Observe that the equation of (1) is from the algebraic point of view a second-order equation in the variable

$Y={V}^{\mathrm{\prime}\mathrm{\prime}}$. So, solving it algebraically in order of

${V}^{\u2033}$, we obtain the equivalent form

${V}^{\u2033}=\frac{-p{x}^{2}\pm \sqrt{{p}^{2}{x}^{4}-4q{x}^{3}(x{V}^{\prime}-V)}}{2{x}^{3}},$

which leads us to consider the equation

${V}^{\u2033}+{H}_{\pm}(x,V,{V}^{\prime})=0,$

where

${H}_{\pm}(x,V,{V}^{\prime})=\frac{p{x}^{2}\mp \sqrt{{p}^{2}{x}^{4}-4q{x}^{3}(x{V}^{\prime}-V)}}{2{x}^{3}}.$

This fact suggests the study of the auxiliary problem

$\{\begin{array}{c}{V}^{\u2033}+H(x,V,{V}^{\prime})=0,\hfill \\ V(c)={V}_{c},\phantom{\rule{2em}{0ex}}V(d)={V}_{d},\hfill \end{array}$

(2)

where

$H(x,V,{V}^{\prime})=\frac{p{x}^{2}-\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}|x{V}^{\prime}-V|}}{2{x}^{3}}.$

The main argument to prove Theorem 2 relies on the method of upper and lower solutions.

We recall that

$\alpha \in {C}^{2}$ is a

*lower solution* of (2) if

$\{\begin{array}{c}{\alpha}^{\mathrm{\prime}\mathrm{\prime}}+H(x,\alpha ,{\alpha}^{\mathrm{\prime}})\ge 0,\hfill \\ \alpha (c)\le {V}_{c},\phantom{\rule{2em}{0ex}}\alpha (d)\le {V}_{d}.\hfill \end{array}$

(3)

Similarly, an

*upper solution* $\beta \in {C}^{2}$ of (2) is defined by reversing the inequalities in (3). A

*solution* of (2) is a function

*u* which is simultaneously a lower and an upper solution. A function

*f* is said to satisfy the Nagumo condition on some given subset

$E\subset I\times {\mathbb{R}}^{2}$ if there exists a positive continuous function

$\phi \in C({\mathbb{R}}_{0}^{+},[\epsilon ,+\mathrm{\infty}[)$,

$\epsilon >0$, such that

$|f(x,y,z)|\le \phi (|z|),\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,y,z)\in E,$

and

${\int}_{0}^{+\mathrm{\infty}}\frac{s}{\phi (s)}\phantom{\rule{0.2em}{0ex}}ds=+\mathrm{\infty}.$

(4)

**Lemma 1**
*Suppose that*
$\frac{{V}_{d}}{d}\le \frac{{V}_{c}}{c}.$

(5)

*Then*:

- 1.
*The function* ${\alpha}_{1}(x)=\frac{{V}_{d}}{d}x$ *is a lower solution of the problem* (2).

- 2.
*If* $k=\sqrt{\frac{q}{{c}^{3}}}$ *is small enough*,

*then the function* ${\alpha}_{2}(x)=\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}$

*is a lower solution of the problem* (2).

- 3.
*The function*
$\beta (x)=\frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}$

*is an upper solution of the problem* (2).

*Proof* 1. If we plug

${\alpha}_{1}(x)=\frac{{V}_{d}}{d}x$ in the first member of the equation in (2), we have that

${\left(\frac{{V}_{d}}{d}x\right)}^{\u2033}+H(x,\frac{{V}_{d}}{d}x,{\left(\frac{{V}_{d}}{d}x\right)}^{\prime})=0+\frac{p{x}^{2}-\sqrt{{p}^{2}{x}^{4}-4q{x}^{3}(\frac{{V}_{d}}{d}x-\frac{{V}_{d}}{d}x)}}{2{x}^{3}}=0,$

and by (5)

${\alpha}_{1}(c)=\frac{{V}_{d}}{d}c\le {V}_{c},\phantom{\rule{2em}{0ex}}{\alpha}_{1}(d)=\frac{{V}_{d}}{d}d={V}_{d}.$

