Existence of solutions of a system of 3D axisymmetric inviscid stagnation flows

Abstract

A system of two integral equations is presented to describe the system of 3D axisymmetric inviscid stagnation flows related to Navier-Stokes equations and existence of its solutions is studied. Utilizing it, we construct analytically the similarity solutions of the 3D system. A nonexistence result is obtained. Previous study was only supported by numerical results.

MSC:34B18.

1 Introduction

The following system of two differential equations arising in the boundary layer problems in fluid mechanics

(1.1)

(1.2)

with boundary conditions

$$\begin{array}{r}f(0)=0,f^{\prime }(0)=0,f^{\prime }(\mathrm{\infty })=1,\\ g(0)=0,g^{\prime }(0)=0,g^{\prime }(\mathrm{\infty })=1\end{array}$$

(1.3)

has been used to describe the system of 3D axisymmetric inviscid stagnation flow [1, 2], which consists of three partial differential equations [2, 3], where λ is a parameter related to the external flow components.

A solution of (1.1)-(1.3) is called a similarity solution and can be used to express the solutions of the 3D system. Regarding the study of (1.1)-(1.3), Howarth [3] presented a numerical study for the case $0<\lambda <1$ which can be applied to the stagnation region of an ellipsoid. Davey [2] investigated numerically the stagnation region near a saddle point ($-1<\lambda <0$). The two-dimensional cases, $\lambda =g=0$ or $\lambda =1$ and $g=f$, and the special cases of the Falkner-Skan equation were solved by Hiemenz [4] and by Homann [5], respectively. Regarding the Falkner-Skan problems, further analytical study can be found in [6–10]. Also, one may refer to recent review of similarity solutions of the Navier-Stokes equations [11].

However, up to now, there has been very little analytical study on the existence of solutions of (1.1)-(1.3).

The main aim of this paper is to study the existence of solutions of (1.1)-(1.3) analytically for the case of $|\lambda |<1$. The method is to present a system of two integral equations and study the existence of its solutions and then use it to construct the solutions of (1.1)-(1.3). Also, a nonexistence result is obtained.

2 A system of two integral equations related to (1.1)-(1.3)

In this section, we present a system of two integral equations to describe a system of (1.1)-(1.3) under suitable conditions, which will be utilized in Section 4.

Let

and

$$\mathrm{\Gamma }=\{(f,g)\in C^3[0,\mathrm{\infty })\times C^3[0,\mathrm{\infty }):f^{\prime }(\eta )\ge 0,g^{″}(\eta )>0,\eta \in [0,\mathrm{\infty })\}.$$

Lemma 2.1 If$(f,g)\in \mathrm{\Gamma }$is a solution of (1.1)-(1.3), then$g^{″}(\mathrm{\infty })=0$.

Proof Since $g^{\prime }(+\mathrm{\infty })=1$, we have

$$\underset{\eta \to \mathrm{\infty }}{lim\hspace{0.17em}inf}{g}^{″}(\eta )=0.$$

(2.1)

Notice that $(f,g)\in \mathrm{\Gamma }$, $f(\eta )={\int }_0^{\eta }{f}^{\prime }(s)ds\ge 0$, $g^{\prime }(\eta )={\int }_0^{\eta }{g}^{″}(s)ds\ge 0$, $g(\eta )={\int }_0^{\eta }{g}^{\prime }(s)ds>0$ and $1>g^{\prime }(\eta )>0$ for $\eta \in (0,+\mathrm{\infty })$.

If $\lambda \ge 0$, we know $g^{‴}(\eta )=-(f(\eta )+\lambda g(\eta ))g^{″}(\eta )-\lambda (1-g^{\mathrm{\prime }2}(\eta ))\le 0$ and then $g^{″}$ is decreasing on $[0,+\mathrm{\infty })$, which implies that ${lim}_{\eta \to \mathrm{\infty }}{g}^{″}(\eta )$ exists. Hence, $g^{″}(\mathrm{\infty })=0$ by (2.1).

If $\lambda <0$, we have $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(0)=-\lambda >0$ by (1.2). By (2.1), there exists ${\eta }_0>0$ such that $g^{″}({\eta }_0)<g^{″}(0)$ and then there exists ${\eta }^{\ast }$ such that $g^{″}({\eta }^{\ast })=max\{g^{″}(\eta ):\eta \in [0,{\eta }_0]\}$. Obviously, ${\eta }^{\ast }\in (0,{\eta }_0]$ by $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(0)>0$. We prove that $g^{″}$ is decreasing on $({\eta }^{\ast },\mathrm{\infty })$.

In fact, if there exist ${\eta }_1,{\eta }_2\in ({\eta }^{\ast },+\mathrm{\infty })$ with ${\eta }_1<{\eta }_2$ such that $g^{″}({\eta }_1)<g^{″}({\eta }_2)$. Let ${\eta }_{\ast }\in [{\eta }^{\ast },{\eta }_2]$ such that $g^{″}({\eta }_{\ast })=min\{g^{″}(\eta ):\eta \in [{\eta }^{\ast },{\eta }_2]\}>0$, then $g^{‴}({\eta }_{\ast })=0$ and $g^{(4)}({\eta }_{\ast })\ge 0$.

Differentiating (1.2) with η, we have

$$g^{(4)}(\eta )=(\lambda g^{\prime }(\eta )-f^{\prime }(\eta ))g^{″}(\eta )-(f(\eta )+\lambda g(\eta ))g^{‴}(\eta ),$$

then

$$g^{(4)}({\eta }_{\ast })=(\lambda g^{\prime }({\eta }_{\ast })-f^{\prime }({\eta }_{\ast }))g^{″}({\eta }_{\ast })<0,$$

a contradiction. Hence, $g^{″}(\eta )$ is decreasing on $({\eta }^{\ast },+\mathrm{\infty })$ and then $g^{″}(\mathrm{\infty })=0$.

