PDE with complex roots of characterictic solution












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I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.



As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
$$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$






pde linear-pde







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    $begingroup$


    I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.



    As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
    $$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$






    pde linear-pde







    share|cite|improve this question









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      0












      0








      0





      $begingroup$


      I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.



      As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
      $$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$






      pde linear-pde







      share|cite|improve this question









      $endgroup$




      I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.



      As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
      $$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$






      pde linear-pde




      pde linear-pde






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      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 19 at 5:27









      Piyush DivyanakarPiyush Divyanakar

      3,338326




      3,338326






















          1 Answer
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          3












          $begingroup$

          $z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.



          I cannot see how you get a so complicated particular solution :
          $$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
          Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?



          Without having the details of your calculus it is not possible to find where a possible mistake occurred.



          There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
          $$z_p=f(x)+g(y)$$
          $$f''(x)+g''(y)=-4pi(x^2+y^2)$$
          $$begin{cases}
          f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
          g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
          end{cases}$$
          No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
          $$z_p=-frac{pi}{3} (x^4+y^4)$$
          $$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
          Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
            $endgroup$
            – Piyush Divyanakar
            Jan 20 at 5:34










          • $begingroup$
            It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
            $endgroup$
            – JJacquelin
            Jan 20 at 7:19











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          $z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.



          I cannot see how you get a so complicated particular solution :
          $$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
          Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?



          Without having the details of your calculus it is not possible to find where a possible mistake occurred.



          There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
          $$z_p=f(x)+g(y)$$
          $$f''(x)+g''(y)=-4pi(x^2+y^2)$$
          $$begin{cases}
          f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
          g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
          end{cases}$$
          No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
          $$z_p=-frac{pi}{3} (x^4+y^4)$$
          $$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
          Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
            $endgroup$
            – Piyush Divyanakar
            Jan 20 at 5:34










          • $begingroup$
            It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
            $endgroup$
            – JJacquelin
            Jan 20 at 7:19
















          3












          $begingroup$

          $z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.



          I cannot see how you get a so complicated particular solution :
          $$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
          Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?



          Without having the details of your calculus it is not possible to find where a possible mistake occurred.



          There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
          $$z_p=f(x)+g(y)$$
          $$f''(x)+g''(y)=-4pi(x^2+y^2)$$
          $$begin{cases}
          f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
          g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
          end{cases}$$
          No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
          $$z_p=-frac{pi}{3} (x^4+y^4)$$
          $$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
          Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
            $endgroup$
            – Piyush Divyanakar
            Jan 20 at 5:34










          • $begingroup$
            It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
            $endgroup$
            – JJacquelin
            Jan 20 at 7:19














          3












          3








          3





          $begingroup$

          $z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.



          I cannot see how you get a so complicated particular solution :
          $$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
          Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?



          Without having the details of your calculus it is not possible to find where a possible mistake occurred.



          There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
          $$z_p=f(x)+g(y)$$
          $$f''(x)+g''(y)=-4pi(x^2+y^2)$$
          $$begin{cases}
          f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
          g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
          end{cases}$$
          No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
          $$z_p=-frac{pi}{3} (x^4+y^4)$$
          $$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
          Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .






          share|cite|improve this answer











          $endgroup$



          $z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.



          I cannot see how you get a so complicated particular solution :
          $$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
          Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?



          Without having the details of your calculus it is not possible to find where a possible mistake occurred.



          There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
          $$z_p=f(x)+g(y)$$
          $$f''(x)+g''(y)=-4pi(x^2+y^2)$$
          $$begin{cases}
          f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
          g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
          end{cases}$$
          No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
          $$z_p=-frac{pi}{3} (x^4+y^4)$$
          $$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
          Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 19 at 8:43

























          answered Jan 19 at 8:33









          JJacquelinJJacquelin

          43.9k21853




          43.9k21853












          • $begingroup$
            I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
            $endgroup$
            – Piyush Divyanakar
            Jan 20 at 5:34










          • $begingroup$
            It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
            $endgroup$
            – JJacquelin
            Jan 20 at 7:19


















          • $begingroup$
            I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
            $endgroup$
            – Piyush Divyanakar
            Jan 20 at 5:34










          • $begingroup$
            It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
            $endgroup$
            – JJacquelin
            Jan 20 at 7:19
















          $begingroup$
          I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
          $endgroup$
          – Piyush Divyanakar
          Jan 20 at 5:34




          $begingroup$
          I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
          $endgroup$
          – Piyush Divyanakar
          Jan 20 at 5:34












          $begingroup$
          It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
          $endgroup$
          – JJacquelin
          Jan 20 at 7:19




          $begingroup$
          It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
          $endgroup$
          – JJacquelin
          Jan 20 at 7:19


















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