Inequality for the $L^2$ norm of a derivative in a Dirichlet problem
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I found some trouble in solving this problem:
given $f$ a continuous function in $[0,1]$ and $uin C^2([0,1])$, such that $u''(x)=f(x)$ on $(0,1)$ and $u(0)=u(1)=0$, prove, $forallepsilon > 0$, the inequality
$||u'||_2^2 leq epsilon||u||_2^2 + frac{1}{4epsilon}||f||_2^2$
where $||.||_2$ is the usual norm on $L^2 ([0,1])$.
First, I rewrote the equation in this way:
$-u(x)u''(x)=-u(x)f(x)$
in order to have
$||u'||_2^2 = -int_{0}^{1}u(x)f(x)dx$.
Then I tried to use a Green function to represent $u$, i.e.
$F(x,y)$=begin{cases} -x(1-y) & text{if $0leq xleq yleq 1$} \ -y(1-x) & text{if $0leq yleq xleq 1$} end{cases}
in order to have $u(x)=int_{0}^{1}F(x,y)f(y)dy$ but I really don't know where that $epsilon$ should come from. I also thought I could use some identity approximation or mollifiers, but I got stuck.
real-analysis integration lebesgue-integral lp-spaces boundary-value-problem
asked Dec 17 '18 at 18:55
LukathLukath
717
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add a comment |
$begingroup$
I found some trouble in solving this problem:
given $f$ a continuous function in $[0,1]$ and $uin C^2([0,1])$, such that $u''(x)=f(x)$ on $(0,1)$ and $u(0)=u(1)=0$, prove, $forallepsilon > 0$, the inequality
$||u'||_2^2 leq epsilon||u||_2^2 + frac{1}{4epsilon}||f||_2^2$
where $||.||_2$ is the usual norm on $L^2 ([0,1])$.
First, I rewrote the equation in this way:
$-u(x)u''(x)=-u(x)f(x)$
in order to have
$||u'||_2^2 = -int_{0}^{1}u(x)f(x)dx$.
Then I tried to use a Green function to represent $u$, i.e.
$F(x,y)$=begin{cases} -x(1-y) & text{if $0leq xleq yleq 1$} \ -y(1-x) & text{if $0leq yleq xleq 1$} end{cases}
in order to have $u(x)=int_{0}^{1}F(x,y)f(y)dy$ but I really don't know where that $epsilon$ should come from. I also thought I could use some identity approximation or mollifiers, but I got stuck.
real-analysis integration lebesgue-integral lp-spaces boundary-value-problem
asked Dec 17 '18 at 18:55
LukathLukath
717
$endgroup$
add a comment |
$begingroup$
I found some trouble in solving this problem:
given $f$ a continuous function in $[0,1]$ and $uin C^2([0,1])$, such that $u''(x)=f(x)$ on $(0,1)$ and $u(0)=u(1)=0$, prove, $forallepsilon > 0$, the inequality
$||u'||_2^2 leq epsilon||u||_2^2 + frac{1}{4epsilon}||f||_2^2$
where $||.||_2$ is the usual norm on $L^2 ([0,1])$.
First, I rewrote the equation in this way:
$-u(x)u''(x)=-u(x)f(x)$
in order to have
$||u'||_2^2 = -int_{0}^{1}u(x)f(x)dx$.
Then I tried to use a Green function to represent $u$, i.e.
$F(x,y)$=begin{cases} -x(1-y) & text{if $0leq xleq yleq 1$} \ -y(1-x) & text{if $0leq yleq xleq 1$} end{cases}
in order to have $u(x)=int_{0}^{1}F(x,y)f(y)dy$ but I really don't know where that $epsilon$ should come from. I also thought I could use some identity approximation or mollifiers, but I got stuck.
real-analysis integration lebesgue-integral lp-spaces boundary-value-problem
asked Dec 17 '18 at 18:55
LukathLukath
717
$endgroup$
I found some trouble in solving this problem:
given $f$ a continuous function in $[0,1]$ and $uin C^2([0,1])$, such that $u''(x)=f(x)$ on $(0,1)$ and $u(0)=u(1)=0$, prove, $forallepsilon > 0$, the inequality
$||u'||_2^2 leq epsilon||u||_2^2 + frac{1}{4epsilon}||f||_2^2$
where $||.||_2$ is the usual norm on $L^2 ([0,1])$.
First, I rewrote the equation in this way:
$-u(x)u''(x)=-u(x)f(x)$
in order to have
$||u'||_2^2 = -int_{0}^{1}u(x)f(x)dx$.
Then I tried to use a Green function to represent $u$, i.e.
$F(x,y)$=begin{cases} -x(1-y) & text{if $0leq xleq yleq 1$} \ -y(1-x) & text{if $0leq yleq xleq 1$} end{cases}
in order to have $u(x)=int_{0}^{1}F(x,y)f(y)dy$ but I really don't know where that $epsilon$ should come from. I also thought I could use some identity approximation or mollifiers, but I got stuck.
real-analysis integration lebesgue-integral lp-spaces boundary-value-problem
real-analysis integration lebesgue-integral lp-spaces boundary-value-problem
asked Dec 17 '18 at 18:55
LukathLukath
717
asked Dec 17 '18 at 18:55
LukathLukath
717
edited Dec 17 '18 at 20:26
user593746
asked Dec 17 '18 at 18:55
LukathLukath
717
asked Dec 17 '18 at 18:55
LukathLukath
717
asked Dec 17 '18 at 18:55
LukathLukath
717
717
add a comment |
add a comment |
1 Answer
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Hint:
$$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
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1
$begingroup$
God, it was just about completing a square... Ockham's razor. Thank you!
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– Lukath
Dec 17 '18 at 19:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Hint:
$$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
$endgroup$
1
$begingroup$
God, it was just about completing a square... Ockham's razor. Thank you!
$endgroup$
– Lukath
Dec 17 '18 at 19:44
add a comment |
$begingroup$
Hint:
$$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
$endgroup$
1
$begingroup$
God, it was just about completing a square... Ockham's razor. Thank you!
$endgroup$
– Lukath
Dec 17 '18 at 19:44
add a comment |
$begingroup$
Hint:
$$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
$endgroup$
Hint:
$$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
answered Dec 17 '18 at 19:10
DylanDylan
12.5k31026
12.5k31026
1
$begingroup$
God, it was just about completing a square... Ockham's razor. Thank you!
$endgroup$
– Lukath
Dec 17 '18 at 19:44
add a comment |
1
$begingroup$
God, it was just about completing a square... Ockham's razor. Thank you!
$endgroup$
– Lukath
Dec 17 '18 at 19:44
1
1
$begingroup$
God, it was just about completing a square... Ockham's razor. Thank you!
$endgroup$
– Lukath
Dec 17 '18 at 19:44
$begingroup$
God, it was just about completing a square... Ockham's razor. Thank you!
$endgroup$
– Lukath
Dec 17 '18 at 19:44
add a comment |
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