This completes the proof. □

Theorem 2.1 If$(f,g)\in \mathrm{\Gamma }$is a solution of (1.1)-(1.2), then

(2.2)

(2.3)

has a solution$(x,y)\in Q$, where$G_{0,1}(t,s)$denotes the Green function for$u^{″}(t)=0$with$u(0)=0$and$u(b)=0$defined by

$$G_{0,b}(t,s)=\{\begin{array}{cc}t(b-s)/b,\hfill & 0\le t\le s\le b,\hfill \\ s(b-t)/b,\hfill & 0\le s\le t\le b.\hfill \end{array}$$

(2.4)

Proof Assume that $(f,g)\in \mathrm{\Gamma }$. Let $\eta :=\eta (t)={(g^{\prime })}^{-1}(t)$ for $t\in [0,1)$ be the inverse function to $t=g^{\prime }(\eta ):[0,\mathrm{\infty })\to [0,1)$. It follows that $g^{\prime }$ is strictly increasing on $[0,+\mathrm{\infty })$ and $\eta (t)={(g^{\prime })}^{-1}(t):[0,1)\to [0,\mathrm{\infty })$ with ${(g^{\prime })}^{-1}(0)=0$, ${lim}_{t\to 1^{-}}{(g^{\prime })}^{-1}(t)=\mathrm{\infty }$. Let $x(t)=g^{″}(\eta )>0$ for $t\in [0,1)$, by Lemma 2.1, $x(1)={lim}_{\eta \to \mathrm{\infty }}{g}^{″}(\eta )=0$. This implies that $x(t)>0$ for $t\in [0,1)$ and x is continuous on $[0,1)$. By Lemma 2.1, we see that x is continuous from the left at 1. Hence, we have $x(t)\in C[0,1]$ and $x(1)=0$, i.e., $x(t)\in Q_1$.

Using the chain rule to $x(t)=g^{″}(\eta )$, we obtain $g^{‴}(\eta )\frac{d\eta }{dt}=x^{\prime }(t)$ and by the inverse function theorem, we have

$$\frac{d\eta }{dt}=\frac{1}{g^{″}(\eta )}=\frac{1}{x(t)}\text{for }t\in [0,1).$$

This, together with $g^{\prime }(\eta )=t$, implies

$$g^{‴}(\eta )=x^{\prime }(t)x(t),\eta ={\int }_0^t\frac{1}{x(s)}ds\text{and}{g}^{\prime }(\eta )\frac{d\eta }{dt}=\frac{t}{x(t)}\text{for }t\in [0,1).$$

Integrating the last equality from 0 to t implies

$$g(\eta (t))={\int }_0^t\frac{s}{x(s)}ds\text{for }t\in [0,1).$$

Let

$$y(t)=f^{\prime }(\eta )=f^{\prime }\left({\int }_0^t\frac{1}{x(s)}ds\right)\text{for }t\in [0,1).$$

Then $y(0)=0$. By $f^{\prime }(\mathrm{\infty })=1$, we know that y is continuous from the left at 1 and then $y(1)=1$.

Notice that $f^{\prime }(\eta )\frac{d\eta }{dt}=\frac{y(t)}{x(t)}$, $t\in [0,1)$, we have $f(\eta )={\int }_0^t\frac{y(s)}{x(s)}ds$.

Differentiating $y(t)$ with t, we have

$$y^{\prime }(t)=f^{″}(\eta )\frac{d\eta }{dt}=\frac{f^{″}(\eta )}{x(t)}\text{for }t\in [0,1).$$

From this, we have $f^{″}(\eta )=y^{\prime }(t)x(t)$ for $\eta \in [0,\mathrm{\infty })$ and $y\in Q_2$.

Differentiating $f^{″}(\eta )$ with t and utilizing $\frac{d\eta }{dt}=\frac{1}{x(t)}$, we have

$$\frac{f^{‴}(\eta )}{x(t)}=y^{″}(t)x(t)+y^{\prime }(t)x^{\prime }(t).$$

Hence,

$$f^{‴}(\eta )=y^{″}(t)x^2(t)+y^{\prime }(t)x(t)x^{\prime }(t).$$

Substituting g, $g^{\prime }$, $g^{″}$, $g^{‴}$ and f into (1.2) implies

$$x^{\prime }(t)=-{\int }_0^t\frac{y(s)+\lambda s}{x(s)}ds+\frac{\lambda (t^2-1)}{x(t)},t\in [0,1).$$

(2.5)

Integrating (2.5) from t to 1, we have

$$\begin{array}{rcl}x(1)-x(t)& =& -{\int }_t^1{\int }_0^{\sigma }\frac{y(s)+\lambda s}{x(s)}dsd\sigma +{\int }_t^1\frac{\lambda (s^2-1)}{x(s)}ds\\ =& {\int }_t^1\frac{\lambda (s^2-1)}{x(s)}ds-{\int }_0^t\left({\int }_t^1\frac{y(s)+\lambda s}{x(s)}d\sigma \right)ds-{\int }_t^1\left({\int }_s^1\frac{y(s)+\lambda s}{x(s)}d\sigma \right)ds\\ =& {\int }_t^1\frac{\lambda (s^2-1)}{x(s)}ds-{\int }_0^t\frac{y(s)+\lambda s}{x(s)}(1-t)ds-{\int }_t^1\frac{(y(s)+\lambda s)(1-s)}{x(s)}ds\\ =& {\int }_t^1\frac{\lambda (s^2-1)-(\lambda s+y(s))(1-s)}{x(s)}ds-(1-t){\int }_0^t\frac{\lambda s+y(s)}{x(s)}ds\\ =& {\int }_t^1\frac{(2\lambda s+\lambda +y(s))(s-1)}{x(s)}ds-(1-t){\int }_0^t\frac{\lambda s+y(s)}{x(s)}ds.\end{array}$$

By $x(1)=0$, then

$$x(t)={\int }_t^1\frac{(2\lambda s+\lambda +y(s))(1-s)}{x(s)}ds+(1-t){\int }_0^t\frac{\lambda s+y(s)}{x(s)}ds.$$

