Phase Separation in the Advective Cahn–Hilliard Equation

Abstract

The Cahn–Hilliard equation is a classic model of phase separation in binary mixtures that exhibits spontaneous coarsening of the phases. We study the Cahn–Hilliard equation with an imposed advection term in order to model the stirring and eventual mixing of the phases. The main result is that if the imposed advection is sufficiently mixing, then no phase separation occurs, and the solution instead converges exponentially to a homogeneous mixed state. The mixing effectiveness of the imposed drift is quantified in terms of the dissipation time of the associated advection–hyperdiffusion equation, and we produce examples of velocity fields with a small dissipation time. We also study the relationship between this quantity and the dissipation time of the standard advection–diffusion equation.

Appendix A. Dissipation Time Bounds of Mixing Vector Fields

In this section, we prove Theorem 3.2. As in Sect. 3, we assume here that M is a smooth, closed, Riemannian manifold with volume 1, and $\Delta$ is the Laplace–Beltrami operator on M. We also fix a divergence-free vector field $u \in L^\infty ( [0, \infty ); C^{2}(M) )$, and let $\theta$ be the solution to the advection–hyperdiffusion equation (1.3) with $\alpha = 2$ on the manifold M, with mean-zero initial data $\theta _0 \in \dot{L}^2(M)$.

The idea behind the proof of Theorem 3.2 is to divide the analysis into two cases. When $\Vert \Delta \theta \Vert _{L^2} / \Vert \theta \Vert _{L^2}$ is large, the energy inequality implies $\Vert \theta \Vert _{L^2}$ decays rapidly. On the other hand, when $\Vert \Delta \theta \Vert _{L^2} / \Vert \theta \Vert _{L^2}$ is small, we use the mixing assumption on u to show that $\Vert \theta \Vert _{L^2}$ still decays rapidly. The outline of the proof is the same as that of Theorem 1.2; however, the proof of the second case is substantially different. We begin by stating two lemmas handling each of the above cases.

Lemma A.1

The solution $\theta$ satisfies the energy inequality

$$\begin{aligned} \partial _t\Vert \theta \Vert _{L^2}^2=-2\gamma \Vert \Delta \theta \Vert _{L^2}^2. \end{aligned}$$

(A.1)

Consequently, if for some $c_0>0$ we have

$$\begin{aligned} \Vert \Delta \theta (t) \Vert _{L^2}^2 \geqslant c_0\Vert \theta (t) \Vert _{L^2}^2, \quad \text { for all } 0\leqslant t \leqslant t_0, \end{aligned}$$

then

$$\begin{aligned} \Vert \theta (t) \Vert _{L^2}^2\leqslant e^{-2\gamma c_0 t}\Vert \theta _0 \Vert _{L^2}^2, \quad \text { for all } 0\leqslant t \leqslant t_0. \end{aligned}$$

(A.2)

Lemma A.2

Let $0 < \lambda _1 \leqslant \lambda _2 \leqslant \cdots$ be the eigenvalues of the Laplacian, where each eigenvalue is repeated according to its multiplicity. Suppose u is weakly mixing with rate function h. There exists positive, finite dimensional constants $\tilde{C}$, $\tilde{c}$ such that for all $\gamma$ sufficiently small the following holds: If $\lambda _N$ is an eigenvalue of the Laplace–Beltrami operator such that¹

$$\begin{aligned} h^{-1}\left( \frac{1}{\tilde{c} \lambda _N^{(d + 4) / 4} } \right) \leqslant \frac{1}{\tilde{C} \lambda _N \sqrt{\gamma } \Vert u \Vert _{C^2}^{1/2} }, \end{aligned}$$

(A.3)

and if

$$\begin{aligned} \Vert \Delta \theta _0 \Vert _{L^2}^2 < \lambda _N^2 \Vert \theta _0 \Vert _{L^2}^2 \end{aligned}$$

