3.1 The stability near the energy minimizer ω

In the class of [ω], ω is the unique harmonic form with respect to g that minimizes the energy functional. When we restrict to the symplectic forms in [ω], the flow (3.1) can be viewed as a gradient flow of the energy functional E(ρ). Hence, it is expected that if one starts with an initial symplectic form ρ ₀ near ω, the flow exists for all time and converges to ω. We justify the expectation in this section, and the key point is the following monotonicity of the normalized energy. Denote

$\begin{align*} E_0=E(\omega )=\int _M \omega \wedge \omega =\int _M \omega \wedge *\omega =\int _M |\omega |^2 \,{\rm dvol}_g. \end{align*}$

For notational simplicity, we will drop the volume elements dvol _g in the integrals when there is no ambiguity. For a closed form α ∊ [ω], we have

$\begin{align*} \int _M |\alpha -\omega |^2 =\int _M |\alpha |^2-2\int _M \alpha \wedge \omega +\int _M |\omega |^2=E(\alpha )-E_0. \end{align*}$

Denote E ₀(α) = E(α) − E ₀. We have the following Poincaré inequality with c ₁ = c ₁(g):

(3.3) $\begin{align*} E_0(\alpha )\leq c_1 \int _M |d^*\alpha |^2 . \end{align*}$

We need the following equivalence for W ^{1, p} norms on the differential forms. For p ≥ 1, one defines the geometric W ^{1, p} norm of a form α by

$\begin{align*} \|\alpha \|_{W^{1, p}}^p=\int _M |\alpha |^p+\int _M |d \alpha |^p+\int _M |d^*\alpha |^p, \end{align*}$

and the analytic W _A ^{1, p} norm is defined as

$\begin{align*} \|\alpha \|_{W_{A}^{1, p}}^p=\int _M |\alpha |^p+\int _M |\nabla \alpha |^p. \end{align*}$

When p = 2, it is standard practice to show that two Sobolev norms are equivalent. A key point we need is that these two norms are equivalent for any p ≥ 1; see [Reference ScottSco95, Corollary 12].

As a consequence of this equivalence, we have the following proposition.

Proposition 3.4. For any close form α ∊ [ω] and p ≥ 1, we have the Poincaré inequality,

(3.4) $\begin{align*} \int _M |\alpha -\omega |^p\leq c_1 \int _M |d^*\alpha |^p. \end{align*}$

Moreover, we have the Sobolev–Poincaré inequality on (M ⁴,ω,g),

(3.5) $\begin{align*} \int _M |\alpha -\omega |^2 \leq c_2 \left (\int _M |d^*\alpha |^{{4}/{3}}\right )^{{3}/{2}}. \end{align*}$

Proof. Given the equivalence of two Sobolev norms on differential forms, the proof is rather standard and we sketch a brief argument. The standard Sobolev inequality implies that

$\begin{align*} \int _M |\alpha -\omega |^2 \leq C \left (\int _M |\nabla (\alpha -\omega )|^{{4}/{3}}\right )^{{3}/{2}}+C\left (\int _M |\alpha -\omega |^{{4}/{3}}\right )^{{3}/{2}}. \end{align*}$

The equivalence of the Sobolev norms implies that, with dα = 0,

$\begin{align*} \int _M |\alpha -\omega |^2 \leq C\left ( \int _M |d^*\alpha |^{{4}/{3}}\right )^{{3}/{2}}+C\left (\int _M |\alpha -\omega |^{{4}/{3}}\right )^{{3}/{2}}. \end{align*}$

Hence, (3.5) follows by the Poincaré inequality (3.4), with p = 4/3. The proof of (3.4) follows by a standard compactness argument and norm equivalence in Proposition 3.3. Suppose otherwise; then there exists a sequence of forms α _k ∊ [ω] such that

$\begin{align*} \int _M |d^*\alpha _k|^p\leq \frac {1}{k} \int _M |\alpha _k-\omega |^p. \end{align*}$

Suppose $\int _M |\alpha _k-\omega |^p=\lambda _k^p$ for each k. We consider β _k = λ _k ⁻¹(α _k −ω). Then β _k is an exact two form, and

$\begin{align*} \int _M |d^*\beta _k|^p\leq \frac {1}{k},\quad \int _M |\beta _k|^p=1. \end{align*}$

We write β _k = dζ _k , where ζ _k ∊ Im(d*) by Hodge decomposition. By the equivalence of the Sobolev norms, β _k ∊ W ^{1, p} with bounded Sobolev norm; note that by the Hodge theory, ζ _k is uniquely determined by β _k and has a uniform W ^{2, p} bound. Hence, β _k converges by subsequence weakly in W ^{1, p} to an exact form β _∞ = dζ _∞. Moreover, the weak convergence implies that

$\begin{align*} \int _M |d^*\beta _\infty |^p\leq \liminf \int _M |d^*\beta _k|^p=0. \end{align*}$

It follows that d*β _∞ = d*dζ _∞ = 0, and hence β _∞ = dζ _∞ = 0 (we have ζ _∞ = 0 as well, as a consequence of ζ _∞ ∊ Im(d*)). However, the Rellich–Kondrachov theorem implies that β _k converges strongly in L ^p . This contradicts the fact that β _k converges strongly in L ^p to β _∞ = 0, since ∥β _k ∥ _{L ^p} = 1.

With the Sobolev–Poincaré inequality, we have the following monotonicity of the normalized energy functional E ₀(ρ) along the flow.

Proposition 3.5. Suppose the flow (3.1) exists in [0, T). For any t < T,

(3.6) $\begin{align*} E_0(\rho (t))\leq E_0(\rho _0) e^{-{t}/{c_2}}. \end{align*}$

Proof. We have, along the flow,

$\begin{align*} \partial _t \int _M |\rho -\omega |^2=\partial _t \int _M |\rho |^2=-\int _M \frac {|d^* \rho |^2}{\sqrt {u}}. \end{align*}$

Apply the Hölder inequality:

$\begin{align*} \int _M fh \leq \left (\int _M f^{{3}/{2}}\right )^{{2}/{3}} \left (\int _M h^3\right )^{{1}/{3}}\quad {\rm with}\,\,f=\frac {|d^*\rho |^{{4}/{3}}}{u^{{1}/{3}}},\quad h=u^{{1}/{3}}. \end{align*}$

We get

$\begin{align*} \left (\int _M |d^*\rho |^{{4}/{3}}\right )^{{3}/{2}}\leq \int _M \frac {|d^* \rho |^2}{\sqrt {u}} \left (\int _M u\right )^{{1}/{2}}= \int _M \frac {|d^* \rho |^2}{\sqrt {u}}. \end{align*}$

By the Sobolev–Poincaré inequality (3.5), we have

$\begin{align*} \int _M |\rho -\omega |^2\leq c_2 \left (\int _M |d^*\rho |^{{4}/{3}}\right )^{{3}/{2}}\leq c_2 \int _M \frac {|d^* \rho |^2}{\sqrt {u}}. \end{align*}$

We get (3.6) as a consequence of

$\begin{align*} \partial _t \int _M |\rho -\omega |^2\leq -\frac {1}{c_2} \int _M |\rho -\omega |^2. \end{align*}$

Now we are ready to state the main theorem in this section.

The proof of Theorem 3.6 follows the standard line of argument given the exponential decay of normalized energy in (3.6). First, we need a more precise statement of short time existence, including the smooth dependence of initial data (finite time stability). The following is a standard fact of quasi-parabolic system applied to (3.1).

Proposition 3.7. Suppose there exists δ ₀ = δ ₀(ω,g), t ₀ = t ₀(ω,g), C ₀ = C ₀(ω,g) and C _k = C _k (ω,g,k), k ≥ 2, for any initial data ρ ₀ satisfying

$\begin{align*} \|\rho _0-\omega \|_{C^{1, \alpha }}\lt \delta _0. \end{align*}$

Then the flow exists over [0,t ₀], and for any t ∊ [0,t ₀],

$\begin{align*} \|\rho -\omega \|_{C^{1, \alpha }}\leq C_0 \|\rho _0-\omega \|_{C^{1, \alpha }}. \end{align*}$

Moreover, for any t ∊ [t ₀/2,t ₀] and k ≥ 2,

$\begin{align*} \|\rho -\omega \|_{C^{k, \alpha }}\leq C_k \|\rho _0-\omega \|_{C^{1, \alpha }}. \end{align*}$

We assume δ ₀ ∊ (0,1) is sufficiently small and denote, for k ≥ 2,

$\begin{align*}V_0=\{\rho \in [\omega ]: \|\rho -\omega \|_{C^{1, \alpha }}\lt \delta _0, \|\rho -\omega \|_{C^{k, \alpha }}\leq C_k\}.\end{align*}$

In particular, for any ρ ∊ V ₀, its potential volume u is closed to 1, such that the flow (3.1) is uniformly parabolic. Now, we fix the constants δ ₀, t ₀, C ₀, C _k and V ₀. For any t ∊ [t ₀/2,t ₀], ρ(t) ∊ V ₀. Since ρ(t ₀) ∊ V ₀, we can get a solution of (3.1) over [t ₀,2t ₀] by applying Proposition 3.7 with initial data ρ(t ₀). For t ∊ [t ₀,2t ₀], we have

$\begin{align*} \|\rho -\omega \|_{C^{1, \alpha }}\leq C_0 \|\rho (t_0)-\omega \|_{C^{1, \alpha }} \leq C_0^2 \|\rho _0-\omega \|_{C^{1, \alpha }} \end{align*}$

and

$\begin{align*} \|\rho -\omega \|_{C^{k, \alpha }}\leq C_k\|\rho (t_0)-\omega \|_{C^{k, \alpha }}\leq C_k^2 \|\rho _0-\omega \|_{C^{1, \alpha }}. \end{align*}$

Now, by choosing ϵ < < δ ₀ sufficiently small, we can get that ρ(t) ∊ V ₀ for t ∊ [t ₀,2t ₀] if∥ρ ₀ − ω∥ _{C ^{1, α}} < ϵ. Applying this repeatedly, we can get finite time stability near ω.

Now we are ready to prove Theorem 3.6.

Proof. Fix T ₀ > > 1 such that the flow exists over [0,T ₀] for the initial data ρ̃ ₀ satisfying ∥ρ̃ ₀ − ω∥ _{C ^{1, α}} < ϵ ₀ = ϵ(T ₀,δ ₀) and ρ(t) stays in V ₀ over [t ₀,T ₀]. Now choose ϵ ₁ < < ϵ ₀ such that the flow exists over [0,T ₁] with T ₁ > > T ₀ and ρ ∊ V ₀ over [t ₀,T ₁] if the initial datasatisfies

$\begin{align*} \|\tilde \rho _0-\omega \|_{C^{1, \alpha }}\lt \epsilon _1. \end{align*}$

To obtain a uniform stability, we apply Proposition 3.6 to get

$\begin{align*} \|\rho (t)-\omega \|^2_{L^2}\leq e^{-{t}/{c_2}} \|\rho _0-\omega \|^2_{L^2}\leq e^{-{t}/{c_2}}. \end{align*}$

Hence, we get that ∥ρ(t) − ω∥ C k, α ≤ C k (since ρ ∊ V 0) and ∥ρ(t) − ω∥ L 2 2 ≤ e −t/c 2 , where c 2 is the Sobolev–Poincaré constant. By the interpolation inequality, we get a uniform estimate for T 1 sufficiently large and t ≥ T 1/2:

(3.7) $\begin{align*} \|\rho (t)-\omega \|_{C^{1, \alpha }}\lt \epsilon _0. \end{align*}$

In particular, ρ(T ₁) satisfies

$\begin{align*} \|\rho (T_1)-\omega \|_{C^{1, \alpha }}\lt \epsilon _0. \end{align*}$

Now we fix T ₁ and take ϵ = ϵ ₁. Extend the flow with initial data ρ(T ₁). Then we obtain the flow solution ρ(t) over [0,T ₁+T ₀] with ρ(t) ∊ V ₀ over [t ₀,T ₁+T ₀]. Since the uniform estimate (3.7) on ρ(t) holds for all t ≥ T ₁/2, by repeating the argument, the flow exists over [0, ∞) with uniformly bounded C ^{k, α} norm, and the energy decay implies that ρ(t) converges to ω in the exponential rate.