Substituting f, $f^{\prime }$, $f^{″}$, $f^{‴}$ and g into (1.1) implies

$$y^{″}(t)x^2(t)+y^{\prime }(t)x(t)x^{\prime }(t)+y^{\prime }(t)x(t){\int }_0^t\frac{\lambda s+y(s)}{x(s)}ds+(1-y^2(t))=0.$$

By ${\int }_0^t\frac{\lambda s+y(s)}{x(s)}ds=\frac{\lambda (t^2-1)}{x(t)}-x^{\prime }(t)$, we have

$$y^{″}(t)+\frac{\lambda (t^2-1)y^{\prime }(t)+(1-y^2(t))}{x^2(t)}=0.$$

Therefore,

$$y(t)={\int }_0^1{G}_{0,1}(t,s)\frac{\lambda (s^2-1)y^{\prime }(s)+(1-y^2(s))}{x^2(s)}ds+t,t\in [0,1),$$

where $G_{0,1}(t,s)$ is defined by (2.4). Hence, $(x,y)$ is a solution of (2.2)-(2.3) in Q. □

3 Positive solutions of the system (2.2)-(2.3)

In this section, we will use the fixed point theorem to study the existence of positive solutions of the system (2.2)-(2.3).

Let

$$\delta =\delta (\lambda )=-\frac{\lambda }{2\lambda +1},\lambda \in (-\frac{1}{3},0].$$

It is easy to verify

$$0<\delta <1\text{if and only if}-\frac{1}{3}<\lambda <0.$$

We define some functions

By computation, $\omega (0)=\frac{1}{36}$, $\omega (-\frac{1}{3})=-\frac{4}{9}$, there exists ${\lambda }_0\in (-\frac{1}{3},0)$ such that $\omega (\lambda )>0$ for $\lambda \in ({\lambda }_0,0]$ and $\omega ({\lambda }_0)=0$.

In order to study the existence of solutions of (2.2)-(2.3) in Q for $\lambda \in ({\lambda }_0,1)$, we denote the norm of the Banach space $C[0,1]\times C^1[0,1]$ by

$$\parallel (x,y)\parallel =\parallel x\parallel +\parallel y\parallel +\parallel y^{\prime }\parallel ,$$

where $\parallel x\parallel =max\{|x(t)|:t\in [0,1]\}$.

Let $(x,y)\in C[0,1]\times C^1[0,1]$ and $n>0$ be a natural number, we define

$$\phi x(t)=max\{x(t),c(t)\},{\phi }_{n}x(t)=max\{x(t),c(t),\frac{1}{n}\},\theta y(t)=max\{y(t),t\},$$

where $c(t)=c_{\lambda }(1-t)$, $t\in [0,1]$,

$$c_{\lambda }=\{\begin{array}{cc}\frac{1}{n},\hfill & \lambda \ge 0,\hfill \\ min\{\sqrt{h(\lambda )},\sqrt{\omega (\lambda )},\frac{(1+\lambda )(1-\delta ){\delta }^2}{4\sigma (\lambda )}\},\hfill & {\lambda }_0<\lambda <0.\hfill \end{array}$$

(3.1)

Notation

and

where $G_{0,1}(t,s)$ is defined by (2.4).

Let $(x,y)\in C[0,1]\times C^1[0,1]$, we define an operator F as follows:

$$F_n(x,y)(t)=(A_n(x,y)(t),B_n(x,y)(t)),$$

where

$$A_n(x,y)(t)=S_n(x,y)(t)+(1-t)T_n(x,y)+\frac{1}{n}.$$

It is easy to verify that ${\phi }_n$, θ are continuous operators from $C[0,1]$ into $C[0,1]$ and ${\phi }_{n}x(t)\ge \frac{1}{n}$, $t\in [0,1]$, we know the following proposition holds:

Lemma 3.1$F_n$is a continuous and compact operator from$C[0,1]\times C^1[0,1]$to$C[0,1]\times C^1[0,1]$.

Lemma 3.2 Let $(\lambda ,z,w)\in (-1,1)\times C[0,1]\times C^1[0,1]$ and $0<\mu \le 1$ such that

(3.2)

(3.3)

Then the following assertions hold:

• (i) $\mu t\le y(t)\le 1$ for $t\in [0,1]$.

• (ii) ${\int }_0^1|y^{\prime }(s)|ds\le 2$ and $V_0^1(y)\le 2$, where $V_0^1(y)$ is a total variation of y on $[0,1]$.

• (iii) If $\mu =1$, then $y(t)$ is increasing on $(0,1)$ and then $\theta y(t)=y(t)$ for $t\in [0,1]$.

Proof We shall use the basic fact: let $u(t)\in C[a,b]\times C^2(a,b)$ and $u(\xi )$ ($\xi \in (a,b)$) be local minimum (maximum), then $u^{″}(\xi )\ge 0$ (≤0).

• (i) If there exists $t_0\in (0,1)$ such that $y(t_0)>1$, by $y(0)=0<\mu =y(1)$, we know that there exists $t_{\ast }\in (0,1)$ such that $y(t_{\ast })=max\{y(t):t\in [0,1]\}>1$. Differentiating (3.3) with t twice, we have

  $$y^{″}(t)=-\mu \frac{h(y)(t)}{{({\phi }_{n}x(t))}^2}.$$

  (3.4)

By $y^{\prime }(t_{\ast })=0$ and (3.4), we have

$$y^{″}(t_{\ast })=-\frac{\mu (1-y^2(t_{\ast }))}{{({\phi }_{n}x(t_{\ast }))}^2}>0,$$

a contradiction. Hence, $y(t)\le 1$ for $t\in (0,1)$.