(A.4)

holds, then we have

$$\begin{aligned} \Vert \theta (t_0) \Vert _{L^2}^2 \leqslant \exp \left( - \frac{\gamma \lambda _N^2 t_0}{4} \right) \Vert \theta _{0} \Vert _{L^2}^2, \end{aligned}$$

(A.5)

at a time $t_0$ given by

$$\begin{aligned} t_0 {\mathop {=}\limits ^{\scriptscriptstyle \hbox {def}}}h^{-1}\Big ( \frac{1}{\tilde{c} \lambda _N^{(d+4)/4}} \Big ) . \end{aligned}$$

(A.6)

If instead u is strongly mixing, then the analog of Lemma A.2 is as follows.

Lemma A.3

Suppose u is strongly mixing with rate function h. There exists a finite dimensional $\tilde{C} > 0$ such that for all $\gamma$ sufficiently small the following holds: If $\lambda _N$ is an eigenvalue of the Laplace–Beltrami operator such that

$$\begin{aligned} 2 h^{-1} \left( \frac{1}{2 \lambda _N} \right) \leqslant \frac{1}{\tilde{C} \lambda _N \sqrt{\gamma } \Vert u \Vert _{C^2}^{1/2} }, \end{aligned}$$

(A.7)

and if (A.4) holds, then (A.5) holds at a time $t_0$ given by

$$\begin{aligned} t_0 {\mathop {=}\limits ^{\scriptscriptstyle \hbox {def}}}2 h^{-1} \left( \frac{1}{2 \lambda _N} \right) . \end{aligned}$$

(A.8)

Finally, for the proof of Theorem 3.2 we need Weyl’s Lemma [see, for instance, (Minakshisundaram and Pleijel 1949)], which describes the asymptotic growth of the eigenvalues of the Laplace–Beltrami operator.

Lemma A.4

(Weyl’s Lemma). Let $0 < \lambda _1 \leqslant \lambda _2 \leqslant \cdots$ be the eigenvalues of the Laplacian, where each eigenvalue is repeated according to its multiplicity. We have

$$\begin{aligned} \lambda _j \approx \frac{4\pi \, \Gamma (\frac{d}{2}+1)^{2/d}}{{{\,\mathrm{vol}\,}}(M)^{2/d}}\,j^{2/d}, \end{aligned}$$

(A.9)

asymptotically as $j \rightarrow \infty$.

Momentarily postponing the proof of Lemmas A.1–A.3, we prove Theorem 3.2.

Proof of Theorem 3.2

For the first assumption, we assume u is weakly mixing with rate function h. Let $\tilde{c}$, $\tilde{C}$ be the constants from Lemma A.2. Note that the intermediate value theorem readily implies the existence of a unique $\lambda _* > 0$ such that

$$\begin{aligned} h^{-1}\left( \frac{1}{\tilde{c} \lambda _*^{(d + 4) / 4} } \right) = \frac{1}{\tilde{C} \lambda _* \sqrt{\gamma } \Vert u \Vert _{C^2}^{1/2} }. \end{aligned}$$

(A.10)

Further, it is easy to see that $\lambda _* \rightarrow \infty$ as $\gamma \rightarrow 0$. Thus, for all sufficiently small $\gamma$, Weyl’s lemma implies $\lambda _{j+1}-\lambda _{j}=o(\lambda _j)$ as $j \rightarrow \infty$. Hence, for all sufficiently large $\lambda _*$, one can always find N large enough such that

$$\begin{aligned} \frac{\lambda _*^2}{2} \leqslant \lambda _N^2 \leqslant \lambda _*^2 . \end{aligned}$$

(A.11)

Now choosing $c_0=\lambda _N^2$ and repeatedly applying Lemmas A.1 and Lemma A.2, we obtain an increasing sequence of times $(t'_k)$, such that $t'_k \rightarrow \infty$, $t_{k+1}'-t_k'\leqslant t_0$, and

$$\begin{aligned} \Vert \theta _s(t_k') \Vert _{L^2}^2 \leqslant \exp \Big (-\frac{\gamma \lambda _N^2 t_k'}{4}\Big )\Vert \theta _{0} \Vert _{L^2}^2. \end{aligned}$$

