Polar laplace equation












-1












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I am really struggling to solve this, the initial condition $u(a,theta)=0$ keeps throwing me off.



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differential-equations pde







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    -1












    $begingroup$


    I am really struggling to solve this, the initial condition $u(a,theta)=0$ keeps throwing me off.



    enter image description here






    differential-equations pde







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I am really struggling to solve this, the initial condition $u(a,theta)=0$ keeps throwing me off.



      enter image description here






      differential-equations pde







      share|cite|improve this question











      $endgroup$




      I am really struggling to solve this, the initial condition $u(a,theta)=0$ keeps throwing me off.



      enter image description here






      differential-equations pde




      differential-equations pde






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 30 '18 at 18:29









      caverac

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      14.2k21130










      asked Nov 30 '18 at 18:27









      Jack LeitchJack Leitch

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          $begingroup$

          Let $u(r,theta)=f(r)g(theta)$. Then the Laplace equation becomes
          $$ f''(r)g(theta)+frac1r f'(r)g(theta)+frac1{r^2}f(r)g''(theta)=0$$
          from which one has
          $$ frac{f''(r)+frac1r f'(r)}{f(r)}=-frac{g''(theta)}{g(theta)}. $$
          Let
          $$ frac{f''(r)+frac1r f'(r)}{f(r)}=K tag{1} $$
          and
          $$-frac{g''(theta)}{g(theta)}=K. tag{2} $$
          Now solving (1) and (2) gives
          $$ f(r)=c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r), g(theta)=d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta). $$
          So
          $$ u(r,theta)=bigg[c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r)bigg] bigg[d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta)bigg]. $$
          Let $u(r,0)=u(r,2pi)=0$ and then one has
          $$ d_1=0, K=frac{n^2}{4},n=1,2,3cdots. $$
          You can choose $K=frac14$ or $K=1$. Let $u(a,theta)=0,u(b,theta)=f(theta)$ and I think you can determine $C_1,C_2$ which is not very hard.






          share|cite|improve this answer









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            $begingroup$

            Let $u(r,theta)=f(r)g(theta)$. Then the Laplace equation becomes
            $$ f''(r)g(theta)+frac1r f'(r)g(theta)+frac1{r^2}f(r)g''(theta)=0$$
            from which one has
            $$ frac{f''(r)+frac1r f'(r)}{f(r)}=-frac{g''(theta)}{g(theta)}. $$
            Let
            $$ frac{f''(r)+frac1r f'(r)}{f(r)}=K tag{1} $$
            and
            $$-frac{g''(theta)}{g(theta)}=K. tag{2} $$
            Now solving (1) and (2) gives
            $$ f(r)=c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r), g(theta)=d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta). $$
            So
            $$ u(r,theta)=bigg[c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r)bigg] bigg[d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta)bigg]. $$
            Let $u(r,0)=u(r,2pi)=0$ and then one has
            $$ d_1=0, K=frac{n^2}{4},n=1,2,3cdots. $$
            You can choose $K=frac14$ or $K=1$. Let $u(a,theta)=0,u(b,theta)=f(theta)$ and I think you can determine $C_1,C_2$ which is not very hard.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let $u(r,theta)=f(r)g(theta)$. Then the Laplace equation becomes
              $$ f''(r)g(theta)+frac1r f'(r)g(theta)+frac1{r^2}f(r)g''(theta)=0$$
              from which one has
              $$ frac{f''(r)+frac1r f'(r)}{f(r)}=-frac{g''(theta)}{g(theta)}. $$
              Let
              $$ frac{f''(r)+frac1r f'(r)}{f(r)}=K tag{1} $$
              and
              $$-frac{g''(theta)}{g(theta)}=K. tag{2} $$
              Now solving (1) and (2) gives
              $$ f(r)=c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r), g(theta)=d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta). $$
              So
              $$ u(r,theta)=bigg[c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r)bigg] bigg[d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta)bigg]. $$
              Let $u(r,0)=u(r,2pi)=0$ and then one has
              $$ d_1=0, K=frac{n^2}{4},n=1,2,3cdots. $$
              You can choose $K=frac14$ or $K=1$. Let $u(a,theta)=0,u(b,theta)=f(theta)$ and I think you can determine $C_1,C_2$ which is not very hard.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $u(r,theta)=f(r)g(theta)$. Then the Laplace equation becomes
                $$ f''(r)g(theta)+frac1r f'(r)g(theta)+frac1{r^2}f(r)g''(theta)=0$$
                from which one has
                $$ frac{f''(r)+frac1r f'(r)}{f(r)}=-frac{g''(theta)}{g(theta)}. $$
                Let
                $$ frac{f''(r)+frac1r f'(r)}{f(r)}=K tag{1} $$
                and
                $$-frac{g''(theta)}{g(theta)}=K. tag{2} $$
                Now solving (1) and (2) gives
                $$ f(r)=c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r), g(theta)=d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta). $$
                So
                $$ u(r,theta)=bigg[c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r)bigg] bigg[d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta)bigg]. $$
                Let $u(r,0)=u(r,2pi)=0$ and then one has
                $$ d_1=0, K=frac{n^2}{4},n=1,2,3cdots. $$
                You can choose $K=frac14$ or $K=1$. Let $u(a,theta)=0,u(b,theta)=f(theta)$ and I think you can determine $C_1,C_2$ which is not very hard.






                share|cite|improve this answer









                $endgroup$



                Let $u(r,theta)=f(r)g(theta)$. Then the Laplace equation becomes
                $$ f''(r)g(theta)+frac1r f'(r)g(theta)+frac1{r^2}f(r)g''(theta)=0$$
                from which one has
                $$ frac{f''(r)+frac1r f'(r)}{f(r)}=-frac{g''(theta)}{g(theta)}. $$
                Let
                $$ frac{f''(r)+frac1r f'(r)}{f(r)}=K tag{1} $$
                and
                $$-frac{g''(theta)}{g(theta)}=K. tag{2} $$
                Now solving (1) and (2) gives
                $$ f(r)=c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r), g(theta)=d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta). $$
                So
                $$ u(r,theta)=bigg[c_1cosh(sqrt Kr)+c_2 isinh(sqrt{K}r)bigg] bigg[d_1cos(sqrt Ktheta)+d_2sin(sqrt Ktheta)bigg]. $$
                Let $u(r,0)=u(r,2pi)=0$ and then one has
                $$ d_1=0, K=frac{n^2}{4},n=1,2,3cdots. $$
                You can choose $K=frac14$ or $K=1$. Let $u(a,theta)=0,u(b,theta)=f(theta)$ and I think you can determine $C_1,C_2$ which is not very hard.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 30 '18 at 20:11









                xpaulxpaul

                22.5k24455




                22.5k24455






























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