Self-Similar Finite-Time Blowups with Smooth Profiles of the Generalized Constantin–Lax–Majda Model

Abstract

We show that the a-parameterized family of the generalized Constantin–Lax–Majda model, also known as the Okamoto–Sakajo–Wunsch model, admits exact self-similar finite-time blowup solutions with interiorly smooth profiles for all $a\le 1$. Depending on the value of a, these self-similar profiles are either smooth on the whole real line or compactly supported and smooth in the interior of their closed supports. The existence of these profiles is proved in a consistent way by considering the fixed-point problem of an a-dependent nonlinear map, based on which detailed characterizations of their regularity, monotonicity, and far-field decay rates are established. Our work unifies existing results for some discrete values of a and also explains previous numerical observations for a wide range of a.

Appendices

Appendix A: Special Functions

1.1 A.1 Special Function F

We define

$$\begin{aligned} F(t):= \frac{t^2-1}{2t}\ln \left| \frac{t+1}{t-1}\right| + 1, \quad t\ge 0. \end{aligned}$$

(A.1)

The derivative of F reads

$$\begin{aligned} F'(t) = \frac{t^2+1}{2t^2}\ln \left| \frac{t+1}{t-1}\right| - \frac{1}{t}. \end{aligned}$$

For $t\in [0,1)$, F(t) and $F'(t)$ have the Taylor expansions

$$\begin{aligned} F(t) = \sum \limits _{n=1}^\infty \frac{2t^{2n}}{4n^2-1},\quad F'(t) = \sum \limits _{n=1}^\infty \frac{4nt^{2n-1}}{4n^2-1}. \end{aligned}$$

For $t\in [0,1)$, F(1/t) and $F'(1/t)$ have the Taylor expansions

$$\begin{aligned} F(1/t) = 2 - \sum \limits _{n=1}^\infty \frac{2t^{2n}}{4n^2-1},\quad F'(1/t) = \sum \limits _{n=1}^\infty \frac{4nt^{2n+1}}{4n^2-1}. \end{aligned}$$

Lemma A.1

The function F defined in (A.1) satisfies

• (1) $F(1/t) = 2-F(t)$, $F'(1/t) = t^2F'(t)$;

• (2) $F\in C([0,+\infty ))$, $F(0) = 0$, $F(1) = 1$, $\lim _{t\rightarrow +\infty }F(t)=2$, $\lim _{t\rightarrow 0}F(t)/t = 0$;

• (3) $F'(0)=0$ and $F'(t)>0$ for $t> 0$.

Proof

Property (1) is straightforward to check. (2) follows from the Taylor expansion of F(t) and property (1). (3) follows from the Taylor expansion of $F'(t)$ and property (1). $\square$

1.2 A.2 Special Function G

We define

$$\begin{aligned} G(t):= \frac{3t^4 - 2t^2 - 1}{8t^3}\ln \left| \frac{t+1}{t-1}\right| + \frac{1}{4t^2} + \frac{7}{12}, \quad t\ge 0. \end{aligned}$$

(A.2)

The derivative of G reads

$$\begin{aligned} G'(t) = \frac{3t^4 + 2t^2 + 3}{8t^4}\ln \left| \frac{t+1}{t-1}\right| - \frac{3t^2+3}{4t^3}. \end{aligned}$$

For $t\in [0,1)$, G(t) and $G'(t)$ have the Taylor expansions

$$\begin{aligned} G(t) = \sum \limits _{n=1}^{+\infty }\frac{4(n+1)t^{2n}}{(2n-1)(2n+1)(2n+3)},\quad G'(t) = \sum \limits _{n=1}^{+\infty }\frac{8n(n+1)t^{2n-1}}{(2n-1)(2n+1)(2n+3)}. \end{aligned}$$

For $t\in [0,1)$, G(1/t) and $G'(1/t)$ have the Taylor expansions

$$\begin{aligned} G(1/t)= & {} \frac{4}{3}-\sum \limits _{n=1}^{+\infty }\frac{4nt^{2n+2}}{(2n-1)(2n+1)(2n+3)},\\ G'(1/t)= & {} \sum \limits _{n=1}^{+\infty }\frac{8n(n+1)t^{2n+3}}{(2n-1)(2n+1)(2n+3)}. \end{aligned}$$

Lemma A.2

The function G defined in (A.2) satisfies

• (1) $G'(1/t) = t^4G'(t)$;

• (2) $G\in C([0,+\infty ))$, $G(0) = 0$, $G(1) = 5/6$, $\lim _{t\rightarrow +\infty }G(t)=4/3$, $\lim _{t\rightarrow 0}G(t)/t=0$;

• (3) $G'(t)\ge 0$ for $t\ge 0$.

