Can I claim that $\nabla \cdot (\delta_{i}(x,y) \nabla u(x,y)) = \delta_{i}(x,y) \Delta u$

Question

Can I claim that $\nabla \cdot (\delta_{i}(x,y) \nabla u(x,y)) = \delta_{i}(x,y) \Delta u$ for \begin{equation*} \delta_{i} (x,y) = \left\{ \begin{array}{ll} a & \mbox{in } [0,0.5] \times [0,0.5) \\ b & \mbox{in } [0,0.5] \times [0.5,1] \\ c & \mbox{in } (0.5,1] \times [0,0.5) \\ d& \mbox{in }(0.5,1] \times [0.5,1] \\ 0& else \end{array} \right. \end{equation*} With a,b,c and d just numbers.

I'm going in circles on this logic.

My first thought was $\nabla \cdot (\delta_{i}(x,y) \nabla u(x,y)) = \nabla \delta_{i}(x,y) \cdot \nabla u + \delta_{i}(x,y) \Delta u $ but I got stuck here since $\nabla \delta_{i}(x,y)$ can't exist because $\delta_{i}(x,y)$ is discontinuous, and therefore non differentiable over the whole interval [0,1].

Then I tried to consider the concept of the weak derivative. I took $\phi \in C_0 ^{\infty} ((0,1))$ and considered the integral

\begin{equation*} \int_0 ^1 \int_0 ^1 (\delta_{i}(x,y) \nabla u(x,y))\nabla \phi \mbox{ dxdy} \end{equation*}

$= \int_{0} ^{0.5} \int_{0} ^{0.5} a \nabla u\nabla \phi \mbox{ dxdy}$ $+\int_{0.5} ^{1} \int_{0} ^{0.5} b \nabla u\nabla \phi \mbox{ dxdy}$ $+ \int_{0} ^{0.5} \int_{0.5} ^{1} c \nabla u\nabla \phi \mbox{ dxdy} $ $+\int_{0.5} ^{1} \int_{0.5} ^{1} d \nabla u\nabla \phi \mbox{ dxdy}$

For each, we essentially have

$ const \int _{Int_y} \int _{Int_x} \nabla u\nabla \phi \mbox{ dxdy} = -const \left( \int _{Int_y} \int _{Int_x} \Delta u \mbox{ }\phi \mbox{ dxdy} - \int _{\partial (Int_x \times Int_y))}(\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$

So we get Original integral $= -a\left(\int_{0} ^{0.5} \int_{0} ^{0.5} \Delta u \mbox{ }\phi \mbox{ dxdy} - \int_{\partial [0,0.5]} \int_{\partial [0,0.5]} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$ $-b \left( \int_{0.5} ^{1} \int_{0} ^{0.5} \Delta u \mbox{ }\phi \mbox{ dxdy}-\int_{\partial [0.5,1]} \int_{\partial [0,0.5]} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$ $-c \left(\int_{0} ^{0.5} \int_{0.5} ^{1} \Delta u \mbox{ }\phi \mbox{ dxdy }- \int_{\partial [0,0.5]}\int_{\partial [0.5,1]} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$ $-d \left(\int_{0.5} ^{1} \int_{0.5} ^{1} \Delta u \mbox{ }\phi \mbox{ dxdy}- \int_{\partial [0.5,1]}\int_{\partial [0.5,1]} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$

We know that $\phi$ has compact support on $[0,1]$ and we know that each boundary $\partial [start,end]$ above is just the collection of the two points $\{start,end\}$ so we get

