A problem in variational calculus












0












$begingroup$


How do you maximize the quotient ||f||/||f'|| of euclidean norms if f is to be a function on [0,1] which vanishes on the boundary?



$||g||^2 = int_0^1g(x)^2textrm{d}x$



I guess $f$ needs to be continuously differentiable for it to make sense or something.






calculus-of-variations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you write down explicitly what the range space and what the norms are?
    $endgroup$
    – Kavi Rama Murthy
    Dec 22 '18 at 11:32










  • $begingroup$
    How about that?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:33
















0












$begingroup$


How do you maximize the quotient ||f||/||f'|| of euclidean norms if f is to be a function on [0,1] which vanishes on the boundary?



$||g||^2 = int_0^1g(x)^2textrm{d}x$



I guess $f$ needs to be continuously differentiable for it to make sense or something.






calculus-of-variations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you write down explicitly what the range space and what the norms are?
    $endgroup$
    – Kavi Rama Murthy
    Dec 22 '18 at 11:32










  • $begingroup$
    How about that?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:33














0












0








0





$begingroup$


How do you maximize the quotient ||f||/||f'|| of euclidean norms if f is to be a function on [0,1] which vanishes on the boundary?



$||g||^2 = int_0^1g(x)^2textrm{d}x$



I guess $f$ needs to be continuously differentiable for it to make sense or something.






calculus-of-variations







share|cite|improve this question











$endgroup$




How do you maximize the quotient ||f||/||f'|| of euclidean norms if f is to be a function on [0,1] which vanishes on the boundary?



$||g||^2 = int_0^1g(x)^2textrm{d}x$



I guess $f$ needs to be continuously differentiable for it to make sense or something.






calculus-of-variations




calculus-of-variations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 12:30







Joel Sjögren

















asked Dec 22 '18 at 10:55









Joel SjögrenJoel Sjögren

2008




2008












  • $begingroup$
    Can you write down explicitly what the range space and what the norms are?
    $endgroup$
    – Kavi Rama Murthy
    Dec 22 '18 at 11:32










  • $begingroup$
    How about that?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:33


















  • $begingroup$
    Can you write down explicitly what the range space and what the norms are?
    $endgroup$
    – Kavi Rama Murthy
    Dec 22 '18 at 11:32










  • $begingroup$
    How about that?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:33
















$begingroup$
Can you write down explicitly what the range space and what the norms are?
$endgroup$
– Kavi Rama Murthy
Dec 22 '18 at 11:32




$begingroup$
Can you write down explicitly what the range space and what the norms are?
$endgroup$
– Kavi Rama Murthy
Dec 22 '18 at 11:32












$begingroup$
How about that?
$endgroup$
– Joel Sjögren
Dec 22 '18 at 12:33




$begingroup$
How about that?
$endgroup$
– Joel Sjögren
Dec 22 '18 at 12:33










1 Answer
1






active

oldest

votes


















1












$begingroup$

Without loss of generality, assume $|g|=1$. Then
$$
begin{align}
0
&=deltaint_0^1g(x)^2,mathrm{d}x\
&=2int_0^1g(x),delta g(x),mathrm{d}xtag1
end{align}
$$
For each $delta g(x)$ that satisfies $(1)$, we want
$$
begin{align}
0
&=deltaint_0^1g'(x)^2,mathrm{d}x\
&=2int_0^1g'(x),delta g'(x),mathrm{d}x\
&=-2int_0^1g''(x),delta g(x),mathrm{d}xtag2
end{align}
$$
To satisfy $(2)$ for all $delta g(x)$ that satisfy $(1)$, we must have
$$
g''(x)=lambda g(x)tag3
$$
for some constant $lambda$.




Explanation of $boldsymbol{(3)}$



Let $lambda$ be so that
$$
int_0^1g(x)g''(x),mathrm{d}x=lambdaint_0^1g(x)^2,mathrm{d}x
$$
Then, since
$$
int_0^1g(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
our condition says that
$$
int_0^1g''(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
Subtract $lambda$ times the former from the latter to get
$$
int_0^1(g''(x)-lambda g(x))^2,mathrm{d}x=0
$$
which implies that $g''(x)=lambda g(x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What is the general rule by which you make the "To satisfy (2)...we must have..." step?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:52