3.2 Shi-type estimates

It is not clear whether the flow will exist for all time or develop a finite time singularity in general. To understand the behavior of a possible finite time singularity, we develop the Shi-type estimate [Reference ShiShi89] for the flow. First, we compute the evolution equations. The evolution equation of ρ reads, in local coordinates,

$\begin{align*} \partial _t \rho _{ij}=\frac {\Delta \rho _{ij}+(\mathcal{R}(\rho ))_{ij}}{\sqrt {u}}+\frac {u_i \rho _{jl, l}-u_j \rho _{il, l}}{2\sqrt {u^3}}. \end{align*}$

In this section, we assume that

(3.8) $\begin{align*} \frac {1}{\sqrt {u}}+|\nabla \rho |^2\leq C_0. \end{align*}$

Since, along the flow, ∥ρ∥ _{L ²(M)} is uniformly bounded, the assumption (3.8) implies that |ρ| is uniformly bounded. To derive the Shi-type estimates, we need only the following structures:

(3.9) $\begin{align*} \partial _t \rho =\frac {\Delta \rho +\mathcal{R}\circ \rho }{\sqrt {u}}+\frac {\rho \circ \nabla \rho \circ \nabla \rho }{\sqrt {u^3}}, \end{align*}$

where A ∘ B describes the possible contraction of A and B. Note that we have

$\begin{align*}\nabla u=\tfrac {1}{2} \langle (*\rho ), \nabla \rho \rangle =\rho \circ \nabla \rho .\end{align*}$

We have the following proposition.

Proposition 3.9. Assuming ( 3.8 ), we have

$\begin{align*} \partial _t |\nabla \rho |^2\leq \frac {\Delta |\nabla \rho |^2-\frac {3}{2}|\nabla^2 \rho |^2}{\sqrt {u}}+C, \end{align*}$

where C is a uniformly bounded constant depending on the metric g, its curvature and derivative of curvature, and the bound in ( 3.8 ). More generally, we have, for k ≥ 2,

(3.10) $\begin{align*} \partial _t |\nabla ^k \rho |^2\leq \frac {\Delta |\nabla ^k \rho |^2-\frac {3}{2}|\nabla ^{k+1} \rho |^2}{\sqrt {u}}+C_k \left (\sum _{i=2}^k |\nabla ^i\rho |^{2{(k+1)}/{i}}+1\right )\!. \end{align*}$

Proof. This is standard practice. With the symbolic notation, we have

$\begin{align*} \partial _t \nabla \rho =\frac {\Delta \nabla \rho }{\sqrt {u}}+\frac {\mathcal{R}\circ \nabla \rho +\nabla \mathcal{R}\circ \rho }{\sqrt {u}}+\frac {\nabla^2 \rho \circ \nabla \rho \circ \rho +\mathcal{R}\circ \rho \circ \rho \circ \nabla \rho }{\sqrt {u^3}}+\frac {\rho \circ \rho \circ \nabla \rho \circ \nabla \rho \circ \nabla \rho }{\sqrt {u^5}}. \end{align*}$

It follows that

(3.11) $\begin{align*} \partial _t |\nabla \rho |^2\leq \frac {\Delta |\nabla \rho |^2-2|\nabla^2\rho |^2}{\sqrt {u}}+C(|\nabla^2\rho |+1)\leq \frac {\Delta |\nabla \rho |^2-\frac {3}{2}|\nabla^2\rho |^2}{\sqrt {u}}+C. \end{align*}$

Note that |∇u| is assumed to be bounded. Hence, we have

$\begin{align*} \partial _t \nabla^2 \rho \leq \frac {\Delta \nabla^2 \rho }{\sqrt {u}}+C (|\nabla^3\rho |+|\nabla^2\rho |^2+|\nabla^2 \rho |+1). \end{align*}$

And, hence,

$\begin{align*} \begin{split} \partial _t |\nabla^2\rho |^2 &\leq \frac {\Delta |\nabla^2 \rho |^2-2|\nabla^3 \rho |^2}{\sqrt {u}}+C|\nabla^2\rho |(|\nabla^3\rho |+|\nabla^2\rho |^2+|\nabla^2 \rho |+1))\\ &\leq \frac {\Delta |\nabla^2 \rho |^2-\frac {3}{2}|\nabla^3 \rho |^2}{\sqrt {u}}+C(|\nabla^2\rho |^3+1). \end{split} \end{align*}$

This proves (3.10) for k = 2. For general k, an inductive computation gives that

$\begin{align*} \partial _t |\nabla ^k\rho |^2\leq \frac {\Delta |\nabla ^k \rho |^2-2|\nabla ^{k+1} \rho |^2}{\sqrt {u}}+C|\nabla ^k\rho |\left (\sum |\nabla ^{i_1}\rho | |\nabla ^{i_2}\rho |\ldots |\nabla ^{i_p}\rho |+1\right )\!, \end{align*}$

where the sum is taken such that i ₁ + i ₂ + ⋯ + i _p ≤ k + 2, with 1 ≤ i _a ≤ k + 1. By applying Young’s inequality,

$\begin{align*} a_1 a_2\ldots a_k\leq \sum \frac {a_i^{p_i}}{p_i},\quad \sum p_i^{-1}=1, \end{align*}$

we can get

$\begin{align*} \partial _t |\nabla ^k\rho |^2\leq \frac {\Delta |\nabla ^k \rho |^2-\frac {3}{2}|\nabla ^{k+1} \rho |^2}{\sqrt {u}}+C_k \left (\sum _{i=2}^k |\nabla ^i\rho |^{{2(k+1)}/{i}}+1\right )\!. \end{align*}$

Now we state the Shi-type estimate.

Proposition 3.10. Suppose ( 3.8) holds over [0,T]. Then we have the uniform bound for ∇ ^k ρ for t ∊ (0, T], k ≥ 2,

(3.12) $\begin{align*} |\nabla ^k\rho |\leq C_k t^{-{(k-1)}/{2}}. \end{align*}$

Such an estimate can be made local. Suppose ( 3.8 ) holds on B _r (p) × [0,T], where B _r (p) is the geodesic ball of the background metric g. Then over B _r/2(p) × [T/2,T], we have

(3.13) $\begin{align*} |\nabla ^{k}\rho |\leq C_k(g, C_0, k, T, r). \end{align*}$

Proof. Given (3.10), the proof of (3.12) follows almost line by line that of He and Li [Reference He and LiHL21, Proposition 2.1]. The local estimate is also standard. Denote Q = (μ+|∇ρ|²)|∇² ρ|², with μ determined later. We compute

$\begin{align*} \begin{split} \partial _t Q&=\partial _t |\nabla \rho |^2 \;|\nabla^2\rho |^2+(\mu +|\nabla \rho |^2) \partial _t |\nabla^2 \rho |^2\\ &\leq \bigg (\frac {\Delta |\nabla \rho |^2-\frac {3}{2}|\nabla^2 \rho |^2}{\sqrt {u}}+C\bigg ) |\nabla^2\rho |^2\\&\quad +(\mu +|\nabla \rho |^2)\bigg (\frac {\Delta |\nabla^2 \rho |^2-\frac {3}{2}|\nabla^3 \rho |^2}{\sqrt {u}}+C(|\nabla^2\rho |^3+1)\bigg )\\ &\leq \frac {\Delta Q}{\sqrt {u}}-\frac {3}{2\sqrt {u}} |\nabla^2\rho |^4-\frac {3\mu }{2\sqrt {u}} |\nabla^3\rho |^2 +C_1(|\nabla^3\rho |\;|\nabla^2\rho |^2+|\nabla^2\rho |^3+1). \end{split} \end{align*}$

Since u ≤ C ₂ = C ₂(C ₀,g) is uniformly bounded by our assumption, by choosing μ sufficiently large, depending on C ₁ and C ₂, we have

(3.14) $\begin{align*} \partial _t Q-\frac {\Delta Q}{\sqrt {u}}\leq -A_0 Q^2+A_1. \end{align*}$

With (3.14) the localization is standard as well. Choose a cut-off function φ supported in B _r (p), and φ = 1 in B _r/2. Denote, for a suitable constant A ₂ = A ₂(r ⁻¹,A ₀,A ₁,C ₀),

$\begin{align*} H=\frac {A_2}{\phi^2}+\frac {1}{A_0 t}+\sqrt {A_1 A_0^{-1}}. \end{align*}$

We compute

$\begin{align*} \partial _t H=-\frac {1}{A_0 t^2},\quad \Delta H=\frac {A_2(-2\phi \Delta \phi+6|\nabla \phi |^2)}{\phi^4},\quad H^2\gt \frac {A_2^2}{\phi^4}+\frac {1}{A_0^2 t^2}+\frac {A_1}{A_0}. \end{align*}$

For x ∊ B _r (p), we have |Δφ| + |∇φ|² ≤ 10r ⁻². Hence, for A ₂ sufficiently large,

$\begin{align*} \partial _t H-\frac {\Delta H}{\sqrt {u}}\gt -A_0 H^2+A_1. \end{align*}$

A maximum principle argument then implies that Q ≤ H over B _r (p) × [0,T]. This proves (3.13) for k = 2. For general k, take Q _k = (μ _k +|∇ ^{k − 1} ρ|²)|∇ ^k ρ|² for sufficiently large μ _k .

Hence, by the Shi-type estimate, if the flow exists for the maximal interval [0, T) with T < ∞, then it develops finite singularities at t = T with $1/\sqrt {u}+|\nabla \rho |^2$ blowing up. Suppose $1/\sqrt {u}$ is bounded; only the Sobolev bound ρ ∊ W ^{1, p} for p > 4 is needed (ρ would be Hölder continuous in particular). This is a well-known technique in the study of harmonic map theory. Even though the argument is standard for specialists, it is technically very involved. Since we do not need this result, we will not pursue such a generalization.

