Using conformal maps to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$












1












$begingroup$


I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.



I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.



Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.



If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.






complex-analysis complex-numbers conformal-geometry mobius-transformation







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    1












    $begingroup$


    I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.



    I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.



    Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.



    If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.






    complex-analysis complex-numbers conformal-geometry mobius-transformation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.



      I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.



      Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.



      If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.






      complex-analysis complex-numbers conformal-geometry mobius-transformation







      share|cite|improve this question









      $endgroup$




      I'm trying to solve a question which asks me to solve the Dirichlet problem on $U = {z : text{Im}z geq 0 }$ on the condition that (where we let $z = x+iy$) $u(x,0) = 0 $ when $ |x| >1 $ and $ u(x,0) = 1 $ when $ |x| < 1$.



      I've already solved the problem for $U = {z : text{Im}z geq 0 }$ with $u(x,0) = 0 $ when $x > 0 $ and $ u(x,0) = 1 $ when $ x< 0$, and I presume there is some way to use a conformal map sending the positive left real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$, but I'm struggling to see how.



      Another result I have from an earlier question is the solution on $U = {x+iy : 0 leq y leq 1 }$, $u(x,0) = 0$, $u(x,1)=1$ as $u(x,y) = y$, but I'm not sure that would be so useful.



      If anyone knows any map which would relate the two I'd really appreciate your help, or if you know of another way to do this using the other result and conformal maps, I'd be really interested to hear that too.






      complex-analysis complex-numbers conformal-geometry mobius-transformation




      complex-analysis complex-numbers conformal-geometry mobius-transformation






      share|cite|improve this question













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      asked Jan 5 at 19:02









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          All conformal maps of the upper half plane are Möbius transformations of the
          form
          $$
          T(z) = frac{az+b}{cz+d}
          $$
          with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
          $$
          T(infty) = -1 , , quad T(0) = 1
          $$
          must hold. Now it should not be too difficult to find that
          $$
          T(z) = frac{1+z}{1-z}
          $$
          satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.






          share|cite|improve this answer











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            $begingroup$

            All conformal maps of the upper half plane are Möbius transformations of the
            form
            $$
            T(z) = frac{az+b}{cz+d}
            $$
            with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
            $$
            T(infty) = -1 , , quad T(0) = 1
            $$
            must hold. Now it should not be too difficult to find that
            $$
            T(z) = frac{1+z}{1-z}
            $$
            satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              All conformal maps of the upper half plane are Möbius transformations of the
              form
              $$
              T(z) = frac{az+b}{cz+d}
              $$
              with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
              $$
              T(infty) = -1 , , quad T(0) = 1
              $$
              must hold. Now it should not be too difficult to find that
              $$
              T(z) = frac{1+z}{1-z}
              $$
              satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                All conformal maps of the upper half plane are Möbius transformations of the
                form
                $$
                T(z) = frac{az+b}{cz+d}
                $$
                with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
                $$
                T(infty) = -1 , , quad T(0) = 1
                $$
                must hold. Now it should not be too difficult to find that
                $$
                T(z) = frac{1+z}{1-z}
                $$
                satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.






                share|cite|improve this answer











                $endgroup$



                All conformal maps of the upper half plane are Möbius transformations of the
                form
                $$
                T(z) = frac{az+b}{cz+d}
                $$
                with $a, b, c, d in Bbb R$, $ad-bc > 0$. We need that $(-infty, 0)$ is mapped to $(-1, 1)$, and since conformal maps preserve the orientation,
                $$
                T(infty) = -1 , , quad T(0) = 1
                $$
                must hold. Now it should not be too difficult to find that
                $$
                T(z) = frac{1+z}{1-z}
                $$
                satisfies all the needs, and allows to transform a solution of your second Dirichlet problem to a solution of your first problem.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 5 at 19:40

























                answered Jan 5 at 19:21









                Martin RMartin R

                27.5k33255




                27.5k33255






























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