This immediately implies

$$\begin{aligned} \tau _2^*(u,\gamma ) \leqslant \frac{8 \ln 2}{\gamma \lambda _N^2}+t_0. \end{aligned}$$

(A.12)

Choosing

$$\begin{aligned} t_* {\mathop {=}\limits ^{\scriptscriptstyle \hbox {def}}}\frac{\sqrt{2}}{\tilde{C} \lambda _* \sqrt{\gamma } \Vert u \Vert _{C^2}^{1/2} }, \end{aligned}$$

and using (A.10), (A.11), and (A.12) yields (3.2) as claimed.

The proof of the second assertion of Theorem 3.2 is almost identical to that of the first assertion. The only change required is to replace Lemma A.2 with A.3. $\square$

It remains to prove Lemmas A.1–A.3.

Proof of Lemma A.1

Multiplying (1.3) by $\theta$, integrating over M, and using the fact that u is divergence-free immediately yield (A.1). The second assertion of Lemma A.1 follows from this and Gronwall’s lemma. $\square$

For Lemmas A.2 and A.3, we will need a standard result estimating the difference between $\theta$ and solutions to the inviscid transport equation.

Lemma A.5

Let $\phi$ be the solution of (3.1) with initial data $\theta _0$. There exists a dimensional constant $C_d$ such that for all $t \geqslant 0$, we have

$$\begin{aligned} \Vert \theta (t)-\phi (t) \Vert _{L^2}^2 \leqslant \sqrt{2\gamma t}\,\Vert \theta _0 \Vert _{L^2} \left( C_d\, \Vert u \Vert _{C^2}\int _0^t\Vert \Delta \theta \Vert _{L^2}^2 \,\mathrm{d}s +\Vert \Delta \theta _0 \Vert _{L^2}^2 \right) ^{1/2}. \end{aligned}$$

(A.13)

Proof

Subtracting (1.3) and (3.1) shows

$$\begin{aligned} \partial _t (\theta -\phi )+u\cdot \nabla (\theta -\phi )+\gamma \Delta ^2\theta =0. \end{aligned}$$

Multiplying this by $\theta (t)-\phi (t)$ and integrating over space and time give

$$\begin{aligned} \Vert \theta (t)-\phi (t) \Vert _{L^2}^2 = -2\gamma \int _0^t \int _M (\theta - \phi ) \Delta ^2 \theta \,\mathrm{d}x\,\mathrm{d}s \leqslant 2\gamma \Vert \theta _0 \Vert _{L^2}\int _0^t\Vert \Delta ^2 \theta \Vert _{L^2}\,\mathrm{d}s. \end{aligned}$$

(A.14)

On the other hand, multiplying (1.3) by $\Delta ^2\theta$ and integrating over M give

$$\begin{aligned} \partial _t \Vert \Delta \theta \Vert _{L^2}^2+2\langle u\cdot \nabla \theta , \Delta ^2 \theta \rangle +2\gamma \Vert \Delta ^2\theta \Vert _{L^2}^2=0. \end{aligned}$$

Integrating the middle term by parts, using the fact that u is divergence-free, and integrating in time yields

$$\begin{aligned} 2\gamma \int _0^t \Vert \Delta ^2\theta \Vert _{L^2}^2\,\mathrm{d}s \leqslant C_d\Vert u \Vert _{C^2}\int _0^t\Vert \Delta \theta \Vert _{L^2}^2\,\mathrm{d}s +\Vert \Delta \theta _0 \Vert _{L^2}^2, \end{aligned}$$

for some dimensional constant $C_d$. Substituting this in (A.14) and using the Cauchy–Schwartz inequality give (A.13) as claimed. $\square$

We now prove Lemma A.2.

