Circular Motion - System of Differential Equations
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The following system of differential equations describes a charged particle in a viscous medium enveloped in an EM field,
$$partial^2_t x(t) = acos(omega t +phi)y +bx +cpartial_t y -dpartial_t x$$
$$partial^2_t y(t) = acos(omega t +phi)x +by -cpartial_t x -dpartial_t y$$
All of the coefficients are real and positive. It's well known there are two circular eigenmotions. Given this, it seems useful to make the change of variables $u(t)=x(t)+i y(t)$. Doing so produces,
$$partial^2_t u(t) = aicos(omega t +phi)u^* +bu -(d+ic)partial_t u$$
In the case of $a=d=0$, a solution of $u(t)=exp(-iw_0t)$ exists with $w_0$ holding two values based on $b$ and $c$. It can be assumed $partial_t x(0)=partial_t y(0)=0$
Any suggestions on how to proceed further? If it makes the problem significantly easier, the case of $d=0$ is also of great interest.
differential-equations systems-of-equations
asked Aug 3 at 22:22
Captain Morgan
33917
add a comment |Â
up vote
2
down vote
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The following system of differential equations describes a charged particle in a viscous medium enveloped in an EM field,
$$partial^2_t x(t) = acos(omega t +phi)y +bx +cpartial_t y -dpartial_t x$$
$$partial^2_t y(t) = acos(omega t +phi)x +by -cpartial_t x -dpartial_t y$$
All of the coefficients are real and positive. It's well known there are two circular eigenmotions. Given this, it seems useful to make the change of variables $u(t)=x(t)+i y(t)$. Doing so produces,
$$partial^2_t u(t) = aicos(omega t +phi)u^* +bu -(d+ic)partial_t u$$
In the case of $a=d=0$, a solution of $u(t)=exp(-iw_0t)$ exists with $w_0$ holding two values based on $b$ and $c$. It can be assumed $partial_t x(0)=partial_t y(0)=0$
Any suggestions on how to proceed further? If it makes the problem significantly easier, the case of $d=0$ is also of great interest.
differential-equations systems-of-equations
asked Aug 3 at 22:22
Captain Morgan
33917
What if you subtract the two equations and solve for the variable $z(t)=x(t)-y(t)$ could it be of any help? Edit: just noticed that the first partial derivative with respect to time don't have the same sign so probably it wouldn't help...maybe choosing a new variable $e=-c$ just for the second equation?
– Davide Morgante
Aug 3 at 22:35
How do you define $rho_1$, $rho_2$, etc?
– Robert Lewis
Aug 3 at 22:41
1
@Robert, I've updated the question to clarify
– Captain Morgan
Aug 3 at 22:44
What's the question? I can only see statements.
– md2perpe
Aug 7 at 16:54
@md2perpe, thanks
– Captain Morgan
Aug 7 at 16:57
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The following system of differential equations describes a charged particle in a viscous medium enveloped in an EM field,
$$partial^2_t x(t) = acos(omega t +phi)y +bx +cpartial_t y -dpartial_t x$$
$$partial^2_t y(t) = acos(omega t +phi)x +by -cpartial_t x -dpartial_t y$$
All of the coefficients are real and positive. It's well known there are two circular eigenmotions. Given this, it seems useful to make the change of variables $u(t)=x(t)+i y(t)$. Doing so produces,
$$partial^2_t u(t) = aicos(omega t +phi)u^* +bu -(d+ic)partial_t u$$
In the case of $a=d=0$, a solution of $u(t)=exp(-iw_0t)$ exists with $w_0$ holding two values based on $b$ and $c$. It can be assumed $partial_t x(0)=partial_t y(0)=0$
Any suggestions on how to proceed further? If it makes the problem significantly easier, the case of $d=0$ is also of great interest.
differential-equations systems-of-equations
asked Aug 3 at 22:22
Captain Morgan
33917
The following system of differential equations describes a charged particle in a viscous medium enveloped in an EM field,
$$partial^2_t x(t) = acos(omega t +phi)y +bx +cpartial_t y -dpartial_t x$$
$$partial^2_t y(t) = acos(omega t +phi)x +by -cpartial_t x -dpartial_t y$$
All of the coefficients are real and positive. It's well known there are two circular eigenmotions. Given this, it seems useful to make the change of variables $u(t)=x(t)+i y(t)$. Doing so produces,
$$partial^2_t u(t) = aicos(omega t +phi)u^* +bu -(d+ic)partial_t u$$
In the case of $a=d=0$, a solution of $u(t)=exp(-iw_0t)$ exists with $w_0$ holding two values based on $b$ and $c$. It can be assumed $partial_t x(0)=partial_t y(0)=0$
Any suggestions on how to proceed further? If it makes the problem significantly easier, the case of $d=0$ is also of great interest.
