Mixing
Chapter 2 Corollary 2.7 of Colding and Minicozzi's minimal surfaces book
I have a question about the last step in the proof of corollary 2.7 in the chapter 2 of Colding and Minicozzi's book A Course in Minimal Surfaces.
Corollary 2.7 If $Sigma^2subsetmathbb{R}^3$ is immersed and minimal, $B_{r_0}^{Sigma}subset Sigma^2$ is an intrinsic disk of radius $r_0$, and $B_{r_0}^{Sigma}cappartialSigma=emptyset$, then
$$something=2(Area(B_{r_0}^{Sigma})-pi r_0^2)le Length(partial B_{r_0}^{Sigma})-2pi r_0^2.$$
The proof of this last inequality is first to notice that $$frac{d^2}{dt^2}Length(partial B_t^{Sigma})ge 0$$ by a previous lemma, therefore
$$tfrac{d}{dt}Length(partial B_t^{Sigma})ge Length(partial B_t^{Sigma})$$
therefore
$$frac{d}{dt}(Length(partial B_t^{Sigma})/t)ge 0.$$
The authors say that the inequality that we want now follows easily. However, I cannot see why. Intuitively the length of $partial B_t^{Sigma}$ should be $c(t)pi t$ and we now know $c(t)$ is increasing.
I guess then we should use the coarea formula somehow to finish it, but I just cannot write it out explicitly. Any help will be appreciated. Thanks in advance!
differential-geometry minimal-surfaces
asked Mar 21 '18 at 10:24
user392347
255
add a comment |
I have a question about the last step in the proof of corollary 2.7 in the chapter 2 of Colding and Minicozzi's book A Course in Minimal Surfaces.
Corollary 2.7 If $Sigma^2subsetmathbb{R}^3$ is immersed and minimal, $B_{r_0}^{Sigma}subset Sigma^2$ is an intrinsic disk of radius $r_0$, and $B_{r_0}^{Sigma}cappartialSigma=emptyset$, then
$$something=2(Area(B_{r_0}^{Sigma})-pi r_0^2)le Length(partial B_{r_0}^{Sigma})-2pi r_0^2.$$
The proof of this last inequality is first to notice that $$frac{d^2}{dt^2}Length(partial B_t^{Sigma})ge 0$$ by a previous lemma, therefore
$$tfrac{d}{dt}Length(partial B_t^{Sigma})ge Length(partial B_t^{Sigma})$$
therefore
$$frac{d}{dt}(Length(partial B_t^{Sigma})/t)ge 0.$$
The authors say that the inequality that we want now follows easily. However, I cannot see why. Intuitively the length of $partial B_t^{Sigma}$ should be $c(t)pi t$ and we now know $c(t)$ is increasing.
I guess then we should use the coarea formula somehow to finish it, but I just cannot write it out explicitly. Any help will be appreciated. Thanks in advance!
differential-geometry minimal-surfaces
asked Mar 21 '18 at 10:24
user392347
255
add a comment |
I have a question about the last step in the proof of corollary 2.7 in the chapter 2 of Colding and Minicozzi's book A Course in Minimal Surfaces.
Corollary 2.7 If $Sigma^2subsetmathbb{R}^3$ is immersed and minimal, $B_{r_0}^{Sigma}subset Sigma^2$ is an intrinsic disk of radius $r_0$, and $B_{r_0}^{Sigma}cappartialSigma=emptyset$, then
$$something=2(Area(B_{r_0}^{Sigma})-pi r_0^2)le Length(partial B_{r_0}^{Sigma})-2pi r_0^2.$$
The proof of this last inequality is first to notice that $$frac{d^2}{dt^2}Length(partial B_t^{Sigma})ge 0$$ by a previous lemma, therefore
$$tfrac{d}{dt}Length(partial B_t^{Sigma})ge Length(partial B_t^{Sigma})$$
therefore
$$frac{d}{dt}(Length(partial B_t^{Sigma})/t)ge 0.$$
The authors say that the inequality that we want now follows easily. However, I cannot see why. Intuitively the length of $partial B_t^{Sigma}$ should be $c(t)pi t$ and we now know $c(t)$ is increasing.
I guess then we should use the coarea formula somehow to finish it, but I just cannot write it out explicitly. Any help will be appreciated. Thanks in advance!
differential-geometry minimal-surfaces
asked Mar 21 '18 at 10:24
user392347
255
I have a question about the last step in the proof of corollary 2.7 in the chapter 2 of Colding and Minicozzi's book A Course in Minimal Surfaces.