• (4) $(4t/3-tG(1/t))' = tF'(1/t)$ for $t\ge 0$.

Proof

Properties (1) is straightforward to check. (2) follows from the Taylor expansions of G(t) and G(1/t). (3) follows from the Taylor expansion of $F'(t)$ and property (1). (4) can be checked straightforwardly by the definitions of G(t) and F(t). $\square$

1.3 A.3 Special Functions $F_i$

Based on the special function F in Appendix A.1, we introduce a series of functions $F_i(t), t\ge 0, i=1,2,3,4$ that appear in the proof of Theorem 4.3:

$$\begin{aligned} \begin{aligned} F_1(t)&:= \int _0^tsF'(s)\, \textrm{d}{s},\\ F_2(t)&:= t^{-1}F_1(t) + tF_1(1/t),\\ F_3(t)&:= \int _0^ts^2F_2(s)\, \textrm{d}{s},\\ F_4(t)&:= t^3\int _0^{1/t}s^5F_3(1/s)\, \textrm{d}{s}. \end{aligned} \end{aligned}$$

It is not hard to check that, for $t>0$,

$$\begin{aligned} F_3'(1/t) = t^{-4}F_3'(t), \end{aligned}$$

which immediately leads to

$$\begin{aligned} F_4(t) = F_4(1/t) = \frac{1}{6}\left( t^{-3}F_3(t) + t^3F_3(1/t)\right) . \end{aligned}$$

Using the Taylor expansions of F and properties of F in Lemma A.1, we can obtain the Taylor expansions of each $F_i$: for $t\in [0,1]$,

$$\begin{aligned} F_1(t)= & {} \sum \limits _{n=1}^\infty \frac{4nt^{2n+1}}{(2n-1)(2n+1)^2},\quad F_1(1/t) = \frac{\pi ^2}{4} - \sum \limits _{n=1}^\infty \frac{4nt^{2n-1}}{(2n-1)^2(2n+1)};\\ F_2(t)= & {} F_2(1/t) = \frac{\pi ^2}{4}t - \sum \limits _{n=1}^\infty \frac{8nt^{2n}}{(2n-1)^2(2n+1)^2};\\ F_3(t)= & {} \frac{\pi ^2}{16}t^4 - \sum \limits _{n=1}^\infty \frac{8nt^{2n+3}}{(2n-1)^2(2n+1)^2(2n+3)},\\ F_3(1/t)= & {} \frac{\pi ^2}{8}t^{-2} + \sum \limits _{n=1}^\infty \frac{8nt^{2n-3}}{(2n-3)(2n-1)^2(2n+1)^2};\\ F_4(t)= & {} F_4(1/t) = \frac{\pi ^2}{32}t + \sum \limits _{n=1}^\infty \frac{8nt^{2n}}{(2n-3)(2n-1)^2(2n+1)^2(2n+3)}. \end{aligned}$$

An elementary calculation shows that $F_4'(t)\ge F_4'(1) = 0$ for $t\in [0,1]$. Hence, the maximum of $F_4(t)$ is achieved at $t=1$ with

$$\begin{aligned} F_4(1) = \frac{\pi ^2}{32} + \sum \limits _{n=1}^\infty \frac{8n}{(2n-3)(2n-1)^2(2n+1)^2(2n+3)} = \frac{\pi ^2}{32} - \frac{1}{6}, \end{aligned}$$

which is used in the proof of Theorem 4.3.

