Enforcing amplitude profile on complex solution of PDE












0












$begingroup$


Consider the following PDE.



$$psi_z = ipsi_{xx}$$



Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be



$$a(x,z)F_x + b(x,z)F_z=0$$



where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.



Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?



What I have done so far is as follows:



Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.



The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.



$$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
$$Aphi_z = A_{xx} - (phi_x)^2A$$



In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.



So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.



Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.



At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).



So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.



Related question, also asked by me: here










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Consider the following PDE.



    $$psi_z = ipsi_{xx}$$



    Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be



    $$a(x,z)F_x + b(x,z)F_z=0$$



    where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.



    Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?



    What I have done so far is as follows:



    Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.



    The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.



    $$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
    $$Aphi_z = A_{xx} - (phi_x)^2A$$



    In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.



    So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.



    Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.



    At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).



    So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.



    Related question, also asked by me: here










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Consider the following PDE.



      $$psi_z = ipsi_{xx}$$



      Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be



      $$a(x,z)F_x + b(x,z)F_z=0$$



      where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.



      Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?



      What I have done so far is as follows:



      Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.



      The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.



      $$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
      $$Aphi_z = A_{xx} - (phi_x)^2A$$



      In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.



      So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.



      Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.



      At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).



      So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.



      Related question, also asked by me: here










      share|cite|improve this question









      $endgroup$




      Consider the following PDE.



      $$psi_z = ipsi_{xx}$$



      Suppose that I can generate the function $F=|psi|^2$ using some other PDE. At this point I am considering this PDE to be



      $$a(x,z)F_x + b(x,z)F_z=0$$



      where by choosing $a$, $b$ and an initial condition $F(x,z=0)$, I generate the desired $F$.



      Once this is done, I want to use the initial PDE as follows. At $z=0$ I want to compute the phase so that the resulting solution will satisfy $|psi|^2=F$. Is there such a method?



      What I have done so far is as follows:



      Suppose that I want a solution such that $F(x,z) = F(x-frac{z^n}{n!})$, for some $nin mathbb{N}$, and as a starting point, consider the case $n = 1$. For this particular case, the choice on $a$ and $b$ is $a = b = 1$.



      The solution of the initial PDE can be considered of this form $psi(x,z) = A(x - z)cdot exp(iphi(x,z))$, and by doing so two differential equations result. I have used the property that $A_z = -A_x$ since $A$ depends on $(x-z)$.



      $$a(x,z) A_x = 2 phi_x A_x + phi_{xx}A$$
      $$Aphi_z = A_{xx} - (phi_x)^2A$$



      In this particular case I mention that those equations can be derived directly by assuming the solution of the above mentioned form ($psi(x,z) = A(x - z)cdot exp(iphi(x,z))$) and then by plugging it in the first PDE, but for more general scenarios I assume that this approach is more general. Coming back to the result, I consider that at $z=0$ I have $a=a_0$. For that value of $a$ I can compute numerically the phase $phi$ and it works. In my particular case that value is $1$.



      So what did I get out of this? The solution given by computing $F$ for the particular case moves in the $(x,z)$ space at an angle $theta$ that is given by $tan(theta) = a/b = 1$. By changing the value of $a$ I also change the angle $theta$. What I don't get is the spread of the solution when using the first PDE to compute the solution for an initial condition that has the amplitude $A$ and phase $phi$ that I have considered/computed, but I have the angle.



      Now, assuming I take $a=z$ and following the same reasoning, when evaluating the two differential equations above at some point $z=0$ in order to compute the phase, the resulting solution of PDE 1 with that phase has a linear trajectory, as in the first case, although the solution given by computing $F$ has a parabolic trajectory. By changing the value of $z$ for some arbitrary value $z_0$, the angle still changes but the trajectory is still linear.



      At this point I considered differentiating $a(x,z)F_x + b(x,z)F_z=0$ one time and then to evaluate at $z=0$ but the computation becomes way to complicated (terms like $A_{xxxx}$ start to appear).



      So coming back to the main question, do you know other approaches that do something similar to what I'm trying to do? Some terminology for googling the issue is welcomed.



      Related question, also asked by me: here







      pde






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 15 at 13:59









      Victor PaleaVictor Palea

      301312




      301312






















          0






          active

          oldest

          votes












          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074464%2fenforcing-amplitude-profile-on-complex-solution-of-pde%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          0






          active

          oldest

          votes








          0






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074464%2fenforcing-amplitude-profile-on-complex-solution-of-pde%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Human spaceflight

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          張江高科駅