If there exists $t_0\in [0,1]$ such that $\mu t_0>y(t_0)$, let $\tau (t)=\mu t-y(t)$, by $\tau (0)=0=\tau (1)$ and $\tau (t_0)>0$, we may assume $t_{\ast }\in (0,1)$ such that $\tau (t_{\ast })=max\{\tau (t):t\in [0,1]\}$. This implies ${\tau }^{\prime }(t_{\ast })=0$, i.e., $y^{\prime }(t_{\ast })=\mu$, and ${\tau }^{″}(t_{\ast })\le 0$. By (3.4) and $\theta y(t_{\ast })=t_{\ast }$, we know

$$h(y)(t_{\ast })=\lambda (t_{\ast }^2-1)\mu +(1-t_{\ast }^2)=(1-\lambda \mu )(1-t_{\ast }^2)>0,$$

then

$${\tau }^{″}(t_{\ast })=-y^{″}(t_{\ast })=\frac{\mu h(y)(t_{\ast })}{{({\phi }_{n}x(t_{\ast }))}^2}>0,$$

a contradiction. Hence, (i) holds.

• (ii) Let $\tilde{t}\in [0,1]$ such that $y(\tilde{t})=max\{y(t):t\in [0,1]\}$ and $\gamma =sup\{\tilde{t}\}$. If $\gamma <1$, we prove that $y(t)$ is increasing on $(0,\gamma )$ and decreasing on $(\gamma ,1)$.

Since $y(0)=0$ and $y(1)=\mu >0$, then $\gamma >0$. Let $\gamma <1$. If there exist $t_1,t_2\in (0,\gamma )$ with $t_1<t_2$ such that $y(t_1)>y(t_2)$, let $t_{\ast }\in (t_1,\gamma )$ such that $y(t_{\ast })=min\{y(t):t\in [t_1,\gamma ]\}$, then $y(t_{\ast })<1$ by (i). From $y^{\prime }(t_{\ast })=0$, $t_{\ast }\le \theta y(t_{\ast })<1$ and (3.4), we know

$$y^{″}(t_{\ast })=-\frac{\mu (1-{(\theta y(t_{\ast }))}^2)}{{({\phi }_{n}x(t_{\ast }))}^2}<0,$$

a contradiction.

If there exist $t_1,t_2\in (\gamma ,1)$ with $t_1<t_2$ such that $y(t_1)<y(t_2)$, let $t_{\ast }\in (\gamma ,t_2)$ such that $y(t_{\ast })=min\{y(t):t\in [\gamma ,t_2]\}$, then $y(t_{\ast })<1$ by (i). Analogously, we know easily

$$y^{″}(t_{\ast })=-\frac{\mu (1-{(\theta y(t_{\ast }))}^2)}{{({\phi }_{n}x(t_{\ast }))}^2}<0,$$

a contradiction. Hence,

$${\int }_0^1|y^{\prime }(s)|ds={\int }_0^{\gamma }{y}^{\prime }(s)ds-{\int }_{\gamma }^1{y}^{\prime }(s)ds=2y(\gamma )-\mu \le 2,$$

and $V_0^1(y)={\int }_0^1|y^{\prime }(s)|ds\le 2$, i.e., (ii) holds.

• (iii) Let $\mu =1$. By (i) and $y(1)=1$, we know $\gamma =1$ and then $y(t)$ is increasing on $(0,1)$ and then $\theta y(t)=y(t)$ for $t\in [0,1]$. Hence, (iii) holds. □

Lemma 3.3[12]

Let E be a Banach space, D be a bounded open set of E and$\theta \in D$, $F:\overline{D}\to E$is compact. If$x\ne \mu Fx$for any$0<\mu <1$and$x\in \partial D$, then F has a fixed point in$\overline{D}$.

Lemma 3.4 Let$\lambda \in (-1,1)$, then F has a fixed point$(x_n,y_n)$in$C[0,1]\times C^1[0,1]$, i.e., there exists$(x_n,y_n)\in C[0,1]\times C^1[0,1]$such that

(3.5)

(3.6)

hold.

Proof Let

$$\mathrm{\Omega }=\{(x,y):(x,y)\in C[0,1]\times C^1[0,1],\parallel (x,y)\parallel <R\},$$

where $R=16n^2$. We prove $(x,y)\ne \mu F(x,y)$ for $0<\mu <1$ and with $\parallel (x,y)\parallel =R$.

In fact, if there exist $(x,y)$ and μ with $\parallel (x,y)\parallel =R$ and $0<\mu <1$ such that $(x,y)\ne \mu F(x,y)$, by Lemma 3.2(i) and (iii), we have $\parallel y\parallel \le 1$.

Since $|\alpha (y)(s)|\le (2|\lambda |+|\lambda |+1)=3|\lambda |+1$ and $|\beta (y)(s)|\le |\lambda |+1$ for $s\in [0,1]$, this, together with $1-t\le 1-s$ for $s\le t$ and ${\phi }_{n}x(t)\ge \frac{1}{n}$, implies

And then $|x(t)|\le |S_n(x,y)(t)|+(1-t)|T_n(x,y)(t)|+1\le 2(2|\lambda |+1)n+1$, i.e., $\parallel x\parallel \le 2(2|\lambda |+1)n+1$.

By (3.3), we have

$$y^{\prime }(t)=-{\int }_0^{t}s\frac{h(y)(s)}{{({\phi }_{n}x(s))}^2}ds+{\int }_t^1(1-s)\frac{h(y)(s)}{{({\phi }_{n}x(s))}^2}ds+1.$$

(3.7)

Noticing that $|h(y)(s)|\le |\lambda ||y^{\prime }(s)|+1$ and ${\phi }_{n}x(s)\ge \frac{1}{n}$ for $s\in [0,1]$, we obtain $|\frac{h(y)(s)}{{({\phi }_{n}x(s))}^2}|\le n^2(|\lambda ||y^{\prime }(s)|+1)$ for $s\in [0,1]$. This, together with (3.7) and Lemma 3.2(ii), implies

$$\begin{array}{rcl}|y^{\prime }(t)|& \le & {\int }_0^1\left|\frac{h(y)(s)}{{({\phi }_{n}x(s))}^2}\right|ds+{\int }_0^1\left|\frac{h(y)(s)}{{({\phi }_{n}x(s))}^2}\right|ds+1\\ \le & 2(2|\lambda |+1)n^2+1,\end{array}$$

i.e., $\parallel y^{\prime }\parallel \le 2(2|\lambda |+1)n^2+1$. Hence,

$$\begin{array}{rcl}\parallel (x,y)\parallel & =& \parallel x\parallel +\parallel y\parallel +\parallel y^{\prime }\parallel \\ \le & 2(2|\lambda |+1)n+1+2(2|\lambda |+1)n^2+1<R,\end{array}$$

a contradiction.