- 2.
Consider now the function

${\alpha}_{2}(x)=\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}.$

Observe first that

${\alpha}_{2}(c)=\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{c}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}={V}_{c}$

and

${\alpha}_{2}(d)=\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{d}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}={V}_{d}.$

On the other hand, we have that

$\begin{array}{rcl}H(x,V,{V}^{\mathrm{\prime}})& =& \frac{p{x}^{2}-\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}|x{V}^{\prime}-V|}}{2{x}^{3}}\\ \ge & \frac{p{x}^{2}-\sqrt{{p}^{2}{x}^{4}}-\sqrt{4q{x}^{3}|x{V}^{\prime}-V|}}{2{x}^{3}}\\ =& -\sqrt{\frac{q}{{x}^{3}}}\sqrt{|x{V}^{\prime}-V|}\\ \ge & -\sqrt{\frac{q}{{c}^{3}}}\sqrt{|x{V}^{\prime}-V|}\\ =& -k\sqrt{|x{V}^{\prime}-V|},\end{array}$

where $k=\sqrt{\frac{q}{{c}^{3}}}$ .

So, if we plug the function

${\alpha}_{2}(x)$ in the first member of the equation in (2), we obtain

$\begin{array}{r}{(\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}})}^{\u2033}\\ \phantom{\rule{2em}{0ex}}+H(x,\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}},{(\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}})}^{\prime})\\ \phantom{\rule{1em}{0ex}}\ge 2\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}-k\sqrt{|2{x}^{2}\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}-\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}-\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}|}\\ \phantom{\rule{1em}{0ex}}=2\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}-k\sqrt{|{x}^{2}\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}-\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}|},\end{array}$

which, for

*k* small enough, is non-negative.

- 3.
Consider now the function

$\beta (x)=\frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}.$

Then, in an analogous way, we plug

$\beta (x)$ in the first member of the equation in (2). We observe that

$\begin{array}{r}{(\frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c})}^{\u2033}\\ \phantom{\rule{2em}{0ex}}+H(x,\frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c},{(\frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c})}^{\prime})\\ \phantom{\rule{1em}{0ex}}=0+\frac{p{x}^{2}-\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}|\frac{{V}_{d}-{V}_{c}}{d-c}x-\frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}|}}{2{x}^{3}}<0\end{array}$

and

$\beta (c)={V}_{c},\phantom{\rule{2em}{0ex}}\beta (d)={V}_{d}.$

So, the three assertions of the lemma are proved. □

**Remark 1** The value

${K}^{\ast}=\frac{2({V}_{d}-{V}_{c})}{({d}^{2}-{c}^{2})\sqrt{{V}_{c}+{V}_{d}}}$ is a suitable upper bound for the possible values that

$k=\sqrt{\frac{q}{{c}^{3}}}$ can take in the above assertion 2. In fact, for

$x\in [c,d]$ and since

$\frac{{V}_{d}}{d}\le \frac{{V}_{c}}{c}$, easy computations show that

$|{x}^{2}({V}_{d}-{V}_{c})-({d}^{2}{V}_{c}-{c}^{2}{V}_{d})|\le ({d}^{2}-{c}^{2})({V}_{d}+{V}_{c}).$

Therefore, if

$k\le {K}^{\ast}$, we have

$\begin{array}{rcl}k& \le & \frac{2({V}_{d}-{V}_{c})}{({d}^{2}-{c}^{2})\sqrt{{V}_{c}+{V}_{d}}}\\ \le & \frac{2\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}}{\sqrt{|{x}^{2}\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}-\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}|}}\phantom{\rule{1em}{0ex}}\text{for all}x\in [c,d],\end{array}$

which is used in the last step of the proof of assertion 2.

Now we state an existence and localisation result for the auxiliary problem (2).

**Theorem 1** *Suppose that* $\frac{{V}_{d}}{d}\le \frac{{V}_{c}}{c}$.

*Then*:

- 1.
*The problem* (2)

*has a solution* *V* *such that* $\frac{{V}_{d}}{d}x\le V(x)\le \frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}.$

- 2.
*If* $k=\sqrt{\frac{q}{{c}^{3}}}$ *is small enough*,

*the problem* (2)

*has a solution* *V* *such that* $\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}\le V(x)\le \frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}.$

- 3.
*If* $\frac{{V}_{d}}{d}=\frac{{V}_{c}}{c}$, *the function* $V(x)=\frac{{V}_{d}}{d}x$ *is a solution of the problem* (2).

