Mixing
Find the extremals of the functional $intsqrt{x^2+y^2}sqrt{1+(y'(x))^2}dx$?
0
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I want to use the polar coordinates $x=rcostheta$ and $y=rsintheta$. After transformation, I get $$int rsqrt{r^2+(r')^2}dtheta.$$
Then, I derived the Euler-Lagrange equation $$2r^2-rr''+3(r')^2=0.$$
I get the solutions to this ODE, but it doesn't match the solutions.
$$theta+C_1=frac{1}{2}arctan(frac{sqrt{r^4-C^2}}{C}), text{ where } C,C_1inmathbb{R}.$$
Solution is $x^2cos(alpha)+2xysin(alpha)-y^2cos(alpha)=beta$, where $alpha$, $beta$ are constants.
polar-coordinates calculus-of-variations
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
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add a comment |
0
$begingroup$
I want to use the polar coordinates $x=rcostheta$ and $y=rsintheta$. After transformation, I get $$int rsqrt{r^2+(r')^2}dtheta.$$
Then, I derived the Euler-Lagrange equation $$2r^2-rr''+3(r')^2=0.$$
I get the solutions to this ODE, but it doesn't match the solutions.
$$theta+C_1=frac{1}{2}arctan(frac{sqrt{r^4-C^2}}{C}), text{ where } C,C_1inmathbb{R}.$$
Solution is $x^2cos(alpha)+2xysin(alpha)-y^2cos(alpha)=beta$, where $alpha$, $beta$ are constants.
polar-coordinates calculus-of-variations
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
$endgroup$
$begingroup$
Well, as the next step you should go back to the rectangular coordinates to see if you can get the proposed solution. I admit to have had some problems with this in the past and I wasn't able to resolve it. As a small contribution, I've tried the alternative way of using [Beltrami's identity][1] as the integrand does not depend on $theta$. I got a first order differential equation ($r'=rsqrt{r^3-c^2}/c^2$) with a solution similar to yours. Maybe someone can be more helpful. [1]: en.wikipedia.org/wiki/Beltrami_identity
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– Marco81
Jan 25 at 13:07
$begingroup$
Shouldn't $int rsqrt{r^2+(r')^2}dtheta$ be $int r(t) sqrt{r'(t)^2+r(t)^2 theta'(t)^2} dt$?
$endgroup$
– md2perpe
Jan 25 at 15:32
$begingroup$
@md2perpe. Not really. If you consider that $dx=costheta dr-sintheta dtheta$ and $dy=drsintheta+rcostheta dtheta$, the $dx^2+dy^2=dr^2+r^2dtheta^2$. In the end $dxsqrt{1+(dy/dx)^2}=dthetasqrt{r^2+(dr/dtheta)^2}$. Obviously $y$ is a function of $x$, $r$ is a function of $theta$, $dy/dx=y'(x)$ and $dr/dtheta=r'(theta)$
$endgroup$
– Marco81
Jan 25 at 16:17
$begingroup$
@Marco81. Yes, I saw later that one gets that integral when parameterizing w.r.t. $theta$.
$endgroup$
– md2perpe
Jan 25 at 17:31
add a comment |
0
0
0
1
$begingroup$
I want to use the polar coordinates $x=rcostheta$ and $y=rsintheta$. After transformation, I get $$int rsqrt{r^2+(r')^2}dtheta.$$
Then, I derived the Euler-Lagrange equation $$2r^2-rr''+3(r')^2=0.$$
I get the solutions to this ODE, but it doesn't match the solutions.
$$theta+C_1=frac{1}{2}arctan(frac{sqrt{r^4-C^2}}{C}), text{ where } C,C_1inmathbb{R}.$$
Solution is $x^2cos(alpha)+2xysin(alpha)-y^2cos(alpha)=beta$, where $alpha$, $beta$ are constants.
polar-coordinates calculus-of-variations
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
$endgroup$
I want to use the polar coordinates $x=rcostheta$ and $y=rsintheta$. After transformation, I get $$int rsqrt{r^2+(r')^2}dtheta.$$
Then, I derived the Euler-Lagrange equation $$2r^2-rr''+3(r')^2=0.$$
I get the solutions to this ODE, but it doesn't match the solutions.
