**Theorem 3.1** *Assume that h*(

*t*)

*and f*(

*t,u*)

*for a* <

*t* <

*b, u*(

*t*) > 0,

*satisfy inequality* (

*2.2*)

*and*$\begin{array}{c}\left(F1\right)\phantom{\rule{2.77695pt}{0ex}}f\left(t,0\right)\equiv 0,\phantom{\rule{2.77695pt}{0ex}}u{f}_{u}\left(t,u\right)>f\left(t,u\right)>0,\\ \left(F2\right){f}_{1}\left(t,u\right)\ge 0,\phantom{\rule{2.77695pt}{0ex}}h\left(t\right)v\left(t\right)+{v}^{\prime}\ge 0,\\ \left(F3\right){h}^{\prime}\left(t\right)\ge 0,\end{array}$

*where* $v\left(t\right)={\int}_{a}^{t}{e}^{-{\int}_{a}^{r}h\left(s\right)ds}dr$, *then problem* (*1.1*) *has at most one positive radial solution*.

**Example 3.1** For the equation

$\Delta u+\frac{A}{|x{|}^{2}}x\cdot \nabla u+{u}^{2}=0,\phantom{\rule{1em}{0ex}}x\in \Omega ,$

where

$-n-1\le A\le -n+1,\Omega :=\left\{x\in {\mathbb{R}}^{n}|\frac{1}{2}<|x|<1\right\},\phantom{\rule{0.3em}{0ex}}n\ge 3.$ Let

*t* = |

*x*|, then

$h\left(t\right)=\frac{A+n-1}{t},\phantom{\rule{1em}{0ex}}v\left(t\right)={\int}_{\frac{1}{2}}^{t}{e}^{-{\int}_{\frac{1}{2}}^{r}\frac{A+n-1}{s}ds}dr.$

A straightforward computation yields

${h}^{\prime}\left(t\right)=-\frac{A+n-1}{{t}^{2}}\ge 0$

and

$h\left(t\right)v\left(t\right)+{v}^{\prime}\left(t\right)={e}^{-\left(A+n-1\right)}\left(t\left(A+n+1\right)-\frac{A+n-1}{4t}\right)\ge 0,\phantom{\rule{1em}{0ex}}t\in \left(\frac{1}{2},1\right).$

Therefore, Theorem 3.1 ensures that there exists at most one positive radial solution.

Before proving our main result, we will do some preliminaries and give some useful lemmas.

Let

*u*(

*t,α*) denote the unique solution of Equation

2.1. If

*α* > 0, then the solution

*u*(

*t,α*) is positive for

*t* slightly larger than

*a*. When it vanishes in (

*a, b*), we define

*b*(

*α*) to be the first zero of

*u*(

*t, α*). More precisely,

*b*(

*α*) is a function of

*α* which has the property that

*u*(

*t, α*) > 0 for

*t* ∈ (

*a, b*(

*α*)) and

*u*(

*b*(

*α*)

*, α*) = 0. Let

*N* denote the set of

*α* > 0 for which the solution

*u*(

*t, α*) has a finite zero

*b*(

*α*). The variation of

*u*(

*t, α*) is defined by

$\varphi \left(t,\alpha \right)=\frac{\partial u\left(t,\alpha \right)}{\partial \alpha}$

and satisfies

${\varphi}^{\u2033}+h\left(t\right){\varphi}^{\prime}+{f}_{u}\left(t,u\right)\varphi =0,\phantom{\rule{1em}{0ex}}\varphi \left(a,\alpha \right)=0,\phantom{\rule{1em}{0ex}}\varphi \left(a,\alpha \right)=1.$

(3.1)

Let

*L* be the linear operator given by

$L\left(\varphi \left(t,\alpha \right)\right)={\varphi}^{\u2033}+h\left(t\right){\varphi}^{\prime}+{f}_{u}\left(t,u\right)\varphi ,\phantom{\rule{1em}{0ex}}a\le t\le b\left(\alpha \right).$

(3.2)

By (2.4), it is easy to show that *u*(*t, α*) has a unique critical point *c*(*α*) in (*a, b*(*α*)), and at this point, *u*(*t, α*) obtains a local maximum value.