*Proof* By the previous lemma, we know already that there are lower and upper solutions for the problem (2). It is also clear that they are well ordered, that is,

${\alpha}_{1}\le \beta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\alpha}_{2}\le \beta .$

So, if the function *H* satisfies the Nagumo condition, the thesis will follow by [1] or [2].

In order to prove assertions 1 and 2, we consider the sets, respectively,

${E}_{1}=\{(x,y,z)\in [c,d]\times {\mathbb{R}}^{2}:\frac{{V}_{d}}{d}x\le y\le \frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}\}$

and

${E}_{2}=\{(x,y,z)\in [c,d]\times {\mathbb{R}}^{2}:\frac{{V}_{d}-{V}_{c}}{{d}^{2}-{c}^{2}}{x}^{2}+\frac{{d}^{2}{V}_{c}-{c}^{2}{V}_{d}}{{d}^{2}-{c}^{2}}\le y\le \frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}\}.$

Function

*H* satisfies the Nagumo condition in

${E}_{1}$ and in

${E}_{2}$. In fact, we have that

$\begin{array}{rl}|H(x,y,z)|& =\left|\frac{p{x}^{2}-\sqrt{{p}^{2}{x}^{4}+4q{x}^{3}|xz-y|}}{2{x}^{3}}\right|\\ \le \frac{p{x}^{2}+\sqrt{4q{x}^{3}|xz-y|}}{2{x}^{3}}\\ \le \frac{p}{2x}+\sqrt{\frac{q}{{x}^{3}}}\sqrt{\frac{{V}_{d}-{V}_{c}}{d-c}x+\frac{d{V}_{c}-c{V}_{d}}{d-c}}+\frac{\sqrt{q}}{x}\sqrt{|z|}\\ \le \frac{p}{2c}+\sqrt{\frac{q}{{c}^{3}}}\sqrt{\frac{{V}_{d}-{V}_{c}}{d-c}d+\frac{d{V}_{c}-c{V}_{d}}{d-c}}+\frac{\sqrt{q}}{c}\sqrt{|z|}.\end{array}$

Thus, for some positive constants

${k}_{1}$ and

${k}_{2}$, we have that

$|H(x,y,z)|\le {k}_{1}+{k}_{2}\sqrt{|z|},\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,y,z)\in {E}_{1}\cup {E}_{2}$

and

${\int}_{0}^{+\mathrm{\infty}}\frac{s}{{k}_{1}+{k}_{2}\sqrt{|s|}}\phantom{\rule{0.2em}{0ex}}ds=+\mathrm{\infty}.$

So, by what was said above, the first two assertions of the thesis hold.

The third assertion follows directly from obvious computations. □

**Proposition 1** *Consider the problem* (2) *and the solution* *V* *given by Theorem* 1. *Then* *V* *is convex and satisfies* $V(x)\ge x{V}^{\mathrm{\prime}}(x)$ *in* $[c,d]$.

*Proof* The convexity of the solution

*V* follows easily by the equation of the problem (2)

${V}^{\u2033}=-H(x,V,{V}^{\prime})=\frac{-p{x}^{2}+\sqrt{p{x}^{4}+4q{x}^{3}|x{V}^{\prime}-V|}}{2{x}^{3}}\ge 0.$

Noting that

${(x{V}^{\prime}(x)-V(x))}^{\prime}=x{V}^{\u2033}(x)\ge 0$

then

$x{V}^{\prime}(x)-V(x)\le d{V}^{\prime}(d)-{V}_{d}.$

But, using Theorem 1, assertion 1, we have that ${V}^{\prime}(d)\le \frac{{V}_{d}}{d}$. In fact, since $\frac{{V}_{d}}{d}x\le V(x)$, it follows that $\frac{{V}_{d}}{d}\ge \frac{V(x)-{V}_{d}}{x-d}$, for $c\le x<d$. Letting $x\to d$, we obtain that $\frac{{V}_{d}}{d}\ge {V}^{\mathrm{\prime}}(d)$.

Then we have

$x{V}^{\prime}(x)-V(x)\le 0.$

□