differential-equations systems-of-equations
asked Aug 3 at 22:22
Captain Morgan
33917
edited Aug 7 at 16:57
asked Aug 3 at 22:22
Captain Morgan
33917
asked Aug 3 at 22:22
Captain Morgan
33917
asked Aug 3 at 22:22
Captain Morgan
33917
33917
What if you subtract the two equations and solve for the variable $z(t)=x(t)-y(t)$ could it be of any help? Edit: just noticed that the first partial derivative with respect to time don't have the same sign so probably it wouldn't help...maybe choosing a new variable $e=-c$ just for the second equation?
– Davide Morgante
Aug 3 at 22:35
How do you define $rho_1$, $rho_2$, etc?
– Robert Lewis
Aug 3 at 22:41
1
@Robert, I've updated the question to clarify
– Captain Morgan
Aug 3 at 22:44
What's the question? I can only see statements.
– md2perpe
Aug 7 at 16:54
@md2perpe, thanks
– Captain Morgan
Aug 7 at 16:57
add a comment |Â
What if you subtract the two equations and solve for the variable $z(t)=x(t)-y(t)$ could it be of any help? Edit: just noticed that the first partial derivative with respect to time don't have the same sign so probably it wouldn't help...maybe choosing a new variable $e=-c$ just for the second equation?
– Davide Morgante
Aug 3 at 22:35
How do you define $rho_1$, $rho_2$, etc?
– Robert Lewis
Aug 3 at 22:41
1
@Robert, I've updated the question to clarify
– Captain Morgan
Aug 3 at 22:44
What's the question? I can only see statements.
– md2perpe
Aug 7 at 16:54
@md2perpe, thanks
– Captain Morgan
Aug 7 at 16:57
What if you subtract the two equations and solve for the variable $z(t)=x(t)-y(t)$ could it be of any help? Edit: just noticed that the first partial derivative with respect to time don't have the same sign so probably it wouldn't help...maybe choosing a new variable $e=-c$ just for the second equation?
– Davide Morgante
Aug 3 at 22:35
What if you subtract the two equations and solve for the variable $z(t)=x(t)-y(t)$ could it be of any help? Edit: just noticed that the first partial derivative with respect to time don't have the same sign so probably it wouldn't help...maybe choosing a new variable $e=-c$ just for the second equation?
– Davide Morgante
Aug 3 at 22:35
How do you define $rho_1$, $rho_2$, etc?
– Robert Lewis
Aug 3 at 22:41
How do you define $rho_1$, $rho_2$, etc?
– Robert Lewis
Aug 3 at 22:41
1
1
@Robert, I've updated the question to clarify
– Captain Morgan
Aug 3 at 22:44
@Robert, I've updated the question to clarify
– Captain Morgan
Aug 3 at 22:44
What's the question? I can only see statements.
– md2perpe
Aug 7 at 16:54
What's the question? I can only see statements.
– md2perpe
Aug 7 at 16:54
@md2perpe, thanks
– Captain Morgan
Aug 7 at 16:57
@md2perpe, thanks
– Captain Morgan
Aug 7 at 16:57
add a comment |Â
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What if you subtract the two equations and solve for the variable $z(t)=x(t)-y(t)$ could it be of any help? Edit: just noticed that the first partial derivative with respect to time don't have the same sign so probably it wouldn't help...maybe choosing a new variable $e=-c$ just for the second equation?
– Davide Morgante
Aug 3 at 22:35
How do you define $rho_1$, $rho_2$, etc?
– Robert Lewis
Aug 3 at 22:41
1
@Robert, I've updated the question to clarify
– Captain Morgan
Aug 3 at 22:44
What's the question? I can only see statements.
– md2perpe
Aug 7 at 16:54
@md2perpe, thanks
– Captain Morgan
Aug 7 at 16:57