  • $begingroup$
    @JoelSjögren: I was writing up the appendix when you commented.
    $endgroup$
    – robjohn♦
    Dec 22 '18 at 13:07










  • $begingroup$
    Where does the premise of the appendix come from?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:39










  • $begingroup$
    Oh ok it's a definition in terms of an arbitrary g.
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:41












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Without loss of generality, assume $|g|=1$. Then
$$
begin{align}
0
&=deltaint_0^1g(x)^2,mathrm{d}x\
&=2int_0^1g(x),delta g(x),mathrm{d}xtag1
end{align}
$$
For each $delta g(x)$ that satisfies $(1)$, we want
$$
begin{align}
0
&=deltaint_0^1g'(x)^2,mathrm{d}x\
&=2int_0^1g'(x),delta g'(x),mathrm{d}x\
&=-2int_0^1g''(x),delta g(x),mathrm{d}xtag2
end{align}
$$
To satisfy $(2)$ for all $delta g(x)$ that satisfy $(1)$, we must have
$$
g''(x)=lambda g(x)tag3
$$
for some constant $lambda$.




Explanation of $boldsymbol{(3)}$



Let $lambda$ be so that
$$
int_0^1g(x)g''(x),mathrm{d}x=lambdaint_0^1g(x)^2,mathrm{d}x
$$
Then, since
$$
int_0^1g(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
our condition says that
$$
int_0^1g''(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
Subtract $lambda$ times the former from the latter to get
$$
int_0^1(g''(x)-lambda g(x))^2,mathrm{d}x=0
$$
which implies that $g''(x)=lambda g(x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What is the general rule by which you make the "To satisfy (2)...we must have..." step?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:52










  • $begingroup$
    @JoelSjögren: I was writing up the appendix when you commented.
    $endgroup$
    – robjohn♦
    Dec 22 '18 at 13:07










  • $begingroup$
    Where does the premise of the appendix come from?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:39










  • $begingroup$
    Oh ok it's a definition in terms of an arbitrary g.
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:41
















1












$begingroup$

Without loss of generality, assume $|g|=1$. Then
$$
begin{align}
0
&=deltaint_0^1g(x)^2,mathrm{d}x\
&=2int_0^1g(x),delta g(x),mathrm{d}xtag1
end{align}
$$
For each $delta g(x)$ that satisfies $(1)$, we want
$$
begin{align}
0
&=deltaint_0^1g'(x)^2,mathrm{d}x\
&=2int_0^1g'(x),delta g'(x),mathrm{d}x\
&=-2int_0^1g''(x),delta g(x),mathrm{d}xtag2
end{align}
$$
To satisfy $(2)$ for all $delta g(x)$ that satisfy $(1)$, we must have
$$
g''(x)=lambda g(x)tag3
$$
for some constant $lambda$.




Explanation of $boldsymbol{(3)}$



Let $lambda$ be so that
$$
int_0^1g(x)g''(x),mathrm{d}x=lambdaint_0^1g(x)^2,mathrm{d}x
$$
Then, since
$$
int_0^1g(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
our condition says that
$$
int_0^1g''(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
Subtract $lambda$ times the former from the latter to get
$$
int_0^1(g''(x)-lambda g(x))^2,mathrm{d}x=0
$$
which implies that $g''(x)=lambda g(x)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What is the general rule by which you make the "To satisfy (2)...we must have..." step?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:52










  • $begingroup$
    @JoelSjögren: I was writing up the appendix when you commented.
    $endgroup$
    – robjohn♦
    Dec 22 '18 at 13:07










  • $begingroup$
    Where does the premise of the appendix come from?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:39










  • $begingroup$
    Oh ok it's a definition in terms of an arbitrary g.
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:41














1












1








1





$begingroup$

Without loss of generality, assume $|g|=1$. Then
$$
begin{align}
0
&=deltaint_0^1g(x)^2,mathrm{d}x\
&=2int_0^1g(x),delta g(x),mathrm{d}xtag1
end{align}
$$
For each $delta g(x)$ that satisfies $(1)$, we want
$$
begin{align}
0
&=deltaint_0^1g'(x)^2,mathrm{d}x\
&=2int_0^1g'(x),delta g'(x),mathrm{d}x\
&=-2int_0^1g''(x),delta g(x),mathrm{d}xtag2
end{align}
$$
To satisfy $(2)$ for all $delta g(x)$ that satisfy $(1)$, we must have
$$
g''(x)=lambda g(x)tag3
$$
for some constant $lambda$.




