Solving a Cauchy Problem using Method of Characteristics

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I came across a partial differential equation (IVP PDE) that I would like to solve:




$$ u(x,0)=0$$




This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:




$$au_x+bu_y=c$$
Such that a, b, and c are constants.




In my particular case, I have:




$$
left{
beginarrayc
a=cos(ky) \
b=1 \
c=ax^2
endarray
right.
$$
Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
Rearranging gives: $$fracdxdy=cos(ky)$$
$$dy=fracduax^2$$
I then canclulated $dx-dy$:
$$du=(ax^2cos(ky)+1)dy-dx$$
Integrate to obtain $u(x,y,a,k)$:
$$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
The question is thus finalised.




This however does not look correct. Am I doing the right thing?



I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.







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    up vote
    2
    down vote

    favorite
    2












    I came across a partial differential equation (IVP PDE) that I would like to solve:




    $$ u(x,0)=0$$




    This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:




    $$au_x+bu_y=c$$
    Such that a, b, and c are constants.




    In my particular case, I have:




    $$
    left{
    beginarrayc
    a=cos(ky) \
    b=1 \
    c=ax^2
    endarray
    right.
    $$
    Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
    Rearranging gives: $$fracdxdy=cos(ky)$$
    $$dy=fracduax^2$$
    I then canclulated $dx-dy$:
    $$du=(ax^2cos(ky)+1)dy-dx$$
    Integrate to obtain $u(x,y,a,k)$:
    $$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
    I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
    The question is thus finalised.




    This however does not look correct. Am I doing the right thing?



    I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.







    share|cite|improve this question






















      up vote
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      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2





      I came across a partial differential equation (IVP PDE) that I would like to solve:




      $$ u(x,0)=0$$




      This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:




      $$au_x+bu_y=c$$
      Such that a, b, and c are constants.




      In my particular case, I have:




      $$
      left{
      beginarrayc
      a=cos(ky) \
      b=1 \
      c=ax^2
      endarray
      right.
      $$
      Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
      Rearranging gives: $$fracdxdy=cos(ky)$$
      $$dy=fracduax^2$$
      I then canclulated $dx-dy$:
      $$du=(ax^2cos(ky)+1)dy-dx$$
      Integrate to obtain $u(x,y,a,k)$:
      $$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
      I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
      The question is thus finalised.




      This however does not look correct. Am I doing the right thing?



      I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.







      share|cite|improve this question












      I came across a partial differential equation (IVP PDE) that I would like to solve:




      $$ u(x,0)=0$$




      This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:




      $$au_x+bu_y=c$$
      Such that a, b, and c are constants.




      In my particular case, I have:




      $$
      left{
      beginarrayc
      a=cos(ky) \
      b=1 \
      c=ax^2
      endarray
      right.
      $$
      Using the Lagrange-Carpit Equations: $$fracdxcos(ky)=fracdy1=fracduax^2$$
      Rearranging gives: $$fracdxdy=cos(ky)$$
      $$dy=fracduax^2$$
      I then canclulated $dx-dy$:
      $$du=(ax^2cos(ky)+1)dy-dx$$
      Integrate to obtain $u(x,y,a,k)$:
      $$u(x,y,a,k)=frac1kax^2sin(ky)+y-x+C_1$$
      I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
      The question is thus finalised.




      This however does not look correct. Am I doing the right thing?



      I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $fracdyds$, $fracdxds$ and $fracdzds$.









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      asked Aug 29 at 10:26









      Hypergeometry

      607




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