**Lemma 3.1** *Assume that* (*F2*) *holds, then ϕ*(*t, α*) > 0 *for all t* ∈ (*a, c*(*α*)).

**Proof**. We introduce a function

$Q\left(t,\alpha \right)=\frac{v\left(t\right)}{{v}^{\prime}\left(t\right)}{u}^{\prime}\left(t,\alpha \right)\ge 0,\phantom{\rule{1em}{0ex}}a\le t\le c\left(\alpha \right),$

where

$v\left(t\right)={\int}_{a}^{t}{e}^{-{\int}_{a}^{r}h\left(s\right)ds}dr$

and accordingly

${v}^{\prime}\left(t\right)={e}^{-{\int}_{a}^{t}h\left(s\right)ds}.$

It is easy to see that

${v}^{\u2033}\left(t\right)+h\left(t\right){v}^{\prime}\left(t\right)=0,\phantom{\rule{1em}{0ex}}t\in \left(a,c\left(\alpha \right)\right).$

Differentiating

*Q(t, α)* with respect to

*t*, we get

${Q}^{\prime}\left(t,\alpha \right)={u}^{\prime}\left(t,\alpha \right)-\frac{v\left(t\right)}{{v}^{\prime}\left(t\right)}f\left(t,u\left(t,\alpha \right)\right)$

and

${Q}^{\u2033}\left(t,\alpha \right)=\left(-h\left(t\right)-\frac{v}{{v}^{\prime}}{f}_{u}\left(t,u\right)\right){u}^{\prime}-\left(2+h\left(t\right)\frac{v\left(t\right)}{{v}^{\prime}\left(t\right)}\right)f\left(t,u\right)-\frac{v}{{v}^{\prime}}{f}_{t}\left(t,u\right).$

Hence, we have

$\begin{array}{lll}\hfill L\left(Q\left(t,\alpha \right)\right)& ={Q}^{\u2033}\left(t,\alpha \right)+h\left(t\right){Q}^{\prime}\left(t,\alpha \right)+{f}_{u}\left(t,u\right)Q\left(t,\alpha \right)\phantom{\rule{2em}{0ex}}& \hfill \\ =-2\left(1+h\left(t\right)\frac{\upsilon \left(t\right)}{{\upsilon}^{\prime}\left(t\right)}\right)f\left(t,u\right)-\frac{\upsilon}{{\upsilon}^{\prime}}{f}_{t}\left(t,u\right).\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

From hypotheses (

*F* 2), we obtain

$L\left(Q\left(t,\alpha \right)\right)\le 0,\phantom{\rule{1em}{0ex}}t\in \left(a,c\left(\alpha \right)\right).$

(3.3)

Since Q(*t*,*α*) > 0 in *t* ∈ (*a*,*c*(*α*)) and inequality (3.3) holds, by the Sturm comparison principle (see [2]), we see that *Q*(*t*,*α*) oscillates faster that *ϕ*(*t*,*α*). Hence, *ϕ*(*t*,*α*) has no zero in *t* ∈ (*a*,*c*(*α*)). From *ϕ*(*a*,*α*) = 0 and *ϕ'*(*a*,*α*) = 1, it follows that *ϕ*(*t, α*) > 0 for all *t* ∈ (*a, c*(*α*)). The proof is complete.

**Remark 3.1** Lemma 3.1 was already proved in [11]. Here we give a simpler proof, directly using Sturm comparison principle.

Now, we present a lemma which has been given to the case *g*(|*x*|) = 0 (see [8]). To make the article as self-contained as possible, we will give a simple proof with a slight modification to [8].