1.4 A.4 Special Functions $(1-x^2 - p)_+ + p$

The functions $f_{m,p}:= (1-x^2 - p)_+ + p$ with $p\in [0,1-\eta /4)$ are a special family in the function set $\mathbb {D}$ that satisfy $b(f_{m,p})/c(f_{m,p}) = (1-p)/3$. In particular, $f_m:= f_{m,0} = (1-x^2)_+$ is the minimal function in $\mathbb {D}$ in the sense that $f(x)\ge f_m(x)$ for all x and all $f\in \mathbb {D}$. We have used the following properties of $f_{m,p}$ in our preceding arguments.

First, we can compute that, for $x\ge 1$,

$$\begin{aligned} {\varvec{T}}(f_{m,p})(x) + b(f_{m,p})&= \frac{1}{\pi }\int _0^{+\infty }f_{m,p}'(y)\cdot y(F(x/y)-2)\, \textrm{d}{y}\\&= \frac{2}{\pi }\int _0^{\sqrt{1-p}}y^2F(y/x)\, \textrm{d}{y} = \frac{2}{\pi }x^3\cdot \int _0^{\sqrt{1-p}/x}t^2F(t)\, \textrm{d}{t} \\&\ge \frac{4}{15\pi }\cdot \frac{(1-p)^{5/2}}{x^2}. \end{aligned}$$

In particular, for $x\ge 1$,

$$\begin{aligned} {\varvec{T}}(f_m)(x) + b(f_m) \ge \frac{4}{15\pi }x^{-2}, \end{aligned}$$

which has been used in the proof of Theorem 4.10 part (1).

Based on the estimates above, we find that for $x\ge 1$,

$$\begin{aligned} {\varvec{R}}_1(f_{m,p})(x)&= {\varvec{T}}_1(f_{m,p})(x) = \left( 1 + \frac{3{\varvec{T}}(f_{m,p})(x)}{c(f_{m,p})}\right) _+ \\&\ge \left( 1 - \frac{3b(f_{m,p})}{c(f_{m,p})} + \frac{4}{5\pi }\cdot \frac{(1-p)^{5/2}}{c(f_{m,p})x^2}\right) _+ \\&= \left( p + \frac{4}{5\pi }\cdot \frac{(1-p)^{5/2}}{c(f_{m,p})x^2}\right) _+ = p + \frac{4}{5\pi }\cdot \frac{(1-p)^{5/2}}{c(f_{m,p})x^2} > p. \end{aligned}$$

However, $f_{m,p}(x) \equiv p$ for $x\ge 1$, which means that $f_{m,p}$ cannot be a fixed point of ${\varvec{R}}_1$. We have used this fact in the proofs of Lemma 4.1 and Lemma 4.2.

Next, we show that $\mu (f_m) = 2Q(f_m)/b(f_m)^2 = \overline{\mu }$, where $\overline{\mu }$ is given in Theorem 4.3, and $\mu (f),Q(f)$ are defined in the proof of this theorem. Owing to the calculations in the proof of Theorem 4.3, for $f\in \mathbb {D}$, we have

$$\begin{aligned} Q(f) = \frac{1}{\pi ^2}\int _0^{+\infty }\int _0^{+\infty }\left( \frac{f'(x)}{x}\right) '\left( \frac{f'(y)}{y}\right) 'x^3y^3F_4(x/y)\, \textrm{d}{x}\, \textrm{d}{y}, \end{aligned}$$

and

$$\begin{aligned} b(f) = \frac{2}{3\pi }\int _0^{+\infty }\left( \frac{f'(y)}{y}\right) 'y^3\, \textrm{d}{y}. \end{aligned}$$

Note that $f'_m(x) = -2x, x\in [0,1)$ and $f'_m(x) = 0, x>1$. We thus have

$$\begin{aligned} \left( \frac{f'(x)}{x}\right) ' = 2\cdot \delta (x-1),\quad x\ge 0, \end{aligned}$$

where $\delta (x)$ is the Dirac function centered at 0. It then follows that

$$\begin{aligned} Q(f_m) = \frac{4}{\pi ^2}F_4(1),\quad b(f_m) = \frac{4}{3\pi }, \end{aligned}$$

and thus

$$\begin{aligned} \mu (f_m) = \frac{2Q(f_m)}{b(f_m)^2} = \frac{9}{2}F_4(1) =: \overline{\mu }. \end{aligned}$$

Recall we have shown in the proof of Theorem 4.3 that $Q(f)\le \overline{\mu }$ for all suitable f. This means that $f_m$ is the maximizer of $\mu (f)$ over the set $\mathbb {D}$.

