Mixing
Complex differential equation for z
$begingroup$
I need help in solving the following differential equation for $z$, a complex number, and $a,A,B,C$ all real:
$$frac{dz}{dt}=4ia(Abar{z}+Bbar{z}^2+Cbar{z}^3)$$
I have tried the ansatz $z=ce^{mt}$ with $c,m$ complex but that gives $m=0$. Also trying $z=r(t)e^{itheta}$ gave me equations I couldn't solve.
For $B=C=0$, the equation is:
$$frac{dz}{dt}=4iaAbar{z}$$
$$implies mc=4iaAbar{c}e^{(m-bar{m})t}$$
This gives $bar{m}=m$ and $m=4iaAbar{c}/c$ If we take $c=re^{itheta}$, $m=4iaAe^{-2itheta}$. Since $m$ has to be real, $cos(2theta)=0$ and thus the system can be solved exactly.
complex-numbers
asked Jan 20 at 0:56
Mani JhaMani Jha
93
$endgroup$
|
show 1 more comment
$begingroup$
I need help in solving the following differential equation for $z$, a complex number, and $a,A,B,C$ all real:
$$frac{dz}{dt}=4ia(Abar{z}+Bbar{z}^2+Cbar{z}^3)$$
I have tried the ansatz $z=ce^{mt}$ with $c,m$ complex but that gives $m=0$. Also trying $z=r(t)e^{itheta}$ gave me equations I couldn't solve.
For $B=C=0$, the equation is:
$$frac{dz}{dt}=4iaAbar{z}$$
$$implies mc=4iaAbar{c}e^{(m-bar{m})t}$$
This gives $bar{m}=m$ and $m=4iaAbar{c}/c$ If we take $c=re^{itheta}$, $m=4iaAe^{-2itheta}$. Since $m$ has to be real, $cos(2theta)=0$ and thus the system can be solved exactly.
complex-numbers
asked Jan 20 at 0:56
Mani JhaMani Jha
93
$endgroup$
$begingroup$
I think you should start with $B=C=0,$ write $z=x+iy$ with real smooth functions $x,y,$ and see what sort of ODE system you get, most likely rotation. Then allow nonzero $B$ in that setting; I think there should be some possible dynamical systems conclusions, but closed form seems optimistic.
$endgroup$
– Will Jagy
Jan 20 at 1:09
$begingroup$
Consider if on the left, you had $doverline{z}/dt$ instead of $dz/dt$, you could just separate variables and integrate. But with conjugation, you lose the comfort of complex numbers and you'll have to split into components. Polar form looks more promising, but as @WillJagy says, start with the simple case to see what works.
$endgroup$
– orion
Jan 20 at 1:14
$begingroup$
The case $B=C=0$ is quite straightforward actually. I have solved it, and either of the mentioned ansatzs work. However even with just B non zero, the trouble is the same.
$endgroup$
– Mani Jha
Jan 20 at 1:19
$begingroup$
@Mani , then edit that into the question.
$endgroup$
– Will Jagy
Jan 20 at 1:20
$begingroup$
@ManiJha I did $B=C=0,$ then $A=a = 1/2,$ so $dot{z}=i bar{z}$ With $z(0) = 1,$ I get $z = cosh t + i sinh t,$ which works. With $z(0) = i,$ I get $z = sinh t + i cosh t.$
$endgroup$
– Will Jagy
Jan 20 at 1:44
|
show 1 more comment
$begingroup$
I need help in solving the following differential equation for $z$, a complex number, and $a,A,B,C$ all real:
$$frac{dz}{dt}=4ia(Abar{z}+Bbar{z}^2+Cbar{z}^3)$$
I have tried the ansatz $z=ce^{mt}$ with $c,m$ complex but that gives $m=0$. Also trying $z=r(t)e^{itheta}$ gave me equations I couldn't solve.
