1 Introduction

In this paper, we consider the Cauchy problem for the following semilinear heat equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{u} - \Delta u = f(u) &{} \qquad \text{ in } (0, \infty ) \times {\mathbb {R}}^{d}, \\ u(0, x) = u_{0}(x)&{} \qquad \text{ in } {\mathbb {R}}^{d}, \end{array}\right. } \end{aligned}$$
(1.1)

where \(u(t, x) : (0, \infty ) \times {\mathbb {R}}^{d} \rightarrow {\mathbb {R}}\), \(d\ge 2\) is the unknown function, f is the nonlinearity and \(u_{0}\in L^q({\mathbb {R}}^d)\) with \(1\le q\le \infty \) is the given initial data.

It is well known that when f is \(C^1({\mathbb {R}},{\mathbb {R}})\), the Cauchy problem (1.1) possesses a unique classical solution if the initial data \(u_0\in L^\infty ({\mathbb {R}}^d)\). For unbounded initial data, this equation has been studied intensively since the pioneering work of Weissler [18]. For instance, in the pure power case i.e. \(f(u) = |u|^{p-1}u\; (p>1)\), equation (1.1) becomes scale invariant, that is, if u(t, x) satisfies (1.1), then so does

$$\begin{aligned} u_{\lambda }(t, x) = \lambda ^{\frac{2}{p-1}}u(\lambda ^{2}t, \lambda x) \end{aligned}$$

for any \(\lambda >0\). Moreover, if one defines the index

$$\begin{aligned} q_{c} = \frac{d(p-1)}{2}, \end{aligned}$$

then the Lebesgue space \(L^{q_{c}}({\mathbb {R}}^d)\) becomes invariant and we have \(\Vert u_{\lambda }\Vert _{L^{q_{c}}} = \Vert u\Vert _{L^{q_{c}}}\) for all \(\lambda >0\). The critical exponent \(q_{c}\) then plays an important role for the well-posedness of the Cauchy problem. Indeed, first recall Weissler’s result [18] concerning the subcritical case \(q > q_{c}\) or the critical case \(q=q_c>1\). Weissler [18] proved that for any \(u_{0} \in L^{q}({\mathbb {R}}^{d})\), there exists a local time \(T = T(u_{0})>0\) and a solution \(u \in C([0, T); L^{q}({\mathbb {R}}^{d})) \cap L_{\text {loc}}^{\infty }((0, T); L^{\infty }({\mathbb {R}}^{d}))\) to (1.1). After that, Brezis and Cazenave [1] proved the unconditional uniqueness of Weissler’s solutions i.e. solution is unique in \( C([0, T); L^{q}({\mathbb {R}}^{d}))\) when the subcritical case \(q>q_c\), \(q\ge p\) or the critical case \(q=q_c>p\). In the supercritical case \(q < q_{c}\), Weissler [18], and Brezis and Cazenave [1] indicated that, for a specific initial data, there is no local solution in any reasonable weak sense. Moreover, Haraux and Weissler [7] proved non-uniqueness of the trivial solution in \(C([0, T); L^{q}({\mathbb {R}}^{d})) \cap L_{\text {loc}}^{\infty } ((0, T); L^{\infty }({\mathbb {R}}^{d}))\) when \(1 + 2/d< p < (d+2)/(d-2)\).

In the critical case \(q = q_{c}\) and \(d \ge 3\), when \(q_{c} > {p}\), Weissler [18] proved the existence of solutions to (1.1) and Brezis and Cazenave [1] obtained the unconditional uniqueness of the solutions. In the double critical case of \(q = q_{c} = p\; (= d/(d-2))\), Weissler [19] proved the conditional well-posedness (uniqueness in a subspace of \(C([0, T); L^{q}({\mathbb {R}}^{d}))\)). In the case where the underlying space is the ball of \({\mathbb {R}}^{d}\) with Dirichlet boundary condition, Ni and Sacks [15] showed that the unconditional uniqueness fails, while Terraneo [17] and Matos and Terraneo [12] extended this result to the entire space \({\mathbb {R}}^{d} \; (d \ge 3)\).

We note that the critical exponent \(q_{c}\) is also important for the blow-up problem (1.1). Let \(u_{0} \in L^{\infty }({\mathbb {R}}^{d})\) and \(T(u_{0})\) be the maximal existence of the time of the classical solution u. It is known that if \(T(u_{0}) < \infty \), the solution u satisfies \(\lim _{t \rightarrow T} \Vert u(t)\Vert _{L^{\infty }} = \infty \) and we say that the solution u blows up in finite time and \(T(u_{0})\) is called the blow-up time of u. In particular, the critical \(L^{q_{c}}\) norm blow-up problem has attracted attention for a long time. Namely, it is a question whether the solution satisfies

$$\begin{aligned} \sup _{t \in (0, T)} \Vert u\Vert _{L^{q_{c}}} = \infty \end{aligned}$$

or not when \(T(u_{0}) < \infty \). Concerning this problem, Giga and Kohn [6] showed that if u is a positive radially decreasing blow-up solution to (1.1), Then, for any neighborhood N of 0 in \({\mathbb {R}}^{d}\), \(\lim _{t \rightarrow T} \Vert u(t)\Vert _{L^{q_{c}, \infty }(N)} = \infty \), where

$$\begin{aligned} \Vert u\Vert _{L^{q_{c}, \infty }(N)} = \sup _{\sigma> 0} \sigma |\left\{ x \in N \mid |u(x)| > \sigma \right\} |^{\frac{1}{q_{c}}} \end{aligned}$$

and \(|\cdot |\) is the Lebesgue measure on \({\mathbb {R}}^{d}\). Since \(\Vert u\Vert _{L^{q_{c}, \infty }(N)} \le \Vert u\Vert _{L^{q_{c}}}\), it follows that \(\lim _{t \rightarrow T} \Vert u(t)\Vert _{L^{q_{c}}} = \infty \). Recently, Mizoguchi and Souplet [14] gave light on this problem and showed that if u is a type I blow-up solution of (1.1), that is, \(\limsup _{t \rightarrow T} (T - t)^{1/(p-1)} \Vert u(t)\Vert _{L^{\infty }} < \infty \), then we have \(\lim _{t \rightarrow T} \Vert u(t)\Vert _{L^{q_{c}}} = \infty \).

Now, let us pay our attention to the two dimensional case. When we consider the two space dimension, we see that any exponent \(1< p < \infty \) is subcritical, and thanks to the result of Weissler [18], we have the local well-posedness of the Cauchy problem in \(L^{q}({\mathbb {R}}^{2})\; (1< p \,{\le }\, q < \infty )\). However, for exponential type nonlinearities, like \(f(u) = \pm u(e^{u^{2}} -1)\), the result of Weissler [18] is not applicable for any Lebesgue space \(L^{q}({\mathbb {R}}^{2}) (1< q < \infty )\). On the other hand, we can show the local well-posedness for \(u_{0} \in L^{\infty }({\mathbb {R}}^{2})\), as we mentioned first. However, \(L^{\infty }({\mathbb {R}}^{2})\) is the subcritical space for the problem (1.1) with \(d=2\) and exponential type nonlinearities. Therefore, one can wonder if there is any notion of criticality in two space dimensions. In this regard, Ibrahim, Jrad, Majdoub and Saanouni [8] have considered the following problem in two space dimensions,

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{u} - \Delta u =f_{0}(u) := \pm u(e^{u^{2}} -1)&{} \qquad \text{ in } (0, \infty ) \times {\mathbb {R}}^{2}, \\ u(0, x) = u_{0}(x)&{} \qquad \text{ in } {\mathbb {R}}^{2}. \end{array}\right. } \end{aligned}$$
(1.2)

The nonlinearity \(f_{0}(u)\) has an energy-critical growth in view of Trudinger-Moser inequality. In [8], the authors proved the local existence and uniqueness in \(C([0,T],H^1({\mathbb {R}}^{2}))\) of the solution to (1.2) with the initial data \(u_{0} \in H^{1}({\mathbb {R}}^{2})\).

On the other hand, it is expected that the problem (1.2) for the heat equation can be solved in spaces which are defined by an integrability of functions such as the Orlicz space as an extension of the class of Lebesgue spaces. Ruf and Terraneo [16] showed the local existence of a solution to (1.2) for small initial data in the Orlicz space \(\text {exp}L^2\) defined by

endowed with Luxemburg norm

$$\begin{aligned} \Vert u\Vert _{\text {exp}L^2}:=\inf \left\{ \lambda >0: \int _{{\mathbb {R}}^2}\exp ( u^2/\lambda ^2)-1\;dx\le 1\right\} . \end{aligned}$$

Subsequently, Ioku [9] has shown that these local solutions are indeed global-in time, and the behavior of \(f_0(u)\sim u^m\) near \(u\sim 0\) with \(m\ge 3\) was important in his result. Later on, Ioku, Ruf and Terraneo [10] proved that there is no local solution do not exist for the specific initial data

$$\begin{aligned} u_*(r)=a (-\log r)^\frac{1}{2} \;\text {for}\; r\le 1,\;\; a\gg 1,\;\text {and}\; u_*(r)=0,\;\text {elsewhere} \end{aligned}$$

that belongs to the Orlicz space, while they showed the local well-posedness (local existence and uniqueness) for initial data in the subclass of the Orlicz space

$$\begin{aligned} \text {exp}L^2_0({\mathbb {R}}^2):=\left\{ u\in L^1_{loc}({\mathbb {R}}^2):\;\text {such that}\; \int _{{\mathbb {R}}^2}\exp (\alpha u^2)-1\;dx<\infty \;\text {for every}\;\alpha >0\right\} \end{aligned}$$

The aim of this paper is to construct an explicit initial data, with neither small nor too large norm in Orlicz space, for which two corresponding distinct solutions exist. The idea is to use the concept of singular solutions as in Ni and Sacks [15]. Before stating our result, let us recall the strategy of the proof of [15]. Ni and Sacks first constructed a singular static solution \(\phi _{*}\) to (1.1) in the unit ball. Namely, \(\phi _{*}\) satisfies the following:

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta \phi = f(\phi ) &{} \qquad \text{ in } B_{1}, \\ \phi = 0 \quad \text{ on } \partial B_{1}, &{}\quad \lim _{x \rightarrow 0} \phi (x) = \infty , \end{array}\right. } \end{aligned}$$
(1.3)

where \(B_{1}\) is the unit ball in \({\mathbb {R}}^{d}\; (d \ge 3)\) and \(f(\phi ) = |\phi |^{p-1}\phi \). Then, they showed that there exists a regular solution \(u_{\text {R}}\) to the Dirichlet problem corresponding to (1.1) with \(u(0, x) = \phi _{*}\). Setting \(u_{\text {S}} = \phi _{*}\), we see that both of \(u_{\text {S}}\) and \(u_{\text {R}}\) are solutions to (1.1), but \(u_{\text {S}} \ne u_{\text {R}}\) because for any \(t>0\), \(u_{\text {R}} (t,\cdot ) \in L^{\infty }(B_{1})\) while \(u_{\text {S}} \notin L^{\infty }(B_{1})\) (the subscripts S and R indicating ‘singular’ and ‘regular’ solutions). For the entire space \({\mathbb {R}}^{d}\), Terraneo [17] constructed a solution \(\phi \in C^{2}({\mathbb {R}}^{d} {\setminus } \{-x_{0}, x_{0}\})\) to the equation

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta \phi = f(\phi ) &{} \qquad \text{ in } {\mathbb {R}}^{d}, \\ \lim _{|x| \rightarrow \infty }\phi (x) = 0 &{}\quad \end{array}\right. } \end{aligned}$$
(1.4)

such that \(\limsup _{x \rightarrow x_{0}} \phi (x) = \infty \) and \(\liminf _{x \rightarrow -x_{0}} \phi (x) = -\infty \), where \(d \ge 3, x_{0} = (1, \ldots , 1)\) and \(f(\phi ) = |\phi |^{p-1} \phi \). However, we cannot apply the method of [15] nor of [17] to two dimensional entire space \({\mathbb {R}}^{2}\) for \(f_{0}(u) = (e^{u^{2}} - 1)u\) directly. Actually, there is no positive solution to the equation (1.4) with \(f(u) = (e^{u^{2}} - 1)u\) (see the proof of Theorem 2.1 below). For this reason, we consider

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{u} - \Delta u = f_{m}(u) &{} \text{ in } (0, \infty ) \times {\mathbb {R}}^{2}, \\ u(0, x) = u_{0}(x) &{} \text{ in } {\mathbb {R}}^{2}, \end{array}\right. } \end{aligned}$$
(1.5)

where our nonlinearity \(f_{m}\), which depends on \(m >0\), satisfies

$$\begin{aligned} \lim _{u\rightarrow \infty } \frac{f_m(u)}{(e^{u^2}-1)u} = 1, \qquad \lim _{u\rightarrow 0} \frac{f_m(u)}{m u} = -1. \end{aligned}$$
(1.6)

See (2.17) below for the precise form of \(f_{m}\). Here, we would like to stress that the essential characterization of the asymptotic form of our nonlinearity \(f_{m}\) at infinity and 0 is just given by (1.6). Let X be the Fréchet space defined as the intersection of the Lebesgue Banach spaces

$$\begin{aligned} X:=\bigcap _{1\le p<\infty } L^p({\mathbb {R}}^2), \end{aligned}$$

endowed with the metric

$$\begin{aligned} \text {dist}(u,v)=\sum _{p=1}^\infty 2^{-p}\frac{\Vert u-v\Vert _{L^p}}{\Vert u-v\Vert _{L^p}+1}. \end{aligned}$$

Recall that the continuity of any map defined on X is equivalent to its continuity on all \(L^p\) spaces with \(1\le p<\infty \).