Appendix B: On the Hilbert Transform

We prove two useful lemmas that exploit properties the Hilbert transform.

Lemma B.1

For any suitable function $\omega$ on $\mathbb {R}$,

$$\begin{aligned} \frac{{\varvec{H}}(\omega )(x)-{\varvec{H}}(\omega )(0)}{x} = {\varvec{H}}\left( \frac{\omega }{x}\right) (x). \end{aligned}$$

As a result,

$$\begin{aligned} \frac{1}{\pi }\int _{\mathbb {R}}\frac{{\varvec{H}}(\omega )(x)\cdot \omega (x)}{x}\, \textrm{d}{x} = -\frac{1}{2}\big ({\varvec{H}}(\omega )(0)\big )^2. \end{aligned}$$

Proof

The first equation follows directly from the definition of the Hilbert transform on the real line. The second equation is derived from the first one as follows:

$$\begin{aligned} \frac{1}{\pi }\int _{\mathbb {R}}\frac{{\varvec{H}}(\omega )\cdot \omega }{x}\, \textrm{d}{x}&= \frac{1}{\pi }\int _{\mathbb {R}}\frac{({\varvec{H}}(\omega )-{\varvec{H}}(\omega )(0))}{x}\omega \, \textrm{d}{x} + {\varvec{H}}(\omega )(0)\cdot \frac{1}{\pi }\int _{\mathbb {R}}\frac{\omega }{x}\, \textrm{d}{x}\\&= \frac{1}{\pi }\int _{\mathbb {R}}{\varvec{H}}\left( \frac{\omega }{x}\right) \omega \, \textrm{d}{x} - \big ({\varvec{H}}(\omega )(0)\big )^2\\&= -\frac{1}{\pi }\int _{\mathbb {R}}\frac{\omega }{x}{\varvec{H}}(\omega )\, \textrm{d}{x} - \big ({\varvec{H}}(\omega )(0)\big )^2. \end{aligned}$$

Rearranging the equation above yields the desired result. $\square$

Lemma B.2

Given a function $\omega$, suppose that $\Vert x^\delta \omega \Vert _{L^{+\infty }(\mathbb {R})} = \sup _{x\in \mathbb {R}}|x|^{\delta }|\omega (x)|<+\infty$ for some $\delta >0$. If $\omega \in H^k_{loc}(A,B)$ for some $A<B$ and some integer $k\ge 0$, then ${\varvec{H}}(\omega )\in H^k_{loc}(A,B)$.

Proof

We first prove a formula for the k-th derivative of ${\varvec{H}}(\omega )$: if $\omega \in H^k_{loc}(A,B)$, then for any $A<a<b<B$ and any $x\in (a,b)$,

$$\begin{aligned} {\varvec{H}}(\omega )^{(k)}(x) = {\varvec{H}}(\chi _{[a,b]}\omega ^{(k)})(x) + g_{a,b}^{(k)}(x) + \sum \limits _{j=0}^{k-1}f_{a,b,j}^{(k-j)}(x), \end{aligned}$$

(B.1)

where the summation is 0 if $k=0$, and

$$\begin{aligned} {\varvec{H}}(\chi _{[a,b]}\omega ^{(k)})(x)= & {} \frac{1}{\pi }P.V.\int _a^b\frac{\omega ^{(k)}(y)}{x-y}\, \textrm{d}{y},\\ g_{a,b}(x):= & {} \frac{1}{\pi }\int _{-\infty }^a\frac{\omega (y)}{x-y}\, \textrm{d}{y} + \int _b^{+\infty }\frac{\omega (y)}{x-y}\, \textrm{d}{y},\\ f_{a,b,j}(x):= & {} \frac{1}{\pi }\left( \omega ^{(j)}(a)\ln |x-a| - \omega ^{(j)}(b)\ln |x-b|\right) ,\quad j=0,1,2,\dots . \end{aligned}$$