For $B=C=0$, the equation is:
$$frac{dz}{dt}=4iaAbar{z}$$
$$implies mc=4iaAbar{c}e^{(m-bar{m})t}$$
This gives $bar{m}=m$ and $m=4iaAbar{c}/c$ If we take $c=re^{itheta}$, $m=4iaAe^{-2itheta}$. Since $m$ has to be real, $cos(2theta)=0$ and thus the system can be solved exactly.
complex-numbers
asked Jan 20 at 0:56
Mani JhaMani Jha
93
$endgroup$
I need help in solving the following differential equation for $z$, a complex number, and $a,A,B,C$ all real:
$$frac{dz}{dt}=4ia(Abar{z}+Bbar{z}^2+Cbar{z}^3)$$
I have tried the ansatz $z=ce^{mt}$ with $c,m$ complex but that gives $m=0$. Also trying $z=r(t)e^{itheta}$ gave me equations I couldn't solve.
For $B=C=0$, the equation is:
$$frac{dz}{dt}=4iaAbar{z}$$
$$implies mc=4iaAbar{c}e^{(m-bar{m})t}$$
This gives $bar{m}=m$ and $m=4iaAbar{c}/c$ If we take $c=re^{itheta}$, $m=4iaAe^{-2itheta}$. Since $m$ has to be real, $cos(2theta)=0$ and thus the system can be solved exactly.
complex-numbers
complex-numbers
asked Jan 20 at 0:56
Mani JhaMani Jha
93
asked Jan 20 at 0:56
Mani JhaMani Jha
93
edited Jan 20 at 1:35
Mani Jha
asked Jan 20 at 0:56
Mani JhaMani Jha
93
asked Jan 20 at 0:56
Mani JhaMani Jha
93
asked Jan 20 at 0:56
Mani JhaMani Jha
93
93
$begingroup$
I think you should start with $B=C=0,$ write $z=x+iy$ with real smooth functions $x,y,$ and see what sort of ODE system you get, most likely rotation. Then allow nonzero $B$ in that setting; I think there should be some possible dynamical systems conclusions, but closed form seems optimistic.
$endgroup$
– Will Jagy
Jan 20 at 1:09
$begingroup$
Consider if on the left, you had $doverline{z}/dt$ instead of $dz/dt$, you could just separate variables and integrate. But with conjugation, you lose the comfort of complex numbers and you'll have to split into components. Polar form looks more promising, but as @WillJagy says, start with the simple case to see what works.
$endgroup$
– orion
Jan 20 at 1:14
$begingroup$
The case $B=C=0$ is quite straightforward actually. I have solved it, and either of the mentioned ansatzs work. However even with just B non zero, the trouble is the same.
$endgroup$
– Mani Jha
Jan 20 at 1:19
$begingroup$
@Mani , then edit that into the question.
$endgroup$
– Will Jagy
Jan 20 at 1:20
$begingroup$
@ManiJha I did $B=C=0,$ then $A=a = 1/2,$ so $dot{z}=i bar{z}$ With $z(0) = 1,$ I get $z = cosh t + i sinh t,$ which works. With $z(0) = i,$ I get $z = sinh t + i cosh t.$
$endgroup$
– Will Jagy
Jan 20 at 1:44
|
show 1 more comment
$begingroup$
I think you should start with $B=C=0,$ write $z=x+iy$ with real smooth functions $x,y,$ and see what sort of ODE system you get, most likely rotation. Then allow nonzero $B$ in that setting; I think there should be some possible dynamical systems conclusions, but closed form seems optimistic.
$endgroup$
– Will Jagy
Jan 20 at 1:09
$begingroup$
Consider if on the left, you had $doverline{z}/dt$ instead of $dz/dt$, you could just separate variables and integrate. But with conjugation, you lose the comfort of complex numbers and you'll have to split into components. Polar form looks more promising, but as @WillJagy says, start with the simple case to see what works.
$endgroup$
– orion
Jan 20 at 1:14
$begingroup$
The case $B=C=0$ is quite straightforward actually. I have solved it, and either of the mentioned ansatzs work. However even with just B non zero, the trouble is the same.
$endgroup$
– Mani Jha
Jan 20 at 1:19
$begingroup$
@Mani , then edit that into the question.
$endgroup$
– Will Jagy
Jan 20 at 1:20
$begingroup$
@ManiJha I did $B=C=0,$ then $A=a = 1/2,$ so $dot{z}=i bar{z}$ With $z(0) = 1,$ I get $z = cosh t + i sinh t,$ which works. With $z(0) = i,$ I get $z = sinh t + i cosh t.$
$endgroup$
– Will Jagy
Jan 20 at 1:44
$begingroup$
I think you should start with $B=C=0,$ write $z=x+iy$ with real smooth functions $x,y,$ and see what sort of ODE system you get, most likely rotation. Then allow nonzero $B$ in that setting; I think there should be some possible dynamical systems conclusions, but closed form seems optimistic.
