# A first order ODE problem

Fix $a>0$ and $b>0$. Does the following ODE $$G(x)^2+2axG(x)G'(x)+2aG'(x)(x-b)=0 \tag{*}$$ have a solution, say, $F(x)$, that satisfies $F(x)>0$ and $F'(x)<0$ on $(b,\infty)$?

I tried to solve it by Mathematica, and it gives $G$ as solution to $$x=e^{-2a\left(\text{log}\,G(x)-\frac{1}{G(x)}\right)}\left(C_0+b(2a)^{2a}\text{Gamma}\left(1-2a,\frac{2a}{G(x)}\right)\right) \tag{**}$$

for some free parameter $C_0$, where Gamma$(\cdot,\cdot)$ is the upper incomplete gamma function. However I'm not sure if this is a proper inverse function or if $G$ given by this equation has domain $(b,\infty)$.

Thanks!!

The answer is yes: indeed, for any $a>0$ and $b>0$, your equation $(**)$ with any $C_0>0$ determines
a solution $G$ to your ODE $(*)$ such that $G>0$ and $G'<0$ on $(b,\infty)$ -- and even on $(0,\infty)$.

Indeed, rewrite $(**)$ as $$x=X(G(x)),\quad X(g):= e^{-2a\left(\text{log}\,g-\frac{1}{g}\right)}\left(C_0+b(2a)^{2a}\text{Gamma}\left(1-2a,\frac{2a}{g}\right)\right).$$ Take any $C_0>0$. Then $X(0+)=\infty$ and $X(\infty-)=0$. Let $$x_1(g):=X'(g)\,\frac{g^2 e^{2 a \left(\log g-\frac{1}{g}\right)}}{g+1}.$$ Then $$x'_1(g)=-\frac{2 a b e^{-\frac{2 a}{g}} g^{2 a}}{(g+1)^2},$$ which is manifestly negative for positive $a,b,g$, so that $X'<0$ on the interval $(0,\infty)$ and
hence the function $X\colon(0,\infty)\to(0,\infty)$ continuously decreases from $\infty$ to $0$ on $(0,\infty)$. So, the inverse function $G:=X^{-1}$ exists and solves the equation $x=X(G(x))$ for $x>0$. Moreover, $G>0$ and $G'=1/(X'\circ G)<0$, as desired.

It remains to verify that this function $G$ is a solution to your ODE $(*)$. To do this, just replace, in $(*)$, $G(x)$ by $g$ and $G'(x)$ by $1/X'(g)$, and then $x$ by $X(g)$, so that $(*)$ is rewritten as $g^2 + 2 a X(g) \frac g{X'(g)} + 2 \frac a{X'(g)}\, (X(g) - b)=0$, which is straightforward to check.

Details of calculations can be seen in the Mathematica notebook and/or its pdf image.

• Thanks for your reply! I'll spend some time to read it. Is it possible, though, to have $F>1$ (I just edited it) instead of $F>0$ in the old post ? Feb 11 '18 at 3:45
• Actually please ignore my last comment. $F>0$ is what I wanted. I got confused. Feb 11 '18 at 3:58
• All right. Actually, if you follow the lines of the proof, you will see that it is impossible to get $F>1$ in any neighborhood of $\infty$. Feb 11 '18 at 4:02
• How did you get Out[99] in the pdf image? I get "Limit[X[g], g -> [Infinity], Assumptions -> True]" when I run the command. Feb 11 '18 at 4:21
• Also I get a much more complicated expression for $x_1'$. Feb 11 '18 at 4:32