 # Proving single solution of initial value problem is increasing Given the Libby Boone 2022-05-01 Answered
Proving single solution of initial value problem is increasing
Given the initial value problem
${y}^{\prime }\left(x\right)=y\left(x\right)-\mathrm{sin}\left\{y\left(x\right)\right\},y\left(0\right)=1$
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Step 1
We use the simple fact that $v\left(y\right)=y-\mathrm{sin}y$ is increasing and positive on $y>0$
The fact that ${y}_{0}\left(x\right)=0$ is a solution implies that $y\left(x\right)>0$ for all x (by uniqueness). Then
${y}^{\prime }\left(x\right)=v\left(y\left(x\right)\right)>0.$
Thus $\underset{x\to -\mathrm{\infty }}{lim}y\left(x\right)$ exists and is nonnegative. If it's not zero, then $y\left(x\right)\ge {y}_{1}>0$ for some fixed ${y}_{1}$ and so
${y}^{\prime }\left(x\right)=v\left(y\left(x\right)\right)\ge v\left({y}_{1}\right)>0.$
That would imply for any $x<0$ (by the mean value theorem and that $y\left(0\right)=1$)
$y\left(x\right)=y\left(x\right)-y\left(0\right)+y\left(0\right)={y}^{\prime }\left({x}_{0}\right)x+1\le v\left({y}_{1}\right)x+1$
this is impossible as $y\left(x\right)>0$ for all x.
###### Not exactly what you’re looking for? Giovanny Howe
Step 1
Let us introduce the new variable z defined by $z=-x$. Then
${y}^{\prime }\left(x\right)=\frac{dy}{dx}$
$=\frac{dy}{dz}\frac{dz}{dx}$
$=-\frac{dy}{dz}$
Therefore, we have the differential equation with the reversed variable direction
$\frac{dy}{dz}=-y+\mathrm{sin}y$
This is in fact the dynamics of a simple damped pendulum. Let $V\left(y\right)=\frac{{y}^{2}}{2}$
$\frac{dV}{dz}=y\frac{dy}{dz}$
$=-{y}^{2}+y\mathrm{sin}y$
and $\frac{dV}{dz}<0$ for all $\frac{y\in \mathbb{R}}{\left\{0\right\}}$ and $\frac{dV}{dz}=0$ at $y=0$. Hence, $\mathbb{R}$ is positively invariant with respect to the ODE, meaning that every solution starting at $z={z}_{0}$ is defined for all $zgiven{z}_{0}$. Let v be the solution of the ODE with $y\left(0\right)=1$. Then, we have that $\underset{z⇒\mathrm{\infty }}{lim}v\left(z\right)=0$. Finally, we conclude that $\underset{x⇒-\mathrm{\infty }}{lim}u\left(x\right)=0$.
We can also show that $v\left(z\right)$ cannot change its sign and monotonically decreases whenever $v\left(z\right)>0$.