3.3 Doubling-time property

In this subsection, we show a doubling-time property for $|\nabla \rho |^2+\sqrt {u}^{-1}$ as follows. Denote

$\begin{align*} \Lambda (x, t)=|\nabla \rho |^2(x, t)+\frac {1}{\sqrt {u(x, t)}}+1,\quad \Lambda (t)=\max _{x} \Lambda (x, t),\quad \Lambda _0=\Lambda (0). \end{align*}$

Proposition 3.11. There exists a uniformly bounded constant C ₀ such that

(3.15) $\begin{align*} \Lambda (t)\leq \frac {\Lambda _0}{\sqrt [7]{1-7C_0 \Lambda _0^7t}} \quad \rm \ for\ \; t\lt (7C_0\Lambda _0^7)^{-1}. \end{align*}$

Proof. Note that ∥ρ∥ _{L ²} is uniformly bounded; max |ρ|(x,t) ≤ CΛ(t) holds for a uniformly bounded constant C depending only on the background geometry. Recall

$\begin{align*} \begin{split} \partial _t \nabla \rho =\frac {\Delta \nabla \rho }{\sqrt {u}}+\frac {\mathcal{R}\circ \nabla \rho +\nabla \mathcal{R}\circ \rho }{\sqrt {u}}+\frac {\nabla^2 \rho \circ \nabla \rho \circ \rho +\mathcal{R}\circ \rho \circ \rho \circ \nabla \rho }{\sqrt {u^3}}+\frac {\rho \circ \rho \circ \nabla \rho \circ \nabla \rho \circ \nabla \rho }{\sqrt {u^5}}. \end{split} \end{align*}$

Hence, comparing (3.11), we compute

(3.16) $\begin{align*} \partial _t |\nabla \rho |^2-\frac {\Delta |\nabla \rho |^2}{\sqrt {u}}\leq C (\Lambda (t))^8. \end{align*}$

For the potential u, we compute

(3.17) $\begin{align*} \partial _t u=\frac {\Delta u}{\sqrt {u}}-\frac {1}{\sqrt {u}} \langle \nabla (*\rho ), \nabla \rho \rangle +\frac {\nabla u\circ \nabla \rho \circ \rho }{\sqrt {u^3}}+\frac {\langle \mathcal{R}(\rho ), *\rho \rangle }{\sqrt {u}}. \end{align*}$

We can derive

(3.18) $\begin{align*} \partial _t u^{-{1}/{2}}-\frac {\Delta u^{-{1}/{2}}}{\sqrt {u}} \leq C(\Lambda (t))^8. \end{align*}$

It follows that, with a uniformly bounded constant C ₀,

(3.19) $\begin{align*} \partial _t \Lambda (x, t)-\frac {\Delta \Lambda (x, t)}{\sqrt {u}}\leq C_0 (\Lambda (t))^8. \end{align*}$

Hence, by the maximum principle, we can deduce that

$\begin{align*} \frac {d\Lambda (t)}{dt}\leq C_0 (\Lambda (t))^8. \end{align*}$

It follows that, for t ∊ [0, (7C ₀Λ ₀⁷)⁻¹), we have

(3.20) $\begin{align*} \Lambda (t)\leq \frac {\Lambda _0}{\sqrt [7]{1-7C_0 \Lambda _0^7t}}. \end{align*}$

If we denote

$\begin{align*} \tilde \Lambda (x, t)= |\rho |^2(x, t)+|\nabla \rho (x, t)|^2+\frac {1}{\sqrt {u(x, t)}}+1, \end{align*}$

we can derive a pointwise inequality of Λ̃(x,t), compared with (3.19):

(3.21) $\begin{align*} \partial _t \tilde \Lambda (x, t)\leq \frac {\Delta \tilde \Lambda (x, t)}{\sqrt {u}}+C_0( \tilde \Lambda (x, t))^8. \end{align*}$

3.4 Long time existence and convergence in W ^{1, 2} assuming the bound of |∇logu|

Here, we assume |∇logu| ≤ C ₀. Note that we assume the normalized condition

$\begin{align*} \int _M u\, {\rm dvol}=\int _M \,{\rm dvol}=1. \end{align*}$

Hence, for any t, there exists x _t ∊ M such that u(x _t ,t) = 1. If |∇logu| ≤ C ₀, then there exists C ₁ = C ₁(C ₀,g) such that

$\begin{align*} -C_1\leq \log u\leq C_1. \end{align*}$

This, in particular, implies that

$\begin{align*} e^{-C_1}\leq u\leq e^{C_1}\quad {\rm and}\quad |\nabla u|\leq C_0 e^{C_1}. \end{align*}$

We have the following proposition.

Proof. We need to be a bit more precise here, given the extra assumptions on u and ∇u. We write the equation as

(3.22) $\begin{align*} \partial _t \rho =\frac {\Delta \rho }{\sqrt {u}}+\frac {\mathcal{R}\circ \rho }{\sqrt {u}}+\nabla (u^{-{1}/{2}})\circ \nabla \rho . \end{align*}$

We compute

(3.23) $\begin{align*} \partial _t |\rho |^2=\frac {\Delta |\rho |^2-2|\nabla \rho |^2}{\sqrt {u}}+\frac {\langle \mathcal{R} (\rho ), \rho \rangle }{\sqrt {u}}+\frac {\nabla \log u \circ \nabla \rho \circ \rho }{\sqrt {u}}. \end{align*}$

Suppose |∇logu| ≤ C ₀, then we have

(3.24) $\begin{align*} \partial _t |\rho |^2-\frac {\Delta |\rho |^2}{\sqrt {u}}\leq \frac {-2|\nabla \rho |^2}{\sqrt {u}} +C(|\rho |^2+|\nabla \rho | |\rho |). \end{align*}$

It follows that |ρ|² ≤ |ρ ₀|² e ^Ct at time t. A straightforward computation gives

(3.25) $\begin{align*} \partial _t |\nabla \rho |^2\leq \frac {\Delta |\nabla \rho |^2-2 |\nabla^2\rho |^2}{\sqrt {u}}+C(|\nabla^2 \rho | |\nabla \rho |+|\nabla \rho |^2+|\nabla \rho | |\rho |+|\nabla^2 u||\nabla \rho |). \end{align*}$

For the potential u, we compute

(3.26) $\begin{align*} \partial _t u=\frac {\Delta u}{\sqrt {u}}-\frac {1}{\sqrt {u}} \langle \nabla (*\rho ), \nabla \rho \rangle +\frac {\nabla u\circ \nabla \rho \circ \rho }{\sqrt {u^3}}+\frac {\langle \mathcal{R}(\rho ), *\rho \rangle }{\sqrt {u}}. \end{align*}$

A straightforward computation also gives, with uniform bounds on u and |∇u|,

(3.27) $\begin{align*} \begin{split} \partial _t |\nabla u|^2&\leq \frac {\Delta |\nabla u|^2-2|\nabla^2 u|^2}{\sqrt {u}}\\ &\quad +C(|\nabla^2 u|+|\nabla^2\rho ||\nabla \rho |+|\nabla^2 u||\nabla \rho |+|\nabla^2\rho |+|\nabla \rho |^2+|\nabla \rho |+|\rho |^2). \end{split} \end{align*}$

Denote f = |∇ρ|² + A|∇u|² + B|ρ|² + 1 with 1 < < A < < B, where the constants A, B depend on the bound on u and ∇u. Then we have, for a uniformly bounded constant C ₁,

(3.28) $\begin{align*} \partial _t f-\frac {\Delta f}{\sqrt {u}}\leq C_1 f. \end{align*}$

A straightforward maximum principle argument implies that

$\begin{align*} |\nabla \rho |^2\leq C_2 e^{C_1 t}. \end{align*}$

Hence, if the flow exists over [0,T] for T < ∞ such that |∇logu| ≤ C ₀, then |∇ρ|² ≤ C ₂ e ^{C ₁ T} . By the Shi-type estimate, we know that the flow can be extended across T. In other words, the flow exists as long as |∇logu| stays bounded.

We have the following convergence result when assuming |∇logu| ≤ C ₀.

Proof. If one obtains a uniform estimate of |∇logu| ≤ C ₀ along the flow, then the long time existence follows Proposition 3.12. The point bounds on |ρ|², |∇ρ|², etc., derived in the last subsection, assuming |∇logu| ≤ C ₀ depends on a time factor e ^Ct which is not bounded when t → ∞. To derive convergence, we use the L ² estimate. Recall that we have

(3.29) $\begin{align*} \partial _t \int _M |\rho -\omega _0|^2 =-\int _M \frac {|d^*\rho |^2}{\sqrt {u}}\leq -C_1 \int _M |d^*\rho |^2. \end{align*}$

We compute, using |∇logu| ≤ C ₀,

(3.30) $\begin{align*} \begin{split} \partial _t \int _M |d^*\rho |^2=&-\int _M \left \langle d^*\rho , d^* d\left ( \frac {d^*\rho }{\sqrt {u}}\right )\right \rangle =-\int _M \left \langle dd^*\rho , d\left ( \frac {d^*\rho }{\sqrt {u}}\right )\right \rangle \\ \leq &-\int _M \frac {|dd^*\rho |^2}{\sqrt {u}}+ C\int _M |dd^*\rho | |d^*\rho |. \end{split} \end{align*}$

Denote

$\begin{align*} Q_1=\int _M |d^*\rho |^2+A_1 \int _M |\rho -\omega _0|^2. \end{align*}$

It follows that, for A ₁ sufficiently large,

(3.31) $\begin{align*} \partial _t Q_1\leq -C_1 \left (\int _M |dd^*\rho |^2+\int _M |d^*\rho |^2\right )\!. \end{align*}$

An application of the Poincaré inequality implies that ∂ _t Q ₁ ≤ − CQ ₁ for some positive constant C. It follows that, for t ∊ [0, ∞),

(3.32) $\begin{align*} \int _M |d^*\rho |^2+A_1 \int _M |\rho -\omega _0|^2\leq A_0 e^{-Ct}. \end{align*}$

When t → ∞, ρ(t) has a uniformly bounded W ^{1, 2} norm, and hence converges strongly in L ² and weakly in W ^{1, 2} to a limit form ρ _∞ with ρ _∞ ∊ [ω ₀], dρ _∞ = 0, ρ _∞ ∊ W ^{1, 2}. The convergence implies that

$\begin{align*} \int _M |d^*\rho _\infty |^2\leq \liminf \int _M |d^*\rho |^2=0,\quad \int _M |\rho _\infty -\omega _0|^2=0. \end{align*}$

It follows that ρ _∞ = ω ₀.

3.5 Smooth convergence

To obtain smooth convergence, we need to obtain the pointwise bound on |ρ| and |∇ρ|. This requires more careful analysis. We have the following proposition.

