find the solution of a differential equation
$begingroup$
we have a particle in laboratory whose equation of motion is a differential equation as
$ddot{x}(s)+2G(x(s),dot{x}(s))=0$
where
$G=frac{-akcos (kx)}{2}(W+(dot{x}-W)^3)$
with $W=(b+asin kx,0,0)$ in which $a$, $b$, and $k$ are real constant values s.t. $0<a<b$ and $a+b<1$.
I also know that
$1=frac{dot{x}(s)}{1+b+asin kx(s)}$.
Trying to find the solution I found
$x(s)=frac{2}{k}arctan(sqrt{1-(frac{a}{1+b})^2}tan ((s+A)frac{k}{2}(1+b)sqrt{1-(frac{a}{1+b})^2})-frac{a}{1+b})$,
where
$A=frac{2}{k(1+b)}frac{1}{sqrt{1-(frac{a}{1+b})^2}}arctanfrac{frac{a}{1+b}}{sqrt{1-(frac{a}{1+b})^2}}$.
Observe that according to the solution, $x(s)$ is periodic with a period of $frac{2pi}{k}$. Moreover, if I am not wrong, its range is $[-frac{pi}{k},frac{pi}{k}]$
Now the problem is that according to the motion of the particle in the laboratory, the range of the is the interval of $[0,infty)$!
So anyone may have any suggestion how to interpret this solution. If there is any other way to rewrite this solution s.t. it is reasonable with the experience. Any suggestion how to deal with this problem?
Thanks a lot!
calculus geometry ordinary-differential-equations
edited Dec 18 '18 at 16:41
Travis
63.8k769151
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
$endgroup$
add a comment |
$begingroup$
we have a particle in laboratory whose equation of motion is a differential equation as
$ddot{x}(s)+2G(x(s),dot{x}(s))=0$
where
$G=frac{-akcos (kx)}{2}(W+(dot{x}-W)^3)$
with $W=(b+asin kx,0,0)$ in which $a$, $b$, and $k$ are real constant values s.t. $0<a<b$ and $a+b<1$.
I also know that
$1=frac{dot{x}(s)}{1+b+asin kx(s)}$.
Trying to find the solution I found
$x(s)=frac{2}{k}arctan(sqrt{1-(frac{a}{1+b})^2}tan ((s+A)frac{k}{2}(1+b)sqrt{1-(frac{a}{1+b})^2})-frac{a}{1+b})$,
where
$A=frac{2}{k(1+b)}frac{1}{sqrt{1-(frac{a}{1+b})^2}}arctanfrac{frac{a}{1+b}}{sqrt{1-(frac{a}{1+b})^2}}$.
Observe that according to the solution, $x(s)$ is periodic with a period of $frac{2pi}{k}$. Moreover, if I am not wrong, its range is $[-frac{pi}{k},frac{pi}{k}]$
Now the problem is that according to the motion of the particle in the laboratory, the range of the is the interval of $[0,infty)$!
So anyone may have any suggestion how to interpret this solution. If there is any other way to rewrite this solution s.t. it is reasonable with the experience. Any suggestion how to deal with this problem?
Thanks a lot!
calculus geometry ordinary-differential-equations
edited Dec 18 '18 at 16:41
Travis
63.8k769151
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
$endgroup$
add a comment |
$begingroup$
we have a particle in laboratory whose equation of motion is a differential equation as
$ddot{x}(s)+2G(x(s),dot{x}(s))=0$
where
$G=frac{-akcos (kx)}{2}(W+(dot{x}-W)^3)$
with $W=(b+asin kx,0,0)$ in which $a$, $b$, and $k$ are real constant values s.t. $0<a<b$ and $a+b<1$.
I also know that
$1=frac{dot{x}(s)}{1+b+asin kx(s)}$.
Trying to find the solution I found
$x(s)=frac{2}{k}arctan(sqrt{1-(frac{a}{1+b})^2}tan ((s+A)frac{k}{2}(1+b)sqrt{1-(frac{a}{1+b})^2})-frac{a}{1+b})$,
where
$A=frac{2}{k(1+b)}frac{1}{sqrt{1-(frac{a}{1+b})^2}}arctanfrac{frac{a}{1+b}}{sqrt{1-(frac{a}{1+b})^2}}$.
Observe that according to the solution, $x(s)$ is periodic with a period of $frac{2pi}{k}$. Moreover, if I am not wrong, its range is $[-frac{pi}{k},frac{pi}{k}]$
Now the problem is that according to the motion of the particle in the laboratory, the range of the is the interval of $[0,infty)$!
So anyone may have any suggestion how to interpret this solution. If there is any other way to rewrite this solution s.t. it is reasonable with the experience. Any suggestion how to deal with this problem?
Thanks a lot!
calculus geometry ordinary-differential-equations
edited Dec 18 '18 at 16:41
Travis
63.8k769151
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
$endgroup$
we have a particle in laboratory whose equation of motion is a differential equation as
$ddot{x}(s)+2G(x(s),dot{x}(s))=0$
where
$G=frac{-akcos (kx)}{2}(W+(dot{x}-W)^3)$
with $W=(b+asin kx,0,0)$ in which $a$, $b$, and $k$ are real constant values s.t. $0<a<b$ and $a+b<1$.
I also know that
$1=frac{dot{x}(s)}{1+b+asin kx(s)}$.
Trying to find the solution I found
$x(s)=frac{2}{k}arctan(sqrt{1-(frac{a}{1+b})^2}tan ((s+A)frac{k}{2}(1+b)sqrt{1-(frac{a}{1+b})^2})-frac{a}{1+b})$,
where
$A=frac{2}{k(1+b)}frac{1}{sqrt{1-(frac{a}{1+b})^2}}arctanfrac{frac{a}{1+b}}{sqrt{1-(frac{a}{1+b})^2}}$.
Observe that according to the solution, $x(s)$ is periodic with a period of $frac{2pi}{k}$. Moreover, if I am not wrong, its range is $[-frac{pi}{k},frac{pi}{k}]$
Now the problem is that according to the motion of the particle in the laboratory, the range of the is the interval of $[0,infty)$!
So anyone may have any suggestion how to interpret this solution. If there is any other way to rewrite this solution s.t. it is reasonable with the experience. Any suggestion how to deal with this problem?
Thanks a lot!
calculus geometry ordinary-differential-equations
calculus geometry ordinary-differential-equations
edited Dec 18 '18 at 16:41
Travis
63.8k769151
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
edited Dec 18 '18 at 16:41
Travis
63.8k769151
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
edited Dec 18 '18 at 16:41
Travis
63.8k769151
edited Dec 18 '18 at 16:41
Travis
63.8k769151
edited Dec 18 '18 at 16:41
Travis
63.8k769151
63.8k769151
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
asked Dec 18 '18 at 14:44
MajidMajid
1,8731926
1,8731926
add a comment |
add a comment |
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