**Lemma 3.2** *Assume α* ∈ *N and f*(*t,u*) *satisfies* (*F* 1)*, then*

(*H* 1) *ϕ*(*t,α*) *vanishes at least once and at most finitely many times in* (*a,b*(*α*)),

(*H2*) *if* 0 < *α*_{1} < *α*_{2}*, and at least one of u*(*t,α*_{1}) *and u*(*t,α*_{2}) *has a finite zero, then they intersect in* (*a*,min{*b*(*α*_{1})*,b*(*α*_{2})}).

**Proof**. We shall prove this by contradiction. Suppose to the contrary that

*ϕ*(

*t, α*) does not vanish in (

*a, b*(

*α*)), then

*ϕ*(

*t, α*) > 0,

*t* ∈ (

*a, b*(

*α*)). Note that

*L*(

*ϕ(t, α*)) = 0, so we have

${\left({e}^{{\int}_{a}^{t}h\left(s\right)ds}{\varphi}^{\prime}\left(t,\alpha \right)\right)}^{\prime}=-{e}^{{\int}_{a}^{t}h\left(s\right)ds}{f}_{u}\left(t,u\left(t,\alpha \right)\right)\varphi \left(t,\alpha \right).$

(3.4)

Using the definition of

*L*, we have

$L\left(u\left(t,\alpha \right)\right)={u}^{\u2033}\left(t,\alpha \right)+h\left(t\right){u}^{\prime}\left(t,\alpha \right)+{f}_{u}\left(t,u\right)u\left(t,\alpha \right).$

Similar to (3.4), we have

${\left({e}^{{\int}_{a}^{t}h\left(s\right)ds}{u}^{\prime}\left(t,\alpha \right)\right)}^{\prime}={e}^{{\int}_{a}^{t}h\left(s\right)ds}L\left(u\left(t,\alpha \right)\right)-{e}^{{\int}_{a}^{t}h\left(s\right)ds}{f}_{u}\left(t,u\left(t,\alpha \right)\right)u\left(t,\alpha \right).$

(3.5)

Multiply both sides of (3.4) by

*u*(

*t, α*) and (3.5) by

*ϕ*(

*t, α*), then subtract the resulting identities and we have

$\begin{array}{c}{\left({e}^{{\int}_{a}^{t}h\left(s\right)ds}\left(\varphi \left(t,\alpha \right){u}^{\prime}\left(t,\alpha \right)-{\varphi}^{\prime}\left(t,\alpha \right)u\left(t,\alpha \right)\right)\right)}^{\prime}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}={e}^{{\int}_{a}^{t}h\left(s\right)ds}\left(u{f}_{u}\left(t,\alpha \right)-f\left(t,u\right)\right)\varphi \left(t,\alpha \right).\end{array}$

(3.6)

By (

*F1*), we have the right side of (3.6) is positive in (

*a, b*(

*α*)). The left side of (3.6) is then a strictly increasing function of

*t* in (3.6). We get

${e}^{{\int}_{a}^{t}h\left(s\right)ds}\left(\varphi \left(t,\alpha \right){u}^{\prime}\left(t,\alpha \right)-{\varphi}^{\prime}\left(t,\alpha \right)u\left(t,\alpha \right)\right)>0\phantom{\rule{1em}{0ex}}\mathsf{\text{at}}\phantom{\rule{2.77695pt}{0ex}}t=b\left(\alpha \right).$

Thus, ${e}^{{\int}_{a}^{b\left(\alpha \right)}h\left(s\right)ds}\varphi \left(b\left(\alpha \right),\alpha \right){u}^{\prime}\left(b\left(\alpha \right),\alpha \right)>0$. However, it contradicts *u*'(*b*(*α*),*α*) < 0 and *ϕ*(*b*(*α*),*α*) ≥ 0.

Since the rest of proof can be completed by the same argument as [8], we omit them.

