charpits equations to solve $u_x u_y=1$












0














$u_x u_y=1$ with initial data $u=0$ and $x+y=1$



Parametrise with $x=s$ and $y=1-s$



diffeq can be written as $F(p,q) = pq - 1 = 0$



Charpits equations give:
$dx/dt = q$,
$dy/dt = p$,
$dp/dt = 0$,
$dq/dt = 0$,
$du/dt = 2pq$



Solving these we get $p=p_0(s)$ and $q=q_0(s)$
and $x = q_0(s)t + x_0 = q_0(s)t + s $
and $y = p_0(s)t + y_0 = p_0(s)t + 1-s $
and $u = 2p_0(s)q_0(s)t$



$F(p_0,q_0) = 0 $ implies that $p_0q_0 = 1$
and $du_0(s)/ds = p_0(dx_0/ds)+ q_0(dy_0/ds)$ implies $p_0 = q_0$



As a result $p_0 = q_0 = -1$ or $1 $ and so $u=2t$



How do i then go about stating where each solution exists and further proving that for a real solution to exist, $[du_0(s) / ds ]^2 geq 4 [dx_0(s)/ds][dy_0(s)/ds]$






differential-equations pde







share|cite|improve this question



























    0














    $u_x u_y=1$ with initial data $u=0$ and $x+y=1$



    Parametrise with $x=s$ and $y=1-s$



    diffeq can be written as $F(p,q) = pq - 1 = 0$



    Charpits equations give:
    $dx/dt = q$,
    $dy/dt = p$,
    $dp/dt = 0$,
    $dq/dt = 0$,
    $du/dt = 2pq$



    Solving these we get $p=p_0(s)$ and $q=q_0(s)$
    and $x = q_0(s)t + x_0 = q_0(s)t + s $
    and $y = p_0(s)t + y_0 = p_0(s)t + 1-s $
    and $u = 2p_0(s)q_0(s)t$



    $F(p_0,q_0) = 0 $ implies that $p_0q_0 = 1$
    and $du_0(s)/ds = p_0(dx_0/ds)+ q_0(dy_0/ds)$ implies $p_0 = q_0$



    As a result $p_0 = q_0 = -1$ or $1 $ and so $u=2t$



    How do i then go about stating where each solution exists and further proving that for a real solution to exist, $[du_0(s) / ds ]^2 geq 4 [dx_0(s)/ds][dy_0(s)/ds]$






    differential-equations pde







    share|cite|improve this question

























      0












      0








      0







      $u_x u_y=1$ with initial data $u=0$ and $x+y=1$



      Parametrise with $x=s$ and $y=1-s$



      diffeq can be written as $F(p,q) = pq - 1 = 0$



      Charpits equations give:
      $dx/dt = q$,
      $dy/dt = p$,
      $dp/dt = 0$,
      $dq/dt = 0$,
      $du/dt = 2pq$



      Solving these we get $p=p_0(s)$ and $q=q_0(s)$
      and $x = q_0(s)t + x_0 = q_0(s)t + s $
      and $y = p_0(s)t + y_0 = p_0(s)t + 1-s $
      and $u = 2p_0(s)q_0(s)t$



      $F(p_0,q_0) = 0 $ implies that $p_0q_0 = 1$
      and $du_0(s)/ds = p_0(dx_0/ds)+ q_0(dy_0/ds)$ implies $p_0 = q_0$



      As a result $p_0 = q_0 = -1$ or $1 $ and so $u=2t$



      How do i then go about stating where each solution exists and further proving that for a real solution to exist, $[du_0(s) / ds ]^2 geq 4 [dx_0(s)/ds][dy_0(s)/ds]$






      differential-equations pde







      share|cite|improve this question













      $u_x u_y=1$ with initial data $u=0$ and $x+y=1$



      Parametrise with $x=s$ and $y=1-s$



      diffeq can be written as $F(p,q) = pq - 1 = 0$



      Charpits equations give:
      $dx/dt = q$,
      $dy/dt = p$,
      $dp/dt = 0$,
      $dq/dt = 0$,
      $du/dt = 2pq$



