Quasilinear generalized parabolic Anderson model equation

Abstract

We present in this note a local in time well-posedness result for the singular 2-dimensional quasilinear generalized parabolic Anderson model equation

$$\begin{aligned} \partial _t u - a(u)\Delta u = g(u)\xi . \end{aligned}$$

The key idea of our approach is a simple transformation of the equation which allows to treat the problem as a semilinear problem. The analysis is done within the elementary setting of paracontrolled calculus.

A Elementary side results

We collect in this Appendix a number of side results, together with their proofs, to make this work self-contained. They can all be found somewhere else in one form or another. As a warm-up, we start with some claims, used in the main body of the text. Recall the classical notation $\Delta _i$ for the Fourier multipliers used in Littlewood-Paley decomposition. We refer to Bahouri et al. [4] textbook for the basics on this subject.

Lemma 9

• 1. For u in the spatial Hölder space $C^\alpha$, we have

  $$\begin{aligned} \big \Vert g (u) \big \Vert _{C^\alpha } \le \Vert g \Vert _{C^1} \big (1 + \Vert u\Vert _{C^\alpha }\big ), \end{aligned}$$

  and

  $$\begin{aligned} \big \Vert g (u) - g (v) \big \Vert _{C^\alpha } \le \Vert g \Vert _{C^2} \big (1 + \Vert u \Vert _{\alpha }\big ) \Vert u - v \Vert _{C^\alpha }. \end{aligned}$$

• 2. For u in the parabolic Hölder space $\mathcal {C}^\alpha$, and $0< \beta \le \alpha$, we have

  $$\begin{aligned} \big \Vert g (u) \big \Vert _{\mathcal {C}^\beta } \lesssim T^{\frac{\alpha - \beta }{2}} \Vert g \Vert _{C^1} \big (1 + \Vert u \Vert _{\mathcal {C}^\alpha }\big ) + \big \Vert g (u_0)\big \Vert _{C^\beta }, \end{aligned}$$

  and if $u_{t=0}=v_{t=0}$, then

  $$\begin{aligned} \big \Vert g (u) - g (v)\big \Vert _{\mathcal {C}^\beta } \lesssim T^{\frac{\alpha - \beta }{2}} \Vert g\Vert _{C^2} \big (1 + \Vert u \Vert _{\mathcal {C}^\alpha }\big ) \Vert u - v\Vert _{\mathcal {C}^\alpha }. \end{aligned}$$

Proof

The Lipschitz estimate of point 1 comes by writing

$$\begin{aligned} \big (g(u)-g(v)\big )(x) - \big (g(u)-g(v)\big )(y) \end{aligned}$$

as the boundary term of the integral of the derivative of the function $s \in [0,1] \mapsto (g(u)-g(v))(y+s(x-y))$.

To see point 2, take any function $h\in \mathcal {C}^\beta$ and start from the following two estimates

$$\begin{aligned} \big \Vert \Delta _i \big \{h(t) - h(0)\big \} \big \Vert _{L^{\infty }} \lesssim \left\{ \begin{array}{l} 2^{- i \alpha } \, \Vert h\Vert _{C_T \mathcal {C}^{\alpha }} \\ T^{\frac{\alpha }{2}} \, \Vert h\Vert _{C^{\alpha / 2}_T L^{\infty }} \end{array} \right. \end{aligned}$$

to get by interpolation of the two upper bounds the estimate

$$\begin{aligned} \Vert h\Vert _{C_T C^\beta } \lesssim T^{\frac{\alpha - \beta }{2}} \Vert h\Vert _{C_T C^\alpha }^\varepsilon \, \Vert h\Vert ^{1 - \varepsilon }_{C_T^\frac{\alpha }{2} L^{\infty }} + \big \Vert h(0)\big \Vert _{C^\beta }. \end{aligned}$$