Definition 1

(Mild solution) We call u a mild solution to (1.5) if the corresponding integral equation

$$\begin{aligned} u(t) = e^{t \Delta } u_{0} + \int _{0}^{t} e^{(t-s)\Delta }f_{m}(u(s))ds \qquad \end{aligned}$$

holds in C([0, T), X), where \(e^{t \Delta }\) denotes the heat semigroup on \({\mathbb {R}}^{2}\).

Then, our main result is the following:

Theorem 1.1

There exist a positive mass \(m_{*}>0\), a nonlinearity \(f_{m_*}\) satisfying (1.6), an initial condition \(\varphi _{*} \in X\) and a time \(T = T(\varphi _{*}) >0\) such that the Cauchy problem (1.5) with \(u_{0} = \varphi _{*}\) has two distinct mild solutions \(u_{S}, u_{R} \in C([0, T), X)\).

Remark 1

  1. (i)

    To prove Theorem 1.1, we construct a singular stationary solution \(\varphi _{*}\) to (1.5). Here, by a singular stationary solution, we mean a time independent function which satisfies the equation (1.5) in the distribution sense on the whole domain and diverges at some points. The result seems to be of independent interest.

  2. (ii)

    We essentially use the condition \(\lim _{u \rightarrow 0}f_{m}(u)/mu = -1\) only for the decay of a singular stationary soliton \(\varphi _{*}\) to (1.5), that is, \(\lim _{|x| \rightarrow \infty } \varphi _{*}(x) = 0\). The other argument in our proof works even without the condition. It is a challenging problem to study whether non-uniqueness still hold without the condition or not.

Remark 2

After completing this paper, it has been brought to our attention that Ioku, Ruf and Terraneo [11] obtained a similar non-uniqueness result for the following two dimensional heat equation on a ball with Dirichlet boundary condition:

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{u} - \Delta u = f(u) &{} \qquad \text{ in } (0, \infty ) \times B_{\rho } \\ u = 0 &{} \qquad \text{ on } (0, \infty ) \times \partial B_{\rho } \end{array}\right. }, \end{aligned}$$
(1.7)

for some \(\rho > 0\), and a nonlinearity f satisfying

$$\begin{aligned} f(u) := {\left\{ \begin{array}{ll} \frac{1}{u^{3}} e^{u^{2}} &{} \qquad \text{ if } |u| > \beta , \\ \alpha u^{2} &{} \qquad \text{ if } |u| \le \beta , \end{array}\right. } \end{aligned}$$

with \(\alpha = e^{5/2} (5/2)^{-5/2}\) and \(\beta = (5/2)^{1/2}\). First, the authors constructed a stationary singular solution \(\varphi _{*}\) of (1.7). Then, they solved the Cauchy problem (1.7) with the initial data \(u_{0}(x) = \mu \varphi _{*}\) for \(\mu > 0\). They proved the following interesting result:

  1. (i)

    If \(\mu < 1\), the equation (1.7) has a unique regular local-in-time solution

  2. (ii)

    If \(\mu = 1\), the equation (1.7) has not only a singular solution \(u_{S} (= \varphi _{*})\) but also a regular solution \(u_{R}\) with the same initial data,

  3. (iii)

    If \(\mu > 1\), the equation has no non-negative solution on any positive time interval.

See [11, Theorem 2.1] for a more precise statement of their trichotomy result. From (ii), we see that non-uniqueness of a solution holds since there exists two distinct solutions \(u_{S}\) and \(u_{R}\) with the same initial data. Thus, the strategy of the proof for the non-uniqueness is same as that of our result. However, in addition to the fact that the settings are different, the ways of constructing the singular and regular solutions are not similar either. For example, their proof of the constructing of the singular stationary solution \(\varphi _{*}\) depended on the fact that the following equation

$$\begin{aligned} - U^{\prime \prime } - \frac{1}{r} U^{\prime } = \frac{1}{U^{3}} e^{U^{2}}, \qquad 0< r < 1 \end{aligned}$$

has an explicit solution \(U = \sqrt{-2 \log r}\). On the other hand, the singular solution we constructed is not explicit. We construct the singular solution in a ball by using the contraction mapping theorem and extend it to the entire space by the shooting method. Moreover, for the construction of the regular solution, they employed the sub-super-solution method, while we employ the contraction mapping theorem.

Remark 3

The problem of uniqueness of solutions for PDEs is a classical and old issue that can be tricky sometimes. It has caught a special attention in the last few years. Among many others, one can refer to the pioneer works of De Lellis and Székelyhidi [4] showing non-uniqueness of very weak solutions to the Euler problem. Their proof relies on convex integration techniques, which more recently, have been upgraded to show non-uniqueness of weak solutions of the Navier-Stokes system thanks to the work of Buckmaster and Vicol [2]. Davila, Del Pino and Wei [3] constructed non-unique weak solutions for the two-dimensional harmonic map, by attaching reverse bubbling. In [5], Germain, Ghoul and Miura investigated the genericity of the non-unique solutions of the supercritical heat flow map.

This paper is organized as follows. In Sect. 2, we construct a singular static soliton \(\varphi _{*}\) to (1.5). To this end, we first prove the existence of a singular soliton \(\phi _{*}\) to (1.3) with \(f(\phi ) = e^{\phi ^{2}}(\phi - 1)\) in the ball in \({\mathbb {R}}^{2}\), following Merle and Peletier [13]. Then, we seek the singular stationary soliton \(\varphi _{*}\) to (1.5) by the shooting method. In Sect. 3, we shall show the existence of a regular solution to (1.5) with \(u|_{t=0} = \varphi _{*}\) by the heat iteration and give a proof of Theorem 1.1. In Appendix A, we show a monotonicity property of solution to the linear heat equation.

1.1 Notation

Throughout the paper, C denotes a positive constant, that does not depend on the parameters, unless otherwise noted and may change from line to line.

2 Construction of singular soliton

2.1 Singular stationary solution on some disk

In this section, we construct a singular solution to the following elliptic equation on a disk

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta u = u(e^{u^{2}} -1 ) \qquad &{} \text{ in } B_{R}, \\ u = 0 \qquad &{} \text{ on } \partial B_{R}, \end{array}\right. } \end{aligned}$$
(2.1)

where \(R > 0\). More precisely, we shall show the following:

Theorem 2.1

There exist \(R_{\infty }>0\) and a singular solution \(u_{\infty } \in C^{\infty }(B_{R_{\infty }} {\setminus } \{0\})\) to (2.1) with \(R = R_{\infty }\) satisfying

$$\begin{aligned} \begin{aligned} u_{\infty }(x)&= \left( -2 \log |x| -2 \log (-\log |x|) -2 \log 2\right) ^{\frac{1}{2}} + O((-\log |x|)^{- \frac{3}{2}} \log (-\log |x|))\qquad \quad \\&\quad \qquad \text{ as } x \rightarrow 0. \end{aligned} \end{aligned}$$
(2.2)

To prove Theorem 2.1, we look for radially symmetric solutions to (2.1). We first pay our attention for \(0 < r = |x| \ll 1\) and employ the following Emden-Fowler transformation:

$$\begin{aligned} y(\rho ) = u(x), \qquad \rho = 2 |\log r|. \end{aligned}$$

Then, we see that the equation in (2.1) is equivalent to the following:

$$\begin{aligned} - \frac{d^{2} y}{d \rho ^{2}} = \frac{e^{-\rho }}{4} y(e^{y^{2}} - 1). \end{aligned}$$
(2.3)

We shall show the following:

Proposition 2.2

There exists \(\Lambda _{\infty } >0\) and a solution \(y_{\infty }(\rho )\) to (2.3) for \(\rho \in [\Lambda _{\infty }, \infty )\) satisfying

$$\begin{aligned} y_{\infty }(\rho ) = (\rho - 2 \log \rho )^{\frac{1}{2}} + O(\rho ^{-\frac{3}{2}} \log \rho ) \qquad \text{ as } \rho \rightarrow \infty . \end{aligned}$$

In order to prove Proposition 2.2, we write

$$\begin{aligned} y(\rho ) = \phi (\rho ) + \eta (\rho ), \qquad \phi (\rho ) := (\rho - 2\log \rho )^{\frac{1}{2}}, \end{aligned}$$
(2.4)

and look for the ODE satisfied by \(\eta \). We have

$$\begin{aligned} \frac{d^{2} y}{d \rho ^{2}} = - \frac{\rho ^{-\frac{3}{2}}}{4} - \frac{1}{4}(\phi ^{-3} - \rho ^{-\frac{3}{2}}) + \phi ^{- 3}\left\{ \frac{1}{\rho } - \frac{1}{\rho ^{2}}\right\} + \phi ^{-1}\rho ^{-2} + \frac{d^{2} \eta }{d \rho ^{2}}. \end{aligned}$$
(2.5)

Now, since

$$\begin{aligned}&y = \rho ^{\frac{1}{2}}\left( 1 - 2\frac{\log \rho }{\rho }\right) ^{\frac{1}{2}} + \eta = \rho ^{\frac{1}{2}} + \rho ^{\frac{1}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \eta , \end{aligned}$$

and

$$\begin{aligned}&e^{y^{2} -\rho } = \rho ^{-2} e^{y^{2} - \phi ^{2}} =\rho ^{-2} \left\{ 1 + (y^{2} - \phi ^{2}) + \left( e^{y^{2} - \phi ^{2}} - 1 - (y^{2} - \phi ^{2})\right) \right\} , \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned} y e^{y^{2}- \rho } =&\, \rho ^{-2}\left\{ \rho ^{\frac{1}{2}} + \rho ^{\frac{1}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \eta \right\} \\&\times \left\{ 1 + (y^{2} - \phi ^{2}) + \left( e^{y^{2} - \phi ^{2}} - 1 - (y^{2} - \phi ^{2})\right) \right\} \\ =&\,\rho ^{-2} \left\{ \rho ^{\frac{1}{2}} + \rho ^{\frac{1}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \eta \right\} \\&+ \rho ^{-2}(y^{2} - \phi ^{2}) \left\{ \rho ^{\frac{1}{2}} + \rho ^{\frac{1}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \eta \right\} \\&+ \rho ^{-2} \left( e^{y^{2} - \phi ^{2}} - 1 - (y^{2} - \phi ^{2})\right) \left\{ { \rho ^{\frac{1}{2}} + \rho ^{\frac{1}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} } + \eta \right\} \\ =&\,\rho ^{-\frac{3}{2}} + \rho ^{-\frac{3}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \rho ^{-2}\eta \\&+ \rho ^{- \frac{3}{2}} (y^{2} - \phi ^{2}) + \rho ^{-\frac{3}{2}}(y^{2} - \phi ^{2}) \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \rho ^{-2}(y^{2} - \phi ^{2}) \eta \\&+ \rho ^{-2} \left( e^{y^{2} - \phi ^{2}} - 1 - (y^{2} - \phi ^{2})\right) \left\{ \rho ^{\frac{1}{2}}(1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} + \eta \right\} . \end{aligned} \end{aligned}$$

In addition, since \( y^{2} - \phi ^{2} = 2\phi \eta + \eta ^{2} = 2 \rho ^{\frac{1}{2}} \eta + \eta ^{2} + 2(\phi - \rho ^{\frac{1}{2}}) \eta , \) we have

$$\begin{aligned} (y^{2} - \phi ^{2})\rho ^{-\frac{3}{2}} = \frac{2 \eta }{\rho } + \rho ^{-\frac{3}{2}}\eta ^{2} + 2 \rho ^{-\frac{3}{2}}(\phi - \rho ^{\frac{1}{2}})\eta . \end{aligned}$$

All this yields that

$$\begin{aligned} \begin{aligned} y e^{y^{2}-\rho } =&\,\rho ^{-\frac{3}{2}} + \rho ^{-\frac{3}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \rho ^{-2}\eta \\&+ 2 \frac{\eta }{\rho } + \rho ^{-\frac{3}{2}}\eta ^{2} + 2 \rho ^{-\frac{3}{2}}(\phi - \rho ^{\frac{1}{2}})\eta \\&+ (y^{2} - \phi ^{2}) \rho ^{-\frac{3}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \rho ^{-2}(y^{2} - \phi ^{2})\eta \\&+ \rho ^{-2} \left( e^{y^{2} - \phi ^{2}} - 1 - (y^{2} - \phi ^{2})\right) \left\{ \rho ^{\frac{1}{2}}(1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} + \eta \right\} . \end{aligned} \end{aligned}$$
(2.6)