By Lemmas 3.1 and 3.3, F has a fixed point $(x_n,y_n)$ in $C[0,1]\times C^1[0,1]$. □

Lemma 3.5 Let$(x_n,y_n)$be in Lemma  3.4, then

• (i) $\{x_n(t)\}$ is bounded on $[0,1]$.

• (ii) $\{x_n^{\prime }(t)\}$ is bounded on $[0,b]$ for any $b\in (\frac{1}{2},1)$.

Proof By Lemma 3.3(i), we know $0\le y_n(t)\le 1$. By (3.5), we have

$$x_n^{\prime }(t)=\frac{-\lambda (1-t^2)}{\phi x_n(t)}-{\int }_0^t\frac{y_n(s)+\lambda s}{\phi x_n(s)}ds,t\in [0,1).$$

(3.8)

• (i) For $\lambda \ge 0$, we know $x_n^{\prime }(t)<0$ for $t\in [0,1]$, i.e., $x_n(t)$ is decreasing in $[0,1]$, by $x_n(1)=\frac{1}{n}$, $\phi x_n(t)=x_n(t)$ for $t\in [0,1]$. By $\alpha (y_n)(t)\ge t$ for $t\in [0,1]$ and (3.5), we have

  $$x_n(t)\ge S_n(x_n,y_n)(t)\ge {\int }_t^1\frac{s}{{\phi }_{n}x(s)}ds\ge \frac{t}{x_n(t)}{\int }_t^1(1-s)ds.$$

  (3.9)

And then $x_n(t)\ge \frac{(1-t)\sqrt{2t}}{2}$ for $t\in [0,1]$. Obviously, $x_n(t)\ge \frac{1-t}{2}$ for $t\in [\frac{1}{2},1]$. This, together with the decrease in $x_n$, implies

$$x_n(t)\ge \frac{1-t}{4}\text{for }t\in [0,1].$$

(3.10)

Let $c^{\ast }(t)=\mu (1-t)$, μ defined by

$$\mu =\{\begin{array}{cc}\frac{1}{4}\hfill & \text{if }\lambda \ge 0,\hfill \\ c_{\lambda }\hfill & \text{if }\lambda <0,\hfill \end{array}$$

where $c_{\lambda }$ defined in (3.1).

It is easy to verify ${\phi }_n{x}_n(t)\ge c^{\ast }(t)$ for $t\in [0,1]$. And then

The last two inequalities imply that $\{x_n(t)\}$ is bounded on $[0,1]$.

• (ii) By (3.8),

  $$|x_n^{\prime }(t)|\le \frac{|\lambda |(1+t)}{\mu }+{\int }_0^t\frac{2}{c^{\ast }(s)}ds,t\in [0,1),$$

we know that $\{x_n^{\prime }(t)\}$ is bounded on $[0,b]$ for any $b\in (\frac{1}{2},1)$. □

Lemma 3.6 Let$(x_n,y_n)$be in Lemma  3.4, then

• (i) $t\le y_n(t)\le 1$ and $y_n(t)$ is increasing in $[0,1]$.

• (ii) $\{y_n^{\prime }(t)\}$ is bounded and equicontinuous in $[0,b]$ for any $b\in (\frac{1}{2},1)$.

Proof

• (i) Lemma 3.2(i) and (iii) imply the desired results.

• (ii) For $b\in (\frac{1}{2},1)$, let $t_b\in [0,b]$ such that $y_n^{\prime }(t_b)=min\{y_n^{\prime }(t):t\in [0,b]\}$. Since $y_n^{\prime }(t)\ge 0$ on $[0,1]$, by Lemma 3.2(ii), $y_n^{\prime }(t_b)b\le {\int }_0^b{y}_n^{\prime }(s)ds\le {\int }_0^1{y}_n^{\prime }(s)ds\le 2$, we obtain $y_n^{\prime }(t_b)\le \frac{2}{b}$.

Differentiating (3.6) with t twice, we have $y_n^{″}(t)=-\frac{h(y_n)(t)}{{({\phi }_n{x}_n(t))}^2}$. Integrating this equality from 0 to $t\le b$, we have

$$y_n^{\prime }(t)-y_n^{\prime }(t_b)=-{\int }_{t_b}^t\frac{h(y_n)(s)}{{({\phi }_n{x}_n(s))}^2}ds.$$

Noticing that $|h(y_n)(t)|\le |\lambda |y_n^{\prime }(t)+1$ and $c^{\ast }(t)\ge c^{\ast }(b)$ for $t\in [0,b]$ and Lemma 3.2(ii), we know

$$|y_n^{\prime }(t)|\le \frac{2|\lambda |+1}{{(c^{\ast }(b))}^2}+|y_n^{\prime }(t_b)|\le \frac{2|\lambda |+1}{{(c^{\ast }(b))}^2}+\frac{2}{b},$$

i.e., $\{y_n^{\prime }(t)\}$ is bounded on $[0,b]$. Let $M_b=sup\{M_n\}$ (where $M_n=max\{y_n^{\prime }(t):t\in [0,b]\}$), we know

$$|y_n^{″}(t)|=\frac{|h(y_n)(t)|}{{(\phi x_n(t))}^2}\le \frac{|\lambda |M_b+1}{{(c^{\ast }(b))}^2}<+\mathrm{\infty }\text{for }0\le t\le b.$$

