charpits method to solve $u_x^2 + yu_y = u$












1














$u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



Charpits equations then give me:
$dx/dt = 2p$,
$dy/dt = y$,
$dp/dt = p$,
$dq/dt = 0 $,
$du/dt = 2p^2 + qy$



How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






differential-equations pde partial-derivative characteristics







share|cite|improve this question





























    1














    $u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



    Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



    Charpits equations then give me:
    $dx/dt = 2p$,
    $dy/dt = y$,
    $dp/dt = p$,
    $dq/dt = 0 $,
    $du/dt = 2p^2 + qy$



    How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






    differential-equations pde partial-derivative characteristics







    share|cite|improve this question



























      1












      1








      1







      $u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



      Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



      Charpits equations then give me:
      $dx/dt = 2p$,
      $dy/dt = y$,
      $dp/dt = p$,
      $dq/dt = 0 $,
      $du/dt = 2p^2 + qy$



      How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






      differential-equations pde partial-derivative characteristics







      share|cite|improve this question















      $u_x^2 + yu_y = u$ subject to $u(x,1) = 1 +x^2 /4$ for $-infty < x < infty $



      Setting $p = du/dx$ and $q = du/dy$, I get $p^2 +yq = u$ and so I am able to write the diffeq as $F(p,q,y,u) = p^2 + qy - u = 0 $



      Charpits equations then give me:
      $dx/dt = 2p$,
      $dy/dt = y$,
      $dp/dt = p$,
      $dq/dt = 0 $,
      $du/dt = 2p^2 + qy$



      How would I go about solving these equations in order to find u in explicit form and then find where it is uniquely defined by the data?






      differential-equations pde partial-derivative characteristics




      differential-equations pde partial-derivative characteristics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 24 at 13:07

























      asked Nov 24 at 13:02









      pablo_mathscobar

      786




      786






















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          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer























          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11











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          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer























          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11
















          2














          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer























          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11














          2












          2








          2






          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.






          share|cite|improve this answer














          You should be able to see that some equations can be solved directly, $y=y_0e^t$, $p=p_0e^t$, $q=q_0$, and then insert into the remaining equations, $$x=x_0+2p_0(e^t-1),$$ $$z=z_0+p_0^2(e^{2t}-1)+q_0y_0(e^t-1).$$



          Now apply $$dz=p,dx+q,dyimplies dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)$$ along the initial curve $x_0(s)=s$, $y_0(s)=1$, $z_0(s)=u(x_0(s),y_0(s))=1+frac{s^2}4$ to get $$frac{x_0}2=p_0cdot 1+q_0cdot 0.$$ Using the original equation $p_0^2+q_0=z_0$ gives $q_0=1$. Thus all collected,
          begin{align}
          x&=x_0y\
          z=u(x,y)&=1+frac{x_0^2}4+frac{x_0^2}4(y^2-1)+(y-1)=frac{x^2}4+y
          end{align}
          etc.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 13:53

























          answered Nov 24 at 13:08









          LutzL

          55.4k42053




          55.4k42053












          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11


















          • I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
            – pablo_mathscobar
            Nov 24 at 13:23












          • Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
            – LutzL
            Nov 24 at 13:39










          • when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
            – pablo_mathscobar
            Nov 24 at 13:44












          • Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
            – LutzL
            Nov 24 at 13:54










          • brilliant thanks
            – pablo_mathscobar
            Nov 24 at 14:11
















          I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
          – pablo_mathscobar
          Nov 24 at 13:23






          I think you assumed that $du/dt=2p2+qx$ and not $du/dt=2p2+qy$ Solving $du/dt=2p2+qy$ is $u = p_0^2 e^2t + q_0y_0e^t + u_0$ If $u_0=u(x,1)=1+x^2/4$, how do I find $p_0$ and $q_0$, is $p_0 = x/2$ and $q_0=0$
          – pablo_mathscobar
          Nov 24 at 13:23














          Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
          – LutzL
          Nov 24 at 13:39




          Corrected. The initial curve has $x_0(s)=s$, $y_0(s)=1$ and $z_0(s)=1+s^2/4$. Now insert into $$dot z_0(s)=p_0(s)dot x_0(s)+q(s)dot y_0(s)$$ to find $p_0(s)=s/2$. Then use the original equation to find $q_0(s)=1$.
          – LutzL
          Nov 24 at 13:39












          when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
          – pablo_mathscobar
          Nov 24 at 13:44






          when solving $dx/dt = 2p_0 e^t$ we get $x = 2p_0e^t + C_1$ since $x_0 = s$ and $p_0 = s/2$ surely this should give $C_1 = 0$ at t=0 and so the equation for x should be $x = 2p_0e^t $ and not $x=x0+2p_0e^t = s + 2p_0e^t$ as you claim
          – pablo_mathscobar
          Nov 24 at 13:44














          Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
          – LutzL
          Nov 24 at 13:54




          Yes, if you integrate $e^t$ you get $(e^t-1)$ etc.
          – LutzL
          Nov 24 at 13:54












          brilliant thanks
          – pablo_mathscobar
          Nov 24 at 14:11




          brilliant thanks
          – pablo_mathscobar
          Nov 24 at 14:11


















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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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