Compute the limit for a harmonic function given two known limits












1












$begingroup$


This problem is from a set of exercises that I have. It states:




Let $uin C(overline{mathbb{R}_+^2})$ be a bounded harmonic function in the upper half plane $mathbb{R}_+^2$, satisfying $u(x,0) to pi$ as $xto infty$ and $u(x,0)to 0$ as $x to -infty$. Compute the limit of $u(rcostheta, rsintheta)$ as $rtoinfty$ for each $0<theta < pi$.




I tried using the Poisson formula for the half plane but didn't manage to get anywhere. I appreciate any input!






real-analysis complex-analysis pde harmonic-functions







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$endgroup$

















    1












    $begingroup$


    This problem is from a set of exercises that I have. It states:




    Let $uin C(overline{mathbb{R}_+^2})$ be a bounded harmonic function in the upper half plane $mathbb{R}_+^2$, satisfying $u(x,0) to pi$ as $xto infty$ and $u(x,0)to 0$ as $x to -infty$. Compute the limit of $u(rcostheta, rsintheta)$ as $rtoinfty$ for each $0<theta < pi$.




    I tried using the Poisson formula for the half plane but didn't manage to get anywhere. I appreciate any input!






    real-analysis complex-analysis pde harmonic-functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      This problem is from a set of exercises that I have. It states:




      Let $uin C(overline{mathbb{R}_+^2})$ be a bounded harmonic function in the upper half plane $mathbb{R}_+^2$, satisfying $u(x,0) to pi$ as $xto infty$ and $u(x,0)to 0$ as $x to -infty$. Compute the limit of $u(rcostheta, rsintheta)$ as $rtoinfty$ for each $0<theta < pi$.




      I tried using the Poisson formula for the half plane but didn't manage to get anywhere. I appreciate any input!






      real-analysis complex-analysis pde harmonic-functions







      share|cite|improve this question











      $endgroup$




      This problem is from a set of exercises that I have. It states:




      Let $uin C(overline{mathbb{R}_+^2})$ be a bounded harmonic function in the upper half plane $mathbb{R}_+^2$, satisfying $u(x,0) to pi$ as $xto infty$ and $u(x,0)to 0$ as $x to -infty$. Compute the limit of $u(rcostheta, rsintheta)$ as $rtoinfty$ for each $0<theta < pi$.




      I tried using the Poisson formula for the half plane but didn't manage to get anywhere. I appreciate any input!






      real-analysis complex-analysis pde harmonic-functions




      real-analysis complex-analysis pde harmonic-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 21:46









      zhw.

      72.2k43175




      72.2k43175










      asked Dec 8 '18 at 20:15









      QuokaQuoka

      1,240212




      1,240212






















          1 Answer
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          $begingroup$

          Hint: Suppose $u(t,0)to 0$ as $|t|to infty,$ with the other hypotheses unchanged. Let $v(z) = u(-1/z).$ Show $v$ is continuous at the origin, with value $0$ there. Show that this implies $lim_{rto infty} u(re^{it})=0$ for all $tin (0,pi).$ Now to the original problem: There is an easy bounded harmonic function $u_0$ that equals $pi$ on the positive real axis and $0$ on the negative real axis. Given $u$ as in your problem, apply the above to $u-u_0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the hint! I tried your approach unfortunately am still getting stuck. I know that showing that $v$ is continuous (and equal to 0) at the origin is equivalent to showing that $lim_{rtoinfty} u(re^{it}) = 0$ for all $0leq tleq pi$. Do you have another hint about how to prove that?
            $endgroup$
            – Quoka
            Dec 9 '18 at 2:10










          • $begingroup$
            Actually that's not an equivalence, it's a $implies$ To show the result about $v,$ recall that $v=P[v|_{mathbb R}].$ And if the boundary function has a point of continuity on the line, then so does $v$ in the upper half plane.
            $endgroup$
            – zhw.
            Dec 9 '18 at 23:48












          • $begingroup$
            Suppose a function $w$ is bounded and harmonic in the half plane $mathbb{R}^2_+$. If $w$ is continuous up to the boundary then I can prove that $w$ is given by the Poisson formula. How can I prove this before showing continuity?
            $endgroup$
            – Quoka
            Dec 10 '18 at 6:08










          • $begingroup$
            Things clicking better now?
            $endgroup$
            – zhw.
            Dec 10 '18 at 20:56










          • $begingroup$
            Yes. Thank you :-)
            $endgroup$
            – Quoka
            Dec 11 '18 at 0:42











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          1 Answer
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          1 Answer
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          active