$$theta+C_1=frac{1}{2}arctan(frac{sqrt{r^4-C^2}}{C}), text{ where } C,C_1inmathbb{R}.$$
Solution is $x^2cos(alpha)+2xysin(alpha)-y^2cos(alpha)=beta$, where $alpha$, $beta$ are constants.
polar-coordinates calculus-of-variations
polar-coordinates calculus-of-variations
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
asked Jan 23 at 22:18
Yuchen DongYuchen Dong
614
614
$begingroup$
Well, as the next step you should go back to the rectangular coordinates to see if you can get the proposed solution. I admit to have had some problems with this in the past and I wasn't able to resolve it. As a small contribution, I've tried the alternative way of using [Beltrami's identity][1] as the integrand does not depend on $theta$. I got a first order differential equation ($r'=rsqrt{r^3-c^2}/c^2$) with a solution similar to yours. Maybe someone can be more helpful. [1]: en.wikipedia.org/wiki/Beltrami_identity
$endgroup$
– Marco81
Jan 25 at 13:07
$begingroup$
Shouldn't $int rsqrt{r^2+(r')^2}dtheta$ be $int r(t) sqrt{r'(t)^2+r(t)^2 theta'(t)^2} dt$?
$endgroup$
– md2perpe
Jan 25 at 15:32
$begingroup$
@md2perpe. Not really. If you consider that $dx=costheta dr-sintheta dtheta$ and $dy=drsintheta+rcostheta dtheta$, the $dx^2+dy^2=dr^2+r^2dtheta^2$. In the end $dxsqrt{1+(dy/dx)^2}=dthetasqrt{r^2+(dr/dtheta)^2}$. Obviously $y$ is a function of $x$, $r$ is a function of $theta$, $dy/dx=y'(x)$ and $dr/dtheta=r'(theta)$
$endgroup$
– Marco81
Jan 25 at 16:17
$begingroup$
@Marco81. Yes, I saw later that one gets that integral when parameterizing w.r.t. $theta$.
$endgroup$
– md2perpe
Jan 25 at 17:31
add a comment |
$begingroup$
Well, as the next step you should go back to the rectangular coordinates to see if you can get the proposed solution. I admit to have had some problems with this in the past and I wasn't able to resolve it. As a small contribution, I've tried the alternative way of using [Beltrami's identity][1] as the integrand does not depend on $theta$. I got a first order differential equation ($r'=rsqrt{r^3-c^2}/c^2$) with a solution similar to yours. Maybe someone can be more helpful. [1]: en.wikipedia.org/wiki/Beltrami_identity
$endgroup$
– Marco81
Jan 25 at 13:07
$begingroup$
Shouldn't $int rsqrt{r^2+(r')^2}dtheta$ be $int r(t) sqrt{r'(t)^2+r(t)^2 theta'(t)^2} dt$?
$endgroup$
– md2perpe
Jan 25 at 15:32
$begingroup$
@md2perpe. Not really. If you consider that $dx=costheta dr-sintheta dtheta$ and $dy=drsintheta+rcostheta dtheta$, the $dx^2+dy^2=dr^2+r^2dtheta^2$. In the end $dxsqrt{1+(dy/dx)^2}=dthetasqrt{r^2+(dr/dtheta)^2}$. Obviously $y$ is a function of $x$, $r$ is a function of $theta$, $dy/dx=y'(x)$ and $dr/dtheta=r'(theta)$
$endgroup$
– Marco81
Jan 25 at 16:17
$begingroup$
@Marco81. Yes, I saw later that one gets that integral when parameterizing w.r.t. $theta$.
