Given the initial value problem

Libby Boone
2022-05-01
Answered

Proving single solution of initial value problem is increasing

Given the initial value problem

${y}^{\prime}\left(x\right)=y\left(x\right)-\mathrm{sin}\left\{y\left(x\right)\right\},y\left(0\right)=1$

Given the initial value problem

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wellnesshaus4n4

Answered 2022-05-02
Author has **24** answers

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.

We use the simple fact that

The fact that

Thus

That would imply for any

this is impossible as

Giovanny Howe

Answered 2022-05-03
Author has **18** answers

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 $\underset{z\Rightarrow \mathrm{\infty}}{lim}v\left(z\right)=0$ . Finally, we conclude that $\underset{x\Rightarrow -\mathrm{\infty}}{lim}u\left(x\right)=0$ .

We can also show that$v\left(z\right)$ cannot change its sign and monotonically decreases whenever $v\left(z\right)>0$ .

Let us introduce the new variable z defined by

Therefore, we have the differential equation with the reversed variable direction

This is in fact the dynamics of a simple damped pendulum. Let

and

We can also show that

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