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          $begingroup$

          Hint: Suppose $u(t,0)to 0$ as $|t|to infty,$ with the other hypotheses unchanged. Let $v(z) = u(-1/z).$ Show $v$ is continuous at the origin, with value $0$ there. Show that this implies $lim_{rto infty} u(re^{it})=0$ for all $tin (0,pi).$ Now to the original problem: There is an easy bounded harmonic function $u_0$ that equals $pi$ on the positive real axis and $0$ on the negative real axis. Given $u$ as in your problem, apply the above to $u-u_0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the hint! I tried your approach unfortunately am still getting stuck. I know that showing that $v$ is continuous (and equal to 0) at the origin is equivalent to showing that $lim_{rtoinfty} u(re^{it}) = 0$ for all $0leq tleq pi$. Do you have another hint about how to prove that?
            $endgroup$
            – Quoka
            Dec 9 '18 at 2:10










          • $begingroup$
            Actually that's not an equivalence, it's a $implies$ To show the result about $v,$ recall that $v=P[v|_{mathbb R}].$ And if the boundary function has a point of continuity on the line, then so does $v$ in the upper half plane.
            $endgroup$
            – zhw.
            Dec 9 '18 at 23:48












          • $begingroup$
            Suppose a function $w$ is bounded and harmonic in the half plane $mathbb{R}^2_+$. If $w$ is continuous up to the boundary then I can prove that $w$ is given by the Poisson formula. How can I prove this before showing continuity?
            $endgroup$
            – Quoka
            Dec 10 '18 at 6:08










          • $begingroup$
            Things clicking better now?
            $endgroup$
            – zhw.
            Dec 10 '18 at 20:56










          • $begingroup$
            Yes. Thank you :-)
            $endgroup$
            – Quoka
            Dec 11 '18 at 0:42
















          1












          $begingroup$

          Hint: Suppose $u(t,0)to 0$ as $|t|to infty,$ with the other hypotheses unchanged. Let $v(z) = u(-1/z).$ Show $v$ is continuous at the origin, with value $0$ there. Show that this implies $lim_{rto infty} u(re^{it})=0$ for all $tin (0,pi).$ Now to the original problem: There is an easy bounded harmonic function $u_0$ that equals $pi$ on the positive real axis and $0$ on the negative real axis. Given $u$ as in your problem, apply the above to $u-u_0.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the hint! I tried your approach unfortunately am still getting stuck. I know that showing that $v$ is continuous (and equal to 0) at the origin is equivalent to showing that $lim_{rtoinfty} u(re^{it}) = 0$ for all $0leq tleq pi$. Do you have another hint about how to prove that?
            $endgroup$
            – Quoka
            Dec 9 '18 at 2:10










          • $begingroup$
            Actually that's not an equivalence, it's a $implies$ To show the result about $v,$ recall that $v=P[v|_{mathbb R}].$ And if the boundary function has a point of continuity on the line, then so does $v$ in the upper half plane.
            $endgroup$
            – zhw.
            Dec 9 '18 at 23:48












          • $begingroup$
            Suppose a function $w$ is bounded and harmonic in the half plane $mathbb{R}^2_+$. If $w$ is continuous up to the boundary then I can prove that $w$ is given by the Poisson formula. How can I prove this before showing continuity?
            $endgroup$
            – Quoka
            Dec 10 '18 at 6:08










          • $begingroup$
            Things clicking better now?
            $endgroup$
            – zhw.
            Dec 10 '18 at 20:56










          • $begingroup$
            Yes. Thank you :-)
            $endgroup$
            – Quoka
            Dec 11 '18 at 0:42














          1












          1








          1





          $begingroup$

          Hint: Suppose $u(t,0)to 0$ as $|t|to infty,$ with the other hypotheses unchanged. Let $v(z) = u(-1/z).$ Show $v$ is continuous at the origin, with value $0$ there. Show that this implies $lim_{rto infty} u(re^{it})=0$ for all $tin (0,pi).$ Now to the original problem: There is an easy bounded harmonic function $u_0$ that equals $pi$ on the positive real axis and $0$ on the negative real axis. Given $u$ as in your problem, apply the above to $u-u_0.$






          share|cite|improve this answer









          $endgroup$



          Hint: Suppose $u(t,0)to 0$ as $|t|to infty,$ with the other hypotheses unchanged. Let $v(z) = u(-1/z).$ Show $v$ is continuous at the origin, with value $0$ there. Show that this implies $lim_{rto infty} u(re^{it})=0$ for all $tin (0,pi).$ Now to the original problem: There is an easy bounded harmonic function $u_0$ that equals $pi$ on the positive real axis and $0$ on the negative real axis. Given $u$ as in your problem, apply the above to $u-u_0.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 22:10









          zhw.zhw.