It follows that if $u_{t=0}=v_{t=0}$, then we have

$$\begin{aligned} \big \Vert g (u) - g (v) \big \Vert _{C_T \mathcal {C}^{\beta }}\lesssim & {} T^{\frac{\alpha -\beta }{2}} \, \big \Vert g (u) - g (v) \big \Vert _{C_T C^\alpha }^{\varepsilon } \big \Vert g (u) - g (v) \big \Vert ^{1 - \varepsilon }_{C_T^{{\alpha /2}} L^{\infty }} \\\lesssim & {} T^{\frac{\alpha - \beta }{2}} \Vert g \Vert _{C^2} \Big \{(1 + \Vert u \Vert _{C_T \mathcal {C}^{\alpha }}) \Vert u - v \Vert _{C_T \mathcal {C}^{\alpha }} \\&\quad +\,\Big (1 + \Vert u\Vert _{C_T^{\alpha /2} L^{\infty }}\Big ) \Vert u - v \Vert _{C_T^{\alpha /2} L^{\infty }}\Big \}. \end{aligned}$$

The Lipschitz estimate of point 2 then comes as a consequence of the inequality

$$\begin{aligned} \big \Vert g (u) - g (v) \big \Vert _{C^{\beta / 2}_T L^{\infty }}\le & {} T^{\frac{\alpha - \beta }{2}} \, \big \Vert g (u) - g (v) \big \Vert _{C^{\alpha / 2}_T L^{\infty }} \\\le & {} T^{\frac{\alpha - \beta }{2}} \Vert g \Vert _{C^2} \Big (1 + \Vert u \Vert _{C^{\alpha / 2}_T L^{\infty }}\Big ) \Vert u - v \Vert _{C^{\alpha / 2}_T L^{\infty }}. \end{aligned}$$

$\square$

The next result is a variation on Gubinelli et al. fundamental “commutator” lemma. It is the first part of Lemma 3. It also happens to be special case of a more general result, proved in Theorem 2 of [6]. Recall that we work with $\alpha >\frac{2}{3}$.

Lemma 10

Let $f, g \in C^\beta$, and $a \in C^\alpha$ and $b \in C^{\alpha - 2}$ be given, such that $\Pi (a,b)$ is a well-defined element of $C^{2\alpha -2}$. Then

$$\begin{aligned} \Big \Vert \Pi \big (\Pi _f a, \Pi _g b\big ) - (f g) \Pi (a,b) \Big \Vert _{C^{2 \alpha + \beta - 2}} \lesssim \Vert f \Vert _{C^\beta } \Vert g \Vert _{C^\beta } \Vert a\Vert _{C^\alpha } \Vert b\Vert _{C^{\alpha - 2}}. \end{aligned}$$

Proof

Denote by $K_{k, x} (z) := K_k (x - z)$ the convolution kernel of the Littlewood-Paley projector $\Delta _k$. We have

$$\begin{aligned} \Delta _k \Big (\Pi \big (\Pi _f a, \Pi _g b\big )\Big ) (x)= & {} \int K_{k, x} (y) \Big (\Pi \big (\Pi _{f (y)}a, \Pi _{g (y)}b\big )\Big )(y)\,dy \\&\quad + \int K_{k, x} (y) \Big (\Pi \big (\Pi _{f - f (y)}a, \Pi _{g (y)}b\big )\Big ) (y)\,dy \\&\quad + \int K_{k, x} (y) \Big (\Pi \big (\Pi _f a, \Pi _{g - g (y)}b\big )\Big ) (y)\,dy \\=: & {} \mathsf{I}_1 + \mathsf{I}_2 + \mathsf{I}_3. \end{aligned}$$

The first term gives

$$\begin{aligned} \Big (\Pi \big (\Pi _{f (y)}a, \Pi _{g (y)}b\big )\Big ) (y)= & {} \sum _{i \sim j} \Delta _i \big (\Pi _{f (y)}a\big )(y) \Delta _j\big (\Pi _{g (y)}b\big )(y) \\= & {} \sum _{i \sim j} f (y) g (y) \big (\Delta _i a\big ) (y) \big (\Delta _j b\big ) (y) {=} f (y) g (y) \, \Pi (a, b) (y) \end{aligned}$$

so it remains to estimate $\mathsf{I}_2$ and $\mathsf{I}_3$ in the spatial Hölder space $C^{2\alpha + \beta - 2}$. We have

$$\begin{aligned} | \mathsf{I}_2 |= & {} \left| \int K_{k, x} (y) \sum _{i \sim j \gtrsim k} \Delta _i\big (\Pi _{f - f (y)}a\big ) (y) g (y) \big (\Delta _j b\big ) (y) \,dy\right| \\\lesssim & {} \Vert f \Vert _{C^\alpha } \Vert a\Vert _{C^\alpha } \Vert g\Vert _{L^{\infty }} \Vert b\Vert _{C^{\alpha - 2}} \int \big |K_{k, x}(y)\big | \left( \sum _{i \gtrsim k} 2^{- (2 \alpha + \beta - 2) i}\right) \,dy. \end{aligned}$$