From (2.3), (2.5) and (2.6), we have

$$\begin{aligned} \begin{aligned}&\frac{\rho ^{-\frac{3}{2}}}{4} + \frac{1}{4}(\phi ^{-3} - \rho ^{-\frac{3}{2}}) - \phi ^{- 3} \left\{ \frac{1}{\rho } - \frac{1}{\rho ^{2}}\right\} - \phi ^{-1} \rho ^{-2} - \frac{d^{2} \eta }{d \rho ^{2}}\\&\quad = \frac{\rho ^{-\frac{3}{2}}}{4} + \frac{\rho ^{-\frac{3}{2}}}{4} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \frac{\rho ^{-2}}{4} \eta \\&\qquad + \frac{1}{2 \rho }\eta + \frac{\rho ^{-\frac{3}{2}}\eta ^{2}}{4} + \frac{1}{2} \rho ^{-\frac{3}{2}}(\phi - \rho ^{\frac{1}{2}})\eta \\&\qquad + \frac{1}{4} (y^{2} - \phi ^{2}) \rho ^{-\frac{3}{2}} \left\{ (1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} + \frac{1}{4} \rho ^{-2}(y^{2} - \phi ^{2}) \eta \\&\qquad + \frac{\rho ^{-2}}{4} \left( e^{y^{2} - \phi ^{2}} - 1 - (y^{2} - \phi ^{2})\right) \left\{ \rho ^{\frac{1}{2}}(1 - 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} + \eta \right\} - \frac{e^{-\rho } y}{4}. \end{aligned} \end{aligned}$$

Namely, \(\eta \) satisfies the following:

$$\begin{aligned} \begin{aligned} 0 = \frac{d^{2} \eta }{d \rho ^{2}} + \frac{1}{2\rho } \eta + f(\rho ) + \sum _{i=1}^{6}g_{i}(\rho , \eta ) - \frac{e^{-\rho } y}{4}, \end{aligned} \end{aligned}$$
(2.7)

where

$$\begin{aligned} f(\rho ) =&- \frac{1}{4}(\phi ^{-3} - \rho ^{-\frac{3}{2}}) + \phi ^{- 3} \left\{ \frac{1}{\rho } - \frac{1}{\rho ^{2}}\right\} + \phi ^{-1} \rho ^{-2}\\&+ \frac{\rho ^{-\frac{3}{2}}}{4} \left\{ \left( 1 - 2 \frac{\log \rho }{\rho }\right) ^{\frac{1}{2}} - 1\right\} , \\ g_{1}(\rho , \eta ) =&\, \frac{\rho ^{-2}}{4} \eta , \quad g_{2}(\rho , \eta ) = \frac{\rho ^{- \frac{3}{2}}}{2}(\phi - \rho ^{\frac{1}{2}})\eta , \quad g_{3}(\rho , \eta ) = \frac{\rho ^{- \frac{3}{2}}}{4} \eta ^{2}, \\ g_{4}(\rho , \eta ) =&\, \frac{1}{4}(y^{2} - \phi ^{2}) \rho ^{-\frac{3}{2}} \left\{ \left( 1 - 2 \frac{\log \rho }{\rho }\right) ^{\frac{1}{2}} - 1\right\} , \quad g_{5}(\rho , \eta ) = \frac{\rho ^{-2}}{4} (y^{2} - \phi ^{2}) \eta , \\ g_{6}(\rho , \eta ) =&\, \frac{\rho ^{-2}}{4} \left( e^{y^{2} - \phi ^{2}} - 1 - (y^{2} - \phi ^{2})\right) \left\{ \rho ^{\frac{1}{2}}\left( 1 - 2\frac{\log \rho }{\rho }\right) ^{\frac{1}{2}} + \eta \right\} . \end{aligned}$$

To solve equation (2.7), we first consider the following linear equation: Footnote 1

$$\begin{aligned} \frac{d^{2} \eta }{d \rho ^{2}} + \left( \frac{1}{2 \rho } + \frac{3}{16 \rho ^{2}} \right) \eta = 0. \end{aligned}$$
(2.8)

Any solution \(\eta \) to (2.8) can be written explicitly as follows: we have

$$\begin{aligned} \eta (\rho ) = A \Phi (\rho ) + B \Psi (\rho ) \end{aligned}$$

for some \(A, B \in {\mathbb {R}}\), where

$$\begin{aligned} \Phi (\rho ) = \rho ^{\frac{1}{4}} \sin ((2 \rho )^{\frac{1}{2}}), \qquad \Psi (\rho ) = \rho ^{\frac{1}{4}} \cos ((2 \rho )^{\frac{1}{2}}). \end{aligned}$$

Namely, \(\Phi \) and \(\Psi \) are form the fundamental system of solutions to (2.8). For a given function F, we seek a solution to the following problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{d^{2} \eta }{d \rho ^{2}} + (\frac{1}{2 \rho } + \frac{3}{16 \rho ^{2}}) \eta + F = 0, &{} \qquad \rho \gg 1, \\ \lim _{\rho \rightarrow \infty } \eta (\rho ) = 0. \end{array}\right. } \end{aligned}$$
(2.9)

By the variation of parameters, we see that (2.9) is equivalent to the following integral equation:

$$\begin{aligned} \eta (\rho )= & {} \int _{\rho }^{\infty } (\Phi (s) \Psi (\rho ) - \Phi (\rho ) \Psi (s)) F(s) ds\\= & {} \int _{\rho }^{\infty } (\rho s)^{\frac{1}{4}} \sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}}) F(s) ds. \end{aligned}$$

By integrating by parts, we can obtain the following:

Lemma 2.3

Let \(\sigma > 1\). Then, we have

Let \(g_{7}(\rho , \eta ) = - \frac{3}{16 \rho ^{2}} \eta \). In order to prove Theorem 2.1, we seek a solution to the following problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{d^{2} \eta }{d \rho ^{2}} + (\frac{1}{2\rho } + \frac{3}{16 \rho ^{2}})\eta + f(\rho ) + \sum _{i=1}^{7}g_{i}(\rho , \eta ) - \frac{e^{-\rho } y}{4} = 0, &{} \qquad \rho \gg 1, \\ \lim _{\rho \rightarrow \infty } \eta (\rho ) = 0. &{} \end{array}\right. }\qquad \qquad \end{aligned}$$
(2.10)

To this end, we prepare several estimates, which are needed later. First, note that

$$\begin{aligned} \phi (\rho ) = \rho ^{\frac{1}{2}} + O(\rho ^{-\frac{1}{2}} \log \rho ). \end{aligned}$$
(2.11)

By (2.11) and elementary calculations, we can obtain the following:

Lemma 2.4

Let \(\phi \) be the function given by (2.4). Then, for sufficiently large \(\rho >0\), we have

$$\begin{aligned} \phi ^{-3} - \rho ^{-3/2}= & {} \,3 \rho ^{-5/2} \log \rho + O(\rho ^{-7/2}\log ^2\rho ), \\ \biggl | \rho ^{-\frac{3}{2}} \left\{ (1- 2 \frac{\log \rho }{\rho })^{\frac{1}{2}} - 1\right\} \biggl |= & {} \,-\rho ^{-5/2}\log \rho + O(\rho ^{-7/2}\log ^2\rho ), \\ \phi ^{-1}\rho ^{-2}= & {} \,\rho ^{-5/2} + O(\rho ^{-7/2} \log \rho ), \\ \phi ^{-3} (\rho ^{-1} - \rho ^{-2})= & {} \, \rho ^{-5/2} + O(\rho ^{-7/2}\log \rho ). \end{aligned}$$

In addition, we have the following estimates on the terms \(g_j\) of equation (2.10). More specifically, we have

Lemma 2.5

Suppose that \(\eta _{1}(\rho ), \eta _{2}(\rho ) = O(\rho ^{-\frac{3}{2}}\log \rho )\). Then, we have

$$\begin{aligned}&|g_{1}(\rho , \eta _{1}) - g_{1}(\rho , \eta _{2})|, |g_{7}(\rho , \eta _{1}) - g_{7}(\rho , \eta _{2})| \le C \rho ^{-2} |\eta _{1} - \eta _{2}|, \\&|g_{2}(\rho , \eta _{1}) - g_{2}(\rho , \eta _{2})|, |g_{4}(\rho , \eta _{1}) - g_{4}(\rho , \eta _{2})|, |g_{6}(\rho , \eta _{1}) - g_{6}(\rho , \eta _{2})| \\&\quad \le C {\rho ^{-2} (\log \rho }) |\eta _{1} - \eta _{2}|, \\&|g_{3}(\rho , \eta _{1}) - g_{3}(\rho , \eta _{2})|, |g_{5}(\rho , \eta _{1}) - g_{5}(\rho , \eta _{2})| \le C{ \rho ^{-3} (\log \rho )} |\eta _{1} - \eta _{2}|. \end{aligned}$$

Proof

From the definitions of \(g_{i}\) and (2.11), we have

$$\begin{aligned}&|g_{1}(\rho , \eta _{1}) - g_{1}(\rho , \eta _{2})|, |g_{7}(\rho , \eta _{1}) - g_{7}(\rho , \eta _{2})| \le C \rho ^{-2} |\eta _{1} - \eta _{2}|, \\&|g_{2}(\rho , \eta _{1}) - g_{2}(\rho , \eta _{2})| \le C \rho ^{-\frac{3}{2}} |\phi - \rho ^{\frac{1}{2}}| |\eta _{1} - \eta _{2}| \le C { \rho ^{-2}(\log \rho )} |\eta _{1} - \eta _{2}|, \\&|g_{3}(\rho , \eta _{1}) - g_{3}(\rho , \eta _{2})| \le C \rho ^{-\frac{3}{2}} |\eta _{1}^{2} - \eta _{2}^{2}| = \rho ^{-\frac{3}{2}}|\eta _{1} + \eta _{2}||\eta _{1} - \eta _{2}| \\&\quad \le C {\rho ^{-3} (\log \rho )} |\eta _{1} - \eta _{2}| {.} \end{aligned}$$

Let \(y_{1}(\rho ) = \phi (\rho ) + \eta _{1}(\rho )\) and \(y_{2}(\rho ) = \phi (\rho ) + \eta _{2}(\rho )\). Then, we obtain

$$\begin{aligned} \begin{aligned} |g_{4}(\rho , \eta _{1}) - g_{4}(\rho , \eta _{2})|&\le C { \rho ^{-\frac{5}{2}} (\log \rho )} |(y_{1}^{2}(\rho ) - \phi ^{2}(\rho )) - (y_{2}^{2}(\rho ) - \phi ^{2}(\rho ))| \\&\le C { \rho ^{-\frac{5}{2}} (\log \rho )} (2\phi |\eta _{1} - \eta _{2}| + |\eta _{1} + \eta _{2}||\eta _{1} - \eta _{2}|) \\&\le C { \rho ^{-2} (\log \rho )}|\eta _{1} - \eta _{2}|. \end{aligned} \end{aligned}$$

Since \(y_{i}^{2} - \phi ^{2} = 2\phi \eta _{i} + \eta _{i}^{2}\) for \(i = 1, 2\), we have

$$\begin{aligned} |y_{i}^{2} - \phi ^{2}| \le |2 \phi + \eta _{i}||\eta _{i}| \le C \rho ^{-1} \log \rho \qquad \text{ for } i= 1, 2. \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned} |g_{5}(\rho , \eta _{1}) - g_{5}(\rho , \eta _{2})| \le&\, C \rho ^{-2} |(y_{1}^{2} - \phi ^{2})\eta _{1} - (y_{2}^{2} - \phi ^{2})\eta _{2}| \\ \le&\, C \rho ^{-2}|(y_{1}^{2} - \phi ^{2})(\eta _{1} - \eta _{2})| + \rho ^{-2} |y_{1}^{2} - y_{2}^{2}||\eta _{2}| \\ \le&\, C {\rho ^{-3}(\log \rho )} |\eta _{1} - \eta _{2}| + \rho ^{-2} {\rho ^{-\frac{3}{2}} (\log \rho )} \rho ^{\frac{1}{2}} |\eta _{1} - \eta _{2}| \\ \le&\, C { \rho ^{-3} (\log \rho )}|\eta _{1} - \eta _{2}|.\\ |g_{6}(\rho , \eta _{1}) - g_{6}(\rho , \eta _{2})| \le&\, C \rho ^{-2} |e^{y_{1}^{2} - \phi ^{2}} - 1 - (y_{1}^{2} - \phi ^{2})||\eta _{1} - \eta _{2}| \\&+ C \rho ^{-2} |(e^{y_{1}^{2} - \phi ^{2}} - 1 - (y_{1}^{2} - \phi ^{2})) \\&- (e^{y_{2}^{2} - \phi ^{2}} - 1 - (y_{2}^{2} - \phi ^{2}))| |\rho ^{\frac{1}{2}} + \eta _{2}| \\ \le&\, C \rho ^{-2} |e^{y_{1}^{2} - \phi ^{2}} - 1 - (y_{1}^{2} - \phi ^{2})| |\eta _{1} - \eta _{2}| \\&+ C\rho ^{-\frac{3}{2}} |(e^{y_{1}^{2} - \phi ^{2}} - 1 - (y_{1}^{2} - \phi ^{2}))\\&- (e^{y_{2}^{2} - \phi ^{2}} - 1 - (y_{2}^{2} - \phi ^{2}))| \\ =:&\, I + II. \end{aligned} \end{aligned}$$

-1pt From (2.1), we have

$$\begin{aligned} I \le C \rho ^{-2} |y_{1}^{2} - \phi ^{2}|^{2} |\eta _{1} - \eta _{2}| \le C \rho ^{-4} (\log \rho )^{2} |\eta _{1} - \eta _{2}|. \end{aligned}$$

-1pt Using the mean value theorem and (2.1), we have

$$\begin{aligned} \begin{aligned}&|(e^{y_{1}^{2} - \phi ^{2}} - 1 - (y_{1}^{2} - \phi ^{2})) - (e^{y_{2}^{2} - \phi ^{2}} - 1 - (y_{2}^{2} - \phi ^{2}))| \\&\quad \le C |\exp [\theta (y_{1}^{2} - \phi ^{2}) + (1-\theta ) (y_{2}^{2} - \phi ^{2})] - 1||y_{1}^{2} - y_{2}^{2}| \\&\quad \le C|\theta (y_{1}^{2} - \phi ^{2}) + (1-\theta )(y_{2}^{2} - \phi ^{2})| |y_{1}^{2} - y_{2}^{2}| \\&\quad \le C { \rho ^{-1} (\log \rho )} (2\phi |\eta _{1} - \eta _{2}| + |\eta _{1} + \eta _{2}| |\eta _{1} - \eta _{2}|) \\&\quad \le C { \rho ^{-\frac{1}{2}} (\log \rho )} |\eta _{1} - \eta _{2}|. \end{aligned} \end{aligned}$$

This yields that \(II \le C { \rho ^{-2} (\log \rho )} |\eta _{1} - \eta _{2}|\). Thus, we see that

$$\begin{aligned} |g_{6}(\rho , \eta _{1}) - g_{6}(\rho , \eta _{2})| \le C { \rho ^{-2} (\log \rho )} |\eta _{1} - \eta _{2}|, \end{aligned}$$

as desired. \(\square \)

We are now in a position to prove Proposition 2.2.