Proof of Lemma A.2

We claim that our choice of $\lambda _N$ and $t_0$ will guarantee

$$\begin{aligned} \int _0^{t_0}\Vert \Delta \theta (s) \Vert _{L^2}^2\,\mathrm{d}s \geqslant \frac{\lambda _N^2t_0\Vert \theta _{0} \Vert _{L^2}^2}{8}. \end{aligned}$$

(A.15)

Once this is established, integrating (A.1) in time immediately yields (A.5).

Thus, to prove Lemma A.2, we only need to prove (A.15). Suppose, for contradiction, the inequality (A.15) does not hold. Letting $P_N :\dot{L}^2(M) \rightarrow \dot{L}^2(M)$ denote the orthogonal projection onto the span of the first N eigenfunctions of the Laplace–Beltrami operator, we observe

$$\begin{aligned} \frac{\lambda _N^2 t_0 \Vert \theta _0 \Vert _{L^2}^2}{8}&> \int _0^{t_0}\Vert \Delta \theta (s) \Vert _{L^2}^2\,\mathrm{d}s \geqslant \lambda _N^2\int _{t_0/2}^{t_0}\Vert (I-P_N)\theta (s) \Vert _{L^2}^2\,\mathrm{d}s \nonumber \\&\geqslant \frac{\lambda _N^2}{2}\int _{t_0/2}^{t_0}\Vert (I-P_N)\phi (s) \Vert _{L^2}^2\,\mathrm{d}s -\lambda _N^2\int _{t_0/2}^{t_0}\Vert (I-P_N)\left( \theta (s)-\phi (s) \right) \Vert _{L^2}^2\,\mathrm{d}s \nonumber \\&\geqslant \frac{\lambda _N^2 t_0}{4}\Vert \theta _{0} \Vert _{L^2}^2-\frac{\lambda _N^2}{2}\int _{t_0/2}^{t_0}\Vert P_N \phi (s) \Vert _{L^2}^2\,\mathrm{d}s -\lambda _N^2\int _{0}^{t_0}\Vert \theta (s)-\phi (s) \Vert _{L^2}^2\,\mathrm{d}s. \end{aligned}$$

(A.16)

We will now bound the last two terms in (A.16).

For the last term in (A.16), we use Lemma A.5 to obtain

$$\begin{aligned}&\int _{0}^{t_0}\Vert \theta (s)-\phi (s) \Vert _{L^2}^2\,\mathrm{d}s \nonumber \\&\quad \leqslant \int _{0}^{t_0} \sqrt{2\gamma s}\,\Vert \theta _0 \Vert _{L^2}\left( C_d \Vert u \Vert _{C^2}\int _0^s\Vert \Delta \theta (t) \Vert _{L^2}^2\,dt+ \Vert \Delta \theta _{0} \Vert _{L^2}^2 \right) ^{1/2}\,\mathrm{d}s\nonumber \\&\quad \leqslant C \sqrt{\gamma }\, t_0^{3/2} \Vert \theta _{0} \Vert _{L^2} \left( \Vert u \Vert _{C^2}\int _0^{t_0}\Vert \Delta \theta (t) \Vert _{L^2}^2\,dt + \Vert \Delta \theta _{0} \Vert _{L^2}^2 \right) ^{1/2} \nonumber \\&\quad \leqslant C \sqrt{\gamma }\,t_0^{3/2}\lambda _N\Vert \theta _{0} \Vert _{L^2}^2 \left( \Vert u \Vert _{C^2}\,t_0 +1 \right) ^{1/2}. \end{aligned}$$

(A.17)

For the last inequality above, we used our assumption that the inequality (A.15) does not hold.