**Lemma 3.3** *If* (*F1*) *and* (*F* 3) *hold, then ϕ*(*b*(*α*)*, α*) ≠ 0.

**Proof**. We shall prove this by contradiction. Suppose to the contrary that

*ϕ*(

*b*(

*α*)

*, α*) = 0. Now, we may as well define

*τ*(

*α*) to be the last zero of

*ϕ*(

*t, α*) in (

*a, b*(

*α*)). By Lemma 3.1, it is easy to get

*c*(

*α*) ≤

*τ*(

*α*), thus

*u*'(

*τ*(

*α*),

*α*) ≤ 0 and

*u*'(

*t, α*) < 0 for all

*t* ∈ (

*τ*(

*α*),

*b*(

*α*)]. We introduce a function

$G\left(t,\alpha \right)={u}^{\prime}\left(t,\alpha \right).$

Differentiating

*G*(

*t, α*) with respect to

*t*, we get

${G}^{\prime}\left(t,\alpha \right)={u}^{\u2033}\left(t,\alpha \right)=-h\left(t\right){u}^{\prime}-f\left(t,u\right)$

and

${G}^{\u2033}\left(t,\alpha \right)=-{h}^{\prime}\left(t\right){u}^{\prime}\left(t,\alpha \right)-h\left(t\right){u}^{\u2033}\left(t,\alpha \right)-{f}_{u}\left(t,u\right){u}^{\prime}-{f}_{t}\left(t,u\right).$

Hence,

$\begin{array}{lll}\hfill L\left(G\left(t,\alpha \right)\right)& ={G}^{\u2033}\left(t,\alpha \right)+h\left(t\right){G}^{\prime}\left(t,\alpha \right)+{f}_{u}\left(t,u\right)G\left(t,\alpha \right)\phantom{\rule{2em}{0ex}}& \hfill \\ =-{h}^{\prime}\left(t\right){u}^{\prime}\left(t,\alpha \right)-h\left(t\right){u}^{\u2033}\left(t,\alpha \right)-{f}_{u}\left(t,u\right){u}^{\prime}\left(t,\alpha \right)-{f}_{t}\left(t,u\right)\phantom{\rule{2em}{0ex}}& \hfill \\ -{h}^{2}\left(t\right){u}^{\prime}\left(t,\alpha \right)-h\left(t\right)f\left(t,u\right)+{f}_{u}\left(t,u\right){u}^{\prime}\left(t,\alpha \right)\phantom{\rule{2em}{0ex}}& \hfill \\ =-{h}^{\prime}\left(t\right){u}^{\prime}\left(t,\alpha \right)+{f}_{t}\left(t,u\right).\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

Hence, we have

${\left({e}^{{\int}_{a}^{t}h\left(s\right)ds}{G}^{\prime}\left(t,\alpha \right)\right)}^{\prime}={e}^{{\int}_{a}^{t}h\left(s\right)ds}L\left(G\left(t,\alpha \right)\right)-{e}^{{\int}_{a}^{t}h\left(s\right)ds}{f}_{u}\left(t,u\left(t,\alpha \right)\right)G\left(t,\alpha \right).$

(3.7)

Similar to the argument of Lemma 3.2, multiply both sides of (3.4) by

*G*(

*t, α*), and (3.7) by

*ϕ*(

*t, α*) then we have

${\left({e}^{{\int}_{a}^{t}h\left(s\right)ds}\left(\varphi \left(t,\alpha \right){G}^{\prime}\left(t,\alpha \right)-{\varphi}^{\prime}\left(t,\alpha \right)G\left(t,\alpha \right)\right)\right)}^{\prime}={e}^{{\int}_{a}^{t}h\left(s\right)ds}L\left(G\left(t,\alpha \right)\right)\varphi \left(t,\alpha \right).$

(3.8)

Note that

*ϕ*(

*b*(

*α*)