$= -a\left(\int_{0} ^{0.5} \int_{0} ^{0.5} \Delta u \mbox{ }\phi \mbox{ dxdy} - \int_{y=0.5} \int_{x=0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$ $-b \left( \int_{0.5} ^{1} \int_{0} ^{0.5} \Delta u \mbox{ }\phi \mbox{ dxdy}-\int_{y=0.5} \int_{x= 0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$ $-c \left(\int_{0} ^{0.5} \int_{0.5} ^{1} \Delta u \mbox{ }\phi \mbox{ dxdy }- \int_{y=0.5}\int_{x=0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$ $-d \left(\int_{0.5} ^{1} \int_{0.5} ^{1} \Delta u \mbox{ }\phi \mbox{ dxdy}-\int_{y=0.5}\int_{x=0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} \right)$

So in order to have $= - \int_0 ^1 \int_0 ^1 \delta_{i}\Delta u \mbox{ } \phi \mbox{ dxdy}$

We have to require that

$a \int_{y=0.5} \int_{x=0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} +b \int_{y=0.5} \int_{x= 0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} +c\int_{y=0.5}\int_{x=0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} +d\int_{y=0.5}\int_{x=0.5} (\nabla u \cdot \vec{n})\phi \mbox{ dS} = 0$

$\rightarrow \nabla u \cdot \vec{n} = 0$ for $x = 0.5$ and $y = 0.5$

So we get that if the condition $\nabla u \cdot \vec{n} = 0$ for $x = 0.5$ and $y = 0.5$ holds

Is this extra Neumann condition what allows me to ensure the solutions on the subdomains coincide on the subdomain borders?

$\nabla \cdot (\delta_{i}(x,y) \nabla u(x,y)) = \delta_{i}(x,y) \Delta u$ in terms of the weak derivative (I think).

My questions are as follows

1. Can I claim that $\nabla \cdot (\delta_{i}(x,y) \nabla u(x,y)) = \delta_{i}(x,y) \Delta u$ from the above?
2. If I wanted to use $u(x,y) = cos(2\pi x)cos(2\pi y)$ would I get \begin{equation*} \nabla \cdot (\delta_{i}(x,y) \nabla u(x,y)) = \left\{ \begin{array}{ll} a (-8 \pi ^2 cos(2\pi x)cos(2\pi y) & \mbox{in } [0,0.5] \times [0,0.5) \\ b (-8 \pi ^2 cos(2\pi x)cos(2\pi y) & \mbox{in } [0,0.5] \times [0.5,1] \\ c (-8 \pi ^2 cos(2\pi x)cos(2\pi y) & \mbox{in } (0.5,1] \times [0,0.5) \\ d (-8 \pi ^2 cos(2\pi x)cos(2\pi y) & \mbox{in }(0.5,1] \times [0.5,1] \\ 0& else \end{array} \right. \end{equation*} I'm really new to this and any insight is greatly appreciated.

Answer 1

You only have $$ \nabla \cdot (\delta \nabla u) = (\nabla \delta) \cdot (\nabla u) + \delta \Delta u $$ at those locations where $\delta$ and $u$ are once and twice differentiable, respectively. At the other locations, the right hand side is simply not defined in the classical sense. Note that this is a stricter requirement than you have on the left hand side: On the left, all you need is that $u$ is differentiable and that $\delta\nabla u$ is differentiable. (In physical terms, that means that the potential $u$ is differentiable, not necessarily continuously so, and that the flux $\delta \nabla u$ has a divergence.)

You can argue that the right hand side above is still well defined if you're willing to allow yourself to do things such as saying that the derivative of the Heaviside function is the Dirac delta function. In other words, if you're willing to engage in calculus with distributions, the right hand side can be made to have sense.

The problem you are considering comes from physics, where typically you have $u$ being a potential (e.g., a fluid pressure in a porous medium, or the electric potential in a body), $\delta \nabla u$ being a flux, and the divergence of the flux being a source term. In these sorts of situations, $u$ frequently has kinks, $\delta$ is discontinuous, and so the right hand side above does not make sense because $\nabla \delta$ does not exist, and neither does $\Delta u$. But the flux, being the product of two discontinuous functions, is still valid as it is continuous and differentiable. So the left hand side makes sense, you just can't apply the product rule.