Proof. The main point is to bound f = |∇ρ|² + A|∇u|² + B|ρ|² + 1, which satisfies (3.28) given the uniform control on |∇logu|. We recall the inequality

(3.33) $\begin{align*} \partial _t f\leq \frac {\Delta f}{\sqrt {u}}+C_1 f. \end{align*}$

We apply the parabolic Moser iteration, which is a standard and very powerful tool for specialists. Since the argument is involved and technically delicate in its nature, we include the details for completeness. For simplicity, we work on the time interval [0,1]. For 2p − 1 > 0 (say take p ≥ (5/8)), we compute

(3.34) $\begin{align*} \begin{split} \partial _t \int _M f^{2p}&\leq 2p\int _M f^{2p-1}\frac {\Delta f}{\sqrt {u}}+2pC_1\int _M f^{2p}\\ &=-2p(2p-1)\int _M f^{2p-2} \frac {|\nabla f|^2}{\sqrt {u}}-2p\int _Mf^{2p-1}\nabla f \nabla (u^{-{1}/{2}})+2p\int _M f^{2p}\\ &\leq -4c_0 p(2p-1)\int _M f^{2p-2}|\nabla f|^2+pC\int _M f^{2p-1} |\nabla f|+2p\int _M f^{2p}\\ &\leq -4c_0\left (2-\frac {1}{p}\right )\int _M |\nabla (f^p)|^2+C\int _M f^{p-1}|\nabla (f^p)|+2p\int _M f^{2p}\\ &\leq -c_0\int _M |\nabla (f^p)|^2+pC \int _M f^{2p}, \end{split} \end{align*}$

where we use the Hölder inequality and Young’s inequality. We use c ₀ to denote a uniform positive constant depending on min u ^−1/2, and C to denote a uniformly bounded positive constant which can vary line by line. Applying a Sobolev inequality to h = f ^p ,

$\begin{align*} \left (\int _M h^4\right )^{{1}/{2}}\leq c_1 \int _M (|\nabla h|^2+h^2), \end{align*}$

we have the following inequality, for $p\in \left [\frac {5}{8}, \infty \right )$ :

(3.35) $\begin{align*} \partial _t \int _M f^{2p}+c_0\left (\int _M f^{4p}\right )^{{1}/{2}}\leq pC \int _M f^{2p}. \end{align*}$

Now, given 0 ≤ a < b ≤ 1, let φ : [0,1] → [0,1] be a Lipschitz function such that

$\begin{align*}\phi (s)=0,\quad s\in [0, a];\quad \phi (s)=1,\quad s\in [b, 1],\end{align*}$

and φ(s) is linear for s ∊ [a,b]. Then

$\begin{align*}\phi {'}(s)=\frac {1}{b-a} ,\quad s\in (a, b);\quad \phi {'}=0,\quad s\in [0, a)\cup (b, 1].\end{align*}$

Taking integration with respect to t over [0,t ₀], for any t ₀ ∊ [a,1], we have

(3.36) $\begin{align*} \int _0^{t_0} \phi^2(t)\partial _t\int _M f^{2p}+c_0 \int _0^{t_0} \phi^2(t) \left (\int _M f^{4p}\right )^{{1}/{2}}\leq pC\int _0^{t_0} \phi^2(t)\int _M f^{2p}. \end{align*}$

Integrating by parts on the first term above, we have

(3.37) $\begin{align*} (\phi^2\int _M f^{2p}) (t_0)+c_0\int _0^{t_0} \phi^2 \left (\int _M f^{4p}\right )^{{1}/{2}}\leq \left (pC+\frac {2}{b-a}\right )\int _a^{t_0}\int _M f^{2p}. \end{align*}$

We have, by (3.37),

(3.38) $\begin{align*} \max _{[b, 1]}\int _M f^{2p}\leq \left (pC+\frac {2}{b-a}\right )\int _a^1\int _M f^{2p} \end{align*}$

and

(3.39) $\begin{align*} \int _{b}^1\left (\int _M f^{4p}\right )^{{1}/{2}}\leq \left (pC+\frac {2}{b-a}\right )\int _a^1\int _M f^{2p}. \end{align*}$

Now, for any t ∊ [b,1], applying the Hölder inequality, we get

(3.40) $\begin{align*} \int _M f^{3p} \leq \left (\int _M f^{4p}\right )^{{1}/{2}}\left (\int _M f^{2p}\right )^{{1}/{2}}\leq \left (\max _{[b, 1]}\int _M f^{2p}\right )^{{1}/{2}}\left (\int _M f^{4p}\right )^{{1}/{2}}. \end{align*}$

It follows that over the space time [b,1] × M,

(3.41) $\begin{align*} \int _b^1\int _M f^{3p}\leq \left (\max _{[b, 1]}\int _M f^{2p}\right )^{{1}/{2}} \int _b^1 \left (\int _M f^{4p}\right )^{{1}/{2}}. \end{align*}$

Applying (3.38) and (3.39) to (3.41), we have

(3.42) $\begin{align*} \left ( \int _b^1\int _M f^{3p}\right )^2\leq \left (pC+\frac {2}{b-a}\right )^3\left (\int _a^1\int _M f^{2p}\right )^3. \end{align*}$

We write the above as

(3.43) $\begin{align*} \|f\|_{L^{3p}([b, 1]\times M)}\leq \left (pC+\frac {2}{b-a}\right )^{{1}/{2p}} \|f\|_{L^{2p}([a, 1]\times M)}. \end{align*}$

In other words, we have

(3.44) $\begin{align*} \log \|f\|_{L^{3p}([b, 1]\times M)}\leq \frac {1}{2p} \log (pC+2(b-a)^{-1})+\log \|f\|_{L^{2p}([a, 1]\times M)}. \end{align*}$

Now, take a sequence of p _i and b _i such that

$\begin{align*} p_i=\frac {1}{2}\left (\frac {3}{2}\right )^{i+1},\quad b_i=\frac {1}{2}-\frac {1}{2^{i+1}},\quad i=0, 1, 2, \ldots \end{align*}$

and denote q _i = 2p _i . For each i, applying (3.44) with p = p _i , a = b _i , b = b _{i + 1}, we get

(3.45) $\begin{align*} \log \|f\|_{L^{q_{i+1}}([b_{i+1}, 1]\times M)}\leq \left (\frac {2}{3}\right )^{i+1}\log (2^{i+2}+Cq_i)+\log \|f\|_{L^{q_i}([b_i, 1]\times M}). \end{align*}$

By letting i → ∞ and the fact that $\sum _i (2/3)^i i\lt \infty$ , (3.45) gives

(3.46) $\begin{align*} \log \|f\|_{L^\infty ([\frac {1}{2}, 1]\times M)}\leq C+\log \|f\|_{L^{{3}/{2}}([0, 1]\times M)}. \end{align*}$

By choosing the constants in a more careful way, we can indeed get that, for any r > 1, since (3.44) holds only for 2p > 1,

(3.47) $\begin{align*} \log \|f\|_{L^\infty ([\frac {1}{2}, 1]\times M)}\leq C_1(r)+\log \|f\|_{L^{r}([0, 1]\times M)}, \end{align*}$

where the constant C ₁ depends on r, and it could blow up when r → 1. To overcome this difficulty, we prove that f is indeed in L ^3/2([t,t+1]×M) for any t ≥ 0. Note that ρ is in W ^{1, 2}(M) and |ρ| is in L ⁴(M). But at the moment, we have no uniform control on $\int _M |\nabla \rho |^3$ for each time slice. Note that, given bounded |∇u|,

$\begin{align*} \int _M f^{{3}/{2}}\leq 2\int _M |\nabla \rho |^3+C \int _M |\rho |^3+C. \end{align*}$

Hence, to show that f ∊ L ^3/2([t,t+1]×M) for each t, it suffices to show that for any t,

(3.48) $\begin{align*} \int _t^{t+1}\int _M |\nabla \rho |^3\leq C. \end{align*}$

To obtain (3.48), we need to use (3.31). Taking the integral over [t,t+1], we get that

$\begin{align*} \int _t^{t+1} \int _M |d^*d\rho |^2+\int _t^{t+1} \int _M |d^*\rho |^2\leq Q_1(t)-Q_1(t+1)\leq C. \end{align*}$

Using the Bochner–Weitzenböck formula, we obtain that

$\begin{align*} \int _t^{t+1} \int _M |\nabla^2\rho |^2+\int _t^{t+1} \int _M |\nabla \rho |^2\leq C. \end{align*}$

By applying the Sobolev inequality on each time slice, we obtain

$\begin{align*} \left (\int _M |\nabla \rho |^4\right )^{{1}/{2}}\leq C\left (\int _M |\nabla^2\rho |^2+|\nabla \rho |^2\right )\!. \end{align*}$

Hence, we have

$\begin{align*} \int _t^{t+1} \left (\int _M |\nabla \rho |^4\right )^{{1}/{2}}\leq C. \end{align*}$

Now, we have, since |∇ρ| ∊ L ²(M),

$\begin{align*} \int _t^{t+1}\int _M |\nabla \rho |^3\leq \int _t^{t+1} \left (\int _M |\nabla \rho |^4\right )^{{1}/{2}} \left (\int _M |\nabla \rho |^2\right )^{{1}/{2}}\leq C \int _t^{t+1}\left (\int _M |\nabla \rho |^4\right )^{{1}/{2}}\leq C. \end{align*}$

This proves (3.48). Applying (3.46) over [t,t+1] and (3.48), we have the uniform control on f over [t+(1/2),t] for any t. This gives the uniform bounds on |ρ| and |∇ρ|, which implies the smooth convergence of ρ → ω ₀ along the flow.

The assumption |∇logu| ≤ C can be weakened considerately if we carefully refine the iteration arguments above.

Proof. First, we note that u is bounded above by assuming |∇u| ∊ L ^{p ₀} . The key is to bound |ρ| and |∇ρ| along the flow via the parabolic Moser iteration, and the strategy is essentially the same as for Proposition 3.14. We use the same notation and highlight only the major difference in the arguments. We have

(3.49) $\begin{align*} \begin{split} \partial _t |\rho |^2 &\leq u^{-{1}/{2}} (\Delta |\rho |^2-2|\nabla \rho |^2)+C(|\nabla u| |\nabla \rho | |\rho |+|\rho |^2)\\ &\leq u^{-{1}/{2}} (\Delta |\rho |^2-|\nabla \rho |^2)+ C(|\nabla u|^2+1) |\rho |^2. \end{split} \end{align*}$

Denote h = |ρ|². We compute

(3.50) $\begin{align*} \begin{split} \partial _t \int _M h^{2p}=&\;2p\int _M h^{2p-1}\partial _t h\\ \leq &\;2p\int _M h^{2p-1} \left (u^{-{1}/{2}} (\Delta h-|\nabla \rho |^2)+ C(|\nabla u|^2+1) h\right )\\ \leq &-c_0 \int _M |\nabla (h^p)|^2+pC\int _M h^{2p} (1+|\nabla u|^2), \end{split} \end{align*}$

which is similar to (3.34); note that the only difference is that we have no pointwise bound on |∇u|, and hence we have an extra term involved with |∇u|² compared with (3.34). The iteration process is almost identical, but it is more involved without the pointwise bound on |∇u|. We proceed as in (3.42) to get

(3.51) $\begin{align*} \left (\int _b^1 \int _M h^{3p}\right )^2\leq \left (pC+\frac {2}{b-a}\right )^3\left (\int _a^1\int _M h^{2p}(1+|\nabla u|^2)\right )^3. \end{align*}$

Since |∇u| ∊ L ^{p ₀/2}(M), we can obtain

(3.52) $\begin{align*} \int _M h^{2p}(1+|\nabla u|^2)\leq \left (\int _M h^{2p q_0}\right )^{{1}/{q_0}}\left (\int _M (1+|\nabla u|^2)^{{p_0}/{2}}\right )^{{2}/{p_0}},\quad \frac {1}{q_0}+\frac {2}{p_0}=1. \end{align*}$

Note that q ₀ < (3/2) and

(3.53) $\begin{align*} \int _a^1 \left (\int _M h^{2p q_0}\right )^{{1}/{q_0}} dt\leq (1-a)^{{p_0}/{2}}\left (\int _a^1 \int _M h^{2pq_0}\right )^{{1}/{q_0}}. \end{align*}$

Hence, we obtain

(3.54) $\begin{align*} \left (\int _b^1 \int _M h^{3p}\right )^2\leq \left (pC+\frac {2}{b-a}\right )^3 C_0 \left (\int _a^1\int _M h^{2pq_0}\right )^{{3}/{q_0}}, \end{align*}$

where C ₀ depends on ∥∇u∥ _{L ^{p 0}} . This implies that

(3.55) $\begin{align*} \log \|h\|_{L^{3p}([b, 1]\times M)}\leq \frac {1}{2p} \left (\log (pC+2(b-a)^{-1})+\log C_0\right )+\log \|h\|_{L^{2pq_0}([a, 1]\times M)}. \end{align*}$

Since q ₀ < (3/2), we can then proceed with the iteration process to bound h over [(1/2),1], and hence to bound |ρ|. The arguments are almost identical from this point, and we skip the details. To bound |∇ρ|, denote again f = |∇ρ|² + A|∇u|² + B|ρ|² + 1. Without the bound on |∇u|, we can obtain

(3.56) $\begin{align*} \partial _t f\leq \frac {\Delta f}{\sqrt {u}}+C f(1+|\nabla u|^2). \end{align*}$

We can bound f following the same argument, given that |∇u| ∊ L ^{p ₀} for p ₀ > 6.

