Method of characteristics for the Beltrami equation when $mu$ is real analytic
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I am reading a proof of the measurable mapping theorem which shows the existence of a solution to the Beltrami equation in the simple case when $mu in L^{infty}(mathbb{C})$ is real analytic (due to Gauss)
begin{equation}
(1-mu(x,y))frac{partial{f}}{partial{x}}+ i(1+mu(x,y)))frac{partial{f}}{partial{y}}=0
tag{4.5.8}
end{equation}
From what I understand, we are looking for a curve $(x(t), y(t))$ which satisfies
$$frac{dx}{dt}=1-mu(x,y)$$ and
$$frac{dy}{dt}=i(1+mu(x,y))$$.
The author then points out that such a curve would satisfy
$$frac{dy}{dx}= ifrac{1+u}{1-u} tag{*}$$
I wanted to understand how one goes from the system of ODE's above to (*). Does $mu$ being real analytic imply that there is a solution to the system above
Furthermore, if there is such a solution, the inverse function theorem tells us that the solution is a local homeomorphism, can we say anything about uniqueness just from the ODE theory?
pde teichmueller-theory
asked Nov 21 at 17:03
user135520
911718
add a comment |
up vote
1
down vote
favorite
I am reading a proof of the measurable mapping theorem which shows the existence of a solution to the Beltrami equation in the simple case when $mu in L^{infty}(mathbb{C})$ is real analytic (due to Gauss)
begin{equation}
(1-mu(x,y))frac{partial{f}}{partial{x}}+ i(1+mu(x,y)))frac{partial{f}}{partial{y}}=0
tag{4.5.8}
end{equation}
From what I understand, we are looking for a curve $(x(t), y(t))$ which satisfies
$$frac{dx}{dt}=1-mu(x,y)$$ and
$$frac{dy}{dt}=i(1+mu(x,y))$$.
The author then points out that such a curve would satisfy
$$frac{dy}{dx}= ifrac{1+u}{1-u} tag{*}$$
I wanted to understand how one goes from the system of ODE's above to (*). Does $mu$ being real analytic imply that there is a solution to the system above
Furthermore, if there is such a solution, the inverse function theorem tells us that the solution is a local homeomorphism, can we say anything about uniqueness just from the ODE theory?
pde teichmueller-theory
asked Nov 21 at 17:03
user135520
911718
I suspect your third equation should read $dy/dt = i(1 + mu(x, y))$, not $dx/dt = i(1 + mu(x, y))$. Could I be right? Cheers!
– Robert Lewis
Nov 21 at 17:32
1
Yes! Thank you!
– user135520
Nov 21 at 17:57
Glad to help out, my friend!
– Robert Lewis
Nov 21 at 18:02
1
Are you asking how $$x' = 1 - mu, quad y' = i(1+mu) implies frac{y'}{x'} = frac{dy}{dt} cdot frac{dt}{dx} = frac{dy}{dx} = frac{i(1+mu)}{1-mu}$$?
– Mattos
Nov 22 at 1:55
Yes, I realized the chain rule here, after posting the question. So I think I understand that such a $y$ exists because $i(1+mu)/(1-mu)$ is something that can be integrated.
– user135520
Nov 22 at 2:05
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading a proof of the measurable mapping theorem which shows the existence of a solution to the Beltrami equation in the simple case when $mu in L^{infty}(mathbb{C})$ is real analytic (due to Gauss)
begin{equation}
(1-mu(x,y))frac{partial{f}}{partial{x}}+ i(1+mu(x,y)))frac{partial{f}}{partial{y}}=0
tag{4.5.8}
end{equation}
From what I understand, we are looking for a curve $(x(t), y(t))$ which satisfies
$$frac{dx}{dt}=1-mu(x,y)$$ and
$$frac{dy}{dt}=i(1+mu(x,y))$$.
The author then points out that such a curve would satisfy
$$frac{dy}{dx}= ifrac{1+u}{1-u} tag{*}$$
I wanted to understand how one goes from the system of ODE's above to (*). Does $mu$ being real analytic imply that there is a solution to the system above
Furthermore, if there is such a solution, the inverse function theorem tells us that the solution is a local homeomorphism, can we say anything about uniqueness just from the ODE theory?
pde teichmueller-theory
asked Nov 21 at 17:03
user135520
911718
I am reading a proof of the measurable mapping theorem which shows the existence of a solution to the Beltrami equation in the simple case when $mu in L^{infty}(mathbb{C})$ is real analytic (due to Gauss)
begin{equation}
(1-mu(x,y))frac{partial{f}}{partial{x}}+ i(1+mu(x,y)))frac{partial{f}}{partial{y}}=0
tag{4.5.8}
end{equation}
From what I understand, we are looking for a curve $(x(t), y(t))$ which satisfies
$$frac{dx}{dt}=1-mu(x,y)$$ and
$$frac{dy}{dt}=i(1+mu(x,y))$$.
