differential equation y''=1/y^m

Question

I'm interested in knowing the asymptotics of solutions to the nonlinear ordinary differential equation \begin{equation*} \begin{array}{ll} y''=1/y^m\tag{*}\\ y(0)=a>0, \text{ and }y'(0)=0. \end{array} \end{equation*}

When $m=3$ $(*)$ has a closed form solution $y=\frac{1}{a}\sqrt{a^4+x^2}$, and $y$ is asymptotic to $\frac{1}{a}|x|$. I have been trying to deduce for which $m$ $(*)$ has a solution which is asymptotically linear (as for $m=3$), and also the constants $c,b$ depending on $m,a$ such that $y$ is asymptotic to $c|x|+b$.

So far I've begun by turning the second order ODE into a first order equation. This follows from letting $z=\frac{dy}{dx}$ and noticing that $(*)$ is equivalent to $z \frac{dz}{dy}=y'' = 1/y^m$, and solving for $z=\frac{dy}{dx}$ to get \begin{equation} (y')^2=\frac{2}{1-m} y^{1-m}+C. \end{equation}

Then we can evaluate $C$ using the initial condition and get for $x\geq 0$ and $y\geq a$ \begin{equation} y' = \left( \frac{2}{m-1} \left(\frac{1}{a^{m-1}} - \frac{1}{y^{m-1}}\right)\right)^{1/2}. \end{equation}

$y$ is a convex function because $y''=1/y^m \geq 1/a^m$, so in particular it is unbounded, so letting $y\to \infty$ in the above expression gives \begin{equation} \lim_{x\to \infty} y'(x) = \left(\frac{2}{(m-1)a^{m-1}}\right)^{1/2} :=c. \end{equation}

This step required that $m>1$, so that's the first constraint on $m$. If $y$ is asymptotic to anything, it will be of the form $y=c|x|+b$ for some constant $b$. Since $0\leq y' \leq c$ it follows that $y$ is bounded above by $c|x|+a$, but it's not clear to me that $y$ is necessarily bounded below by a linear function $c|x|+b$ for some $b$. I've tried get asymptotic bounds on the integral

\begin{equation} \int_a^y \left( \frac{2}{m-1} \left(\frac{1}{a^{m-1}} - \frac{1}{t^{m-1}}\right)\right)^{1/2} \,dt \end{equation}

as $y\to \infty$, but I haven't had much success. If anyone has any suggestions for how to prove the solution $y$ is bounded below by $c|x|+b$ for some $b$ I'd be very grateful.

Update: I haven't proved this yet, but I've made progress by showing that $y$ cannot be asymptotic to $cx-ln(x)$. This would mean that there exists $\alpha>\beta>0$ such that as $x\to \infty$ \begin{equation} cx-\alpha\, ln(x) \leq y(x) \leq cx-\beta \,ln(x). \end{equation} Because $f$ is convex this implies a bound on the derivative. \begin{equation} c-\frac{\alpha}{x} \leq y'(x) \leq c-\frac{\beta}{x}. \end{equation} And then you can show from $(*)$ that $y'''<0$ so $y'$ is concave and this implies a bound on the second derivative. \begin{equation} \frac{\alpha}{x^2} \leq y''(x) \leq \frac{\beta}{x^2}. \end{equation}

Then if $m>2$ we get the following inequality for large enough $x$:

\begin{equation} \frac{1}{y^m} \leq \frac{1}{(cx-ln(x))^m} < \frac{\alpha}{x^2} \leq y''. \end{equation}

This provides us with a contradiction so if $m>2$ then $y$ cannot be asymptotic to $cx-ln(x)$. But this is only one example, and doesn't stop $y$ from being some other sublinear function. One could probably prove a similar thing for $y$ asymptotic to $cx-ln(ln(x))$, but this method will never completely prove that $y$ is asymptotically linear.