One might ask whether it is sufficient for long time existence to assume u ⁻¹ + ∥∇u∥ _{L ^p} ≤ C for any p > 4. The following technical question would be much more difficult and one would expect new technical ingredients to prove such a statement.

Question 3.16. Does the flow exist for all time when assuming

$\begin{align*} u+u^{-1}+\|\nabla u\|_{L^4}\leq C? \end{align*}$

Remark 3.17. The conformal factor $\sqrt {u}$ can be replaced by any power u ^r for r ∊ (0, (1/2)]. For such r, the metric g _r = u ^r g has total volume

$\begin{align*} V_r=\int _M u^{2r} dv\leq \int _M dv. \end{align*}$

The results about stability and long time existence also hold true for r ∊ (0, (1/2)]. But for r > (1/2), because of the Sobolev embedding, we will not have a straightforward exponential energy decay. For stability near ω, one needs extra input.

3.6 Examples and discussions

In this section, we discuss some examples. First, we consider an example on T ⁴ = S ¹ × S ¹ × S ¹ × S ¹. Let ω ₀ = dx ₀ ∧ dx ₁ + dx ₂ ∧ dx ₃ and $g_0=\sum dx_i\otimes dx_i$ . Suppose ρ ₀ = u ₀(x ₁,x ₂)dx ₁ ∧ dx ₂ + dx ₃ ∧ dx ₄ for a positive function u ₀ : T ² = S ¹ × S ¹ → ℝ₊ satisfying

(3.57) $\begin{align*} \int _{T^2} u_0 dx_1\wedge dx_2=\int _{T^2} dx_1\wedge dx_2. \end{align*}$

Clearly, dρ ₀ = 0 and (3.57) imply that ρ ₀ ∊ [ω ₀]. A straightforward computation implies that the flow (3.1) reduces to the parabolic flow of u(t) : T ² → ℝ₊,

$\begin{align*} \partial _t u=\frac {\Delta u}{\sqrt {u}}-\frac {|\nabla u|^2}{2\sqrt {u^3}}. \end{align*}$

Note that one can write the equation as

(3.58) $\begin{align*} \partial _t u=2\Delta ({u^{\frac {1}{2}}}). \end{align*}$

This is an example of the fast diffusion equation. Since u ₀ is assumed to be positive, the maximum principle implies that u stays uniformly bounded and uniformly positive. By the Krylov–Safonov estimate, u has a uniformly bounded Hölder norm. Hence, (3.58) has a unique long time smooth solution with uniformly bounded Hölder norms by the standard parabolic theory. The gradient flow nature implies that u converges smoothly to u _∞ ≡ 1.

This extends to the example (M ⁴,ω ₀=ω ₁⊕ω ₂) = (Σ ₁,ω ₁) × (Σ ₂,ω ₂) for two closed surfaces with two volume forms ω ₁, ω ₂. Suppose ρ ₀ ∊ [ω ₀] of the form ρ ₀ = u ₀ ω ₁ ⊕ ω ₂ for u ₀ : Σ ₁ → ℝ₊ such that

$\begin{align*} \int _{\Sigma _1} u_0 \omega _1=\int _{\Sigma _1} \omega _1. \end{align*}$

The flow (3.1) is then reduced to the fast diffusion equation in (3.58) in the classical setting. The computation is almost identical. The example above does not extend to the case when ρ ₀ = v ₀(x ₁,x ₂)dx ₁ ∧ dx ₂ + w ₀(x ₃,x ₄)dx ₃ ∧ dx ₄ satisfying v ₀, w ₀ > 0, and

$\begin{align*} \int _{T^2} v_0 dx_1\wedge dx_2=\int _{T^2} w_0 dx_3\wedge dx_4=\int _{T^2} dx_1\wedge dx_2=\int _{T^2} dx_3\wedge dx_4. \end{align*}$

In this case, ρ ₀ ∊ [ω ₀], but the product structure is not preserved along the flow (3.1). Note that in this example, ρ ₀ is isotropic to ω ₀ in an obvious way.

Problem 3.18. Does the flow (3.1) exist for all time and converge to ω ₀ on T ⁴ in the above example, with the initial symplectic form

$\begin{align*}\rho _0=v_0(x_1, x_2) dx_1\wedge dx_2+w_0(x_3, x_4) dx_3\wedge dx_4?\end{align*}$

The product example above might give some indication of how to choose the metric g̃ = g(Q,·) and the corresponding Hodge star $\tilde *$ . Suppose the matrices A and B are defined as ρ = g(A·,·) and * ρ = g(B·,·). For g ₀ and ρ ₀ given above on T ⁴, we have

Hence, a = − A ² = diag(v ²,v ²,w ²,w ²) and b = − B ² = diag(w ²,w ²,v ²,v ²). One can defineQ = fa ^r or fb ^r for some positive function f and some power r. For the conformal metric, we have $Q=\sqrt {u} I$ . For the metric g _ρ , Q is given by u ⁻¹ a. Neither choice of $Q=\sqrt {u}I$ nor Q = a (or Q = u ⁻¹ a) would preserve the product structure. In particular, the flow becomes complicated for these choices. But if we choose Q = b (or b ^r for r > 0), then the flow preserves the product structure such that

$\begin{align*} \partial _t v=\Delta \left (-\frac {1}{v}\right ),\quad \partial _t w=\Delta \left (-\frac {1}{w}\right )\!. \end{align*}$

If we choose $Q=\sqrt {b}$ , then the flow equation preserves the product structure such that

$\begin{align*} \partial _t v=\Delta (\log v),\quad \Delta w=\Delta (\log w). \end{align*}$

For such a choice, the flow equation becomes the classical fast diffusion equation and the flow exists for all time and converges to ω ₀ on T ⁴ by the standard theory of parabolic equations.

Problem 3.19. Let (M,ω,g) be an almost Kähler manifold of dimension four. For each symplectic form ρ ∊ [ω], we define the metric g̃ = g(b·,·). Consider the flow

$\begin{align*} \partial _t \rho =d(\tilde * d*\rho ). \end{align*}$

Does the flow exist for all time?

We will analyze the different Hodge flows in the next section. For the second example, we consider ρ ₀ ∊ [ω ₀] on (T ⁴,ω ₀,g ₀), with

$\begin{align*} \rho _0=dx_1\wedge dx_2+dx_3\wedge dx_4+d\theta \quad {\rm with}\; \theta =a(x_1, x_2)dx_3+b(x_1, x_2) dx_4, \end{align*}$

where a, b are two functions on S ¹ × S ¹ (hence, periodic functions on ℝ × ℝ). The nondegenerate condition implies that

$\begin{align*} u_0=1-a_{x_1} b_{x_2}+a_{x_2} b_{x_1}\gt 0. \end{align*}$

Note that for this example, ρ ₀ is isotropic to ω ₀ via ρ _s = ω ₀ + sdθ for s ∊ [0,1], with ρ _s nondegenerate for each s. The linear Hodge Laplacian flow reduces to the heat equations for a, b separately:

$\begin{align*} \partial _t a=\Delta a=a_{x_1x_1}+a_{x_2x_2},\quad \partial _t b=\Delta b=b_{x_1x_1}+b_{x_2x_2}. \end{align*}$

However, along the Hodge Laplacian flow, the positivity condition on u = 1 − a _{x 1} b _{x 2} + a _{x 2} b _{x 1} might not hold. This fact should be well known for experts. Since we cannot find a direct reference, we construct an explicit example as follows.

Example 3.20. We consider the linear Hodge flow for ρ = dx ∧ dy + dz ∧ dw + d(a(x,y)dz+b(x,y)dw) on T ⁴. The flow is reduced to the heat equation on T ²,

$\begin{align*} \partial _t a=\Delta a,\quad \partial _t b=\Delta b, \end{align*}$

where a and b are 2π periodic functions for x and y, respectively. Take a(x,y) = a(x), and suppose a _x = f(x) while b(x,y) = h(x)e(y). Along the linear flow, f(x), h(x), e(y) are all periodic functions (functions on S ¹) satisfying the heat equation

$\begin{align*} \partial _t f=f_{xx},\quad \partial _t h=h_{xx}\quad {\rm and}\,\; \partial _t\, e=e_{yy}. \end{align*}$

Denote e _y = c(y). For the initial data, u ₀(x,y) = 1 − f ₀(x)h ₀(x)c ₀(y) and u(x,y,t) = 1 − f(x,t)h(x,t)c(y,t), since a(x,y,t) = f(x,t) is independent of y. There is a restriction on f and c such that

$\begin{align*} \int _0^{2\pi } f_0(x)\, dx=\int _0^{2\pi } c_0(y)\, dy=0. \end{align*}$

We choose f ₀, h ₀ such that f ₀ = 0 for x ∊ [0,π] and f ₀(x) = sin 2x for x ∊ [π,2π], and h ₀(x) = 0 for x ∊ [π,2π] and h ₀(x) = sin (2x) for x ∊ [0,π]. Hence, u ₀(x,y) = 1 for any (x,y) ∊ S ¹ × S ¹. Let f(x,t) and h(x,t) solve the heat equation with initial data f ₀ and h ₀, respectively. Clearly, f(x,t)h(x,t) is not identically zero for all t > 0, and f(x,t) and h(x,t) both converge to zero when t → ∞. Let max f(x,1)h(x,1) = 1/A > 0. Let c(y,t) = A ₀ e ^−t sin y. Then u(x,y,1) < 0 for some point if A ₀ > Ae. By choosing A ₀ sufficiently large depending on t ₀, u(x,y,t ₀) can become negative for arbitrarily small t ₀ > 0.