$endgroup$
– md2perpe
Jan 25 at 17:31
$begingroup$
Well, as the next step you should go back to the rectangular coordinates to see if you can get the proposed solution. I admit to have had some problems with this in the past and I wasn't able to resolve it. As a small contribution, I've tried the alternative way of using [Beltrami's identity][1] as the integrand does not depend on $theta$. I got a first order differential equation ($r'=rsqrt{r^3-c^2}/c^2$) with a solution similar to yours. Maybe someone can be more helpful. [1]: en.wikipedia.org/wiki/Beltrami_identity
$endgroup$
– Marco81
Jan 25 at 13:07
$begingroup$
Well, as the next step you should go back to the rectangular coordinates to see if you can get the proposed solution. I admit to have had some problems with this in the past and I wasn't able to resolve it. As a small contribution, I've tried the alternative way of using [Beltrami's identity][1] as the integrand does not depend on $theta$. I got a first order differential equation ($r'=rsqrt{r^3-c^2}/c^2$) with a solution similar to yours. Maybe someone can be more helpful. [1]: en.wikipedia.org/wiki/Beltrami_identity
$endgroup$
– Marco81
Jan 25 at 13:07
$begingroup$
Shouldn't $int rsqrt{r^2+(r')^2}dtheta$ be $int r(t) sqrt{r'(t)^2+r(t)^2 theta'(t)^2} dt$?
$endgroup$
– md2perpe
Jan 25 at 15:32
$begingroup$
Shouldn't $int rsqrt{r^2+(r')^2}dtheta$ be $int r(t) sqrt{r'(t)^2+r(t)^2 theta'(t)^2} dt$?
$endgroup$
– md2perpe
Jan 25 at 15:32
$begingroup$
@md2perpe. Not really. If you consider that $dx=costheta dr-sintheta dtheta$ and $dy=drsintheta+rcostheta dtheta$, the $dx^2+dy^2=dr^2+r^2dtheta^2$. In the end $dxsqrt{1+(dy/dx)^2}=dthetasqrt{r^2+(dr/dtheta)^2}$. Obviously $y$ is a function of $x$, $r$ is a function of $theta$, $dy/dx=y'(x)$ and $dr/dtheta=r'(theta)$
$endgroup$
– Marco81
Jan 25 at 16:17
$begingroup$
@md2perpe. Not really. If you consider that $dx=costheta dr-sintheta dtheta$ and $dy=drsintheta+rcostheta dtheta$, the $dx^2+dy^2=dr^2+r^2dtheta^2$. In the end $dxsqrt{1+(dy/dx)^2}=dthetasqrt{r^2+(dr/dtheta)^2}$. Obviously $y$ is a function of $x$, $r$ is a function of $theta$, $dy/dx=y'(x)$ and $dr/dtheta=r'(theta)$
$endgroup$
– Marco81
Jan 25 at 16:17
$begingroup$
@Marco81. Yes, I saw later that one gets that integral when parameterizing w.r.t. $theta$.
$endgroup$
– md2perpe
Jan 25 at 17:31
$begingroup$
@Marco81. Yes, I saw later that one gets that integral when parameterizing w.r.t. $theta$.
$endgroup$
– md2perpe
Jan 25 at 17:31
add a comment |
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$begingroup$
Well, as the next step you should go back to the rectangular coordinates to see if you can get the proposed solution. I admit to have had some problems with this in the past and I wasn't able to resolve it. As a small contribution, I've tried the alternative way of using [Beltrami's identity][1] as the integrand does not depend on $theta$. I got a first order differential equation ($r'=rsqrt{r^3-c^2}/c^2$) with a solution similar to yours. Maybe someone can be more helpful. [1]: en.wikipedia.org/wiki/Beltrami_identity
$endgroup$
– Marco81
Jan 25 at 13:07
$begingroup$
Shouldn't $int rsqrt{r^2+(r')^2}dtheta$ be $int r(t) sqrt{r'(t)^2+r(t)^2 theta'(t)^2} dt$?
$endgroup$
– md2perpe
Jan 25 at 15:32
$begingroup$
@md2perpe. Not really. If you consider that $dx=costheta dr-sintheta dtheta$ and $dy=drsintheta+rcostheta dtheta$, the $dx^2+dy^2=dr^2+r^2dtheta^2$. In the end $dxsqrt{1+(dy/dx)^2}=dthetasqrt{r^2+(dr/dtheta)^2}$. Obviously $y$ is a function of $x$, $r$ is a function of $theta$, $dy/dx=y'(x)$ and $dr/dtheta=r'(theta)$
$endgroup$
– Marco81
Jan 25 at 16:17
$begingroup$
@Marco81. Yes, I saw later that one gets that integral when parameterizing w.r.t. $theta$.
$endgroup$
– md2perpe
Jan 25 at 17:31