Proof of Proposition 2.2

Note that (2.7) is equivalent to the following integral equation:

$$\begin{aligned} \eta (\rho ) = {\mathcal {T}}[\eta ](\rho ), \end{aligned}$$

in which

$$\begin{aligned} {\mathcal {T}}[\eta ](\rho ) = \int _{\rho }^{\infty } (\rho s)^{\frac{1}{4}}\sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}}) F(s, \eta ) ds, \end{aligned}$$

where

$$\begin{aligned} F(\rho , \eta ) = f(\rho ) + \sum _{i=1}^{7} g_{i}(\rho , \eta ) - \frac{e^{-\rho }}{4} y. \end{aligned}$$

Fix \(\Lambda >0\) sufficiently large and let X be a space of continuous functions on \([\Lambda , \infty )\) equipped with the following norm:

$$\begin{aligned} \Vert \xi \Vert = \sup \left\{ |\rho |^{\frac{3}{2}}(\log \rho )^{-1} |\xi (\rho )| \mid \rho \ge \Lambda \right\} . \end{aligned}$$

Lemmas 2.3 and 2.4 insure that there exists a constant \(C_{*}>0\) such that

$$\begin{aligned} \biggl | \int _{\rho }^{\infty } (\rho s)^{\frac{1}{4}}\sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}})f(s) ds\biggl | \le C_{*} \rho ^{-\frac{3}{2}} \log \rho . \end{aligned}$$
(2.12)

For this constant \(C_{*}>0\), define the space

$$\begin{aligned} \Sigma = \left\{ \xi \in X \mid \Vert \xi \Vert \le 3 C_{*} \right\} , \end{aligned}$$

and we shall first show that \({\mathcal {T}}\) maps \(\Sigma \) to itself. From Lemma 2.5, we have

$$\begin{aligned} \biggl |\sum _{i=1}^{7}g_{i}(\rho , \eta )\biggl | \le C \rho ^{-2} \log \rho |\eta | \le C \rho ^{-\frac{7}{2}}(\log \rho )^{2}, \end{aligned}$$

which yields that

$$\begin{aligned} \begin{aligned}&\biggl | \int _{\rho }^{\infty } (\rho s)^{\frac{1}{4}}\sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}}) \sum _{i=1}^{7}g_{i}(\rho , \eta ) ds \biggl | \le C \int _{\rho }^{\infty } \rho ^{\frac{1}{4}} s^{\frac{1}{4}} s^{-\frac{7}{2}} (\log s)^{2} ds \\&\quad \le C \rho ^{\frac{1}{4}} \int _{\rho }^{\infty } s^{-3} ds \le C \rho ^{-\frac{7}{4}}. \end{aligned} \end{aligned}$$

Therefore, we can take \(\Lambda >0\) sufficiently large so that

$$\begin{aligned} \biggl | \int _{\rho }^{\infty } (\rho s)^{\frac{1}{4}} \sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}}) \sum _{i=1}^{7}g_{i}(\rho , \eta ) ds \biggl | \le C_{*} \rho ^{-\frac{3}{2}} \log \rho \end{aligned}$$
(2.13)

for \(\rho \ge \Lambda \). We can easily find that for sufficiently large \(\Lambda >0\), we have

$$\begin{aligned} \biggl | \int _{\rho }^{\infty } (\rho s)^{\frac{1}{4}}\sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}}) e^{- s}y(s) ds \biggl | \le C e^{-\frac{\rho }{2}} \le C_{*} \rho ^{-\frac{3}{2}} \log \rho \end{aligned}$$
(2.14)

for \(\rho \ge \Lambda \). By (2.12), (2.13) and (2.14), we obtain

$$\begin{aligned} |{\mathcal {T}}[\eta ](\rho )| \le 3 C_{*} \rho ^{-\frac{3}{2}} \log \rho \end{aligned}$$

for \(\eta \in \Sigma \). This yields that \({\mathcal {T}}[\eta ] \in \Sigma \).

Next, we shall show that \({\mathcal {T}}\) is a contraction mapping. For \(\eta _{1}, \eta _{2} \in \Sigma \), we have

$$\begin{aligned} \begin{aligned} |{\mathcal {T}}[\eta _{1}](\rho ) - {\mathcal {T}}[\eta _{2}](\rho )|&\le \,\sum _{i=1}^{7} \int _{\rho }^{\infty } |(\rho s)^{\frac{1}{4}}\sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}})| |g_{i}(s, \eta _{1}) - g_{i}(s, \eta _{2})| ds \\&\qquad \!+\! \int _{\rho }^{\infty } |(\rho s)^{\frac{1}{4}}\sin ((2\rho )^{\frac{1}{2}} \!-\! (2s)^{\frac{1}{2}})| e^{-s}|\eta _{1} \!-\! \eta _{2}| ds. \end{aligned} \end{aligned}$$

From Lemma 2.5 follows the estimate

$$\begin{aligned} \begin{aligned} |{\mathcal {T}}[\eta _{1}](\rho ) - {\mathcal {T}}[\eta _{2}](\rho )| \le&\,\, C \int _{\rho }^{\infty } |(\rho s)^{\frac{1}{4}}\sin ((2\rho )^{\frac{1}{2}} - (2s)^{\frac{1}{2}})| s^{-2} \log s|\eta _{1} - \eta _{2}| ds\\&+ C \Vert \eta _{1} - \eta _{2}\Vert e^{-\rho /2} \\ \le&\,\, C \int _{\rho }^{\infty } s^{\frac{1}{4}} \rho ^{\frac{1}{4}}s^{-\frac{7}{2}}(\log s)^{2} \Vert \eta _{1} - \eta _{2}\Vert ds + C \Vert \eta _{1} - \eta _{2}\Vert e^{-\rho }\\ \le&\,\, C \Vert \eta _{1} - \eta _{2}\Vert \rho ^{-2} (\log \rho )^{2}. \end{aligned} \end{aligned}$$

This yields that

$$\begin{aligned} \Vert {\mathcal {T}}[\eta _{1}](\rho ) - {\mathcal {T}}[\eta _{2}](\rho )\Vert \le C \rho ^{-\frac{1}{4}} \Vert \eta _{1} - \eta _{2}\Vert < \frac{1}{2} \Vert \eta _{1} - \eta _{2}\Vert \end{aligned}$$

for sufficiently large \(\rho > 0\). Thus, we find that \({\mathcal {T}}\) is a contraction mapping. This completes the proof. \(\square \)

Proof of Theorem 2.1

Let \(u_{\infty }(r) = y_{\infty }(\rho )\), where \(y_{\infty }(\rho )\) is the solution to (2.3) obtained by Proposition 2.2. We note that \(u_{\infty }\) satisfies the following:

$$\begin{aligned} - \frac{d^{2} u_{\infty }}{d r^{2}} - \frac{1}{r} \frac{d u_{\infty }}{d r} = u_{\infty } (e^{u_{\infty }^{2}} - 1)\qquad \text{ for } r \in (0, R_{\infty }), \end{aligned}$$
(2.15)

where \(R_{\infty } = e^{- \Lambda _{\infty }/2}\). Since \(u_{\infty }\) is a solution to the ordinary differential equation (2.15), we can extend \(u_{\infty }\) in the positive direction of r as long as \(u_{\infty }\) remains bounded. We claim that \(u_{\infty }\) has a zero at some point. Suppose the contrary that \(u_{\infty }(r) >0\) for all \(0< r < \infty \). Then, we see that \(u_{\infty }\) is monotone decreasing. Indeed, if not, there exists a local minimum point \(r_{*} \in (0, \infty )\). It follows that \(\partial ^{2}_{r} u_{\infty }(r_{*}) \ge 0\) and \(\partial _{r} u_{\infty }(r_{*}) = 0\). Then, from the equation (2.15), we obtain

$$\begin{aligned} 0 \le \frac{d^{2} u_{\infty }}{d r^{2}}(r_{*}) = - u_{\infty }(r_{*}) (e^{u_{\infty }^{2}(r_{*})} - 1) < 0, \end{aligned}$$

which is a contradiction.

Since \(u_{\infty }\) is positive and monotone decreasing, there exists a constant \(C_{\infty } \ge 0\) such that \(u_{\infty }(r) \rightarrow C_{\infty }\) as \(r \rightarrow \infty \). To prove that \(C_{\infty } = 0\), suppose the contrary that \(C_{\infty } > 0\). This together with (2.15) yields that

$$\begin{aligned} 0 = \lim _{r \rightarrow \infty } \left( \frac{d^{2} u_{\infty }}{d r^{2}}(r) + \frac{1}{r} \frac{d u_{\infty }}{d r} (r)\right) = - \lim _{r \rightarrow \infty } (e^{u_{\infty }^{2}(r)} - 1)u_{\infty }(r) < 0, \end{aligned}$$

which is absurd. Therefore, we see that \(C_{\infty } = 0\), that is, \(\lim _{r \rightarrow \infty }u_{\infty }(r) = 0\). Multiplying (2.15) by r and integrating the resulting equation from 0 to r yields that

$$\begin{aligned} -r \frac{d u_{\infty }}{d r}(r) = \int _{0}^{r} s u_{\infty } (e^{u_{\infty }^{2}} - 1) ds > 0. \end{aligned}$$

This yields that for any \(R>0\), there exists a constant \(C_{1} >0\) such that \(- d u_{\infty }/dr(r) \ge C_{1}/r\) for all \(r > R\). It follows that

$$\begin{aligned} u_{\infty }(r) - u_{\infty }(R) = \int _{R}^{r} \frac{d u_{\infty }}{d s}(s) ds \le - C_{1} \int _{R}^{r} \frac{1}{s} ds. \end{aligned}$$

Letting \(r \rightarrow \infty \), we have

$$\begin{aligned} - u_{\infty }(R) = \lim _{r \rightarrow \infty }(u_{\infty }(r) - u_{\infty }(R)) \le - { C_{1}} \lim _{r \rightarrow \infty } \int _{R}^{r} \frac{1}{s} ds = -\infty , \end{aligned}$$

which is a contradiction. Therefore, there exists \(R_{\infty } > 0\) such that \(u_{\infty }(r)\) has a zero at \(r = R_{\infty }\).