      Solving these we get $p=p_0(s)$ and $q=q_0(s)$
      and $x = q_0(s)t + x_0 = q_0(s)t + s $
      and $y = p_0(s)t + y_0 = p_0(s)t + 1-s $
      and $u = 2p_0(s)q_0(s)t$



      $F(p_0,q_0) = 0 $ implies that $p_0q_0 = 1$
      and $du_0(s)/ds = p_0(dx_0/ds)+ q_0(dy_0/ds)$ implies $p_0 = q_0$



      As a result $p_0 = q_0 = -1$ or $1 $ and so $u=2t$



      How do i then go about stating where each solution exists and further proving that for a real solution to exist, $[du_0(s) / ds ]^2 geq 4 [dx_0(s)/ds][dy_0(s)/ds]$






      differential-equations pde




      differential-equations pde






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 at 14:30









      pablo_mathscobar

      736




      736






















          1 Answer
          1






          active

          oldest

          votes


















          0














          For general initial conditions $z_0(s)=u(x_0(s),y_0(s))$ you get the system of equations
          $$
          p_0(s)q_0(s)=1\
          p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=dot z_0(s)
          $$
          which looks almost like the Viete's equations for a quadratic equation. Multiply the first equation with $dot x_0dot y_0$ to get that $p_0(s)dot x_0(s)$ and $q_0(s)dot y_0(s)$ are the root pair of
          $$
          r^2-dot z_0cdot r + dot x_0dot y_0=0.
          $$
          Now check the discriminant for a condition to get real solutions.






          share|cite|improve this answer





















          • Cheers, ill try that. Is the rest of my answer correct?
            – pablo_mathscobar
            Nov 24 at 14:54










          • Yes, and as $0>-1$, the condition is satisfied.
            – LutzL
            Nov 24 at 14:57










          • Hi in struggling to see how you got the equation involving r^2, how did you come to that part
            – pablo_mathscobar
            Nov 24 at 16:14










          • By Viete. $(r-r_1)(r-r_2)=r^2-(r_1+r_2)r+r_1r_2$.
            – LutzL
            Nov 24 at 16:36











          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%2f3011621%2fcharpits-equations-to-solve-u-x-u-y-1%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          For general initial conditions $z_0(s)=u(x_0(s),y_0(s))$ you get the system of equations
          $$
          p_0(s)q_0(s)=1\
          p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=dot z_0(s)
          $$
          which looks almost like the Viete's equations for a quadratic equation. Multiply the first equation with $dot x_0dot y_0$ to get that $p_0(s)dot x_0(s)$ and $q_0(s)dot y_0(s)$ are the root pair of
          $$
          r^2-dot z_0cdot r + dot x_0dot y_0=0.
          $$
          Now check the discriminant for a condition to get real solutions.






          share|cite|improve this answer





















          • Cheers, ill try that. Is the rest of my answer correct?
            – pablo_mathscobar
            Nov 24 at 14:54










          • Yes, and as $0>-1$, the condition is satisfied.
            – LutzL
            Nov 24 at 14:57










          • Hi in struggling to see how you got the equation involving r^2, how did you come to that part
            – pablo_mathscobar
            Nov 24 at 16:14










          • By Viete. $(r-r_1)(r-r_2)=r^2-(r_1+r_2)r+r_1r_2$.
            – LutzL
            Nov 24 at 16:36
















          0














          For general initial conditions $z_0(s)=u(x_0(s),y_0(s))$ you get the system of equations
          $$
          p_0(s)q_0(s)=1\
          p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=dot z_0(s)
          $$
          which looks almost like the Viete's equations for a quadratic equation. Multiply the first equation with $dot x_0dot y_0$ to get that $p_0(s)dot x_0(s)$ and $q_0(s)dot y_0(s)$ are the root pair of
          $$
          r^2-dot z_0cdot r + dot x_0dot y_0=0.
          $$
          Now check the discriminant for a condition to get real solutions.






          share|cite|improve this answer





















          • Cheers, ill try that. Is the rest of my answer correct?
            – pablo_mathscobar
            Nov 24 at 14:54