To estimate the second term on the right of (A.16), let $e_j$ denote the eigenfunction of the Laplace–Beltrami operator corresponding to the eigenvalue $\lambda _j$. Now

$$\begin{aligned} \int _{t_0/2}^{t_0}\Vert P_N\phi (s) \Vert _{L^2}^2\,\mathrm{d}s \leqslant \sum _{j = 1}^N \int _0^{t_0} |\langle \phi (s), e_j \rangle |^2 \, \mathrm{d}s \leqslant t_0 h^2(t_0) \Vert \phi _0 \Vert _{H^1}^2 \sum _{j= 1}^N \lambda _j . \end{aligned}$$

Using Weyl’s lemma (A.9) and the assumption (A.4), we see

$$\begin{aligned} \int _{t_0/2}^{t_0}\Vert P_N\phi (s) \Vert _{L^2}^2\,\mathrm{d}s \leqslant C t_0 h^2(t_0) \Vert \phi _0 \Vert _{L^2}^2 \lambda _N^{(d + 4) /2}, \end{aligned}$$

(A.18)

for some constant $C = C(M)$.

We now let $C_1$ be the larger of the constants appearing in (A.17) and (A.18). Using these two inequalities in (A.16) shows

$$\begin{aligned} \tfrac{1}{8} > \tfrac{1}{4} - C_1 \lambda _N \sqrt{\gamma t_0} \left( 1 + t_0 \Vert u \Vert _{C^2} \right) ^{1/2} - C_1 \lambda _N^{(d + 4) / 2} h^2(t_0) . \end{aligned}$$

(A.19)

If we choose $\tilde{c} \geqslant \sqrt{16 C_1}$, then by equation (A.6) the last term on the right is at most 1/16. Next, when $\gamma$ is sufficiently small we will have $t_0 \Vert u \Vert _{C^2} \geqslant 1$. Thus, if $\tilde{C} \geqslant 16 \sqrt{2} C_1$ and $\lambda _N$ is the largest eigenvalue for which (A.3) holds, then the second term above is also at most 1/16. This implies $1/8 > 1/8$, which is the desired contradiction. $\square$

The proof of Lemma A.3 is very similar to that of Lemma A.2.

Proof of Lemma A.3

Follow the proof of Lemma A.2 until (A.18). Now, to estimate the second term on the right of (A.16), the strongly mixing property of u gives

$$\begin{aligned} \int _{t_0/2}^{t_0}\Vert P_N\phi (s) \Vert _{L^2}^2\,\mathrm{d}s&\leqslant \lambda _N\int _{t_0/2}^{t_0}\Vert \phi (s) \Vert _{H^{-1}}^2\,\mathrm{d}s \leqslant \lambda _N \int _{t_0/2}^{t_0}h^2(s) \Vert \theta _{0} \Vert _{H^1}^2\,\mathrm{d}s \nonumber \\&\leqslant \frac{t_0}{2}\lambda _N h^2(t_0/2)\,\Vert \Delta \theta _0 \Vert _{L^2}\Vert \theta _0 \Vert _{L^2} \leqslant \frac{t_0}{2}\lambda _N^{2}h^2(t_0/2)\, \Vert \theta _{0} \Vert _{L^2}^2. \end{aligned}$$

(A.20)

Above, the last inequality followed from interpolation and the assumption (A.4).

Now let $C_1$ be the constant appearing in (A.17). Using (A.17) and (A.20) in (A.16) implies

$$\begin{aligned} \tfrac{1}{8} > \tfrac{1}{4} - C_1 \lambda _N \sqrt{\gamma t_0} \left( 1 + t_0 \Vert u \Vert _{C^2} \right) ^{1/2} - \tfrac{1}{4} {\lambda _N^2 } h^2(t_0/2). \end{aligned}$$

If $t_0$ is defined by (A.8), then the last term above is at most 1/16. Moreover, if $\tilde{C} = 2^{9/2}\, C_1$ and $\lambda _N$ is the largest eigenvalue of the Laplace–Beltrami operator satisfying (A.7), then the second term above is also at most 1/16. This again forces $1/8 > 1/8$, which is our desired contradiction. $\square$