Corollary 2.7 If $Sigma^2subsetmathbb{R}^3$ is immersed and minimal, $B_{r_0}^{Sigma}subset Sigma^2$ is an intrinsic disk of radius $r_0$, and $B_{r_0}^{Sigma}cappartialSigma=emptyset$, then
$$something=2(Area(B_{r_0}^{Sigma})-pi r_0^2)le Length(partial B_{r_0}^{Sigma})-2pi r_0^2.$$
The proof of this last inequality is first to notice that $$frac{d^2}{dt^2}Length(partial B_t^{Sigma})ge 0$$ by a previous lemma, therefore
$$tfrac{d}{dt}Length(partial B_t^{Sigma})ge Length(partial B_t^{Sigma})$$
therefore
$$frac{d}{dt}(Length(partial B_t^{Sigma})/t)ge 0.$$
The authors say that the inequality that we want now follows easily. However, I cannot see why. Intuitively the length of $partial B_t^{Sigma}$ should be $c(t)pi t$ and we now know $c(t)$ is increasing.
I guess then we should use the coarea formula somehow to finish it, but I just cannot write it out explicitly. Any help will be appreciated. Thanks in advance!
differential-geometry minimal-surfaces
differential-geometry minimal-surfaces
asked Mar 21 '18 at 10:24
user392347
255
asked Mar 21 '18 at 10:24
user392347
255
asked Mar 21 '18 at 10:24
user392347
255
asked Mar 21 '18 at 10:24
user392347
255
asked Mar 21 '18 at 10:24
user392347
255
255
add a comment |
add a comment |
1 Answer
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I integrated the equation:
$int_0^{r_0} tfrac{d}{dt}Length(partial B_t^Sigma)geq int_0^{r_0}Length(partial B_t^Sigma)$
Observed that right hand side is equal to $Area(B_{r_0}^Sigma)$
By integrating by parts left hand side we get:
$r_0Length(partial B_t^Sigma)- Area(B_{r_0}^Sigma)geq Area(B_{r_0}^Sigma)$
and hence we conclude by subtracting from both equations $2pi r_0^2$
answered Oct 24 '18 at 15:27
F.T.
12
add a comment |
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1 Answer
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I integrated the equation:
$int_0^{r_0} tfrac{d}{dt}Length(partial B_t^Sigma)geq int_0^{r_0}Length(partial B_t^Sigma)$
Observed that right hand side is equal to $Area(B_{r_0}^Sigma)$
By integrating by parts left hand side we get:
$r_0Length(partial B_t^Sigma)- Area(B_{r_0}^Sigma)geq Area(B_{r_0}^Sigma)$
and hence we conclude by subtracting from both equations $2pi r_0^2$
answered Oct 24 '18 at 15:27
F.T.
12
add a comment |
I integrated the equation:
$int_0^{r_0} tfrac{d}{dt}Length(partial B_t^Sigma)geq int_0^{r_0}Length(partial B_t^Sigma)$
Observed that right hand side is equal to $Area(B_{r_0}^Sigma)$
By integrating by parts left hand side we get:
$r_0Length(partial B_t^Sigma)- Area(B_{r_0}^Sigma)geq Area(B_{r_0}^Sigma)$
and hence we conclude by subtracting from both equations $2pi r_0^2$
answered Oct 24 '18 at 15:27
F.T.
12
add a comment |
I integrated the equation:
$int_0^{r_0} tfrac{d}{dt}Length(partial B_t^Sigma)geq int_0^{r_0}Length(partial B_t^Sigma)$
Observed that right hand side is equal to $Area(B_{r_0}^Sigma)$
By integrating by parts left hand side we get:
$r_0Length(partial B_t^Sigma)- Area(B_{r_0}^Sigma)geq Area(B_{r_0}^Sigma)$
and hence we conclude by subtracting from both equations $2pi r_0^2$
answered Oct 24 '18 at 15:27
F.T.
12
I integrated the equation:
$int_0^{r_0} tfrac{d}{dt}Length(partial B_t^Sigma)geq int_0^{r_0}Length(partial B_t^Sigma)$
Observed that right hand side is equal to $Area(B_{r_0}^Sigma)$
By integrating by parts left hand side we get:
$r_0Length(partial B_t^Sigma)- Area(B_{r_0}^Sigma)geq Area(B_{r_0}^Sigma)$
and hence we conclude by subtracting from both equations $2pi r_0^2$
answered Oct 24 '18 at 15:27
F.T.
12
edited Nov 21 '18 at 8:46
answered Oct 24 '18 at 15:27
F.T.
12
answered Oct 24 '18 at 15:27
F.T.
12
answered Oct 24 '18 at 15:27
F.T.
12
12
add a comment |
add a comment |
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