The conformal flow (3.1) is reduced to the system

(3.59) $\begin{align*} \partial _t a=\frac {\Delta a}{\sqrt {u}},\quad \partial _t b=\frac {\Delta b}{\sqrt {u}}, \end{align*}$

and the factor $1/\sqrt {u}$ forces the positivity of u before developing a singularity.

Proof. This is a straightforward computation and the key point is that d*θ = 0. Suppose a, b are unique solutions of (3.59). Then

$\begin{align*} \partial _t \rho =d(\partial _t \theta )=d\left (\frac {\Delta a\; dx_3+\Delta b\; dx_4}{\sqrt {u}}\right )\!. \end{align*}$

Since d*θ = 0, we have

$\begin{align*}-d*\rho =-d^* d\theta =\Delta \theta =\Delta a \;dx_3+\Delta b \; dx_4.\end{align*}$

Hence, we have

$\begin{align*} \partial _t \rho =-d\left (\frac {d^*\rho }{\sqrt {u}}\right )\!. \end{align*}$

The system (3.59) is a quasilinear parabolic system for (a,b) on T ², together with a positivity restriction u = 1 − a _{x 1} b _{x 2} + a _{x 2} b _{x 1} > 0.

It is not clear at the moment whether the flow (3.1) would develop a singularity or not when n = 4. The study of the two special examples above might give a hint for this problem.

3.7 Singularity and soliton

When n = 6, the corresponding geometric flow would have to develop singularities in general. In dimension 6, we choose r = 1/3 to consider the flow

$\begin{align*} \partial _t \rho =-d\left (\frac {d^*\rho }{u^{{1}/{3}}}\right )\!. \end{align*}$

With almost identical arguments as for n = 4, one can show that if |∇logu| stays uniformly bounded along the flow, then the flow exists for all time and converges to ω strongly in L ². (If one assumes further that |∇ρ| is bounded, then the convergence would be in C ^∞ topology). One expects that, in general, the symplectic forms in a cohomology class [ω] are not path-connected. A particular example on T ² × S ² × S ² is provided by McDuff. In this case, the flow has to develop singularities. The flow can be viewed as a gradient flow, and the normalized energy decreases strictly along the flow (except at ω); it would be interesting to analyze the formation of singularities along the flow. We pose a concrete question on formation of singularities.

One considers the solitons on ℝ⁴. A particular type is

(3.60) $\begin{align*} d\left (\frac {d^*\rho _0}{\sqrt {u_0}}\right )=d(i_X\rho _0), \end{align*}$

where X is a Killing vector field of the Euclidean metric. Let ψ _t be the flow generated by X. Then ρ(t) = ψ _t *ρ ₀ satisfies the flow equation on ℝ⁴,

$\begin{align*} \partial _t \rho =-d\left (\frac {d^*\rho }{\sqrt {u}}\right )\!. \end{align*}$

We can describe a family of special solutions to (3.60) as follows. Let

$\begin{align*} \rho _0= a(x, y) dx\wedge dy+dw\wedge dz. \end{align*}$

Choose X = (V ¹,V ²,V ³,V ⁴) to be a constant vector field. Then (3.60) is reduced to

(3.61) $\begin{align*} \left (\frac {a_x}{\sqrt {a}}\right )_x+\left (\frac {a_y}{\sqrt {a}}\right )_y+V^1 a_x+V^2 a_y=0. \end{align*}$

It would be an interesting question to classify solutions to (3.60).

4.1 The linear Hodge flow h _ij = δ _ij

When Q _ij = h _ij = δ _ij , the flow is the Hodge flow

$\begin{align*} \partial _t \rho =\Delta \rho . \end{align*}$

The standard theory shows that the flow exists for all time with uniform control and converges to ω ₀ at exponential rate. But the potential u does not remain positive in general. In the previous section, we construct an example, for any given time t ₀ > 0, in which one can choose an initial symplectic form ρ ₀ with u ₀ = 1, but along the flow u can become negative at t ₀. Nevertheless, the linear Hodge flow can indicate some key difficulties in the study of nonlinear Hodge flow, in particular for conformal Hodge flows. We have

$\begin{align*} \begin{split} \partial _t |\rho |^2 &=\Delta |\rho |^2-2|\nabla \rho |^2, \partial _t u=\Delta u-\langle \nabla \rho , \nabla (*\rho )\rangle ,\\ \partial _t |\rho ^{+}|^2&=\Delta |\rho ^{+}|^2-2|\nabla \rho ^{+}|^2, \partial _t |\rho ^{-}|^2=\Delta |\rho ^{-}|^2-2|\nabla \rho ^{-}|^2. \end{split} \end{align*}$

The eigenvalues of A are $\pm \sqrt {-1} \lambda _1, \pm \sqrt {-1} \lambda _2$ , which are determined by

$\begin{align*} \lambda _1=\frac {1}{\sqrt {2}} (|\rho ^{+}|+|\rho ^{-}|),\quad \lambda _2=\frac {1}{\sqrt {2}} (|\rho ^{+}|-|\rho ^{-}|). \end{align*}$

Note that λ ₁ and λ ₂ are smooth functions at points where ρ ⁻ ≠ 0. We compute

$\begin{align*} \begin{split} \partial _t \lambda _1=\Delta \lambda _1-\frac {|\nabla \rho ^{+}|^2-|\nabla |\rho ^{+}||^2}{\sqrt {2}|\rho ^{+}|}-\frac {|\nabla \rho ^{-}|^2-|\nabla |\rho ^{-}||^2}{\sqrt {2}|\rho ^{-}|},\\ \partial _t \lambda _2=\Delta \lambda _2-\frac {|\nabla \rho ^{+}|^2-|\nabla |\rho ^{+}||^2}{|\rho ^{+}|}+\frac {|\nabla \rho ^{-}|^2-|\nabla |\rho ^{-}||^2}{|\rho ^{-}|}. \end{split} \end{align*}$

We denote, at points where ρ ⁻ ≠ 0,

(4.3) $\begin{align*} J=\frac {|\nabla \rho ^{+}|^2-|\nabla |\rho ^{+}||^2}{\sqrt {2}|\rho ^{+}|},\quad K=\frac {|\nabla \rho ^{-}|^2-|\nabla |\rho ^{-}||^2}{\sqrt {2}|\rho ^{-}|}. \end{align*}$

The maximum principle argument implies the larger eigenvalue λ ₁ is bounded above, but fails to preserve the positivity of the smaller eigenvalue λ ₂. (Strictly speaking, our computation only works at points where λ ₁ ≠ λ ₂, but with modifications the argument can work at points where λ ₁ = λ ₂ as well). Note that at the minimum of λ ₂, we have

$\begin{align*} \nabla |\rho ^{+}|=\nabla |\rho ^{-}|. \end{align*}$

At the first time, λ ₂ = 0, and hence |ρ ⁺| = |ρ ⁻| as well; the failure to preserve the positivity (nonnegativity) of λ ₂ lies in the fact that at such points (the first time λ ₂ = 0), the following control does not hold:

(4.4) $\begin{align*} |\nabla \rho ^{+}|\leq |\nabla \rho ^{-}|. \end{align*}$

Similar consideration of the evolution equation on u leads to the same observation: that at the point where u = 0 for the first time, (4.4) does not hold (it has to fail in general). At the maximum point of u ⁻¹|ρ|², we have

(4.5) $\begin{align*} |\rho ^{-}|^2 \nabla |\rho ^+|^2=|\rho ^{+}|^2\nabla |\rho ^{-}|^2. \end{align*}$

However, u ⁻¹|ρ|² fails to be bounded above along the flow, which shows that the following control at maximum point of u ⁻¹|ρ|² fails as well (when u is positive):

(4.6) $\begin{align*} |\rho ^{-}|^2 |\nabla \rho ^{+}|^2\leq |\rho ^{+}|^2 |\nabla \rho ^{-}|^2. \end{align*}$

The failure of positivity of u along the linear Hodge flow lies in the difference between (4.5) and (4.6) from a technical perspective. Now, we compute the evolution equations of λ ₁ and λ ₂ with the different choices of h _ij .

4.2 The case when h _ij = fδ _ij

The evolution equation reads

$\begin{align*} \partial _t \rho _{ij}=f \Delta \rho _{ij}+f_i \rho _{jl,l}-f_{j}\rho _{il,l}. \end{align*}$

Hence, we compute

$\begin{align*} \begin{split} \partial _t |\rho |^2&=f(\Delta |\rho |^2-2|\nabla \rho |^2)+2f_{i}\rho _{ij}\rho _{jl,l},\\ \partial _t u &= f(\Delta u-|\nabla \rho ^{+}|^2+|\nabla \rho ^{-}|^2)+f_{i}(*\rho )_{ij}\rho _{jl,l},\\ \partial _t |\rho ^{+}|^2&=f(\Delta |\rho ^{+}|^2-2|\nabla \rho ^{+}|^2)+2f_i \rho ^{+}_{ij} \rho _{jl, l},\\ \partial _t |\rho ^{-}|^2&=f(\Delta |\rho ^{-}|^2-2|\nabla \rho ^{-}|^2)+2f_i \rho ^{-}_{ij} \rho _{jl, l}.\\ \end{split} \end{align*}$

The conformal factor f can take the form of u ^−r for some r > 0, or something like u ⁻¹|ρ|². The structure of the evolution equation along the flow indicates the almost identical technical difficulty of preserving the uniform positivity of u (or λ ₂), since the maximum principle argument fails similarly as for the linear Hodge flow. It is still possible that the flow exists for all time with f = u ^−r , but it is hard to obtain the uniform positivity of u along the flow via a direct maximum principle argument.

