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(x)=y(x)sin{y(x)},y(0)=1
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Answers (2)

wellnesshaus4n4
Answered 2022-05-02 Author has 24 answers
Step 1
We use the simple fact that v(y)=ysiny is increasing and positive on y>0
The fact that y0(x)=0 is a solution implies that y(x)>0 for all x (by uniqueness). Then
y(x)=v(y(x))>0.
Thus limxy(x) exists and is nonnegative. If it's not zero, then y(x)y1>0 for some fixed y1 and so
y(x)=v(y(x))v(y1)>0.
That would imply for any x<0 (by the mean value theorem and that y(0)=1)
y(x)=y(x)y(0)+y(0)=y(x0)x+1v(y1)x+1
this is impossible as y(x)>0 for all x.
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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(x)=dydx
=dydzdzdx
=dydz
Therefore, we have the differential equation with the reversed variable direction
dydz=y+siny
This is in fact the dynamics of a simple damped pendulum. Let V(y)=y22
dVdz=ydydz
=y2+ysiny
and dVdz<0 for all yR{0} and dVdz=0 at y=0. Hence, R is positively invariant with respect to the ODE, meaning that every solution starting at z=z0 is defined for all zgivenz0. Let v be the solution of the ODE with y(0)=1. Then, we have that limzv(z)=0. Finally, we conclude that limxu(x)=0.
We can also show that v(z) cannot change its sign and monotonically decreases whenever v(z)>0.
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