We prove this formula with induction. The base case $k=0$ is trivial:

$$\begin{aligned} {\varvec{H}}(\omega )(x) = \frac{1}{\pi }P.V.\int _{-\infty }^{+\infty }\frac{\omega (y)}{x-y}\, \textrm{d}{y} = \frac{1}{\pi }P.V.\int _a^b\frac{\omega (y)}{x-y}\, \textrm{d}{y} + g_{a,b}(x). \end{aligned}$$

Now suppose that (B.1) is true for some integer $k\ge 0$, we need to show that it is then also true for $k+1$. Under the assumption that $\omega \in H^{k+1}_{loc}(A,B)$, we can use integration by parts to rewrite the first term on the right-hand side of (B.1) as

$$\begin{aligned} {\varvec{H}}(\chi _{[a,b]}\omega ^{(k)})(x)&= \frac{1}{\pi }P.V.\int _a^b\frac{\omega ^{(k)}(y)}{x-y}\, \textrm{d}{y} \\&= -\frac{1}{\pi }\omega ^{(k)}(y)\ln |x-y|\Big |_{y=a}^{y=b} + \frac{1}{\pi }\int _a^b\omega ^{(k+1)}\ln |x-y|\, \textrm{d}{y}\\&= f_{a,b,k}(x) + \frac{1}{\pi }\int _a^b\omega ^{(k+1)}\ln |x-y|\, \textrm{d}{y}. \end{aligned}$$

Note that $\omega ^{(k)}(a)$ and $\omega ^{(k)}(b)$ are finite because $\omega \in H^{k+1}_{loc}(A,B)$. It then follows from the inductive assumption that, for $x\in (a,b)$,

$$\begin{aligned} {\varvec{H}}(\omega )^{(k+1)}(x)&= \left( {\varvec{H}}(\chi _{[a,b]}\omega ^{(k)})(x)\right) ' + \left( g_{a,b}^{(k)}(x) + \sum \limits _{j=0}^{k-1}f_{a,b,j}^{(k-j)}(x)\right) '\\&= f_{a,b,k}'(x) + \left( \frac{1}{\pi }\int _a^b\omega ^{(k+1)}\ln |x-y|\, \textrm{d}{y}\right) ' \\&\quad + \left( g_{a,b}^{(k)}(x) + \sum \limits _{j=0}^{k-1}f_{a,b,j}^{(k-j)}(x)\right) '\\&= f_{a,b,k}'(x) + \frac{1}{\pi }P.V.\int _a^b\frac{\omega ^{(k+1)}}{x-y}\, \textrm{d}{y} \\&\quad + g_{a,b}^{(k+1)}(x) + \sum \limits _{j=0}^{k-1}f_{a,b,j}^{(k+1-j)}(x)\\&= \frac{1}{\pi }P.V.\int _a^b\frac{\omega ^{(k+1)}}{x-y}\, \textrm{d}{y} + g_{a,b}^{(k+1)}(x) + \sum \limits _{j=0}^kf_{a,b,j}^{(k+1-j)}(x). \end{aligned}$$

Hence, (B.1) is also true for $k+1$. This completes the induction.

We then use (B.1) to prove the lemma. Note that under the assumptions of the lemma, it is easy to see that $g_{a,b}(x)$ and $f_{a,b,j}(x), j=0,1,\dots , k-1,$ are infinitely smooth in the interior of (a, b), and thus $g_{a,b},f_{a,b,j}\in H^k_{loc}(a,b)$. As for the first term on the right-hand side of (B.1), we have

$$\begin{aligned} \left\| {\varvec{H}}(\chi _{[a,b]}\omega ^{(k)})\right\| _{L^2([a,b])}\le & {} \left\| {\varvec{H}}(\chi _{[a,b]}\omega ^{(k)})\right\| _{L^2(\mathbb {R})}\\= & {} \left\| \chi _{[a,b]}\omega ^{(k)}\right\| _{L^2(\mathbb {R})} = \Vert \omega \Vert _{\dot{H}^k([a,b])}<+\infty . \end{aligned}$$

Therefore, (B.1) implies that ${\varvec{H}}(\omega )\in H^k_{loc}(a,b)$. Since this is true for any $A<a<b<B$, we immediately have ${\varvec{H}}(\omega )\in H^k_{loc}(A,B)$. $\square$