(\Omega)$ with $f \in L^2(\Omega)$ has regularity $\operatorname{div}(k \nabla u) \in L^2(\Omega)$. Comparing the integral equation for the weak solution with the condition I wrote above for $h \in L^2(\Omega)$ to represent the distributional derivative of $\operatorname{div}(k \nabla u)$, you can see that $h = - f$ and if $f \in L^2(\Omega)$ thus $\operatorname{div}(k \nabla u) \in L^2(\Omega)$. The regularity $k \in L^\infty(\Omega)$ is here only used such that $v = k \nabla u \in L^2(\Omega)$ (and of course for less regular $k$ the integral equation for the weak solutions is no longer well-defined for $\varphi \in H^1_0(\Omega))$). This argument of course only works because $u$ is not any function in $H^1(\Omega)$ but one that satisfies the integral equation, which gives you the regularity $\operatorname{div}(k \nabla u) \in L^2(\Omega)$ "for free" if your right-hand side $f$ has that regularity.

As a last note, the space of vector fields $v \in (L^2(\Omega))^d$ with $\operatorname{div} v \in L^2(\Omega)$ is typically denoted by $H(\operatorname{div}; \Omega)$, thus $k \nabla u \in H(\operatorname{div}; \Omega)$ is the minimal regularity for $\operatorname{div}(k \nabla u)$ to be well-defined.