Finally, we shall show that \(u_{\infty }\) is a solution (2.1) in a distribution sense. Let \(\eta (\rho )\) be the solution to (2.10), obtained in the proof of Proposition 2.2. Since \(\eta (\rho ) = O(\rho ^{-3/2} \log \rho )\), we see that \(u_{\infty }\) satisfies

$$\begin{aligned} u_{\infty }(r)= & {} \left( -2 \log r -2 \log (-\log r) -2 \log 2 \right) ^{\frac{1}{2}} + O((-\log r)^{- \frac{3}{2}} \log (-\log r))\\&\text{ as } r \rightarrow 0. \end{aligned}$$

This yields that there exists a constant \(C>0\) such that

$$\begin{aligned} u_{\infty } e^{u_{\infty }^{2}} \le C r^{-2} (-\log r)^{-\frac{3}{2}} \qquad \text{ for } \text{ sufficiently } \text{ small } r > 0. \end{aligned}$$
(2.16)

This together with the monotonicity implies that \(u_{\infty } e^{u_{\infty }^{2}} \in L^{1}_{\text {loc}}(B_{R_{\infty }})\). Therefore, we see that \(u_{\infty }\) is a distributional solution to (2.1). This completes the proof. \(\square \)

2.2 Singular soliton by the shooting

Let \(R\in (0,R_\infty )\) be the unique point such that \(u_\infty (R)=2\). For each \(m\ge 0\), we put

$$\begin{aligned} f_{m}(s) := s(e^{s^2}-1)-m\chi (s)s, \end{aligned}$$
(2.17)

where \(\chi \in C^\infty ({\mathbb {R}})\) is a cut-off function satisfying \(\chi (t)=1\) for \(|t|\le 1, \chi (t)=0\) for \(|t|\ge 2\), and \(t\chi '(t)\le 0\) for all \(t\in {\mathbb {R}}\). Consider a family of radial ODE’s with a parameter \(m\ge 0\):

$$\begin{aligned} {\left\{ \begin{array}{ll} - \varphi _{*}^{\prime \prime } - \frac{\varphi _{*}^{\prime }}{r} = f_m(\varphi _{*}), &{}(r>R)\\ \varphi _{*}(R)=u_\infty (R)=2,\ \varphi _{*}^{\prime }(R)= u_\infty ^{\prime }(R), \end{array}\right. } \end{aligned}$$
(2.18)

where the prime mark denotes the differentiation with respect to r. Let \(\phi _m\) be the unique solution of the above. We shall show that there exists \(m_{*}>0\) such that \(\phi _{m_{*}}(r)\searrow 0\) as \(r\rightarrow \infty \). To this end, we show the following:

Proposition 2.6

Let \(m\ge 0\) and \(\phi _{m}\) be the solution to (2.18). There exists \(m_{S}>0\) and \(m_{L}>0\) such that if \(m \in [0, m_{S})\), \(\phi _{m}\) has a zero in \((R, \infty )\), and if \(m \in (m_{L}, \infty )\), \(\phi _{m}(r)\) is positive for all \(r \ge R\).

First, we define an energy density function \(E_{m}:[R,\infty )\rightarrow {\mathbb {R}}\) by

$$\begin{aligned} E_{m}(r) := \frac{(\phi _{m}^{\prime }(r))^{2}}{2} + F_m(\phi _m(r)), \end{aligned}$$

where \(F_m\) is the nonlinear potential energy defined by

$$\begin{aligned} F_m(u):=\int _0^u f_m(s)ds= \frac{e^{u^2} - 1 - u^2}{2} - m \int _{0}^{u} \chi (s) s ds, \end{aligned}$$

which enjoys the standard superquadratic condition:

$$\begin{aligned}&0\le G_m(u):=uF_m'(u)-2F_m(u)=\displaystyle \sum _{k=2}^\infty \frac{(k-1)}{k!}u^{2k} + 2 m\displaystyle \int _0^u(\chi (s)-\chi (u))sds,\nonumber \\ \end{aligned}$$
(2.19)

and, thanks to the monotonicity of \(\chi \),

$$\begin{aligned} G_m(u)=0\quad \text{ if } \text{ and } \text{ only } \text{ if }\quad u=0. \end{aligned}$$
(2.20)

It follows from (2.18) that

$$\begin{aligned} E_{m}^{\prime }(r) = - \frac{(\phi _{m}^{\prime }(r))^{2}}{r} \le 0. \end{aligned}$$
(2.21)

Thus, \(E_{m}(r)\) is a non-increasing function of r. Using (2.21), we shall show the following:

Lemma 2.7

Let \(m>0\) and \(\phi _{m}\) be a solution to (2.18). Let \(s\ge R\), \(E_{m}(s) <0\) and \(\phi _{m}(s)>0\). Then, we have \(\phi _{m}(r) >0\) for all \(r \in (s, \infty )\).

Proof

Suppose the contrary. Then there exists \(z \in (s, \infty )\) such that \(\phi _{m}(z) = 0\). This together with (2.21) yields that

$$\begin{aligned} 0> E_{m}(s) \ge E_{m}(z) = \frac{(\phi _{m}'(z))^{2}}{2} \ge 0, \end{aligned}$$

which is a contradiction. \(\square \)

We are now in a position to prove Proposition 2.6.

Proof of Proposition 2.6

We note that \(m = 0\), \(\phi _{0}(r) = u_{\infty }\) has a zero at \(r = R_{\infty }\). Then, by the continuity of \(\phi _m\) with respect to m, we see that \(\phi _{m}(r)\) still has a zero if \(m>0\) is sufficiently small.

On the other hand, we have

$$\begin{aligned} E_m(R)=\frac{(u_\infty '(R))^2}{2}+\frac{e^4-5}{2}-m\int _0^2\chi (s)sds < 0 \end{aligned}$$

for large \(m>0\), then Lemma 2.7 implies that \(\phi _m(r)>0\) for all \(r>R\). \(\square \)

We put

$$\begin{aligned} m_{*} = \inf \left\{ m>0 \mid \phi _{m} (r)>0 \quad \text{ on } r > R \right\} . \end{aligned}$$
(2.22)

We see from Proposition 2.6 that \(0< m_{*} < \infty \). We extend \(\phi _{m_{*}}\) by defining \(\phi _{m_{*}}(r) = u_{\infty }(r)\) for \(r \in (0, R)\) (still denoted by the same symbol). Then, we have the following:

Theorem 2.8

Let \(m_{*} >0\) be the number defined by (2.22). Then, \(\phi _{m_{*}}\) is a singular positive radial solution to the following elliptic equation

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta \phi = f_{m_{*}}(\phi ) &{} \qquad \text{ in } {\mathbb {R}}^{2}, \\ \lim _{|x| \rightarrow \infty }\phi (x) = 0, &{} \end{array}\right. } \end{aligned}$$
(2.23)

where \(f_{m}(s)\) is defined by (2.17). Moreover, \(\phi _{m_{*}}\) is strictly decreasing in the radial direction. Moreover, for any \(m \in (0, m_{*})\) there exists \(C_{m} \in (0, \infty )\) such that

$$\begin{aligned} \phi _{m_{*}}(r)+|\phi _{m_{*}}^{\prime }(r)| \le C_{m} e^{-\sqrt{m} r} \qquad \text{ for } \text{ all } r \ge R. \end{aligned}$$
(2.24)

Proof

To prove Theorem 2.8, it suffices to show that \(\phi _{m_{*}}(r) > 0, \phi _{m_{*}}^{\prime }(r) < 0\) for all \(r > 0\), \(\lim _{r \rightarrow \infty } \phi _{m_{*}}(r) = 0\) and (2.24).

First, we shall show that \(\phi _{m_{*}}(r) >0\) on \(r \in (R, \infty )\). By definition of \(m_*\), there exists a sequence \(m_{n} \searrow m_{*}\) such that \(\phi _{m_{n}}(r) >0\) for all \(r> R\) and \(n \in {\mathbb {N}}\). Then \(\phi _{m_*}(r)=\lim _{n\rightarrow \infty }\phi _{m_n}(r) \ge 0\) for all \(r> R\). If \(\phi _{m_*}(r)=0\) at some \(r>R\), then \(\phi _{m_*}'(r)=0\) and so \(\phi _{m_*}\equiv 0\) by the ODE, a contradiction. Hence \(\phi _{m_*}(r)>0\) for all \(r>R\).

Next, we claim that

$$\begin{aligned} E_{m_{*}}(r) \ge 0 \qquad \text{ for } \text{ all } r \in (R, \infty ). \end{aligned}$$
(2.25)

Suppose the contrary that there exists \(R_{*}>0\) such that \(E_{m_{*}}(R_{*}) < 0\). Then the continuity for m yields that \(\phi _m(r)>0\) on \(R\le r\le R_*\) and \(E_{m}(R_{*}) < 0\) when \(m\in (0,m_*)\) is close enough to \(m_*\). Then Lemma 2.7 implies \(\phi _m(r)>0\) for \(r\ge R_*\), hence for all \(r>R\), contradicting the definition of \(m_*\). Hence we have (2.25).

Next, we shall show that \(\phi _{m_{*}}^{\prime }(r) < 0\) for all \(r > R\). Suppose the contrary and let \(s>R\) be the first zero of \(\phi _{m_*}'\). Then we have \(0=\phi _{m_*}'(s)\le \phi _{m_*}''(s)=-f_{m_*}(\phi _{m_*}(s))\), \(0\le E_{m_*}(s)=F_{m_*}(\phi _{m_*}(s))\), and so \(G_{m_*}(\phi _{m_*}(s))\le 0\), contradicting (2.19).

Therefore \(\phi _{m_*}'(r)<0<\phi _{m_*}(r)\) for all \(r>R\), so \(\phi _{m_*}(r)\searrow \exists C_*\in [0,2)\) and \(\phi _{m_*}'(r)\rightarrow 0\) as \(r\rightarrow \infty \). Then we have \(0\le \lim _{r\rightarrow \infty }E_{m_*}(r)=F_{m_*}(C_*)\) and

$$\begin{aligned} \phi _{m_*}''(r)=-\phi _{m_*}'(r)/r-f_{m_*}(\phi _{m_*}(r))\rightarrow -f_{m_*}(C_*) \quad (r\rightarrow \infty ), \end{aligned}$$

which has to be 0 because \(\phi _{m_*}'(r)\rightarrow 0\). Hence \(G_{m_*}(C_*)\le 0\) and so \(C_*=0\) by (2.19).

Next, let \(m\in (0,m_*)\). Then \(\phi _{m_*}(r)\searrow 0\) together with the definition of \(f_m\) implies

$$\begin{aligned} \phi _{m_*}''(r) = -\phi _{m_*}'(r)/r - f_{m_*}(\phi _{m_*}(r))> m \phi _{m_*}(r) > 0 \end{aligned}$$

for sufficiently large \(r>R\). Hence \(e^{\sqrt{m}r}(\sqrt{m}-\partial _r)\phi _{m_*}(r)\) is decreasing for large r, which implies the desired exponential decay.

Finally, it follows from (2.16) and (2.24) that \(f_{m_{*}} \in L^{1}({\mathbb {R}}^{2})\). Thus, we see that \(\phi _{m_{*}}\) is a distributional solution to (2.23). This completes the proof. \(\square \)

Remark 4

Let \(u_{\infty }\) be a solution to (2.1), which is obtained in Theorem 2.1. Since \(\phi _{m_{*}} (r) = u_{\infty }(r)\) for \(r \in (0, R]\), we see from (2.2) that \(\phi _{m_{*}}\) satisfies

$$\begin{aligned} \phi _{m_{*}}(r)= & {} \left( -2 \log r -2 \log (-\log r) -2 \log 2 \right) ^{\frac{1}{2}} \nonumber \\&+ O((-\log r)^{- \frac{3}{2}} \log (-\log r)) \quad \text{ as } r \rightarrow 0. \end{aligned}$$
(2.26)

Moreover, we have, by (2.10), Lemmas 2.3 and 2.4, that \(|d^{2} \eta /d \rho ^{2} (\rho )| \le C \rho ^{-5/2} \log \rho \). Thus, by integrating, we see that \(|d \eta /d \rho (\rho )| \le C \rho ^{-3/2} \log \rho \). Thus, \(\phi _{m_{*}}^{\prime }\) satisfies

$$\begin{aligned} \phi _{m_{*}}^{\prime }(r)= & {} - \left( -2 \log r -2 \log (-\log r) -2 \log 2\right) ^{-\frac{1}{2}} \left( \frac{1}{r} + \frac{1}{r\log r} \right) \nonumber \\&+ O((-\log r)^{- 3/2} \log (-\log r)) \quad \text{ as } r \rightarrow 0. \end{aligned}$$
(2.27)

3 Regular solution by the heat iteration

In what follows, we denote \(\phi _{m_{*}}\), which is obtained in Theorem 2.8, by \(\varphi _{*}\). Now we are ready to construct a regular solution to (1.5) with initial data \(\varphi _*\).

Theorem 3.1

Let \(u_0:=e^{t\Delta }\varphi _{*}\). Then for any \(t>0\), \(u_0(t)\) is bounded on \({\mathbb {R}}^2\). Moreover, there exists a small \(T>0\), and a solution \(u_{R}\) to (1.5) with \(u_{R}|_{t=0} = \varphi _{*}\) satisfying

$$\begin{aligned} { |\log t|^{1/2}(u_{R} - u_0) \in L^\infty ([0, T) \times {\mathbb {R}}^{2}). } \end{aligned}$$

Note that \(\varphi _{*}\in C^\infty ({\mathbb {R}}^2{\setminus }\{0\})\) is a positive radial function satisfying the asymptotic form,

$$\begin{aligned} \varphi _{*}(r)= & {} (\rho - 2\log \rho )^{1/2}+O(\rho ^{-3/2}\log \rho )\nonumber \\= & {} (\rho -2\log \rho +O(\rho ^{-1}\log \rho ))^{1/2} \quad (\rho \rightarrow \infty ), \end{aligned}$$
(3.1)

where, as before, \(\rho :=|\log r^2|=2|\log r|\gg 1\). The above two equivalent expressions of remainder will be frequently switched in the following computations.

