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.










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    2












    $begingroup$


    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.










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      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.










      share|cite|improve this question











      $endgroup$




      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






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      edited Dec 17 '18 at 20:26







      user593746

















      asked Dec 17 '18 at 18:55









      LukathLukath

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          $begingroup$

          Hint:



          $$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$






          share|cite|improve this answer









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            $begingroup$
            God, it was just about completing a square... Ockham's razor. Thank you!
            $endgroup$
            – Lukath
            Dec 17 '18 at 19:44











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          1 Answer
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          1 Answer
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          active

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          $begingroup$

          Hint:



          $$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            God, it was just about completing a square... Ockham's razor. Thank you!
            $endgroup$
            – Lukath
            Dec 17 '18 at 19:44
















          1












          $begingroup$

          Hint:



          $$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$






          share|cite|improve this answer









          $endgroup$









          • 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








          1





          $begingroup$

          Hint:



          $$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$ epsilon u^2 + frac{1}{4epsilon} f^2 + uf = left( sqrt{epsilon} u + frac{1}{2sqrt{epsilon}}f right)^2 $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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














          • 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


















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