The author then points out that such a curve would satisfy
$$frac{dy}{dx}= ifrac{1+u}{1-u} tag{*}$$
I wanted to understand how one goes from the system of ODE's above to (*). Does $mu$ being real analytic imply that there is a solution to the system above
Furthermore, if there is such a solution, the inverse function theorem tells us that the solution is a local homeomorphism, can we say anything about uniqueness just from the ODE theory?
pde teichmueller-theory
pde teichmueller-theory
asked Nov 21 at 17:03
user135520
911718
asked Nov 21 at 17:03
user135520
911718
edited Nov 21 at 17:57
asked Nov 21 at 17:03
user135520
911718
asked Nov 21 at 17:03
user135520
911718
asked Nov 21 at 17:03
user135520
911718
911718
I suspect your third equation should read $dy/dt = i(1 + mu(x, y))$, not $dx/dt = i(1 + mu(x, y))$. Could I be right? Cheers!
– Robert Lewis
Nov 21 at 17:32
1
Yes! Thank you!
– user135520
Nov 21 at 17:57
Glad to help out, my friend!
– Robert Lewis
Nov 21 at 18:02
1
Are you asking how $$x' = 1 - mu, quad y' = i(1+mu) implies frac{y'}{x'} = frac{dy}{dt} cdot frac{dt}{dx} = frac{dy}{dx} = frac{i(1+mu)}{1-mu}$$?
– Mattos
Nov 22 at 1:55
Yes, I realized the chain rule here, after posting the question. So I think I understand that such a $y$ exists because $i(1+mu)/(1-mu)$ is something that can be integrated.
– user135520
Nov 22 at 2:05
add a comment |
I suspect your third equation should read $dy/dt = i(1 + mu(x, y))$, not $dx/dt = i(1 + mu(x, y))$. Could I be right? Cheers!
– Robert Lewis
Nov 21 at 17:32
1
Yes! Thank you!
– user135520
Nov 21 at 17:57
Glad to help out, my friend!
– Robert Lewis
Nov 21 at 18:02
1
Are you asking how $$x' = 1 - mu, quad y' = i(1+mu) implies frac{y'}{x'} = frac{dy}{dt} cdot frac{dt}{dx} = frac{dy}{dx} = frac{i(1+mu)}{1-mu}$$?
– Mattos
Nov 22 at 1:55
Yes, I realized the chain rule here, after posting the question. So I think I understand that such a $y$ exists because $i(1+mu)/(1-mu)$ is something that can be integrated.
– user135520
Nov 22 at 2:05
I suspect your third equation should read $dy/dt = i(1 + mu(x, y))$, not $dx/dt = i(1 + mu(x, y))$. Could I be right? Cheers!
– Robert Lewis
Nov 21 at 17:32
I suspect your third equation should read $dy/dt = i(1 + mu(x, y))$, not $dx/dt = i(1 + mu(x, y))$. Could I be right? Cheers!
– Robert Lewis
Nov 21 at 17:32
1
1
Yes! Thank you!
– user135520
Nov 21 at 17:57
Yes! Thank you!
– user135520
Nov 21 at 17:57
Glad to help out, my friend!
– Robert Lewis
Nov 21 at 18:02
Glad to help out, my friend!
– Robert Lewis
Nov 21 at 18:02
1
1
Are you asking how $$x' = 1 - mu, quad y' = i(1+mu) implies frac{y'}{x'} = frac{dy}{dt} cdot frac{dt}{dx} = frac{dy}{dx} = frac{i(1+mu)}{1-mu}$$?
– Mattos
Nov 22 at 1:55
Are you asking how $$x' = 1 - mu, quad y' = i(1+mu) implies frac{y'}{x'} = frac{dy}{dt} cdot frac{dt}{dx} = frac{dy}{dx} = frac{i(1+mu)}{1-mu}$$?
– Mattos
Nov 22 at 1:55
Yes, I realized the chain rule here, after posting the question. So I think I understand that such a $y$ exists because $i(1+mu)/(1-mu)$ is something that can be integrated.
– user135520
Nov 22 at 2:05
Yes, I realized the chain rule here, after posting the question. So I think I understand that such a $y$ exists because $i(1+mu)/(1-mu)$ is something that can be integrated.
– user135520
Nov 22 at 2:05
add a comment |
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I suspect your third equation should read $dy/dt = i(1 + mu(x, y))$, not $dx/dt = i(1 + mu(x, y))$. Could I be right? Cheers!
– Robert Lewis
Nov 21 at 17:32
1
Yes! Thank you!
– user135520
Nov 21 at 17:57
Glad to help out, my friend!
– Robert Lewis
Nov 21 at 18:02
1
Are you asking how $$x' = 1 - mu, quad y' = i(1+mu) implies frac{y'}{x'} = frac{dy}{dt} cdot frac{dt}{dx} = frac{dy}{dx} = frac{i(1+mu)}{1-mu}$$?
– Mattos
Nov 22 at 1:55
Yes, I realized the chain rule here, after posting the question. So I think I understand that such a $y$ exists because $i(1+mu)/(1-mu)$ is something that can be integrated.
– user135520
Nov 22 at 2:05