We need a precise estimate or asymptotic behavior around \(t,r\rightarrow 0\) of the iteration sequence. Consider the first (or zeroth) iteration

$$\begin{aligned} u_0:=e^{t\Delta }\varphi _{*} = \frac{1}{4\pi t}\int _{{\mathbb {R}}^2}e^{-\frac{|x-y|^2}{4t}}\varphi _{*}(y)dy = \int _{{\mathbb {R}}^2} \frac{e^{-|z|^2/4}}{4\pi }\varphi _{*}(x-\sqrt{t}z)dz. \end{aligned}$$

We shall show the following:

Lemma 3.2

There exists \(\varepsilon >0\) such that if \(\max \{t, |x|^{2}\} < \varepsilon ^{2}\), we have

$$\begin{aligned} u_{0}(t, x) \le \min \{\varphi _{*}(\sqrt{t}), \varphi _{*}(x)\} + O(|\log t|^{-\frac{1}{2}}). \end{aligned}$$
(3.2)

Proof

We take \(\varepsilon >0\) sufficiently small so that for any \(r < \varepsilon \), \(\varphi _{*}(r)\) satisfies (3.1), and consider the region \((t, x) \in (0, \infty ) \times {\mathbb {R}}^{2}\) satisfying \(\max \{t, |x|^{2}\} < \varepsilon ^{2}\). Put

$$\begin{aligned} \ell :=-\log t=|\log t| \gg 1,\quad \nu :=|x|^2/t\in [0,\infty ). \end{aligned}$$

We shall estimate \(u_{0}\) by dividing the space-time region \(\max \{t, |x|^{2}\} < \varepsilon ^{2}\) into the three subregions: inside the parabolic cylinder \(\nu \le 1\), \(\nu > 8 \log \ell \gg 1\) and \(1\le \nu \le 8 \log \ell \).

First, we consider the region \(\nu \le 1\). It follows from Lemma A.1 that

$$\begin{aligned} \Vert u_0(t,\cdot )\Vert _{L^\infty _x} = u_0(t, 0) = \int _{{\mathbb {R}}^2} \frac{e^{-|z|^2/4}}{4\pi }\varphi _{*}(\sqrt{t}z)dz. \end{aligned}$$

Thus, by (3.1), we obtain

$$\begin{aligned} \Vert u_0(t,\cdot )\Vert _{L^\infty _x}= & {} \int _{{\mathbb {R}}^2} \frac{e^{-|z|^2/4}}{4\pi }\varphi _{*}(\sqrt{t}z)dz \nonumber \\= & {} \int _0^{t} e^{-s/4}(-\log (ts) - 2\log |\log (ts)| + O(1))^{1/2}ds/4 \nonumber \\&+ \int _{t}^{\frac{\varepsilon ^{2}}{t}} e^{-s/4}(-\log (ts) - 2\log |\log (ts)| + O(1))^{1/2}ds/4 \nonumber \\&+ \int _{\frac{\varepsilon ^2}{t}}^{\infty } e^{- \frac{s}{4}} \varphi _{*}(\sqrt{ts})ds/4 \nonumber \\=: & {} \mathrm{I} + \mathrm{J} + \mathrm{K}. \end{aligned}$$
(3.3)

For \(0< s < t\), we have \(ts>s^{2}\). This yields via an integration by parts

$$\begin{aligned} \mathrm{I} \le C \int _0^{t} |2 \log s|^{1/2}ds \le C t \ell ^{1/2} \le C\ell ^{-1/2}. \end{aligned}$$
(3.4)

For \(t< s < \frac{\varepsilon ^2}{t}\), we have \(|\log s| <\ell \), so the integrand is expanded

$$\begin{aligned} (-\log (ts) - 2\log |\log (ts)|+O(1))^{1/2}= & {} (\ell -2\log \ell - \log s +O(1))^{1/2}\nonumber \\= & {} (\ell -2\log \ell )^{1/2} + O(\ell ^{-1/2}{\langle \log s \rangle }),\nonumber \\ \end{aligned}$$
(3.5)

where we set \({\langle \log s \rangle } = \sqrt{1 + |\log s|^{2}}\). Since \(\varphi _{*}(r)\) is monotone decreasing in \(r > 0\), we have, by (3.1) and (3.5), that

$$\begin{aligned} \varphi _{*}(\sqrt{ts}) \le (\ell -2\log \ell )^{1/2} + O(\ell ^{-1/2}{\langle \log s \rangle }) \end{aligned}$$

for \(s \ge \frac{\varepsilon ^{2}}{t}\). Integration against \(e^{-s/4}\) yields

$$\begin{aligned} \mathrm{J} + \mathrm{K}\le & {} \int _{t}^{\infty } e^{-s/4} ((\ell -2\log \ell )^{1/2} + O(\ell ^{-1/2}{\langle \log s \rangle })ds/4 \nonumber \\\le & {} (\ell -2\log \ell )^{1/2} + O(\ell ^{-1/2}). \end{aligned}$$
(3.6)

Estimates (3.4) and (3.6) together with (3.3) imply

$$\begin{aligned} u_0(t, 0)-(\ell -2\log \ell )^{1/2} \le C \ell ^{-1/2}. \end{aligned}$$

Namely, we have

$$\begin{aligned} u_{0} \le \varphi _{*}(\sqrt{t}) + O(\ell ^{-1/2}). \end{aligned}$$
(3.7)

This is enough in the region \(\nu \le 1\).

Before focusing on the two remaining regions, let \(0<\delta \le \varepsilon \) be a small parameter and set

$$\begin{aligned} B:=\{z\in {\mathbb {R}}^2 \mid |x-\sqrt{t}z|<\delta |x|\}. \end{aligned}$$

Decomposing \(u_{0}\) as

$$\begin{aligned} u_{0}(t, x) = \int _{B} \frac{e^{-|z|^2/4}}{4\pi }\varphi _{*}(x-\sqrt{t}z)dz + \int _{B^c} \frac{e^{-|z|^2/4}}{4\pi }\varphi _{*}(x-\sqrt{t}z)dz =: u_{0}^{I} + u_{0}^{X},\nonumber \\ \end{aligned}$$
(3.8)

and writing

$$\begin{aligned} \displaystyle u_0^X =\int _{B^C}\frac{e^{-|z|^2/4}}{4\pi }[\varphi _*(x) + \varphi _{*}(x-\sqrt{t}z)-\varphi _*(x)]dz. \end{aligned}$$
(3.9)

First, we consider \(z \in B^{C}\) satisfying \(|x - \sqrt{t}z| \le |x|\). One can apply the mean value theorem, and use (2.27) to write

$$\begin{aligned} |\varphi _{*}(|x-\sqrt{t}z|)-\varphi _{*}(|x|)| \le \sup _{\delta |x|\le |y|\le |x|} |\varphi _{*}^{\prime }(|y|)| \big ||x - \sqrt{t}z| - |x|\big | \le C_{\delta } \frac{\sqrt{t}|z|}{r|\log r|^{1/2}}\nonumber \\ \end{aligned}$$
(3.10)

for \(z \in B^{C}\) satisfying \(|x - \sqrt{t}z| \le |x|\). Here, \(C_{\delta }\) is a positive constant which depends on \(\delta \). Next, we consider \(z\in B^C\) satisfying \(|x - \sqrt{t}z| \ge |x|\). Since \(\varphi _{*}\) is monotone decreasing in \(r = |x|\), we see that

$$\begin{aligned} \varphi _{*}(x - \sqrt{t}z) \le \varphi _{*}(x). \end{aligned}$$
(3.11)

for \(z\in B^C\) satisfying \(|x - \sqrt{t}z| \ge |x|\). From (3.9)–(3.11), we conclude that

$$\begin{aligned} u_0^X\le & {} \int _{B^C}\frac{e^{-|z|^2/4}}{4\pi } \left[ \varphi _{*}(x)+ { C_{\delta } \frac{\sqrt{t}|z|}{r|\log r|^{1/2}}}dz\right] \nonumber \\= & {} \varphi _{*}(x)(1-\theta _B) + { C_{\delta } \sqrt{t/r^2}|\log r|^{-1/2}, } \end{aligned}$$
(3.12)

where we set

$$\begin{aligned} \theta _B:=\int _B \frac{e^{-|z|^2/4}}{4\pi }dz \in (0,1). \end{aligned}$$

Now, for \(z\in B\), it follows that

$$\begin{aligned} |\sqrt{t} z| \ge |x| - |x - \sqrt{t}z| \ge (1 - \delta )|x|. \end{aligned}$$
(3.13)

Thus, we have \(|z|^2 \ge \nu (1- \delta )^{2}\), and so

$$\begin{aligned} u_0^I =\int _B \frac{e^{-|z|^2/4}}{4\pi }\varphi _{*}(x-\sqrt{t}z)dz \le \frac{e^{-\nu (1-\delta )^{2}/4 }}{4\pi t} \int _{|y|^2<t\delta ^2\nu } \varphi _{*}(y)dy. \end{aligned}$$
(3.14)

Second, we estimate \(u_{0}^I\) in the region \(\nu > 8 \log \ell \gg 1\). For \(0<a\ll 1\) we have

$$\begin{aligned} \frac{1}{\pi } \int _{|y|^2<a}\varphi _{*}(y)dy= & {} \int _0^a (-\log s - 2\log |\log s| + O(|\log s|^{-1}\log |\log s|))^{1/2}ds \nonumber \\= & {} \int _{|\log a|}^\infty (t-2\log t +O(t^{-1}\log t))^{1/2} e^{-t}dt\nonumber \\= & {} a\left[ (|\log a|-2\log |\log a|)^{1/2}\right. \nonumber \\&+\left. \frac{1}{2}|\log a|^{-1/2}+O(|\log a|^{-3/2}\log |\log a|)\right] \nonumber \\= & {} a\left[ |\log a|-2\log |\log a|+1+O(|\log a|^{-1}\log |\log a|)\right] ^{1/2},\nonumber \\ \end{aligned}$$
(3.15)

by partial integration on \(e^{-t}\). Note that \(t \delta ^{2} \nu = \delta |x|^{2} \le \delta \varepsilon ^{2} \ll 1\). Plugging the formula (3.15) into (3.14) yields

$$\begin{aligned} u_0^I \le \frac{e^{-\nu (1-\delta )^{2}/4}}{4\pi }\delta ^2\nu (|\log (t\delta ^2\nu )|-2\log |\log (t\delta ^2\nu )|+O(1))^{1/2}. \end{aligned}$$
(3.16)

In the region \(\nu >8\log \ell \gg 1\), we have \(\ell = |\log t| < e^{\nu /8}\), yielding

$$\begin{aligned} |\log (t\delta ^2\nu )|-2\log |\log (t\delta ^2\nu )| + O(1)\le & {} 2 |\log (t\delta ^2\nu )| \nonumber \\\le & {} 2 \left( |\log t| + 2|\log \delta | + |\log \nu |\right) \nonumber \\\le & {} 2 \left( e^{\nu /8} + 2 |\log \delta | + |\log \nu | \right) \nonumber \\\le & {} 4 \left( e^{\nu /8} + 2 |\log \delta | \right) . \end{aligned}$$
(3.17)

If we take \(\delta > 0\) sufficiently small so that \(- (1-\delta )^{2}/4 +1/16 < -1/6\), we have, by (3.16) and (3.17), that

$$\begin{aligned} \begin{aligned} u_{0}^{I}&\le \frac{e^{-\nu (1-\delta )^{2}/4}}{\pi } \delta ^2\nu e^{\nu /16} (1 + 2 e^{- \nu /8} |\log \delta |)^{1/2} \\&\le \nu e^{-\nu /6} \delta ^{2} (1 + 2 |\log \delta |)^{1/2} \\&\le \nu e^{-\nu /6}. \end{aligned} \end{aligned}$$

Thus, we get the first intermediate estimate \(u_0^I \le \nu e^{- \nu /6}\). Moreover, using the fact that \( |\log t| < e^{\nu /8}\) again, we obtain

$$\begin{aligned} |2 \log r| = - \log r^{2} = - \log (\nu t) = - \log \nu + |\log t| \le e^{\nu /8}, \end{aligned}$$

and therefore,

$$\begin{aligned} u_0^I \ll |\log r|^{-1/2}\nu ^{-1/2}. \end{aligned}$$
(3.18)

Finally, observe that in the same region \(\nu >8\log \ell \gg 1\), we have \(r^{2} \gg t\) and

$$\begin{aligned} r^{2} |\log r| = -\frac{r^{2} \log r^{2}}{2} \gg - \frac{t \log t}{2} = \frac{t \ell }{2}. \end{aligned}$$

This together with (3.8), (3.12) and (3.18) yields that for \(\nu >8\log \ell \gg 1\), we have obtained

$$\begin{aligned} u_0 - \varphi _{*}(x) \le C_{\delta } |\log r|^{-1/2}\nu ^{-1/2} \ll \ell ^{-1/2}. \end{aligned}$$
(3.19)