Since $2 \alpha + \beta - 2 > 0$ we obtain in the end

$$\begin{aligned} | \mathsf{I}_2 | \lesssim 2^{- (2 \alpha + \beta - 2) k} \Vert f \Vert _{C^\alpha } \Vert a\Vert _{C^\alpha } \Vert g \Vert _{L^{\infty }} \Vert b \Vert _{C^{\alpha - 2}}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} | \mathsf{I}_3 | \lesssim 2^{- (2 \alpha + \beta - 2) k} \Vert f \Vert _{L^{\infty }} \Vert a\Vert _{C^\alpha } \Vert g \Vert _{C^\beta } \Vert b\Vert _{C^{\alpha - 2}}, \end{aligned}$$

which completes the proof.$\square$

Lemma 11

Given $f \in C^{2 \beta }$, $g \in C^\beta$, and $h \in C^{\alpha - 2}$, with regularity exponents in (0, 2), we have

$$\begin{aligned} \big \Vert f \, \Pi _g h - \Pi _{f g} h \big \Vert _{C^{\beta + \alpha - 2}} \lesssim \Vert f\Vert _{C^{2 \beta }} \Vert g \Vert _{C^\beta } \Vert h \Vert _{C^{\alpha - 2}}. \end{aligned}$$

Proof

We have

$$\begin{aligned} f \, \Pi _g h= & {} \Pi _f (\Pi _g h) + \Pi _{\Pi _g h} (f) + \Pi \big (f, \Pi _gh\big ) \\= & {} \big (\Pi _f (\Pi _g h) - \Pi _{f g}h\big ) + \Pi _{f g}h + \Pi _{\Pi _g h}(f) + \Pi \big (f, \Pi _g h\big ), \end{aligned}$$

where

$$\begin{aligned} \big \Vert \Pi _f \big (\Pi _g h\big ) - \Pi _{f g}h \big \Vert _{C^{\beta + \alpha - 2}} \lesssim \Vert f \Vert _{L^{\infty }}\, \Vert g \Vert _{C^\beta } \, \Vert h \Vert _{{C^{\alpha - 2}}}, \end{aligned}$$

due to Proposition 23 of [6], and

$$\begin{aligned} \big \Vert \Pi \big (f, \Pi _g h\big )\big \Vert _{C^{2\beta + \alpha - 2}}\lesssim & {} \Vert f \Vert _{C^{2\beta }} \,\Vert g \Vert _{L^{\infty }}\, \Vert h \Vert _{C^{\alpha - 2}} \\ \big \Vert \Pi _{\Pi _g h} f \big \Vert _{C^{2\alpha - 2}}\lesssim & {} \Vert f \Vert _{C^\alpha } \, \Vert g\Vert _{L^{\infty }} \,\Vert h \Vert _{C^{\alpha - 2}}. \end{aligned}$$

The next proposition gives the second part of Lemma 3, recall $\mathscr {L}^0 = \partial _t - a(u_0)\Delta$.$\square$

Proposition 12

Given $u'\in \mathcal {C}^\beta$, we have the following continuity result for a commutator

$$\begin{aligned} \Big \Vert \mathscr {L}^0 \big (\overline{\Pi }_{u'} X\big ) - \Pi _{a \big (u^T_0\big ) u'} (-\Delta X) \Big \Vert _{C_T\mathcal {C}^{\alpha + \beta - 2}} \lesssim \big (1 + T^{- \frac{2\beta -\alpha }{2}} \Vert u_0\Vert _{C^\alpha }\big ) \Vert u' \Vert _{\mathcal {C}^\beta } \Vert X \Vert _{C^\alpha }. \end{aligned}$$