Answer 1

$$y''=\frac{1}{y^m}$$ $$2y''y'=\frac{2y'}{y^m}$$ Case $m\neq 1$ : $$(y')^2= -\frac{2}{(m-1)y^{m-1}}+c_1$$ With conditions $y(0)=a$ and $y'(0)=0$ : $\quad c_1=\frac{2}{(m-1)a^{m-1}}$ $$y'=\pm\sqrt{-\frac{2}{(m-1)y^{m-1}}+c_1}$$ $$dy=\pm\sqrt{-\frac{2}{(m-1)y^{m-1}}+c_1}\:dx$$ $$\frac{dy}{\pm\sqrt{-\frac{2}{(m-1)y^{m-1}}+c_1}}=dx$$ $$\int \frac{dy}{ \sqrt{-\frac{2}{(m-1)y^{m-1}}+c_1}} =\pm x+c_2$$

Integration leads to the general solution expressed on implicit form : $$\frac{1}{c_1} \sqrt{-\frac{2}{(m-1)y^{m-1}}+c_1} \:\: _2F_1\left(A\:,\: B \:;\: C \:;\: Y(y)\right)= \pm x+c_2$$ $_2F_1(A,B;C;Y)$ is the Gauss hypergeometric function with : $\begin{cases} A=1 \\ B=\frac{1}{2}-\frac{1}{m-1} \\ C=1-\frac{1}{m-1} \\ Y(y)=\frac{2}{c_1(m-1)y^{m-1}} \end{cases}$

$$ \sqrt{\frac{1-Y(y)}{c_1}} \:\: _2F_1\left(A\:,\: B \:;\: C \:;\: Y(y)\right)= \pm x+c_2$$

The condition $y(0)=a$ determines $c_2$ from the above solution with $x=0$ and $y=a$ in it.

For some integer values of $m$, the hypergeometric function might be reduced to simpler functions.

Asymptotic behaviour :

$y\to\infty\quad$ then $\quad Y\to 0\quad$ the asypmtotic expansion of the hypergeometric function is : $$_2F_1(A,B;C;Y)=1+\frac{AB}{C}Y+O(Y^2)$$ This allows to derive the asymptotic Relationship between $Y$ and $x$ and then between $y$ and $x$

Answer 2

As suggested by Did, we start by making the substitution \begin{equation} y(x) = a\,cosh(z(x))^{2/(m-1)}. \end{equation}

Taking a derivative and plugging it into the nonlinear first order equation gives \begin{equation} z'(x) = \sqrt{\frac{m-1}{2a^{m+1}}}\,cosh(z(x))^{(m-1)/2}. \end{equation}

If we introduce the constants $c=\sqrt{\frac{2}{(m-1)a^{m-1}}}$ and $\alpha = \frac{2}{m-1}$. Then the integrated form of this equation is

\begin{equation} \int_0^z cosh(t)^\alpha\,dt = \frac{c}{\alpha a}x +d, \end{equation} for some constant $d$, to be determined. Now we restrict to the case when $m\geq 3$, which forces $0<\alpha \leq 1$. This means that we have inequalities $(a+b)^\alpha < a^\alpha + b^\alpha$ for $a,b>0$. Likewise we have the inequalities $(a-b)^\alpha\geq a^\alpha - b^\alpha$ for $a\geq b>0$. Using this on the function $cosh(t)^\alpha$ we get

\begin{align} \frac{c}{a \alpha}x + d &\leq \int_0^z \frac{1}{2^\alpha} (e^{\alpha t} + e^{-\alpha t})\,dt\\ &\leq \frac{1}{\alpha 2^\alpha} (e^{\alpha z}-e^{-\alpha z}). \end{align}

At this point we can evaluate at the initial condition $z(0)=0$ to get that $d=0$ and we get a string of inequalities, using the above inequalities once more, to get

\begin{align} \frac{c}{a}x &\leq \frac{1}{2^\alpha} (e^{\alpha z}-e^{-\alpha z})\\ &\leq \frac{1}{2^\alpha}(e^z-e^{-z})^\alpha\\ &=(cosh^2(z)-1)^\frac{\alpha}{2}\\ &\leq cosh(z)^\alpha = \frac{y}{a}. \end{align}

This implies that $y(x)$ is bounded below by $c|x|$ when $m\geq 3$. This doesn't find the largest value of $b$ such that $c|x|+b\leq y(x)$, but it does guarantee $b\geq 0$ for $m\geq 3$. And it doesn't rule out the fact that $y$ could be bounded below by $c|x|+b$ for $m<3$.