This implies that $\{y_n^{\prime }(t)\}$ is equicontinuous on $[0,b]$. □

Theorem 3.1 There exists $(x,y)\in C[0,1]\times (C[0,1]\cap C^1[0,1))$ such that

(3.11)

(3.12)

hold, where

Proof Let $(x_n,y_n)$ be in Lemma 3.4, by Lemma 3.5(ii) and (iii), we know that $\{x_n(t)\}$ is bounded and equicontinuous on $[0,b]$ for any $b\in (\frac{1}{2},1)$. Letting $b=1-\frac{1}{k}$ ($k=3,4,\dots$), utilizing the diagonal principle and the Arzela-Ascoli theorem, we know that there exists a subsequence $\{x_{n_k}(t)\}$ of $\{x_n(t)\}$ and $x(t)\in C[0,1)$ such that $x_{n_k}(t)$ converges to $x(t)$ for $t\in [0,1)$. Without loss of generality, we assume that $\{x_{n_k}(t)\}$ is itself of $\{x_n(t)\}$.

By Lemma 3.6, we know that $\{y_n^{\prime }(t)\}$ is bounded and equicontinuous on $[0,b]$ for any $b\in (\frac{1}{2},1)$ and then $\{y_n(t)\}$ is bounded and equicontinuous on $[0,b]$. Let $b=1-\frac{1}{k}$ ($k=3,4,\dots$), the diagonal principle and the Arzela-Ascoli theorem imply that there exist y and $y_0$ in $C[0,1)$ and two subsequences $\{y_{n_k}(t)\}$ and $\{y_{n_i}^{\prime }(t)\}$ with $\{y_{n_i}(t)\}\subseteq \{y_{n_k}(t)\}\subseteq \{y_n(t)\}$ such that $y_{n_k}(t)$ converges to $y(t)$ for $t\in [0,1)$ with $y(1)=1$ and $y_{n_i}^{\prime }(t)$ converges to $y_0(t)$ for each $t\in [0,1)$. For the sake of convenience, we assume that $\{y_{n_i}(t)\}$ and $\{y_{n_k}(t)\}$ are itself of $\{y_n(t)\}$. By $y_n(t)={\int }_0^t{y}_n^{\prime }(s)ds$, we obtain $y(t)={\int }_0^t{y}_0(s)ds$ and then $y_0(t)=y^{\prime }(t)$ for $t\in [0,1)$.

Since

$\alpha (y_n)(t)$ converges to $\alpha (y)(t)$ and $\beta (y_n)(t)$ converges to $\beta (y)(t)$ for $t\in [0,1)$, by the Lebesgue dominated theorem (the dominated function $F(s)=\frac{3|\lambda |+1}{c^{\ast }}$, $s\in [0,1])$, we have that $(x,y)$ satisfies (3.11) and $x\in Q_1$.

Fix $t\in (0,1)$ and choose $b\in (0,1)$ such that $t\le b$, then

$$y_n(t)={\int }_0^b{G}_{0,b}\frac{h(y_n)(s)}{{(\phi x_n(s))}^2}ds+\frac{t}{b}{y}_n(b)\text{for }t\in [0,b].$$

Noticing that $|h(y_n)(s)|\le |\lambda ||y_n^{\prime }(s)|+1\le |\lambda |M_b+1$ and $h(y_n)(s)$ converges to $h(y)(s)$ for $s\in [0,b]$, by the Lebesgue dominated theorem (the dominated function $F(s)=\frac{M_b+1}{{(c^{\ast }(b))}^2}$ on $\in [0,b]$), we have

$$y(t)={\int }_0^b{G}_{0,b}\frac{h(y)(s)}{{(\phi x(s))}^2}ds+\frac{t}{b}y(b)\text{for }t\in [0,b].$$

Differentiating the last equality twice, we know

$$y^{″}(t)=-\frac{h(y)(t)}{{(\phi x(t))}^2}\text{for }t\in [0,1).$$

By (i), we know $t\le y(t)\le 1$ and ${lim}_{t\to 1}y(t)=1=y(1)$ and then $y\in C[0,1]\cap C^1[0,1)$. This, together with (2.4), implies that $y(t)$ satisfies (3.12). Clearly, $(x,y)\in Q$. □

Theorem 3.2 For$\lambda \in ({\lambda }_0,1)$, the system (2.2)-(2.3) has at least a solution$(x,y)$in Q.

Proof Let $(x,y)$ in Theorem 3.1. It is clear that we only prove $\phi x(t)=x(t)$. If $\lambda \ge 0$, by (3.10), we obtain $x(t)\ge \frac{1-t}{4}$ for $t\in [0,1]$ and then $\phi x(t)=x(t)$. Next, we prove $x(t)\ge c_{\lambda }(1-t)$ for $t\in [0,1]$ for ${\lambda }_0<\lambda <0$.

Let $\gamma \in [0,1]$ such that $M=\phi x(\gamma )=max\{\phi x(t):t\in [0,1]\}$, then

$$\begin{array}{rcl}M& \ge & x(\delta )={\int }_{\delta }^1\frac{(2\lambda s+\lambda +y(s))(1-s)}{\phi x(s)}+(1-\delta ){\int }_0^{\delta }\frac{\lambda s+y(s)}{\phi x(s)}ds\\ \ge & {\int }_{\delta }^1\frac{(2\lambda s+\lambda +s)(1-s)}{max\{M,c_{\lambda }\}}ds+(1-\delta ){\int }_0^{\delta }\frac{\lambda s+s}{max\{M,c_{\lambda }\}}ds\\ =& \frac{1}{max\{M,c_{\lambda }\}}{\int }_{\delta }^1(2\lambda s+\lambda +s)(1-s)ds+(1-\delta ){\int }_0^{\delta }(\lambda s+s)ds\\ =& \frac{h(\lambda )}{max\{M,c_{\lambda }\}}.\end{array}$$

From this and $c_{\lambda }\le \sqrt{h(\lambda )}$, we obtain $M\ge \sqrt{h(\lambda )}$ and $x(\gamma )=\phi x(\gamma )=M$.