Proof

Recall first from Lemma 5.1 in [11] that

$$\begin{aligned} \big \Vert \overline{\Pi }_{u'} (\Delta X) - \Pi _{u'} (\Delta X) \big \Vert _{C_T C^{\alpha + \beta - 2}} \lesssim \Vert u' \Vert _{C^{\beta / 2}_TL^{\infty }} \Vert X \Vert _{C^\alpha }. \end{aligned}$$

(12)

Then we write

$$\begin{aligned} \mathscr {L}^0 \big (\overline{\Pi }_{u'}X\big ) - \Pi _{a \big (u^T_0\big ) u'} (-\Delta X)= & {} \Big \{\mathscr {L}^0 \big (\overline{\Pi }_{u'}X\big ) + a \big (u^T_{^{} 0}\big ) \overline{\Pi }_{u'} (\Delta X)\Big \} \\&\quad +\,a \big (u^T_0\big ) \Big \{\Pi _{u'} (\Delta X) - \overline{\Pi }_{u'} (\Delta X)\Big \} \\&\quad +\,\Big \{\Pi _{a \big (u^T_0\big ) u'} (\Delta X) - a \big (u^T_0\big ) \Pi _{u'} (\Delta X)\Big \}, \end{aligned}$$

and observe that the second term on the right hand side can be estimated with inequality (12) to obtain

$$\begin{aligned} \Big \Vert a \big (u^T_0\big ) \Big \{\Pi _{u'} (\Delta X) - \overline{\Pi }_{u'} (\Delta X)\Big \}\Big \Vert _{C^{\alpha + \beta - 2}} \lesssim \big (1 + \Vert u_0 \Vert _{\alpha }\big ) \Vert u'\Vert _{C^{\beta / 2}_T L^{\infty }} \Vert X \Vert _{C^\alpha }, \end{aligned}$$

and that the third term can be taken care of by Lemma 11 to obtain

$$\begin{aligned} \big \Vert \Pi _{a \big (u^T_0\big ) u'} (\Delta X) - a \big (u^T_0\big ) \Pi _{u'} (\Delta X) \big \Vert _{C^{\alpha + \beta - 2}} \lesssim \left( 1 + T^{- \frac{2\beta -\alpha }{2}} \Vert u_0\Vert _{C^\alpha }\right) \Vert u' \Vert _{\mathcal {C}^\beta } \Vert X \Vert _{C^\alpha }. \end{aligned}$$

We now estimate the first term. Since X does not depend on t,

$$\begin{aligned} \partial _t \big (\overline{\Pi }_{u'} X\big ) = \sum _i \partial _t (S_{i - 1} Q_i u') \Delta _i X. \end{aligned}$$

Since the spatial Fourier transform of $\partial _t \big (S_{i - 1} Q_i u'\big ) \Delta _i X$ is localized in an annulus of size $2^i$, we obtain from estimate (32) in [11] that

$$\begin{aligned} \big \Vert \partial _t (S_{i - 1} Q_i u')\big \Vert _{C_T L^{\infty }}= & {} \big \Vert \partial _t (Q_i S_{i - 1} u')\big \Vert _{C_T L^{\infty }} \\\lesssim & {} 2^{- (\beta - 2) i} \big \Vert S_{i - 1} u' \big \Vert _{C^{\beta / 2}_T L^{\infty }} \\\lesssim & {} 2^{- (\beta - 2) i} \Vert u' \Vert _{C^{\beta / 2}_T L^{\infty }}, \end{aligned}$$

so

$$\begin{aligned} \big \Vert \partial _t (\overline{\Pi }_{u'}X\big )\big \Vert _{C_T C^{\alpha + \beta - 2}} \lesssim \Vert u' \Vert _{\mathcal {C}^\beta } \Vert X \Vert _{C^\alpha }. \end{aligned}$$

(13)

We have, on the other hand,

$$\begin{aligned} \Delta \big (\overline{\Pi }_{u'}X\big ) - \overline{\Pi }_{u'} (\Delta X) = \overline{\Pi }_{\Delta u}X - 2 \overline{\Pi }_{\nabla u'} (\nabla X), \end{aligned}$$

with

$$\begin{aligned} \big \Vert Q_i S_{i - 1} \Delta u' \big \Vert _{C_T L^{\infty }} \le \big \Vert S_{i - 1} \Delta u' \big \Vert _{C_T L^{\infty }} \lesssim 2^{- (\beta - 2) i} \Vert u' \Vert _{C_T C^\beta } \end{aligned}$$

and

$$\begin{aligned} \big \Vert Q_i S_{i - 1} \nabla u' \big \Vert _{C_T L^{\infty }} \lesssim 2^{- (\beta - 1) i} \Vert u' \Vert _{C_T C^\beta }. \end{aligned}$$

