Tricomi equation canonical form and solution












1












$begingroup$


Consider the Tricomi equation



$$u_{xx}+xu_{yy}=0$$



With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:



$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$










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$endgroup$












  • $begingroup$
    This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
    $endgroup$
    – Autolatry
    Nov 18 '14 at 14:52
















1












$begingroup$


Consider the Tricomi equation



$$u_{xx}+xu_{yy}=0$$



With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:



$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$










share|cite|improve this question











$endgroup$












  • $begingroup$
    This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
    $endgroup$
    – Autolatry
    Nov 18 '14 at 14:52














1












1








1


1



$begingroup$


Consider the Tricomi equation



$$u_{xx}+xu_{yy}=0$$



With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:



$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$










share|cite|improve this question











$endgroup$




Consider the Tricomi equation



$$u_{xx}+xu_{yy}=0$$



With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:



$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$







pde linear-pde






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edited Dec 12 '18 at 18:26









Harry49

7,42331340




7,42331340










asked Nov 18 '14 at 14:42









Joemak BoblinkeJoemak Boblinke

895




895












  • $begingroup$
    This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
    $endgroup$
    – Autolatry
    Nov 18 '14 at 14:52


















  • $begingroup$
    This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
    $endgroup$
    – Autolatry
    Nov 18 '14 at 14:52
















$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52




$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52










1 Answer
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$begingroup$

In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
$$
u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
$$

The functions $r,q$ satisfy the Beltrami differential equations
$$
r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
$$

Particular solutions of the Tricomi equation can be found here.





Further reading: p. 162-163 of



R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234






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    $begingroup$

    In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
    $$
    u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
    $$

    The functions $r,q$ satisfy the Beltrami differential equations
    $$
    r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
    $$

    Particular solutions of the Tricomi equation can be found here.





    Further reading: p. 162-163 of



    R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
      $$
      u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
      $$

      The functions $r,q$ satisfy the Beltrami differential equations
      $$
      r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
      $$

      Particular solutions of the Tricomi equation can be found here.





      Further reading: p. 162-163 of



      R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
        $$
        u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
        $$

        The functions $r,q$ satisfy the Beltrami differential equations
        $$
        r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
        $$

        Particular solutions of the Tricomi equation can be found here.





        Further reading: p. 162-163 of



        R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234






        share|cite|improve this answer











        $endgroup$



        In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
        $$
        u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
        $$

        The functions $r,q$ satisfy the Beltrami differential equations
        $$
        r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
        $$

        Particular solutions of the Tricomi equation can be found here.





        Further reading: p. 162-163 of



        R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 12 '18 at 18:32

























        answered Dec 12 '18 at 13:10









        Harry49Harry49

        7,42331340




        7,42331340






























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