          • Yes, and as $0>-1$, the condition is satisfied.
            – LutzL
            Nov 24 at 14:57










          • Hi in struggling to see how you got the equation involving r^2, how did you come to that part
            – pablo_mathscobar
            Nov 24 at 16:14










          • By Viete. $(r-r_1)(r-r_2)=r^2-(r_1+r_2)r+r_1r_2$.
            – LutzL
            Nov 24 at 16:36














          0












          0








          0






          For general initial conditions $z_0(s)=u(x_0(s),y_0(s))$ you get the system of equations
          $$
          p_0(s)q_0(s)=1\
          p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=dot z_0(s)
          $$
          which looks almost like the Viete's equations for a quadratic equation. Multiply the first equation with $dot x_0dot y_0$ to get that $p_0(s)dot x_0(s)$ and $q_0(s)dot y_0(s)$ are the root pair of
          $$
          r^2-dot z_0cdot r + dot x_0dot y_0=0.
          $$
          Now check the discriminant for a condition to get real solutions.






          share|cite|improve this answer












          For general initial conditions $z_0(s)=u(x_0(s),y_0(s))$ you get the system of equations
          $$
          p_0(s)q_0(s)=1\
          p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=dot z_0(s)
          $$
          which looks almost like the Viete's equations for a quadratic equation. Multiply the first equation with $dot x_0dot y_0$ to get that $p_0(s)dot x_0(s)$ and $q_0(s)dot y_0(s)$ are the root pair of
          $$
          r^2-dot z_0cdot r + dot x_0dot y_0=0.
          $$
          Now check the discriminant for a condition to get real solutions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 14:52









          LutzL

          55.3k42053




          55.3k42053












          • Cheers, ill try that. Is the rest of my answer correct?
            – pablo_mathscobar
            Nov 24 at 14:54










          • Yes, and as $0>-1$, the condition is satisfied.
            – LutzL
            Nov 24 at 14:57










          • Hi in struggling to see how you got the equation involving r^2, how did you come to that part
            – pablo_mathscobar
            Nov 24 at 16:14










          • By Viete. $(r-r_1)(r-r_2)=r^2-(r_1+r_2)r+r_1r_2$.
            – LutzL
            Nov 24 at 16:36


















          • Cheers, ill try that. Is the rest of my answer correct?
            – pablo_mathscobar
            Nov 24 at 14:54










          • Yes, and as $0>-1$, the condition is satisfied.
            – LutzL
            Nov 24 at 14:57










          • Hi in struggling to see how you got the equation involving r^2, how did you come to that part
            – pablo_mathscobar
            Nov 24 at 16:14










          • By Viete. $(r-r_1)(r-r_2)=r^2-(r_1+r_2)r+r_1r_2$.
            – LutzL
            Nov 24 at 16:36
















          Cheers, ill try that. Is the rest of my answer correct?
          – pablo_mathscobar
          Nov 24 at 14:54




          Cheers, ill try that. Is the rest of my answer correct?
          – pablo_mathscobar
          Nov 24 at 14:54












          Yes, and as $0>-1$, the condition is satisfied.
          – LutzL
          Nov 24 at 14:57




          Yes, and as $0>-1$, the condition is satisfied.
          – LutzL
          Nov 24 at 14:57












          Hi in struggling to see how you got the equation involving r^2, how did you come to that part
          – pablo_mathscobar
          Nov 24 at 16:14




          Hi in struggling to see how you got the equation involving r^2, how did you come to that part
          – pablo_mathscobar
          Nov 24 at 16:14












          By Viete. $(r-r_1)(r-r_2)=r^2-(r_1+r_2)r+r_1r_2$.
          – LutzL
          Nov 24 at 16:36




          By Viete. $(r-r_1)(r-r_2)=r^2-(r_1+r_2)r+r_1r_2$.
          – LutzL
          Nov 24 at 16:36


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3011621%2fcharpits-equations-to-solve-u-x-u-y-1%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

          Index of /

          This page is only for reference, If you need detailed information, please check here
          Read more

          Tribalistas

          Image
            Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
          Read more

          Listed building

          Image
          Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
          Read more