Altogether, this gives

$$\begin{aligned}&\big \Vert \Delta (\overline{\Pi }_{u'}X) - \overline{\Pi }_{u'} (\Delta X) \big \Vert _{C_T C^{\alpha + \beta - 2}}\nonumber \\&= \Vert \overline{\Pi }_{\Delta u} (X) - 2 \overline{\Pi }_{\nabla u'} (\nabla X) \Vert _{C_T C^{\alpha + \beta - 2}} \lesssim \Vert u' \Vert _{\mathcal {C}^\beta } \Vert X \Vert _{C^\alpha }, \end{aligned}$$

(14)

so we deduce from (13) and (14) that

$$\begin{aligned}&\Big \Vert \mathscr {L}^0 \big (\overline{\Pi }_{u'} (X)\big ) - a \big (u^T_0\big ) \overline{\Pi }_{u'} (-\Delta X) \Big \Vert _{C_T C^{\alpha + \beta - 2}} \\&\quad = \Big \Vert \partial _t \big (\overline{\Pi }_{u'}X\big ) - a \big (u^T_0\big ) \Big \{\Delta (\overline{\Pi }_{u'} (X)) - \overline{\Pi }_{u'} (\Delta X)\Big \} \Big \Vert _{C_T C^{\alpha + \beta - 2}} \\&\quad \lesssim \big (1 + \Vert a \big (u^T_0\big ) \Vert _{C^\beta }\big ) \Vert u' \Vert _{\mathcal {C}^\beta } \Vert X \Vert _{C^\alpha }, \end{aligned}$$

which concludes the proof.$\square$

Lemma 13

Given $f \in \mathcal {C}^\beta$ and $g \in C^\alpha$, we have

$$\begin{aligned} \big \Vert \overline{\Pi }_fg \big \Vert _{\mathcal {C}^\alpha } \lesssim \Vert f \Vert _{\mathcal {C}^\beta } \Vert g \Vert _{C^\alpha }. \end{aligned}$$

Proof

Let work here with the canonical heat operator $\mathscr {L} := \partial _t - \Delta$. We have from the classical Schauder estimates

$$\begin{aligned} \big \Vert \overline{\Pi }_fg\big \Vert _{\mathcal {C}^\alpha } \lesssim \big \Vert \overline{\Pi }_{f(0)}g\big \Vert _{C^\alpha } + \big \Vert \mathscr {L}\big (\overline{\Pi }_fg\big )\big \Vert _{C_T C^{\alpha - 2}}, \end{aligned}$$

and the rough bound $\big \Vert \overline{\Pi }_{f(0)}g\big \Vert _{C^\alpha } \le \Vert f\Vert _{C_T L^{\infty }} \Vert g\Vert _{C^\alpha }$. Next, we write, with $\mathscr {L}g = -\Delta g$,

$$\begin{aligned} \mathscr {L} \big (\overline{\Pi }_fg\big ) = \Big \{\mathscr {L}\big (\overline{\Pi }_fg\big ) - \overline{\Pi }_f (-\Delta g)\Big \} - \overline{\Pi }_f (\Delta g), \end{aligned}$$

and use commutator Lemma 5.1 of [11] to get

$$\begin{aligned} \Big \Vert \mathscr {L}\big (\overline{\Pi }_fg\big ) - \overline{\Pi }_f (-\Delta g) \Big \Vert _{C_T\mathcal {C}^{\beta + \alpha - 2}} \lesssim \Vert f \Vert _{\mathcal {C}^\beta }\Vert g \Vert _{C^\alpha }, \end{aligned}$$

and

$$\begin{aligned} \Big \Vert \overline{\Pi }_f (-\Delta g)\Big \Vert _{C_T C^{\alpha - 2}} \lesssim \Vert f \Vert _{C_T L^{\infty }} \Vert g \Vert _{C^\alpha }. \end{aligned}$$

$\square$