$endgroup$
– Will Jagy
Jan 20 at 1:09
$begingroup$
I think you should start with $B=C=0,$ write $z=x+iy$ with real smooth functions $x,y,$ and see what sort of ODE system you get, most likely rotation. Then allow nonzero $B$ in that setting; I think there should be some possible dynamical systems conclusions, but closed form seems optimistic.
$endgroup$
– Will Jagy
Jan 20 at 1:09
$begingroup$
Consider if on the left, you had $doverline{z}/dt$ instead of $dz/dt$, you could just separate variables and integrate. But with conjugation, you lose the comfort of complex numbers and you'll have to split into components. Polar form looks more promising, but as @WillJagy says, start with the simple case to see what works.
$endgroup$
– orion
Jan 20 at 1:14
$begingroup$
Consider if on the left, you had $doverline{z}/dt$ instead of $dz/dt$, you could just separate variables and integrate. But with conjugation, you lose the comfort of complex numbers and you'll have to split into components. Polar form looks more promising, but as @WillJagy says, start with the simple case to see what works.
$endgroup$
– orion
Jan 20 at 1:14
$begingroup$
The case $B=C=0$ is quite straightforward actually. I have solved it, and either of the mentioned ansatzs work. However even with just B non zero, the trouble is the same.
$endgroup$
– Mani Jha
Jan 20 at 1:19
$begingroup$
The case $B=C=0$ is quite straightforward actually. I have solved it, and either of the mentioned ansatzs work. However even with just B non zero, the trouble is the same.
$endgroup$
– Mani Jha
Jan 20 at 1:19
$begingroup$
@Mani , then edit that into the question.
$endgroup$
– Will Jagy
Jan 20 at 1:20
$begingroup$
@Mani , then edit that into the question.
$endgroup$
– Will Jagy
Jan 20 at 1:20
$begingroup$
@ManiJha I did $B=C=0,$ then $A=a = 1/2,$ so $dot{z}=i bar{z}$ With $z(0) = 1,$ I get $z = cosh t + i sinh t,$ which works. With $z(0) = i,$ I get $z = sinh t + i cosh t.$
$endgroup$
– Will Jagy
Jan 20 at 1:44
$begingroup$
@ManiJha I did $B=C=0,$ then $A=a = 1/2,$ so $dot{z}=i bar{z}$ With $z(0) = 1,$ I get $z = cosh t + i sinh t,$ which works. With $z(0) = i,$ I get $z = sinh t + i cosh t.$
$endgroup$
– Will Jagy
Jan 20 at 1:44
|
show 1 more comment
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$begingroup$
I think you should start with $B=C=0,$ write $z=x+iy$ with real smooth functions $x,y,$ and see what sort of ODE system you get, most likely rotation. Then allow nonzero $B$ in that setting; I think there should be some possible dynamical systems conclusions, but closed form seems optimistic.
$endgroup$
– Will Jagy
Jan 20 at 1:09
$begingroup$
Consider if on the left, you had $doverline{z}/dt$ instead of $dz/dt$, you could just separate variables and integrate. But with conjugation, you lose the comfort of complex numbers and you'll have to split into components. Polar form looks more promising, but as @WillJagy says, start with the simple case to see what works.
$endgroup$
– orion
Jan 20 at 1:14
$begingroup$
The case $B=C=0$ is quite straightforward actually. I have solved it, and either of the mentioned ansatzs work. However even with just B non zero, the trouble is the same.
$endgroup$
– Mani Jha
Jan 20 at 1:19
$begingroup$
@Mani , then edit that into the question.
$endgroup$
– Will Jagy
Jan 20 at 1:20
$begingroup$
@ManiJha I did $B=C=0,$ then $A=a = 1/2,$ so $dot{z}=i bar{z}$ With $z(0) = 1,$ I get $z = cosh t + i sinh t,$ which works. With $z(0) = i,$ I get $z = sinh t + i cosh t.$
$endgroup$
– Will Jagy
Jan 20 at 1:44