Thus it remains to estimate \(u_0^I\) in the region

$$\begin{aligned} 1 \le \nu =\frac{|x|^2}{t} < 8\log \ell . \end{aligned}$$

Let \(\zeta :=z-x/\sqrt{t}\). On \(z\in B\), we have \(|\zeta |<\delta \sqrt{\nu }\ll \ell \). We shall show that

$$\begin{aligned}&\varphi _{*}(x-\sqrt{t}z)=\varphi _{*}(\sqrt{t}\zeta ) \le (\ell - 2\log \ell )^{1/2} + C\ell ^{-1/2} \langle \log |\zeta | \rangle \end{aligned}$$
(3.20)

for some constant \(C >0\), which is independent of \(\nu \) and \(\delta \). We first consider the region \(1 < |\zeta |\). Since \(1 < |\zeta | \ll \ell \), we have \(\ell ^{-1} \log |\zeta | < \ell ^{-1} \log \ell \ll 1\). Thus, by (3.1) and the Taylor expansion, we have

$$\begin{aligned} \varphi _{*}(x-\sqrt{t}z)= & {} \varphi _{*}(\sqrt{t}\zeta )\nonumber \\\le & {} (-\log t - 2\log |\log t|-\log |\zeta |^2 -2\log |1-\ell ^{-1}\log |\zeta |^2|+C)^{1/2}\nonumber \\\le & {} (\ell - 2\log \ell + C \langle \log |\zeta | \rangle )^{1/2} \nonumber \\\le & {} (\ell - 2\log \ell )^{1/2} (1 + C \ell ^{-1} \langle \log |\zeta | \rangle )^{1/2}\nonumber \\\le & {} (\ell - 2\log \ell )^{1/2} (1+C\ell ^{-1}\langle \log |\zeta | \rangle ) \nonumber \\\le & {} (\ell - 2\log \ell )^{1/2} + C\ell ^{-1/2} \langle \log |\zeta | \rangle . \end{aligned}$$
(3.21)

Next, we consider the region \(|\zeta | \le 1\). Since \(- \log |\zeta |^{2} \ge 0\), we have

$$\begin{aligned} - \log |\zeta |^{2} - 2 \log |1-\ell ^{-1}\log |\zeta |^2| \le -2 \log |\zeta |^{2}. \end{aligned}$$

Using this together with (3.1), we obtain

$$\begin{aligned} \varphi _{*}(x-\sqrt{t}z)\le & {} (-\log t - 2\log |\log t|-\log |\zeta |^2-2\log |1-\ell ^{-1}\log |\zeta |^2|+C)^{1/2} \nonumber \\\le & {} (\ell - 2\log \ell + C(- \log |\zeta | + 1))^{1/2} \nonumber \\\le & {} (\ell - 2\log \ell )^{1/2} + C (\ell - 2\log \ell )^{-1/2} (- \log |\zeta | + 1)\nonumber \\\le & {} (\ell - 2\log \ell )^{1/2} + C\ell ^{-1/2} \langle \log |\zeta | \rangle . \end{aligned}$$
(3.22)

From (3.21) and (3.22), we see that (3.20) holds.

Moreover, it follows from (3.13) that \(|z|^{2} \ge \nu (1 - \delta )^{2}\). This yields that

$$\begin{aligned} \int _{B} e^{- \frac{|z|^{2}}{4}} \langle \log |\zeta | \rangle dz\le & {} \int _{|\zeta | \le \delta \sqrt{\nu }} e^{- \nu (1-\delta )^{2}/4} \langle \log |\zeta | \rangle d \zeta \le e^{- \nu (1-\delta )^{2}/4} \delta ^{2} \nu \langle \log \delta ^{2} \nu \rangle \nonumber \\\le & {} C \end{aligned}$$
(3.23)

for some constant \(C > 0\), which is independent of \(\nu \) and \(\delta \). It follows from (3.23) and (3.20) that

$$\begin{aligned} u_0^I= & {} \int _{B} \frac{e^{-|z|^{2}}/4}{4\pi } \varphi _{*}(x - \sqrt{t}z) dz \nonumber \\\le & {} (\ell -2\log \ell )^{1/2} \int _B\frac{e^{-|z|^2/4}}{4\pi } dz +O(\ell ^{-1/2})\nonumber \\\le & {} \varphi _{*}(\sqrt{t})\int _B\frac{e^{-|z|^2/4}}{4\pi } dz +O(\ell ^{-1/2}). \end{aligned}$$
(3.24)

Hence, by (3.8), (3.12) and (3.24), we have

$$\begin{aligned} u_0 \le \varphi _{*}(x) + (\varphi _{*}(\sqrt{t})-\varphi _{*}(x))\int _B \frac{e^{-|z|^2/4}}{4\pi } dz + O(\ell ^{-1/2}), \end{aligned}$$

where the remainder on B is estimated by

$$\begin{aligned} \begin{aligned} \varphi _{*}(\sqrt{t})-\varphi _{*}(x) \le&\, C \frac{-\log \sqrt{t}+\log r}{\ell ^{1/2}} \le C \frac{\log \nu }{\ell ^{1/2}}, \\ \int _B e^{-|z|^2/4}dz \le&\, e^{-\nu (1-\delta )^{2}/4}\nu \le C e^{-\nu /8}. \end{aligned} \end{aligned}$$

Here, we have used (3.13) in the second inequality. This yields that

$$\begin{aligned} u_{0} \le \varphi _{*}(x) + O(\ell ^{-\frac{1}{2}}). \end{aligned}$$
(3.25)

From (3.7), (3.19) and (3.25), we see that (3.2) holds. \(\square \)

From Lemma 3.2, we have obtained, denoting \(\rho :=|\log r^2|\),

(3.26)

One can use the radial monotonicity of \(u_0\) (cf. Lemma A.1) to extend, to all \(x\in {\mathbb {R}}^2\), the bounds of the functions \(u_0e^{u_0^2}\) and \(u_0^2e^{u_0^2}\) as follows:

$$\begin{aligned} t<\varepsilon ^{2} \implies u_0 \le \sqrt{\ell }, \quad&u_0e^{u_0^2} \le C [(t+r^2)^{-1}+\varepsilon ^{-2}]|\log \min \{t + r^{2}, \varepsilon ^{2}\}|^{-3/2}=:F_0,\nonumber \\&u_0^2e^{u_0^2} \le C [(t+r^2)^{-1}+\varepsilon ^{-2}]| \log \min \{t + r^{2}, \varepsilon ^{2}\}|^{-1}=:F_0'. \end{aligned}$$
(3.27)

We shall show the bound of the function \(u_{0} e^{u_{0}^{2}}\). From the monotonicity of \(u_{0}\), it is enough to consider the region \(0< r < \varepsilon ^{2}\) only.

Note that we have \(\ell ^{-3/2}, \rho ^{-3/2} < |\log (t + r^{2})|^{-3/2}\) for \(\max (t, r^{2}) < \varepsilon ^{2}\). Moreover, it follows that \(\min (\frac{1}{t}, \frac{1}{r^{2}}) < 2(t + r^{2})^{-1}\). These together with (3.26) yield that

$$\begin{aligned} u_0 e^{u_0^2} \le C \min \left( \frac{1}{t}, \frac{1}{r^{2}}\right) |\log (t + r^{2})|^{-3/2} \le C (t + r^{2})^{-1} |\log (t + r^{2})|^{-3/2}.\nonumber \\ \end{aligned}$$
(3.28)

We first consider the region \(t + r^{2} < \varepsilon ^{2}\). It follows from (3.28) that

$$\begin{aligned} u_0 e^{u_0^2}&\le C (t + r^{2})^{-1} |\log \min \{t + r^{2}, \varepsilon ^{2}\}|^{-3/2} \nonumber \\&\le C [(t + r^{2})^{-1} + \varepsilon ^{-2}] |\log \min \{t + r^{2}, \varepsilon ^{2}\}|^{-3/2}. \end{aligned}$$
(3.29)

Next, we consider the region \(\varepsilon ^{2} \le t + r^{2} < 2\varepsilon ^{2}\). Since the function \(s^{-1} |\log s|^{-3/2}\) is decreasing for sufficiently small \(s>0\), we have, by (3.28), that

$$\begin{aligned} u_{0} e^{u_{0}^{2}}&\le C (t + r^{2})^{-1}|\log (t + r^{2})|^{-3/2} \le C \varepsilon ^{-2}|\log \varepsilon ^{2}|^{-3/2}\nonumber \\&\le C [(t + r^{2})^{-1} + \varepsilon ^{-2}] |\log \min \{t + r^{2}, \varepsilon ^{2}\}|^{-3/2}. \end{aligned}$$
(3.30)

From (3.29) and (3.30), we see that \(u_{0}e^{u_{0}^{2}} \le C F_{0}\) for \(t < \varepsilon ^{2}\). We can obtain the bound of the function \(u_{0}^{2} e^{u_{0}^{2}}\) similarly. Thus, (3.27) holds.

Hence by the mean value theorem, for any functions \(v_0\) and \(v_1\), we have, for \(t<{ \varepsilon ^{2}}\), that

$$\begin{aligned} \sqrt{\ell }(|v_0|+|v_1|)\le & {} C \implies |f_{0}(u_0+v_0)-f_{0}(u_0+v_1)|\nonumber \\\le & {} C u_0^2e^{u_0^2}|v_0-v_1| \nonumber \\\le & {} C F_0'|v_0-v_1|, \end{aligned}$$
(3.31)

where \(f_{0}(u) = u (e^{u^{2}} -1)\).

To estimate the second iteration, we prepare the following:

Lemma 3.3

Let \(\alpha > 0\) and \(0< \varepsilon < 1\). For any \((t, r) \in (0, \varepsilon ^{2}) \times (0, \infty )\), there exists a positive constant \(C_{*}\) such that

$$\begin{aligned} \int _0^t e^{(t-s)\Delta }\frac{1}{s+r^2}|\log \min \{s + r^{2}, \varepsilon ^{2}\}|^{- \alpha }ds\le & {} C_{*} \ell ^{-\alpha }, \end{aligned}$$
(3.32)
$$\begin{aligned} \int _0^t e^{(t-s)\Delta }|\log \min \{s + r^{2}, \varepsilon ^{2}\}|^{- \alpha }ds\le & {} C_{*} t \ell ^{-\alpha } \end{aligned}$$
(3.33)

Proof

From Lemma A.1, we have

$$\begin{aligned} \begin{aligned}&\int _0^t e^{(t-s)\Delta }\frac{1}{s+r^2}|\log \min \{s + r^{2}, \varepsilon ^{2}\}|^{- \alpha } ds \\&\quad \le \int _{0}^{t} \int _{0}^{\infty } \frac{re^{- \frac{r^{2}}{4(t-s)}}}{2(t-s)(s + r^{2})} |\log \min \{s + r^{2}, \varepsilon ^{2}\}|^{- \alpha } dr ds. \end{aligned} \end{aligned}$$

Then, the integral is estimated using the following formula

$$\begin{aligned} \begin{aligned}&\int _0^t e^{(t-s)\Delta }\frac{1}{s+r^2} |\log \min \{s + r^{2}, \varepsilon ^{2}\}|^{- \alpha } ds \\&\quad \le \int _0^t \int _0^\infty \frac{e^{-\frac{\sigma }{4s}}}{4s(t-s+\sigma )} |\log \min \{t - s + \sigma , \varepsilon ^{2}\}|^{- \alpha } d\sigma ds \\&\quad =\int _0^\infty \int _0^t \frac{e^{-\eta /4}}{t-s+s\eta } |\log \min \{t -s+{ s\eta }, \varepsilon ^{2}\}|^{- \alpha }ds \frac{d\eta }{4} \end{aligned} \end{aligned}$$

where the variables are changed by \(s\rightarrow t-s\), \(r^2\rightarrow \sigma \rightarrow s\eta \). Then, we split the integral as follows.

$$\begin{aligned} \begin{aligned}&\int _0^\infty \int _0^t \frac{e^{-\eta /4}}{t-s+s\eta } |\log \min \{t-s+s\eta , \varepsilon ^{2}\}|^{- \alpha }ds \frac{d\eta }{4} \\&\quad = \int _0^{1/\sqrt{t}}\int _0^{t} \frac{e^{-\eta /4}}{t-s+s\eta } |\log \min \{t-s+s\eta , \varepsilon ^{2}\}|^{- \alpha }ds \frac{d\eta }{4} \\&\qquad + \int _{1/\sqrt{t}}^{\infty } \int _0^t \frac{e^{-\eta /4}}{t-s+s\eta } |\log \min \{t-s+s\eta , \varepsilon ^{2}\}|^{- \alpha }ds \frac{d\eta }{4} =: I + II. \end{aligned} \end{aligned}$$

We first estimate I. Since \(t -s + s \eta \le \sqrt{t}\) for \(0< s < t\) and \(0< \eta < 1/\sqrt{t}\), we have

$$\begin{aligned} I&\le \int _0^{1/\sqrt{t}}\int _0^{t} \frac{e^{-\eta /4}}{t-s+s\eta } |\log \min \{\sqrt{t}, \varepsilon ^{2}\}|^{- \alpha }ds \frac{d\eta }{4} \nonumber \\&\le C \ell ^{-\alpha } \int _{0}^{1/\sqrt{t}} \int _{0}^{t} \frac{e^{-\eta /4}}{t-s+s\eta } ds d\eta = C \ell ^{-\alpha } \int _{0}^{1/\sqrt{t}} e^{-\eta /4}\frac{\log \eta }{\eta -1} d\eta \le C \ell ^{-\alpha }. \end{aligned}$$
(3.34)