4.3 The case when h _ij = u ⁻¹ a _ij

Here, we choose g _ρ = u ⁻¹ g(A·,A·), which is the metric used in the Donaldson geometric flow. Correspondingly, we have

$\begin{align*} Q_{ij}=h_{ij}=\frac {a_{ij}}{u}. \end{align*}$

Recall a _ij = ρ _ip ρ _jp , b _ij = (*ρ) _ip (*ρ) _jp and a _ij + b _ij = |ρ|² δ _ij . We compute

(4.7) $\begin{align*} \begin{split} \partial _t |\rho |^2 &=\left (\frac {|\rho |^2 \rho _{kj}}{u}-(*\rho )_{kj}\right )\Delta \rho _{kj}+2\rho _{ij} h_{ik, j} \rho _{kl, l},\\ \partial _t u & =\frac {1}{2} \rho _{kj}\Delta \rho _{kj} +(*\rho )_{ij} h_{ik, j} \rho _{kl,l}. \end{split} \end{align*}$

We compute, with x = |ρ ⁻|²/|ρ ⁺|²,

(4.8) $\begin{align*} \begin{aligned} \partial _t |\rho ^{+}|^2&= \frac {1+x}{1-x} \Delta |\rho ^{+}|^2+\frac {2}{1-x}\Delta |\rho ^{-}|^2-\frac {2(1+x)}{1-x}|\nabla \rho ^{+}|^2-\frac {4}{1-x} |\nabla \rho ^{-}|^2\\ &\quad +\!\left (\frac {b_{ik}}{u} +\delta _{ik}\right ) \rho _{ij, j}\rho _{kl,l}+\left (\frac {|\rho |^2}{u}\right )_j \rho _{kj}\rho _{kl,l},\\ \partial _t |\rho ^{-}|^2&= \frac {2x}{1-x} \Delta |\rho ^{+}|^2+\frac {1+x}{1-x} \Delta |\rho ^{-}|^2-\frac {4x}{1-x}|\nabla \rho ^{+}|^2-\frac {2(1+x)}{1-x} |\nabla \rho ^{-}|^2\\ &\quad +\!\left (\frac {b_{ik}}{u} -\delta _{ik}\right ) \rho _{ij, j}\rho _{kl,l}+\left (\frac {|\rho |^2}{u}\right )_j \rho _{kj}\rho _{kl,l}. \end{aligned} \end{align*}$

We compute, at points where |ρ ⁻| ≠ 0,

(4.9) $\begin{align*} \begin{aligned} \partial _t \lambda _1&=\frac {\lambda _1}{\lambda _2}\Delta \lambda _1-\frac {\lambda _1}{\lambda _2}(J+K)+\frac {\lambda _1 b_{ik}-u\lambda _2 \delta _{ik}}{u(\lambda _1^2-\lambda _2^2)} \rho _{ij, j}\rho _{kl,l}+\left (\frac {\lambda _1^2+\lambda _2^2}{\lambda _1\lambda _2}\right )_j \frac {\lambda _1\rho _{kj}\rho _{kl,l}}{\lambda _1^2-\lambda _2^2},\\ \partial _t \lambda _2&=\frac {\lambda _2}{\lambda _1}\Delta \lambda _2+\frac {\lambda _2}{\lambda _1}(K-J)+\frac {\lambda _1u\delta _{ik}-\lambda _2 b_{ik}}{u(\lambda _1^2-\lambda _2^2)} \rho _{ij, j}\rho _{kl,l}-\left (\frac {\lambda _1^2+\lambda _2^2}{\lambda _1\lambda _2}\right )_j \frac {\lambda _2\rho _{kj}\rho _{kl,l}}{\lambda _1^2-\lambda _2^2}. \end{aligned} \end{align*}$

We analyze (4.9) at points where λ ₁ or λ ₂ obtains its maximum or minimum, respectively. Let λ ₁(p,t) = max _{x ∊ M} λ ₁(x,t). Then at the point p, we have

$\begin{align*} \nabla \lambda _1=0,\quad \nabla^2 \lambda _1\leq 0. \end{align*}$

This implies that ∇|ρ ⁺| + ∇|ρ ⁻| = 0. We compute

$\begin{align*} \frac {\langle \rho ^{+}, \nabla \rho ^{+}\rangle }{|\rho ^{+}|}+\frac {\langle \rho ^{-}, \nabla \rho ^{-}\rangle }{|\rho ^{-}|}=0. \end{align*}$

We can choose a coordinate at p such that ρ ₁₂ = λ ₁ and ρ ₃₄ = λ ₂ and

$\begin{align*}\rho _{13}=\rho _{24}=\rho _{23}=\rho _{14}=0.\end{align*}$

This implies that ∇ρ ₁₂ = ∇λ ₁ = 0, ∇λ ₂ = ∇ρ ₃₄ at p. We compute

$\begin{align*} J+K=\frac {\lambda _1}{\lambda _1^2-\lambda _2^2} (|\nabla \rho _{13}|^2+|\nabla \rho _{14}|^2+|\nabla \rho _{23}|^2+|\nabla \rho _{24}|^2)+\frac {2\lambda _2}{\lambda _1^2-\lambda _2^2} (\nabla \rho _{13}\nabla \rho _{24}-\nabla \rho _{23}\nabla \rho _{14}), \end{align*}$

while for

$\begin{align*} \left (\frac {\lambda _1^2+\lambda _2^2}{\lambda _1\lambda _2}\right )_j \frac {\lambda _1\rho _{kj}\rho _{kl,l}}{\lambda _1^2-\lambda _2^2}=&\frac {\lambda _1}{\lambda _2^2} (\rho _{34, 1}\rho _{2l, l}-\rho _{34, 2}\rho _{1l, l})+\frac {\rho _{34,3}\rho _{4l, l}-\rho _{34, 4}\rho _{3l, l}}{\lambda _2},\\ \frac {\lambda _1 b_{ik}-u\lambda _2 \delta _{ik}}{u(\lambda _1^2-\lambda _2^2)} \rho _{ij, j}\rho _{kl,l}=&frac {1}{\lambda _2} [(\rho _{3j, j})^2+(\rho _{4j, j})^2]. \end{align*}$

It seems to be hard to bound λ ₁ via a direct maximum principle argument since the term (λ ₁/λ ₂ ²)(ρ _{34, 1} ρ _{2l, l} −ρ _{34, 2} ρ _{1l, l} ) has the highest order when λ ₂ < < 1 and this term does not have the favorite sign. We denote

$\begin{align*} \partial _t \lambda _2=\frac {\lambda _2}{\lambda _1}\Delta \lambda _2+\mathcal{R} \end{align*}$

and analyze the behavior at the minimum of λ ₂, where λ ₂(q,t) = min λ ₂(·,t). At the point q, we choose a coordinate such that ρ ₁₂ = λ ₁ and ρ ₃₄ = λ ₂ and

$\begin{align*}\rho _{13}=\rho _{24}=\rho _{23}=\rho _{14}=0.\end{align*}$

At q, ∇λ ₂ = 0 = ∇ρ ₃₄, ∇λ ₁ = ∇ρ ₁₂. We are not able to prove that ℛ ≥ 0 at q.

4.4 The case when h _ij = u ⁻² a _ij

We have

$\begin{align*} \rho _{ij} h_{ik}=\frac {|\rho |^2\delta _{kj}}{u^2}-\frac {(*\rho )_{kj}}{u},\quad (*\rho )_{ij}h_{ik}=\frac {\rho _{kj}}{u}. \end{align*}$

We compute

(4.10) $\begin{align*} \begin{aligned} \partial _t |\rho |^2&=\left (\frac {|\rho |^2\rho _{kj}}{u^2}-\frac {(*\rho )_{kj}}{u} \right ) \Delta \rho _{kj}+2\left (\frac {|\rho |^2\rho _{kj}}{u^2}-\frac {(*\rho )_{kj}}{u}\right )_j \rho _{kl, l}-2h_{ik} \rho _{ij, j}\rho _{kl,l},\\ \partial _t u &=\frac {\rho _{kj}}{2u} \Delta \rho _{kj}+\left (\frac {\rho _{kj}}{u}\right )_j \rho _{kl, l}. \end{aligned} \end{align*}$

We compute

$\begin{align*} \begin{split} \partial _t (|\rho |^2+2u)=&\left (\frac {|\rho |^2+u}{u^2} \rho _{kj}-\frac {(*\rho )_{kj}}{u}\right )\Delta \rho _{kj}+2\left (\frac {|\rho |^2+u}{u^2} \rho _{kj}-\frac {(*\rho )_{kj}}{u}\right )_j \rho _{kl,l}-2h_{ik} \rho _{ij, j}\rho _{kl,l},\\ \partial _t (|\rho |^2-2u)=&\left (\frac {|\rho |^2-u}{u^2} \rho _{kj}-\frac {(*\rho )_{kj}}{u}\right )\Delta \rho _{kj}+2\left (\frac {|\rho |^2-u}{u^2} \rho _{kj}-\frac {(*\rho )_{kj}}{u}\right )_j \rho _{kl,l}-2h_{ik} \rho _{ij, j}\rho _{kl,l}. \end{split} \end{align*}$

We compute

(4.11) $\begin{align*} \begin{aligned} \partial _t |\rho ^{+}|^2&=\frac {|\rho |^2}{2u^2}\Delta |\rho ^{+}|^2+\frac {|\rho |^2+2u}{2u^2}\Delta |\rho ^{-}|^2-\frac {|\rho |^2}{u^2} |\nabla \rho ^{+}|^2-\frac {|\rho |^2+2u}{u^2}|\nabla \rho ^{-}|^2\\ &\quad +\!\left (\frac {|\rho |^2+u}{u^2} \rho _{kj}-\frac {(*\rho )_{kj}}{u}\right )_j \rho _{kl,l}-h_{ik} \rho _{ij, j}\rho _{kl,l},\\ \partial _t |\rho ^{-}|^2&=\frac {|\rho |^2-2u}{2u^2}\Delta |\rho ^{+}|^2+\frac {|\rho |^2}{2u^2}\Delta |\rho ^{-}|^2-\frac {|\rho |^2-2u}{u^2} |\nabla \rho ^{+}|^2-\frac {|\rho |^2}{u^2}|\nabla \rho ^{-}|^2\\ &\quad +\!\left (\frac {|\rho |^2-u}{u^2} \rho _{kj}-\frac {(*\rho )_{kj}}{u}\right )_j \rho _{kl,l}-h_{ik} \rho _{ij, j}\rho _{kl,l}. \end{aligned} \end{align*}$

It follows that

(4.12) $\begin{align*} \begin{aligned} \partial _t \lambda _1&=\frac {\Delta \lambda _1}{\lambda _2^2}-\frac {1}{\lambda _2^2} (J+K)+\frac {\lambda _1 b_{ik}-u\lambda _2 \delta _{ik}}{u^2(\lambda _1^2-\lambda _2^2)} \rho _{ij, j}\rho _{kl,l}+P_1,\\ \partial _t \lambda _2&=\frac {\Delta \lambda _2}{\lambda _1^2}-\frac {1}{\lambda _1^2} (K-J)+\frac {\lambda _1u\delta _{ik}-\lambda _2 b_{ik}}{u^2(\lambda _1^2-\lambda _2^2)} \rho _{ij, j}\rho _{kl,l}+P_2, \end{aligned} \end{align*}$

where we denote

$\begin{align*} \begin{split} P_1&=\frac {\lambda _{1, j}\rho _{kl, l}}{\lambda _1(\lambda _1^2-\lambda _2^2)}\left (\frac {(*\rho )_{kj}}{\lambda _2}-\frac {\rho _{kj}}{\lambda _1}\right )+\frac {\lambda _{2, j}\rho _{kl, l}}{\lambda _2(\lambda _1^2-\lambda _2^2)}\left (\frac {(*\rho )_{kj}}{\lambda _2}+\frac {\rho _{kj}}{\lambda _1}-\frac {2\lambda _1}{\lambda _2^2}\rho _{kj}\right )\!,\\ P_2&=\frac {\lambda _{1, j}\rho _{kl, l}}{\lambda _1(\lambda _1^2-\lambda _2^2)}\left (\frac {(*\rho )_{kj}}{\lambda _2}+\frac {\rho _{kj}}{\lambda _2}-\frac {2\lambda _2}{\lambda _1^2}\rho _{kj}\right )+\frac {\lambda _{2, j}\rho _{kl, l}}{\lambda _2(\lambda _1^2-\lambda _2^2)}\left (\frac {(*\rho )_{kj}}{\lambda _2}-\frac {\rho _{kj}}{\lambda _1}\right )\!. \end{split} \end{align*}$