*, α*) = 0, thus integrating both sides of (3.8) from

*τ*(

*α*) to

*b*(

*α*), we obtain

$\begin{array}{c}\left(-{e}^{{\int}_{a}^{b\left(\alpha \right)}h\left(s\right)ds}{\varphi}^{\prime}\left(b\left(\alpha \right),\alpha \right)G\left(b\left(\alpha \right),\alpha \right)\right)-\left(-{e}^{{\int}_{a}^{\tau \left(\alpha \right)}h\left(s\right)ds}{\varphi}^{\prime}\left(\tau \left(\alpha \right),\alpha \right)G\left(\tau \left(\alpha \right),\alpha \right)\right)\\ ={\int}_{\tau \left(\alpha \right)}^{b\left(\alpha \right)}{e}^{{\int}_{a}^{t}h\left(s\right)ds}L\left(G\left(t,\alpha \right)\right)\varphi \left(t,\alpha \right)dt.\end{array}$

(3.9)

Since

*τ*(

*α*) to be the last zero of

*ϕ*(

*t, α*) in (

*a, b*(

*α*)), the behavior of

*ϕ*(

*t, α*) in (

*τ*(

*α*)

*, b*(

*α*)) can be classified into two cases as follows:

- (i)
*ϕ* (*t, α*) > 0 in (*τ*(*α*)*, b*(*α*)), then the left side of (3.9) is negative, but by (*F* 3) the right side is positive.

It is impossible.

- (ii)
*ϕ* (*t, α*) < 0 in (*τ*(*α*)*, b*(*α*)), then the left side of (3.9) is positive, but by (*F* 3) the right side is negative.

It is also impossible. The proof is complete.

**The proof of Theorem 3.1** We will prove it as a standard process. Assume that

*N* is a nonempty set, otherwise we have nothing to prove. Let

*α* ∈

*N*, then

*u*(

*b*(

*α*)

*, α*) = 0. It is easy to see that

*u'*(

*b*(

*α*)

*, α*) ≤ 0. If

*u'*(

*b*(

*α*)

*, α*) = 0, then the assumption

*f*(

*t*, 0) ≡ 0 for all

*t* ≥ 0, and the uniqueness of solution of initial value problems for ordinary differential equations imply that

*u*(

*t, α*) ≡ 0 for all

*t* ∈ [

*a, b*(

*α*)], which contradicts the initial condition of

*u*(

*t, α*). Hence, we have

${u}^{\prime}\left(b\left(\alpha \right),\alpha \right)<0,$

(3.10)

and the implicit function theorem implies that *b*(*α*) is well-defined as a function of *α* in *N* and *b*(*α*) ∈ *C*^{1}(*N*). Furthermore, it follows from (3.10) that *N* is an open set. By Lemma 3.2, we have *N* is an open interval (see [8]).

Differentiate both sides of the identity

*u*(

*b*(

*α*)

*, α*) = 0 with respect to

*α*, we obtain

${u}^{\prime}\left(b\left(\alpha \right),\alpha \right){b}^{\prime}\left(\alpha \right)+\varphi \left(b\left(\alpha \right),\alpha \right)=0.$

From above Lemma 3.3, we have *ϕ*(*b*(*α*)*,α*) ≠ 0. Thus, *b*'(*α*) ≠ 0, *α* ∈ *N*. It means that *b*'(*α*) does not change sign, i.e., *b*(*α*) is monotone. The proof is complete.

**Remark 3.2** Actually, if the functions

*f*(|

*x*|,

*u*) and

*g*(|

*x*|) satisfy some suitable conditions, it is not difficult to get the existence of positive radial solutions to the Dirichlet boundary value problem (1.1). We just need that for Equation

2.1, the functions

*f*(|

*x*|,

*u*) and

*g*(|

*x*|) satisfy inequality (2.2) and

${\int}_{a}^{b}{e}^{-{\int}_{a}^{s}h\left(\xi \right)d\xi}\left(\alpha -{\int}_{a}^{s}{e}^{{\int}_{a}^{\xi}h\left(r\right)dr}f\left(\xi ,u\right)d\xi \right)ds=0.$

However, it seems that these assumptions are too strict.