Let $S(t)=S(x,y)(t)$ and $S=max\{S(t):t\in [0,1]\}$, we prove

$$S\le \sqrt{\frac{3+5\lambda }{3}}.$$

(3.13)

By $\alpha (y)(0)=\lambda <0$ and $\alpha (y)(1)=3\lambda +1>0$, there exists $t_0\in (0,1)$ such that $\alpha (y)(t_0)=0$. Since $\alpha {(y)}^{″}(t)=y^{″}(t)\le 0$ for $t\in [0,1]$, i.e., $\alpha (y)(t)$ is concave down on $[0,1]$, then $\alpha (y)(s)\le 0$ for $s\in [0,t_0]$ and $\alpha (y)(s)\ge 0$ for $s\in [t_0,1]$. Hence, $S=S(t_0)$.

By (3.11), we have

$$\phi x(t)\ge x(t)\ge S(t)\text{for }t\in [t_0,1],$$

we know

$$S(t)(-S^{\prime }(t))=\frac{S(t)\alpha (y)(t)}{\phi x(t)}\le 2\lambda t+\lambda +1\text{for }t\in [t_0,1].$$

Integrating the last inequality from $t_0$ to 1 and utilizing $S(1)=0$, we have

$$\frac{S^2(t_0)}{2}\le {\int }_{t_0}^1(2\lambda s+\lambda +1)(1-s)ds\le {\int }_0^1(2\lambda s+\lambda +1)(1-s)ds=\frac{3+5\lambda }{6}.$$

Hence, (3.13) holds.

By $x^{\prime }(0)>0$, $x(\delta )>0$ and $x(1)=0$, we have $0<\gamma <1$ and $x^{\prime }(\gamma )=0$, then

$$0=x^{\prime }(\gamma )=-\frac{\lambda (1-{\gamma }^2)}{\phi x(\gamma )}-{\int }_0^{\gamma }\frac{\lambda s+y(s)}{\phi x(s)}ds,$$

i.e.,

$${\int }_0^{\gamma }\frac{\lambda s+y(s)}{\phi x(s)}ds=-\frac{\lambda (1-{\gamma }^2)}{\phi x(\gamma )}.$$

Hence,

$$(1-\gamma )T(x,y)(\gamma )=-\frac{\lambda (1-\gamma )(1-{\gamma }^2)}{\phi x(\gamma )}\le -\frac{\lambda }{M}.$$

This, together with (3.13), implies

$$M=x(\gamma )=S(x,y)(\gamma )+(1-\gamma )T(x,y)(\gamma )\le \sqrt{\frac{3+5\lambda }{3}}-\frac{\lambda }{M},$$

i.e.,

$$M\le \frac{\sqrt{\frac{3+5\lambda }{3}}+\sqrt{\frac{3-7\lambda }{3}}}{2}=\sigma (\lambda ).$$

Since $\alpha (y)(t)\ge 2\lambda t+\lambda +t\ge 0$ for $t\in [\delta ,1]$, we have

$$\begin{array}{rcl}x(t)& \ge & (1-t)T(x,y)(t)\ge (1-t)T(x,y)(\delta )\\ \ge & (1-t){\int }_0^{\delta }\frac{\lambda s+s}{\sigma (\lambda )}ds\ge \frac{(\lambda +1){\delta }^2}{2\sigma (\lambda )}(1-t),t\in [\delta ,1].\end{array}$$

And then $x(t)\ge c_{\lambda }(t)$ for $t\in [\delta ,1]$.

Finally, we prove $x(t)\ge c_{\lambda }$ for $t\in [0,\delta ]$.

In fact, if there exists $t\in [0,\delta ]$ such that $x(t)<c_{\lambda }$, by $x(\delta )>c_{\lambda }$, there exists $t^{\prime }\in (0,\delta )$ such that $x(t)>c_{\lambda }$ for $t\in (t^{\prime },\delta ]$ and $x(t^{\prime })=c_{\lambda }$.

From

$$x(\delta )=S(x,y)(\delta )+(1-\delta )T(x,y)(\delta ),$$

$S(x,y)(\delta )\ge {\int }_{\delta }^1\frac{2\lambda s+\lambda +s}{\sigma (\lambda )}ds$ and $T(x,y)(\delta )\ge {\int }_0^{\delta }\frac{\lambda s+s}{\sigma (\lambda )}ds$, we obtain

$$x(\delta )\ge \frac{h(\lambda )}{\sigma (\lambda )}.$$

By (3.11), we have

$$x^{\prime }(t)=-\frac{\lambda (1-t^2)}{x(t)}-{\int }_0^t\frac{\lambda s+y(s)}{\phi x(s)}ds\le -\frac{\lambda (1-t^2)}{x(t)},t\in [t^{\prime },\delta ],$$

i.e., $x(t)x^{\prime }(t)\le -\lambda (1-t^2)$, $t\in [t^{\prime },\delta ]$. Integrating this inequality from $t^{\prime }$ to δ, we have

$$\frac{x^2(\delta )-c_{\lambda }^2}{2}\le {\int }_{t^{\prime }}^{\delta }-\lambda (1-s^2)ds<{\int }_0^{\delta }-\lambda (1-s^2)ds$$

and then $c_{\lambda }^2>x^2(\delta )+2{\int }_0^{\delta }\lambda (1-s^2)ds\ge \frac{h^2(\lambda )}{{\sigma }^2(\lambda )}-2l(\lambda )=\omega (\lambda )$, a contradiction.

This completes the proof. □

4 Existence of solutions of (1.1)-(1.3)

In this section, we use positive solutions obtained in Theorem 3.2 to construct the solutions of (1.1)-(1.3) in Γ.

Theorem 4.1 For$\lambda \in ({\lambda }_0,1)$, the system (1.1)-(1.3) has at least a solution$(f,g)\in \mathrm{\Gamma }$.