4.5 The case when h _ij = u ⁻¹ b _ij

Here, we choose

$\begin{align*} Q_{ij}=h_{ij}=\frac {b_{ij}}{u}. \end{align*}$

We have

$\begin{align*} \rho _{ij} h_{ik}=(*\rho )_{kj},\quad (*\rho )_{ij} h_{ik}=\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{u}. \end{align*}$

We compute

(4.13) $\begin{align*} \begin{aligned} \partial _t |\rho |^2&=(*\rho )_{kj}\Delta \rho _{kj}-\frac {2b_{ik}}{u}\rho _{ij, j}\rho _{kl,l},\\ \partial _t u&=\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{2u} \Delta \rho _{kj}+\left (\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{u}\right )_j \rho _{kl, l}. \end{aligned} \end{align*}$

Hence, we have

(4.14) $\begin{align*} \begin{aligned} \partial _t |\rho ^{+}|^2&=\frac {(|\rho |^2+u)(*\rho )_{kj}-u\rho _{kj}}{2u} \Delta \rho _{kj}-\frac {b_{ik}}{u}\rho _{ij, j}\rho _{kl,l}+\left (\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{u}\right )_j \rho _{kl, l},\\ \partial _t |\rho ^{-}|^2&=\frac {(u-|\rho |^2)(*\rho )_{kj}+u\rho _{kj}}{2u} \Delta \rho _{kj}-\frac {b_{ik}}{u}\rho _{ij, j}\rho _{kl,l}-\left (\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{u}\right )_j \rho _{kl, l}. \end{aligned} \end{align*}$

Hence, we have

(4.15) $\begin{align*} \begin{aligned} \partial _t \lambda _1=\frac {\lambda _2}{\lambda _1}\Delta \lambda _1-\frac {\lambda _2}{\lambda _1} (J+K)-\frac {\lambda _1 b_{ik}-u\lambda _2 \delta _{ik}}{u(\lambda _1^2-\lambda _2^2)} \rho _{ij, j}\rho _{kl,l}-R_1,\\ \partial _t \lambda _2=\frac {\lambda _1}{\lambda _2}\Delta \lambda _2-\frac {\lambda _1}{\lambda _2} (K-J)+\frac {\lambda _2 b_{ik}-u\lambda _1 \delta _{ik}}{u(\lambda _1^2-\lambda _2^2)} \rho _{ij, j}\rho _{kl,l}+R_2, \end{aligned} \end{align*}$

where we denote

$\begin{align*} \begin{split} R_1=\left (\frac {\lambda _1^2+\lambda _2^2}{\lambda _1\lambda _2}\right )_j \frac {\lambda _2 (*\rho )_{kj}\rho _{kl, l}}{\lambda _1^2-\lambda _2^2}, \quad R_2=\left (\frac {\lambda _1^2+\lambda _2^2}{\lambda _1\lambda _2}\right )_j \frac {\lambda _1 (*\rho )_{kj}\rho _{kl, l}}{\lambda _1^2-\lambda _2^2}. \end{split} \end{align*}$

4.6 The case when h _ij = u ⁻² b _ij

Here, we choose

(4.16) $\begin{align*} Q_{ij}=b_{ij},\quad h_{ij}=\frac {b_{ij}}{u^2}. \end{align*}$

We compute

$\begin{align*} \rho _{ij} h_{ik}=\frac {(*\rho )_{kj}}{u},\quad (*\rho )_{ij} h_{ik}=\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{u^2}. \end{align*}$

Hence, we have

(4.17) $\begin{align*} \begin{aligned} \partial _t |\rho |^2&=\frac {(*\rho )_{kj}}{u} \Delta \rho _{kj}+2 \left (\frac {(*\rho )_{kj}}{u} \right )_{,\; j}\rho _{kl, l}-2 h_{ik}\rho _{ip, p}\rho _{kl, l},\\ \partial _t (2u)&=\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{u^2} \Delta \rho _{kj}+2\left (\frac {|\rho |^2(*\rho )_{kj}-u\rho _{kj}}{u^2} \right )_{,\; j}\rho _{kl, l},\\ \partial _t (|\rho |^2+2u)&=R_{kj}\Delta \rho _{kj}+2R_{kj,\,j} \rho _{kl, l}-2h_{ik}\rho _{ip, p}\rho _{kl, l}, \\ \partial _t (|\rho |^2-2u)&S_{kj}\Delta \rho _{kj}+2S_{kj,\;j} \rho _{kl, l}-2h_{ik}\rho _{ip, p}\rho _{kl, l}, \end{aligned} \end{align*}$

where we denote

$\begin{align*} R_{kj}=\frac {(|\rho |^2+u)(*\rho )_{kj} -u\rho _{kj}}{ u^2},\quad S_{kj}=\frac {\rho _{kj}+(*\rho )_{kj}}{u}-\frac {|\rho |^2}{u^2} (*\rho )_{kj}. \end{align*}$

Recall that

$\begin{align*} \lambda _1= \frac {\sqrt {|\rho |^2+2u}}{2}+\frac {\sqrt {|\rho |^2-2u}}{2}, \quad \lambda _2=\frac {\sqrt {|\rho |^2+2u}}{2}-\frac {\sqrt {|\rho |^2-2u}}{2}. \end{align*}$

Note that λ ₁ and λ ₂ are both smooth functions when |ρ ⁻|² ≠ 0. We compute

(4.18) $\begin{align} \partial _t \lambda _1&\frac {\partial _t (|\rho |^2+2u)}{4\sqrt {|\rho |^2+2u}}+\frac {\partial _t (|\rho |^2-2u)}{4\sqrt {|\rho |^2-2u}}\notag \\ &= \left (\frac {R_{kj}\Delta \rho _{kj}}{4\sqrt {|\rho |^2+2u}}+\frac {S_{kj}\Delta \rho _{kj}}{4\sqrt {|\rho |^2-2u}}\right ) + \left (\frac {R_{kj, j}\rho _{kl, l}}{2\sqrt {|\rho |^2+2u}}+\frac {S_{kj, j}\rho _{kl, l}}{2\sqrt {|\rho |^2-2u}}\right ) \notag \\ &\quad -\!\left (\frac {1}{2\sqrt {|\rho |^2+2u}}+\frac {1}{2\sqrt {|\rho |^2-2u}}\right ) h_{ik} \rho _{ip, p}\rho _{kl,l}, \end{align}$

and similarly

(4.19) $\begin{align} \partial _t \lambda _2&=\frac {\partial _t (|\rho |^2+2u)}{4\sqrt {|\rho |^2+2u}}-\frac {\partial _t (|\rho |^2-2u)}{4\sqrt {|\rho |^2-2u}}\notag \\ &= \left (\frac {R_{kj}\Delta \rho _{kj}}{4\sqrt {|\rho |^2+2u}}-\frac {S_{kj}\Delta \rho _{kj}}{4\sqrt {|\rho |^2-2u}}\right ) + \left (\frac {R_{kj, j}\rho _{kl, l}}{2\sqrt {|\rho |^2+2u}}-\frac {S_{kj, j}\rho _{kl, l}}{2\sqrt {|\rho |^2-2u}}\right ) \notag \\ &\quad -\!\left (\frac {1}{2\sqrt {|\rho |^2+2u}}-\frac {1}{2\sqrt {|\rho |^2-2u}}\right ) h_{ik} \rho _{ip, p}\rho _{kl,l}. \end{align}$

By (4.18), we compute, using ρ = ρ ⁺ + ρ ⁻ and * ρ = ρ ⁺ − ρ ⁻,

(4.20) $\begin{align} \partial _t \lambda _1&\frac {\Delta \lambda _1}{\lambda _1^2}-\frac {1}{\lambda _1^2}(J+K)-\frac {\lambda _1 h_{ik}\rho _{ip,p}\rho _{kl,l}}{\lambda _1^2-\lambda _2^2}+\frac {|\rho _{kl,l}|^2}{\lambda _1(\lambda _1^2-\lambda _2^2)}\notag \\ &\quad +\frac {\lambda _{1, j}\rho _{kl, l}}{\lambda _1 (\lambda _1^2-\lambda _2^2)}\left ( \left (\frac {2\lambda _2}{\lambda _1^2}-\frac {1}{\lambda _2}\right )(*\rho )_{kj}-\frac {\rho _{kj}}{\lambda _1}\right )+\frac {\lambda _{2, j} \rho _{kl, l}\left ({(*\rho )_{kj}}/{\lambda _2}-{\rho _{kj}}/{\lambda _1}\right )}{\lambda _2(\lambda _1^2-\lambda _2^2)}, \end{align}$

and, similarly, we compute

(4.21) $\begin{align} \partial _t \lambda _2&=\frac {\Delta \lambda _2}{\lambda _2^2}+\frac {1}{\lambda _2^2}(K-J)+\frac {\lambda _2 h_{ik}\rho _{ip,p}\rho _{kl,l}}{\lambda _1^2-\lambda _2^2}-\frac {|\rho _{kl,l}|^2}{\lambda _2(\lambda _1^2-\lambda _2^2)}\notag \\ &\quad +\frac {\lambda _{2, j}\rho _{kl, l}}{\lambda _2 (\lambda _1^2-\lambda _2^2)}\left (\frac {\rho _{kj}}{\lambda _2}+\frac {(*\rho )_{kj}}{\lambda _1}-\frac {2\lambda _1(*\rho )_{kj}}{\lambda _2^2}\right )+\frac {\lambda _{1, j} \rho _{kl, l}\left ({\rho _{kj}}/{\lambda _2}-{(*\rho )_{kj}}/{\lambda _1}\right )}{\lambda _1(\lambda _1^2-\lambda _2^2)}. \end{align}$

When ρ = λ ₁(x ₁,x ₂)dx ₁ ∧ dx ₂ + λ ₂(x ₃,x ₄)dx ₃ ∧ dx ₄, we have already seen that the nonlinear Hodge flow with h _ij = u ⁻² b _ij preserves the structure and satisfies

(4.22) $\begin{align*} \partial _t \lambda _1=\Delta _{x_1, x_2} (-\lambda _1^{-1}), \partial _t \lambda _2=\Delta _{x_3, x_4}(-\lambda _2^{-1}). \end{align*}$

It is a direct computation to check that (4.22) coincides with (4.20) and (4.21). For example, J = K = 0 and ρ _{1l, l} = ∂ _{x 2} λ ₁, ρ _{2l, l} = − ∂ _{x 1} λ ₁, ρ _{3l, l} = ∂ _{x 4} λ ₂, ρ _{4l, l} = − ∂ _{x 3} λ ₂. Note that (4.22) is a fast diffusion equation for λ ₁ and λ ₂, respectively. The maximum principle indeed applies to show that λ ₁, λ ₂ remain uniformly bounded and uniformly positive. By the parabolic theory, the flow exists for all time with uniform control, and it will converge to ω = dx ₁ ∧ dx ₂ + dx ₃ ∧ dx ₄. However, the maximum principle argument does not apply directly to (4.20) and (4.21) for the general case. On the other hand, (4.21) might still suggest that λ ₂ stays uniformly positive along the flow. When λ ₂ approaches zero, the leading term λ ₂ ⁻² Δλ ₂ is the same as for the fast diffusion equation and it might force λ ₂ to stay uniformly positive. Nevertheless, it is clear that the system is really complicated, and it remains a substantial challenge to address these difficulties.