Explanation of $boldsymbol{(3)}$



Let $lambda$ be so that
$$
int_0^1g(x)g''(x),mathrm{d}x=lambdaint_0^1g(x)^2,mathrm{d}x
$$
Then, since
$$
int_0^1g(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
our condition says that
$$
int_0^1g''(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
Subtract $lambda$ times the former from the latter to get
$$
int_0^1(g''(x)-lambda g(x))^2,mathrm{d}x=0
$$
which implies that $g''(x)=lambda g(x)$.






share|cite|improve this answer











$endgroup$



Without loss of generality, assume $|g|=1$. Then
$$
begin{align}
0
&=deltaint_0^1g(x)^2,mathrm{d}x\
&=2int_0^1g(x),delta g(x),mathrm{d}xtag1
end{align}
$$
For each $delta g(x)$ that satisfies $(1)$, we want
$$
begin{align}
0
&=deltaint_0^1g'(x)^2,mathrm{d}x\
&=2int_0^1g'(x),delta g'(x),mathrm{d}x\
&=-2int_0^1g''(x),delta g(x),mathrm{d}xtag2
end{align}
$$
To satisfy $(2)$ for all $delta g(x)$ that satisfy $(1)$, we must have
$$
g''(x)=lambda g(x)tag3
$$
for some constant $lambda$.




Explanation of $boldsymbol{(3)}$



Let $lambda$ be so that
$$
int_0^1g(x)g''(x),mathrm{d}x=lambdaint_0^1g(x)^2,mathrm{d}x
$$
Then, since
$$
int_0^1g(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
our condition says that
$$
int_0^1g''(x)overbrace{(g''(x)-lambda g(x))}^{delta g(x)},mathrm{d}x=0
$$
Subtract $lambda$ times the former from the latter to get
$$
int_0^1(g''(x)-lambda g(x))^2,mathrm{d}x=0
$$
which implies that $g''(x)=lambda g(x)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 13:05

























answered Dec 22 '18 at 12:43









robjohn♦robjohn

271k27316643




271k27316643












  • $begingroup$
    What is the general rule by which you make the "To satisfy (2)...we must have..." step?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:52










  • $begingroup$
    @JoelSjögren: I was writing up the appendix when you commented.
    $endgroup$
    – robjohn♦
    Dec 22 '18 at 13:07










  • $begingroup$
    Where does the premise of the appendix come from?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:39










  • $begingroup$
    Oh ok it's a definition in terms of an arbitrary g.
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:41


















  • $begingroup$
    What is the general rule by which you make the "To satisfy (2)...we must have..." step?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 12:52










  • $begingroup$
    @JoelSjögren: I was writing up the appendix when you commented.
    $endgroup$
    – robjohn♦
    Dec 22 '18 at 13:07










  • $begingroup$
    Where does the premise of the appendix come from?
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:39










  • $begingroup$
    Oh ok it's a definition in terms of an arbitrary g.
    $endgroup$
    – Joel Sjögren
    Dec 22 '18 at 13:41
















$begingroup$
What is the general rule by which you make the "To satisfy (2)...we must have..." step?
$endgroup$
– Joel Sjögren
Dec 22 '18 at 12:52




$begingroup$
What is the general rule by which you make the "To satisfy (2)...we must have..." step?
$endgroup$
– Joel Sjögren
Dec 22 '18 at 12:52












$begingroup$
@JoelSjögren: I was writing up the appendix when you commented.
$endgroup$
– robjohn♦
Dec 22 '18 at 13:07




$begingroup$
@JoelSjögren: I was writing up the appendix when you commented.
$endgroup$
– robjohn♦
Dec 22 '18 at 13:07












$begingroup$
Where does the premise of the appendix come from?
$endgroup$
– Joel Sjögren
Dec 22 '18 at 13:39




$begingroup$
Where does the premise of the appendix come from?
$endgroup$
– Joel Sjögren
Dec 22 '18 at 13:39












$begingroup$
Oh ok it's a definition in terms of an arbitrary g.
$endgroup$
– Joel Sjögren
Dec 22 '18 at 13:41




$begingroup$
Oh ok it's a definition in terms of an arbitrary g.
$endgroup$
– Joel Sjögren
Dec 22 '18 at 13:41


















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