Next, we estimate II. We note that \(|\log \min \{t - s + s\eta , \varepsilon ^{2}\}|^{- \alpha } \le (- \log \varepsilon ^{2})^{-\alpha }\) and \(t -s + s\eta \ge t\) for \(0< s < t\) and \(\eta \ge 1/\sqrt{t}\). Therefore, we have

$$\begin{aligned} II \le C (- \log \varepsilon ^{2})^{-\alpha } \int _{1/\sqrt{t}}^{\infty } e^{- \eta /4}\int _0^t \frac{1}{t} ds d\eta \le C e^{- \frac{1}{4\sqrt{t}}}. \end{aligned}$$
(3.35)

From (3.34) and (3.35), we obtain (3.32). We can obtain (3.33) by the similar argument above. This completes the proof. \(\square \)

Using Lemma 3.3, we shall show the following:

Lemma 3.4

Let \(\varepsilon >0\) be given by Lemma 3.3. For any space-time function v on \((0,\varepsilon ^{2})\times {\mathbb {R}}^2\), let

$$\begin{aligned} D[v]:=\int _0^t e^{(t-s)\Delta }f_0((u_0+v)(s))ds. \end{aligned}$$

Then, there exists a positive constant \(C_{0}\) such that

$$\begin{aligned}&|D[0]| \le C_{0} \ell ^{-3/2}. \end{aligned}$$
(3.36)

Moreover, for any \(C_1\in (0,\infty )\), there exists \(C_2\in (0,\infty )\) such that for any \(v_0\) and \(v_1\) satisfying

$$\begin{aligned} \sup _{0<t<\varepsilon }|\log t|^{1/2}\Vert v_j(t)\Vert _{L^{\infty }} \le C_1 \quad (j=0,1), \end{aligned}$$

and for any \(\alpha { >} -1\), we have

$$\begin{aligned} |D[v_0]-D[v_1]| \le C_2 \ell ^{-1-\alpha }\sup _{0<s<t}|\log s|^\alpha \Vert v_0(s)-v_1(s) \Vert _{L^\infty }. \end{aligned}$$
(3.37)

Proof

(3.27) and Lemma 3.3 yield

$$\begin{aligned} \begin{aligned} |D[0]|&\le \int _0^t e^{(t-s)\Delta }|f_0(u_0(s))|ds \le C \int _0^t e^{(t-s)\Delta }F_0(s)ds \\&\le C \ell ^{-3/2} + \varepsilon ^{-2} t \ell ^{-3/2} \le C \ell ^{-3/2}. \end{aligned} \end{aligned}$$

Similarly, (3.31) and Lemma 3.3 yield

$$\begin{aligned} \begin{aligned} |D[v_0]-D[v_1]|&\le \, \int _0^t e^{(t-s)\Delta }|f_0(u_0+v_0))-f_0(u_0+v_1)|ds \\&\le \, C \int _0^t e^{(t-s)\Delta }F_0'(s)\Vert v_1(s)-v_0(s)\Vert _\infty ds \\&\le \, C \int _0^t e^{(t-s)\Delta }[(s+r^2)^{-1}+\varepsilon ^{-2}] |\log \min \{s + r^{2}, \varepsilon ^{2}\}|^{-1-\alpha } { ds} \\&\quad \times \sup _{0<s<t}|\log s|^\alpha \Vert v_0(s)-v_1(s)\Vert _{L^{\infty }} \\&\le \, C \ell ^{-1-\alpha }\sup _{0<s<t}|\log s|^\alpha \Vert v_0(s)-v_1(s)\Vert _{L^{\infty }} \end{aligned} \end{aligned}$$

for any \(\alpha { >} -1\). Noting that the constants are independent of \(v_0,v_1,\alpha ,t\), we arrive at the desired conclusion. \(\square \)

We are now in a position to prove Theorem 3.1

Proof of Theorem 3.1

We put

$$\begin{aligned} E[v] := \int _{0}^{t} e^{(t-s)\Delta } L(u_{0} + v)(s)ds, \qquad L(u) := m_{*} \chi (u)u \end{aligned}$$

and \(I[v] := D[v] - E[v]\). Then, we are naturally led to consider the mapping \(v\mapsto I[v]\) for v in the following set

$$\begin{aligned} B_T^{1/2}:=\{v \in C([0,T]\times {\mathbb {R}}^2) \mid \Vert v\Vert _{X_T^{1/2}}:=\sup _{0<t<T}\ell ^{1/2}\Vert v(t)\Vert _{L^{\infty }} \le 1\} \end{aligned}$$

for some constant \(T\in (0, { \varepsilon ^{2}})\) to be determined, which is a closed ball of a Banach space with the norm \(X_T^{1/2}\). The estimates on \(D[\cdot ]\) in (3.36) and (3.37) with \(C_1=1\) imply

$$\begin{aligned}&\Vert D[v_0]-D[v_1]\Vert _{X_T^{1/2}} \le C_2|\log T|^{-1}\Vert v_0-v_1\Vert _{X_T^{1/2}}, \\&\Vert D[v_0]\Vert _{X_T^{1/2}}\le \Vert D[0]\Vert _{X_T^{1/2}}+\Vert D[0]-D[v_0]\Vert _{X_T^{1/2}} \le (C_0+C_2)|\log T|^{-1} \end{aligned}$$

for any \(v_0,v_1\in B_T^{1/2}\). Moreover, we can easily obtain

$$\begin{aligned} \begin{aligned}&\Vert E[v]\Vert _{L^{\infty }} \le t\Vert L\Vert _{L^{\infty }}, \\&\left\| \int _0^t e^{(t-s)\Delta }(L(v_0)-L(v_1))(s)ds\right\| _{L^{\infty }} \le t\Vert L\Vert _{Lip}\Vert v_0-v_1\Vert _{L^{\infty }}. \end{aligned} \end{aligned}$$

Hence if \(T>0\) is small enough then \(v\mapsto I[v]\) is a contraction mapping on \(B_T^{1/2}\), so there is a unique fixed point \(v\in B_T^{1/2}\). Then \(u=u_0+v\) is a local (mild) solution on \(0<t<T\) of

$$\begin{aligned} {\dot{u}} - \Delta u + m_{*} \chi (u) u = f_0(u), \quad u(0)=\varphi _{*}. \end{aligned}$$

This completes the proof. \(\square \)

We can prove Theorem 1.1 immediately from Theorems 2.8 and 3.1 .

Proof of Theorem 1.1

We also denote \(\phi _{m_{*}}\), which is the stationary singular soliton to (1.5) obtained in Theorem 2.8, by \(\varphi _{*}\). Let \(u_{S}(t) = \varphi _{*}\) and \(u_{R}(t)\) be the regular solution to (1.5), obtained in Theorem 3.1. We see that \(u_{S}(0) = u_{R}(0) = \varphi _{*}\).

We shall show that \(u_{S}\) belongs to C([0, T), X) and becomes a mild solution for each \(T > 0\). As we mentioned in the proof of Theorem 2.8, \(\varphi _{*}\) satisfies (2.23) in a distributional sense and \(f_{m_{*}}(\varphi _{*}) \in L^{1}({\mathbb {R}}^{2})\). It follows from the \(L^{p}-L^{q}\) estimate of the heat kernel (see e.g. [19, Proposition 1]) that for \(1 \le p < \infty \), we obtain

$$\begin{aligned} \begin{aligned}&\left\| \int _{0}^{t} e^{\Delta (t-s)}f_{m_{*}}( \varphi _{*})ds \right\| _{L^{p}} \le \int _{0}^{t}\Vert e^{\Delta (t-s)}f_{m_{*}}( \varphi _{*})\Vert _{L^{p}} ds \\&\quad \le C \int _{0}^{t} (t-s)^{- (1 - \frac{1}{p})} \Vert f_{m_{*}}( \varphi _{*})\Vert _{L^{1}} ds \le C \Vert f_{m_{*}}( \varphi _{*})\Vert _{L^{1}} < \infty . \end{aligned} \end{aligned}$$

Thus \(u_{S}(t) = \varphi _{*} \in C([0, T), X)\) satisfies the Duhamel formula:

$$\begin{aligned} u(t) = e^{t \Delta } \varphi _{*} + \int _{0}^{t} e^{(t-s)\Delta }f_{m_{*}}(u(s))ds \qquad \text{ in } C([0, T), X). \end{aligned}$$
(3.38)

Next, we shall show that \(u_{R}\) belongs to C([0, T), X) for sufficiently small \(T > 0\) and becomes a mild solution. The fixed point \(v \in B_{T}^{1/2}\) of I can be obtained as the limit of \(v_{n} = I[v_{n-1}]\) starting from \(v_{0} = 0\). Then for \(T > 0\) small enough, there exists \(M >0\) such that \(\sup _{n \in {\mathbb {N}}} |\log t|^{1/2}\Vert v_{n}(t)\Vert _{L^{\infty }} \le M\) and \(\lim _{n \rightarrow \infty }v_{n}(t) = v(t)\) in \(L^{\infty }({\mathbb {R}}^{2})\) for \(0< t < T\). Let \(u_{n} = u_{0} + v_{n}\). From the estimate (3.26) on \(u_{0}\) together with Lemma A.1 and that \(\Vert v_{n}(t)\Vert _{L^{\infty }} \le M \ell ^{-1/2}\), we obtain

$$\begin{aligned} |u_{n}|^{2} \le (|u_{0}| + |v_{n}|)^{2} \le \ell - \frac{3}{2} \log \ell \end{aligned}$$

uniformly for \(0< t < T \ll 1\), \(x \in {\mathbb {R}}^{2}\) and \(n \in {\mathbb {N}}\). Hence, there exists a constant \(C_{m_{*}} > 0\)

$$\begin{aligned} |f_{m_{*}}(u_{n})| \le C_{m_{*}} |u_{n}| e^{u_{n}^{2}} \le C_{m_{*}} t^{-1} \ell ^{-3/2} |u_{n}|, \end{aligned}$$
(3.39)

where \(f_{m}\) is defined by (2.17). Suppose that \(\sup _{0< s< T} \Vert u_{n}(s)\Vert _{L^{p}} < \infty \) for any \(p \in [1, \infty )\). Using (3.39) in the Duhamel formula

$$\begin{aligned} u_{n+1}(t) = u_{0}(t) + v_{n+1}(t) = u_{0}(t) + \int _{0}^{t} e^{(t-s)\Delta } f_{m_{*}}(u_{n}(s)) ds, \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned} \Vert u_{n+1}(t)\Vert _{L^{p}}&\le \Vert \varphi _{*}\Vert _{L^{p}} + \int _{0}^{t} \Vert f_{m_{*}}(u_{n}(s))\Vert _{L^{p}} ds \\&\le \Vert \varphi _{*}\Vert _{L^{p}} + C_{m_{*}} \int _{0}^{t} s^{-1} |\log s|^{-3/2} \Vert u_{n}(s)\Vert _{L^{p}} ds \\&\le \Vert \varphi _{*}\Vert _{L^{p}} + 2 C_{m_{*}} |\log T|^{-1/2} \sup _{0< s < T}\Vert u_{n}(s)\Vert _{L^{p}} \end{aligned} \end{aligned}$$
(3.40)

for \(0< t < T \ll 1\). Here, we have used the \(L^{p}\) contraction estimate of the heat kernel (see e.g. [19, Proposition 1]) in the first inequality. Thus, we obtain by the induction argument

$$\begin{aligned} \sup _{0< s < T}\Vert u_{n}(s)\Vert _{L^{p}} \le \Vert \varphi _{*}\Vert _{L^{p}} (1 + 3 C_{m_{*}} |\log T|^{-1/2}) \end{aligned}$$
(3.41)

for \(n \in {\mathbb {N}}\). Since, for \(0< t < T \ll 1\),

$$\begin{aligned} \lim _{n \rightarrow \infty } u_{n}(t) = \lim _{n \rightarrow \infty } (u_{0} + v_{n}(t)) = u_{0} + v = u_{R}(t) \qquad \text{ in } L^{\infty } ({\mathbb {R}}^{2}) , \end{aligned}$$

the above uniform bound (3.41) together with Fatou’s lemma implies that,

$$\begin{aligned} \Vert u_{R}(t)\Vert _{L^{p}} \le \liminf _{n \rightarrow \infty } \Vert u_{n}(t)\Vert _{L^{p}} \le \Vert \varphi _{*}\Vert _{L^{p}} (1 + 3 C_{m_{*}} |\log T|^{-1/2}) \end{aligned}$$

for \(0< t < T \ll 1\). In addition, the same estimate as in (3.40) implies that \(f_{m_{*}}(u_{R}) \in L^{1} ([0, T), L^{p}({\mathbb {R}}^{2}))\) and by the Duhamel formula, \(u_{R} \in C([0, T), L^{p}({\mathbb {R}}^{2}))\). Thus, we conclude that \(u_{R} \in C([0, T), L^{p}({\mathbb {R}}^{2}))\) for all \(p \in [1, \infty )\). That is \(u_R\in C([0,T), X)\). Then, we see that \(u_{R}\) also satisfies (3.38).

For all \(t \in (0, T)\), we have \(u_{S}(t) = \varphi _{*} \notin L^{\infty } ({\mathbb {R}}^{2})\) while \(u_{R}(t) \in L^{\infty }({\mathbb {R}}^{2})\). This implies that \(u_{S}(t) \ne u_{R}(t)\) for all \(t \in (0, T)\). This completes the proof. \(\square \)