Harnack's inequality












3












$begingroup$


PDE Evans, 2nd edition, Exercise 2.7




Use Poisson's formula for the ball to prove
$$r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}}u(0) le u(x) le r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}}u(0)$$
whenever $u in C^2(B^0(0,r)) cap C(partial B(0,r))$ is positive and harmonic in $B^0(0,r) subset mathbb{R}^n$. This is an explicit form of Harnack's inequality.




My proof attempt is in the answer, but I would like to know if it's correct.






proof-verification pde







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    PDE Evans, 2nd edition, Exercise 2.7




    Use Poisson's formula for the ball to prove
    $$r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}}u(0) le u(x) le r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}}u(0)$$
    whenever $u in C^2(B^0(0,r)) cap C(partial B(0,r))$ is positive and harmonic in $B^0(0,r) subset mathbb{R}^n$. This is an explicit form of Harnack's inequality.




    My proof attempt is in the answer, but I would like to know if it's correct.






    proof-verification pde







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      PDE Evans, 2nd edition, Exercise 2.7




      Use Poisson's formula for the ball to prove
      $$r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}}u(0) le u(x) le r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}}u(0)$$
      whenever $u in C^2(B^0(0,r)) cap C(partial B(0,r))$ is positive and harmonic in $B^0(0,r) subset mathbb{R}^n$. This is an explicit form of Harnack's inequality.




      My proof attempt is in the answer, but I would like to know if it's correct.






      proof-verification pde







      share|cite|improve this question











      $endgroup$




      PDE Evans, 2nd edition, Exercise 2.7




      Use Poisson's formula for the ball to prove
      $$r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}}u(0) le u(x) le r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}}u(0)$$
      whenever $u in C^2(B^0(0,r)) cap C(partial B(0,r))$ is positive and harmonic in $B^0(0,r) subset mathbb{R}^n$. This is an explicit form of Harnack's inequality.




      My proof attempt is in the answer, but I would like to know if it's correct.






      proof-verification pde




      proof-verification pde






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 8 '17 at 22:50









      Parisina

      839922




      839922










      asked Jun 6 '14 at 17:13









      CookieCookie

      8,723123782




      8,723123782






















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          $begingroup$

          Triangle inequality and reverse triangle inequality assert that
          $$r-|x|le|x-y|le r+|x|$$ for all $x in B^0(0,r)$ and $y in partial B(0,r)$ (which means $|y|=r$).



          Now we appeal to Poisson's formula: $$u(x) = frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{|x-y|^n} dS(y)$$



          Thus, begin{align}
          u(x) &le frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r-|x|)^n} dS(y) \
          &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
          &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} u(0) end{align}
          and begin{align}
          u(x) &ge frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r+|x|)^n} dS(y) \
          &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
          &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} u(0) end{align}
          Therefore, Harnack's inequality has been established, as desired.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Triangle inequality and reverse triangle inequality assert that
            $$r-|x|le|x-y|le r+|x|$$ for all $x in B^0(0,r)$ and $y in partial B(0,r)$ (which means $|y|=r$).



            Now we appeal to Poisson's formula: $$u(x) = frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{|x-y|^n} dS(y)$$



            Thus, begin{align}
            u(x) &le frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r-|x|)^n} dS(y) \
            &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
            &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} u(0) end{align}
            and begin{align}
            u(x) &ge frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r+|x|)^n} dS(y) \
            &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
            &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} u(0) end{align}
            Therefore, Harnack's inequality has been established, as desired.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Triangle inequality and reverse triangle inequality assert that
              $$r-|x|le|x-y|le r+|x|$$ for all $x in B^0(0,r)$ and $y in partial B(0,r)$ (which means $|y|=r$).



              Now we appeal to Poisson's formula: $$u(x) = frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{|x-y|^n} dS(y)$$



              Thus, begin{align}
              u(x) &le frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r-|x|)^n} dS(y) \
              &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
              &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} u(0) end{align}
              and begin{align}
              u(x) &ge frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r+|x|)^n} dS(y) \
              &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
              &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} u(0) end{align}
              Therefore, Harnack's inequality has been established, as desired.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Triangle inequality and reverse triangle inequality assert that
                $$r-|x|le|x-y|le r+|x|$$ for all $x in B^0(0,r)$ and $y in partial B(0,r)$ (which means $|y|=r$).



                Now we appeal to Poisson's formula: $$u(x) = frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{|x-y|^n} dS(y)$$



                Thus, begin{align}
                u(x) &le frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r-|x|)^n} dS(y) \
                &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
                &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} u(0) end{align}
                and begin{align}
                u(x) &ge frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r+|x|)^n} dS(y) \
                &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
                &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} u(0) end{align}
                Therefore, Harnack's inequality has been established, as desired.






                share|cite|improve this answer











                $endgroup$



                Triangle inequality and reverse triangle inequality assert that
                $$r-|x|le|x-y|le r+|x|$$ for all $x in B^0(0,r)$ and $y in partial B(0,r)$ (which means $|y|=r$).



                Now we appeal to Poisson's formula: $$u(x) = frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{|x-y|^n} dS(y)$$



                Thus, begin{align}
                u(x) &le frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r-|x|)^n} dS(y) \
                &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
                &= r^{n-2} frac{r+|x|}{(r-|x|)^{n-1}} u(0) end{align}
                and begin{align}
                u(x) &ge frac{r^2-|x|^2}{n alpha(n)r}int_{partial B(0,r)} frac{g(y)}{(r+|x|)^n} dS(y) \
                &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} frac{1}{n alpha(n)r^{n-1}} int_{partial B(0,r)} g(y) , dS(y) \
                &= r^{n-2} frac{r-|x|}{(r+|x|)^{n-1}} u(0) end{align}
                Therefore, Harnack's inequality has been established, as desired.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jun 20 '14 at 18:34

























                answered Jun 6 '14 at 17:30









                CookieCookie

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