# Integration of Cahn-Hilliard-Oono equation

I am currently reading a paper on the Cahn-Hilliard-Oono equation with Neumann boundary conditions:

\begin{align} \frac{\partial u}{\partial t} + \epsilon u &+ \Delta^2 u - \Delta f(u) = 0, \quad \epsilon \geq 0,\\ \frac{\partial u}{\partial n} &= \frac{\partial \Delta u}{\partial n} = 0 \quad \text{on } \Gamma,\\ &u\lvert_{t=0} = u_0 \end{align}

where $$\Omega \subset \mathbb{R}^3$$ is bounded and regular with boundary $$\Gamma$$. $$n$$ denotes the unit outer normal to $$\Gamma$$ and $$f(u) = u^3-u$$.

Now the author claims that, when integrating the Cahn-Hilliard-Oono equation over $$\Omega$$, we receive:

\begin{align} \frac{d \langle u \rangle}{dt} + \epsilon \langle u \rangle = 0, \end{align} where \begin{align} \langle \cdot \rangle = \frac{1}{\text{Vol}(\Omega)} \int_{\Omega} \cdot \, dx. \end{align}

I am feeling really stupid at the moment, because I cannot comprehend whats happening here. Where does the term $$\frac{1} {\text{Vol}({\Omega})}$$ come from to begin with?

I would have computed: \begin{align} \int_{\Omega} \frac{\partial u}{\partial t} + \epsilon u &+ \Delta^2 u - \Delta f(u) \, dx = \frac{d}{dt} \int_{\Omega} u \, dx + \epsilon \int_{\Omega} u \,dx \end{align}

Aditionally, I am wondering why the last to terms in the above equation are equal to zero. I tried to justify this with the divergence theorem, but I absolutely messed up:

\begin{align} \int_{\Omega} \Delta^2 u \,dx = \int_{\Omega} \text{div}(\nabla(\Delta u)) \,dx = \int_{\Gamma} \nabla(\Delta u)\cdot n \, d\Gamma - \int_{\Omega} \nabla 1 \cdot \nabla (\Delta u) dx = 0\\ \int_{\Omega} \Delta f(u) \,dx =\int_{\Omega} \text{div}(\nabla(f(u)) \, dx = \int_{\Gamma} \nabla f(u) \cdot n \, d\Gamma - \int_{\Omega} \nabla 1\cdot \nabla f(u) \, dx = \int_{\Gamma} f'(u) \nabla u \cdot n \, d\Gamma \end{align}

Can someone help me out?

• $\langle \cdot \rangle$ just denotes the spatial average of a function and so you need to divide by the length/area/volume when in $1$/$2$/$n$ dimensions respectively. And computing the averaged expressions as you did is exactly what to do; for example, you have $$\int_{\Omega} (\Delta^{2} u) \ dx = \int_{\Gamma} \underbrace{(\nabla \Delta u \cdot n)}_{= \ \partial_{n} \Delta u} \ d \Gamma - \int_{\Omega} \underbrace{(\nabla 1)}_{= \ 0} \cdot (\nabla \Delta u) \ dx = 0$$ Keep going with the other expression. Mar 15, 2023 at 11:58
• @MatthewCassell, thanks for your answer. So did I understood you correctly: Integrating over $\Omega$ leads to $\frac{d}{dt} \int_{\Omega} u \,dx + \epsilon \int_{\Omega} u \,dx$ and afterwards, I have to divide by volume to obtain the spatial average. Or ist the spatial average something that is directly linked to the integration? Mar 15, 2023 at 12:12
• @MatthewCassell, and concerning the last expression. I am stuck at the term $\int_{\Gamma} ( f'(u) \nabla u) \cdot n \, d\Gamma$, since I know nothing about $\nabla u$. Mar 15, 2023 at 12:16
• No, when you take the average of the equation by definition of the average you are computing $(\text{vol} \Omega)^{-1} \int_{\Omega} (\cdot)$, not just the integral component, though the result comes out the same if you integrate the entire expression and then multiply everything by $(\text{vol} \Omega)^{-1}$. Then by linearity of the integral, you have $$\langle u_{t} + \epsilon u + \dots \rangle = \langle u_{t} \rangle + \epsilon \langle u \rangle + \dots$$ For the second question $$(f'(u) \nabla u) \cdot n = f'(u) (\nabla u \cdot n) = f'(u) \partial_{n} u$$ Mar 15, 2023 at 12:26
• @MatthewCassell, okay then I just didn't get that the author was computing the mean of the function. Thanks a lot! Mar 15, 2023 at 13:17

\begin{align} \int_{\Omega}\Delta^2 u \, dx &= \int_{\Omega} \text{div}(\nabla \Delta u)\, dx \\ &= \int_{\Gamma} \underbrace{(\nabla \Delta u) \cdot n}_{= \ \partial_n \Delta u \ = \ 0 \text{ on } \Gamma} \, d\Gamma - \int_{\Omega} \underbrace{\nabla 1}_{= \ 0} \cdot (\nabla \Delta u) \, dx \\ &= 0 \end{align}
\begin{align} \int_{\Omega} \Delta f(u) \, dx &= \int_{\Omega} \text{div} (\nabla(f(u)) \,dx \\ &= \int_{\Gamma} \nabla f(u) \cdot n \, d\Gamma - \int_{\Omega} \underbrace{\nabla 1}_{= \ 0} \cdot \nabla f(u) \, dx \\ &= \int_{\Gamma} f'(u) \underbrace{(\nabla u \cdot n)}_{= \ \partial_{n} u \ = \ 0 \text{ on } \Gamma} \, d\Gamma \\ &= 0. \end{align}