Concavity of $log u(x)$












-1












$begingroup$


Let $u>0$ be a solution of the following equation
$$begin{cases}-[|u'(x)|^{p-2}u'(x)]'=lambda_1cdot u(x)^{p-1}&,x in (a,b)\u(a)=u(b)=0 end{cases},$$
where $lambda_1$ is the minimum of the Rayleigh quotient and $p>1.$



Prove that the function $log u(x)$ is concave on $(a,b).$






calculus pde convex-analysis spectral-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – DaveNine
    Dec 16 '18 at 1:43










  • $begingroup$
    To make the second derivative of "log". but the question is : can I use the case when only $p=2?$
    $endgroup$
    – Andrew
    Dec 16 '18 at 3:38
















-1












$begingroup$


Let $u>0$ be a solution of the following equation
$$begin{cases}-[|u'(x)|^{p-2}u'(x)]'=lambda_1cdot u(x)^{p-1}&,x in (a,b)\u(a)=u(b)=0 end{cases},$$
where $lambda_1$ is the minimum of the Rayleigh quotient and $p>1.$



Prove that the function $log u(x)$ is concave on $(a,b).$






calculus pde convex-analysis spectral-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – DaveNine
    Dec 16 '18 at 1:43










  • $begingroup$
    To make the second derivative of "log". but the question is : can I use the case when only $p=2?$
    $endgroup$
    – Andrew
    Dec 16 '18 at 3:38














-1












-1








-1





$begingroup$


Let $u>0$ be a solution of the following equation
$$begin{cases}-[|u'(x)|^{p-2}u'(x)]'=lambda_1cdot u(x)^{p-1}&,x in (a,b)\u(a)=u(b)=0 end{cases},$$
where $lambda_1$ is the minimum of the Rayleigh quotient and $p>1.$



Prove that the function $log u(x)$ is concave on $(a,b).$






calculus pde convex-analysis spectral-theory







share|cite|improve this question









$endgroup$




Let $u>0$ be a solution of the following equation
$$begin{cases}-[|u'(x)|^{p-2}u'(x)]'=lambda_1cdot u(x)^{p-1}&,x in (a,b)\u(a)=u(b)=0 end{cases},$$
where $lambda_1$ is the minimum of the Rayleigh quotient and $p>1.$



Prove that the function $log u(x)$ is concave on $(a,b).$






calculus pde convex-analysis spectral-theory




calculus pde convex-analysis spectral-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 22:19









AndrewAndrew

346




346








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – DaveNine
    Dec 16 '18 at 1:43










  • $begingroup$
    To make the second derivative of "log". but the question is : can I use the case when only $p=2?$
    $endgroup$
    – Andrew
    Dec 16 '18 at 3:38














  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – DaveNine
    Dec 16 '18 at 1:43










  • $begingroup$
    To make the second derivative of "log". but the question is : can I use the case when only $p=2?$
    $endgroup$
    – Andrew
    Dec 16 '18 at 3:38








1




1




$begingroup$
What have you tried?
$endgroup$
– DaveNine
Dec 16 '18 at 1:43




$begingroup$
What have you tried?
$endgroup$
– DaveNine
Dec 16 '18 at 1:43












$begingroup$
To make the second derivative of "log". but the question is : can I use the case when only $p=2?$
$endgroup$
– Andrew
Dec 16 '18 at 3:38




$begingroup$
To make the second derivative of "log". but the question is : can I use the case when only $p=2?$
$endgroup$
– Andrew
Dec 16 '18 at 3:38










1 Answer
1






active

oldest

votes


















1












$begingroup$

It is easy to see that $lambda_1>0$. From the first equation, one has
$$ -(p-2)|u'(x)|^{p-3}frac{u'(x)}{|u'(x)|}u''(x)u'(x)-|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1} $$
or
$$ -(p-1)|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1}. $$
from which it is easy to see
$$ u''(x)=-frac{lambda_1}{p-1}frac{u(x)^{p-1}}{|u'(x)|^{p-2}}<0. tag{1} $$
So by (1), one has
begin{eqnarray*} (ln u(x))''&=&left(frac{1}{u(x)}u'(x)right)'\
&=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
&=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
&<&0.
end{eqnarray*}
Namely $ln u(x)$ is concave.






share|cite|improve this answer









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    $begingroup$

    It is easy to see that $lambda_1>0$. From the first equation, one has
    $$ -(p-2)|u'(x)|^{p-3}frac{u'(x)}{|u'(x)|}u''(x)u'(x)-|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1} $$
    or
    $$ -(p-1)|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1}. $$
    from which it is easy to see
    $$ u''(x)=-frac{lambda_1}{p-1}frac{u(x)^{p-1}}{|u'(x)|^{p-2}}<0. tag{1} $$
    So by (1), one has
    begin{eqnarray*} (ln u(x))''&=&left(frac{1}{u(x)}u'(x)right)'\
    &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
    &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
    &<&0.
    end{eqnarray*}
    Namely $ln u(x)$ is concave.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It is easy to see that $lambda_1>0$. From the first equation, one has
      $$ -(p-2)|u'(x)|^{p-3}frac{u'(x)}{|u'(x)|}u''(x)u'(x)-|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1} $$
      or
      $$ -(p-1)|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1}. $$
      from which it is easy to see
      $$ u''(x)=-frac{lambda_1}{p-1}frac{u(x)^{p-1}}{|u'(x)|^{p-2}}<0. tag{1} $$
      So by (1), one has
      begin{eqnarray*} (ln u(x))''&=&left(frac{1}{u(x)}u'(x)right)'\
      &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
      &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
      &<&0.
      end{eqnarray*}
      Namely $ln u(x)$ is concave.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It is easy to see that $lambda_1>0$. From the first equation, one has
        $$ -(p-2)|u'(x)|^{p-3}frac{u'(x)}{|u'(x)|}u''(x)u'(x)-|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1} $$
        or
        $$ -(p-1)|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1}. $$
        from which it is easy to see
        $$ u''(x)=-frac{lambda_1}{p-1}frac{u(x)^{p-1}}{|u'(x)|^{p-2}}<0. tag{1} $$
        So by (1), one has
        begin{eqnarray*} (ln u(x))''&=&left(frac{1}{u(x)}u'(x)right)'\
        &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
        &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
        &<&0.
        end{eqnarray*}
        Namely $ln u(x)$ is concave.






        share|cite|improve this answer









        $endgroup$



        It is easy to see that $lambda_1>0$. From the first equation, one has
        $$ -(p-2)|u'(x)|^{p-3}frac{u'(x)}{|u'(x)|}u''(x)u'(x)-|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1} $$
        or
        $$ -(p-1)|u'(x)|^{p-2}u''(x)=lambda_1u(x)^{p-1}. $$
        from which it is easy to see
        $$ u''(x)=-frac{lambda_1}{p-1}frac{u(x)^{p-1}}{|u'(x)|^{p-2}}<0. tag{1} $$
        So by (1), one has
        begin{eqnarray*} (ln u(x))''&=&left(frac{1}{u(x)}u'(x)right)'\
        &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
        &=&-frac{1}{u(x)^2}|u'(x)|^2+frac{1}{u(x)}u''(x)\
        &<&0.
        end{eqnarray*}
        Namely $ln u(x)$ is concave.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 0:13









        xpaulxpaul

        23.1k24455




        23.1k24455






























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