Proof Let $\lambda \in ({\lambda }_0,1)$, by Theorem 3.2, the system (2.2)-(2.3) has at least a solution $(x,y)$ in Q. By $x(t)\ge c_{\ast }(t)$ and (2.2), we know

$$\begin{array}{rcl}x(t)& \le & {\int }_t^1\frac{(1-s)(3|\lambda |+1)}{c_{\ast }(s)}ds+(1-t){\int }_0^t\frac{|\lambda |+1}{c_{\ast }(s)}ds\\ \le & \frac{1}{c_{\ast }}({\int }_t^1(3|\lambda |+1)ds+(1-t){\int }_0^t\frac{1+|\lambda |}{1-s}ds)\\ \le & \frac{1}{c_{\ast }}(3|\lambda |+1-(1+|\lambda |)ln(1-t))(1-t).\end{array}$$

Let $u(t)=\frac{1}{c_{\ast }}(3|\lambda |+1-(1+|\lambda |)ln(1-t))$, $du=\frac{1+|\lambda |}{c_{\ast }(1-t)}dt$ and then

$${\int }_0^1\frac{1}{z(s)}ds\ge {\int }_0^1\frac{1}{u(s)(1-s)}ds=\frac{c_{\ast }}{1+|\lambda |}{\int }_0^{\mathrm{\infty }}\frac{du}{u}=\mathrm{\infty },$$

we have ${\int }_0^1\frac{1}{x(s)}ds=\mathrm{\infty }$.

Let

$$\eta :=\eta (t)={\int }_0^t\frac{1}{x(s)}ds,0\le t<1.$$

(4.1)

Then $\eta (t)$ is strictly increasing on $[0,1)$ and

$$\eta (0)=0,\eta (1-0)={\int }_0^1\frac{1}{x(s)}ds=+\mathrm{\infty }.$$

Let $t=h(\eta )$ be the inverse function to $\eta =\eta (t)$, we define the function

$$g(\eta )={\int }_0^{\eta }h(s)ds,f(\eta )={\int }_0^{\eta }y(h(s))ds,0\le \eta <+\mathrm{\infty }.$$

Then

$$g^{\prime }(\eta )=h(\eta ),g(0)=0,g^{\prime }(0)=0,g^{\prime }(\mathrm{\infty })=1$$

and

$$f^{\prime }(\eta )=y(h(\eta )),f(0)=0,f^{\prime }(0)=0,f^{\prime }(\mathrm{\infty })=1.$$

From (4.1), we have

$$\eta =\eta (g^{\prime }(\eta ))={\int }_0^{g^{\prime }(\eta )}\frac{1}{x(s)}ds,0\le \eta <+\mathrm{\infty }.$$

(4.2)

Differentiating (4.2) with respect to η, we have

$$g^{″}(\eta )=x(g^{\prime }(\eta ))=x(t),0\le \eta <+\mathrm{\infty }.$$

(4.3)

Then $g^{″}(\eta )>0$ for $0\le \eta <+\mathrm{\infty }$.

Differentiating (4.3) with respect to η, we have

$$g^{‴}(\eta )=x^{\prime }(g^{\prime }(\eta )),g^{″}(\eta )=x^{\prime }(t)x(t),0\le t<1.$$

(4.4)

Differentiating (2.2) with respect to t, we have

$$x^{\prime }(t)=-{\int }_0^t\frac{\lambda s+y(s)}{x(s)}ds+\frac{-\lambda (1-t^2)}{x(t)},0\le t<1.$$

(4.5)

By setting $s=g^{\prime }(\sigma )$ and utilizing $t=g^{\prime }(\eta )$ and (4.3), we have

$$\begin{array}{rcl}{\int }_0^t\frac{\lambda s+y(s)}{x(s)}ds& =& {\int }_0^{g^{\prime }(\eta )}\frac{\lambda s+y(s)}{x(s)}ds\\ =& {\int }_0^{\eta }(f^{\prime }(\sigma )+\lambda g^{\prime }(\sigma ))d\sigma =f(\eta )+\lambda g(\eta ).\end{array}$$

(4.6)

By (4.3), (4.4), (4.5) and (4.6), we have

$$g^{‴}=-(f+\lambda g)g^{″}+\lambda (g^{\mathrm{\prime }2}-1).$$

By (4.1), we have $\frac{dt}{d\eta }=x(t)$. Differentiating $f^{\prime }(\eta )$ with respect to η, we have

$$f^{″}(\eta )=y^{\prime }(t)\frac{dt}{d\eta }=y^{\prime }(t)x(t),f^{‴}(\eta )=y^{″}(t)x^2(t)+y^{\prime }(t)x^{\prime }(t)x(t).$$

Differentiating (2.3) with t twice and combining (4.5) and (4.6), we obtain

This completes the proof. □

Remark 4.1 For $\lambda <-1$, by Theorem 1 [2], (1.1)-(1.3) has no solution such that ${lim}_{\eta \to \mathrm{\infty }}{g}^{\prime }(\eta )=1$ with $|g^{\prime }(\eta )|<1$ for $\eta \ge {\eta }_0$, ${\eta }_0\ge 0$ is a constant.

Utilizing the system (2.2)-(2.3), we know easily that (1.1)-(1.3) has no solution in Γ for $\lambda \le -1$.

In fact, if (1.1)-(1.3) has a solution $(f,g)\in \mathrm{\Gamma }$ for some $\lambda \le -1$, by Theorem 2.1, then (1.1)-(1.3) has a solution in $(x,y)\in Q$. Noticing that

$$\alpha (y)(t)=2\lambda t+\lambda +y(t)\le 2\lambda t+\lambda +1<0\text{for }t\in (0,1),$$

we know

$$g^{″}(0)=x(0)={\int }_0^1\frac{\alpha (y)(s)(1-s)}{\phi x(s)}ds<0,$$

a contradiction.

This research uses integrals of equations to investigate the existence of solutions of the 3D axisymmetric inviscid stagnation flows related to Navier-Stokes equations and supplies a gap of analytical study in this field.