          72.2k43175




          72.2k43175












          • $begingroup$
            Thanks for the hint! I tried your approach unfortunately am still getting stuck. I know that showing that $v$ is continuous (and equal to 0) at the origin is equivalent to showing that $lim_{rtoinfty} u(re^{it}) = 0$ for all $0leq tleq pi$. Do you have another hint about how to prove that?
            $endgroup$
            – Quoka
            Dec 9 '18 at 2:10










          • $begingroup$
            Actually that's not an equivalence, it's a $implies$ To show the result about $v,$ recall that $v=P[v|_{mathbb R}].$ And if the boundary function has a point of continuity on the line, then so does $v$ in the upper half plane.
            $endgroup$
            – zhw.
            Dec 9 '18 at 23:48












          • $begingroup$
            Suppose a function $w$ is bounded and harmonic in the half plane $mathbb{R}^2_+$. If $w$ is continuous up to the boundary then I can prove that $w$ is given by the Poisson formula. How can I prove this before showing continuity?
            $endgroup$
            – Quoka
            Dec 10 '18 at 6:08










          • $begingroup$
            Things clicking better now?
            $endgroup$
            – zhw.
            Dec 10 '18 at 20:56










          • $begingroup$
            Yes. Thank you :-)
            $endgroup$
            – Quoka
            Dec 11 '18 at 0:42


















          • $begingroup$
            Thanks for the hint! I tried your approach unfortunately am still getting stuck. I know that showing that $v$ is continuous (and equal to 0) at the origin is equivalent to showing that $lim_{rtoinfty} u(re^{it}) = 0$ for all $0leq tleq pi$. Do you have another hint about how to prove that?
            $endgroup$
            – Quoka
            Dec 9 '18 at 2:10










          • $begingroup$
            Actually that's not an equivalence, it's a $implies$ To show the result about $v,$ recall that $v=P[v|_{mathbb R}].$ And if the boundary function has a point of continuity on the line, then so does $v$ in the upper half plane.
            $endgroup$
            – zhw.
            Dec 9 '18 at 23:48












          • $begingroup$
            Suppose a function $w$ is bounded and harmonic in the half plane $mathbb{R}^2_+$. If $w$ is continuous up to the boundary then I can prove that $w$ is given by the Poisson formula. How can I prove this before showing continuity?
            $endgroup$
            – Quoka
            Dec 10 '18 at 6:08










          • $begingroup$
            Things clicking better now?
            $endgroup$
            – zhw.
            Dec 10 '18 at 20:56










          • $begingroup$
            Yes. Thank you :-)
            $endgroup$
            – Quoka
            Dec 11 '18 at 0:42
















          $begingroup$
          Thanks for the hint! I tried your approach unfortunately am still getting stuck. I know that showing that $v$ is continuous (and equal to 0) at the origin is equivalent to showing that $lim_{rtoinfty} u(re^{it}) = 0$ for all $0leq tleq pi$. Do you have another hint about how to prove that?
          $endgroup$
          – Quoka
          Dec 9 '18 at 2:10




          $begingroup$
          Thanks for the hint! I tried your approach unfortunately am still getting stuck. I know that showing that $v$ is continuous (and equal to 0) at the origin is equivalent to showing that $lim_{rtoinfty} u(re^{it}) = 0$ for all $0leq tleq pi$. Do you have another hint about how to prove that?
          $endgroup$
          – Quoka
          Dec 9 '18 at 2:10












          $begingroup$
          Actually that's not an equivalence, it's a $implies$ To show the result about $v,$ recall that $v=P[v|_{mathbb R}].$ And if the boundary function has a point of continuity on the line, then so does $v$ in the upper half plane.
          $endgroup$
          – zhw.
          Dec 9 '18 at 23:48






          $begingroup$
          Actually that's not an equivalence, it's a $implies$ To show the result about $v,$ recall that $v=P[v|_{mathbb R}].$ And if the boundary function has a point of continuity on the line, then so does $v$ in the upper half plane.
          $endgroup$
          – zhw.
          Dec 9 '18 at 23:48














          $begingroup$
          Suppose a function $w$ is bounded and harmonic in the half plane $mathbb{R}^2_+$. If $w$ is continuous up to the boundary then I can prove that $w$ is given by the Poisson formula. How can I prove this before showing continuity?
          $endgroup$
          – Quoka
          Dec 10 '18 at 6:08




          $begingroup$
          Suppose a function $w$ is bounded and harmonic in the half plane $mathbb{R}^2_+$. If $w$ is continuous up to the boundary then I can prove that $w$ is given by the Poisson formula. How can I prove this before showing continuity?
          $endgroup$
          – Quoka
          Dec 10 '18 at 6:08












          $begingroup$
          Things clicking better now?
          $endgroup$
          – zhw.
          Dec 10 '18 at 20:56




          $begingroup$
          Things clicking better now?
          $endgroup$
          – zhw.
          Dec 10 '18 at 20:56












          $begingroup$
          Yes. Thank you :-)
          $endgroup$
          – Quoka
          Dec 11 '18 at 0:42




          $begingroup$
          Yes. Thank you :-)
          $endgroup$
          – Quoka
          Dec 11 '18 at 0:42


















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