Quasilinear PDE using method of characteristics












0












$begingroup$


The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$



I first find that



$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$



Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$



Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$



Applying the initial conditions yields
$A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$



This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,



$A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$



Is this the correct solution?



Thanks!






pde characteristics







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$



    I first find that



    $frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$



    Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$



    Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$



    Applying the initial conditions yields
    $A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$



    This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,



    $A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$



    Is this the correct solution?



    Thanks!






    pde characteristics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$



      I first find that



      $frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$



      Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$



      Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$



      Applying the initial conditions yields
      $A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$



      This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,



      $A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$



      Is this the correct solution?



      Thanks!






      pde characteristics







      share|cite|improve this question











      $endgroup$




      The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$



      I first find that



      $frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$



      Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$



      Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$



      Applying the initial conditions yields
      $A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$



      This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,



      $A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$



      Is this the correct solution?



      Thanks!






      pde characteristics




      pde characteristics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 22:00









      Harry49

      8,38931344




      8,38931344










      asked Dec 11 '18 at 16:40









      Brad ScottBrad Scott

      1548




      1548






















          2 Answers
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          active

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          1












          $begingroup$

          No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.



          You can solve the first and last quotient as
          $$
          x,dx = -duimplies 2u+x^2=c_1
          $$
          This you can then insert into the other equations.
          $$
          0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
          $$
          Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
          $$
          ϕ(2u+x^2)=2xu+frac23x^3-y^2.
          $$




          With the initial conditions $u=y$, $x=0$ this reduces to
          $$
          ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
          $$
          so that the surface follows the equation
          $$
          -(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
          $$
          which is a quadratic equation for the value of $u$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            how do you know which equations to solve?
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 17:39










          • $begingroup$
            You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:43












          • $begingroup$
            You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:49





















          1












          $begingroup$

          $$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
          First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
          $$u+frac{x^2}{2}=c_1$$
          Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :



          $c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$



          $(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$



          $$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
          General solution on the form of implicite equation :
          $$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
          $F$ is an arbitrary function to be determined according to the boundary condition.



          Condition $u(0,y)=y$ :



          $0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$



          The function $F$ is determined :
          $$F(X)=-frac{X^2}{2}$$
          We put it into the general solution where $X=u+frac{x^2}{2}$
          $$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
          Solving for $u$ :
          $$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$



          This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 19:03












          • $begingroup$
            In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 21:55












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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.



          You can solve the first and last quotient as
          $$
          x,dx = -duimplies 2u+x^2=c_1
          $$
          This you can then insert into the other equations.
          $$
          0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
          $$
          Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
          $$
          ϕ(2u+x^2)=2xu+frac23x^3-y^2.
          $$




          With the initial conditions $u=y$, $x=0$ this reduces to
          $$
          ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
          $$
          so that the surface follows the equation
          $$
          -(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
          $$
          which is a quadratic equation for the value of $u$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            how do you know which equations to solve?
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 17:39










          • $begingroup$
            You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:43












          • $begingroup$
            You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:49


















          1












          $begingroup$

          No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.



          You can solve the first and last quotient as
          $$
          x,dx = -duimplies 2u+x^2=c_1
          $$
          This you can then insert into the other equations.
          $$
          0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
          $$
          Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
          $$
          ϕ(2u+x^2)=2xu+frac23x^3-y^2.
          $$




          With the initial conditions $u=y$, $x=0$ this reduces to
          $$
          ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
          $$
          so that the surface follows the equation
          $$
          -(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
          $$
          which is a quadratic equation for the value of $u$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            how do you know which equations to solve?
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 17:39










          • $begingroup$
            You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:43












          • $begingroup$
            You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:49
















          1












          1








          1





          $begingroup$

          No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.



          You can solve the first and last quotient as
          $$
          x,dx = -duimplies 2u+x^2=c_1
          $$
          This you can then insert into the other equations.
          $$
          0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
          $$
          Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
          $$
          ϕ(2u+x^2)=2xu+frac23x^3-y^2.
          $$




          With the initial conditions $u=y$, $x=0$ this reduces to
          $$
          ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
          $$
          so that the surface follows the equation
          $$
          -(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
          $$
          which is a quadratic equation for the value of $u$.






          share|cite|improve this answer











          $endgroup$



          No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.



          You can solve the first and last quotient as
          $$
          x,dx = -duimplies 2u+x^2=c_1
          $$
          This you can then insert into the other equations.
          $$
          0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
          $$
          Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
          $$
          ϕ(2u+x^2)=2xu+frac23x^3-y^2.
          $$




          With the initial conditions $u=y$, $x=0$ this reduces to
          $$
          ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
          $$
          so that the surface follows the equation
          $$
          -(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
          $$
          which is a quadratic equation for the value of $u$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 18:03

























          answered Dec 11 '18 at 17:37









          LutzLLutzL

          60.1k42057




          60.1k42057












          • $begingroup$
            how do you know which equations to solve?
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 17:39










          • $begingroup$
            You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:43












          • $begingroup$
            You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:49




















          • $begingroup$
            how do you know which equations to solve?
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 17:39










          • $begingroup$
            You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:43












          • $begingroup$
            You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
            $endgroup$
            – LutzL
            Dec 11 '18 at 17:49


















          $begingroup$
          how do you know which equations to solve?
          $endgroup$
          – Brad Scott
          Dec 11 '18 at 17:39




          $begingroup$
          how do you know which equations to solve?
          $endgroup$
          – Brad Scott
          Dec 11 '18 at 17:39












          $begingroup$
          You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
          $endgroup$
          – LutzL
          Dec 11 '18 at 17:43






          $begingroup$
          You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
          $endgroup$
          – LutzL
          Dec 11 '18 at 17:43














          $begingroup$
          You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
          $endgroup$
          – LutzL
          Dec 11 '18 at 17:49






          $begingroup$
          You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
          $endgroup$
          – LutzL
          Dec 11 '18 at 17:49













          1












          $begingroup$

          $$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
          First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
          $$u+frac{x^2}{2}=c_1$$
          Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :



          $c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$



          $(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$



          $$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
          General solution on the form of implicite equation :
          $$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
          $F$ is an arbitrary function to be determined according to the boundary condition.



          Condition $u(0,y)=y$ :



          $0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$



          The function $F$ is determined :
          $$F(X)=-frac{X^2}{2}$$
          We put it into the general solution where $X=u+frac{x^2}{2}$
          $$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
          Solving for $u$ :
          $$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$



          This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 19:03












          • $begingroup$
            In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 21:55
















          1












          $begingroup$

          $$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
          First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
          $$u+frac{x^2}{2}=c_1$$
          Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :



          $c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$



          $(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$



          $$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
          General solution on the form of implicite equation :
          $$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
          $F$ is an arbitrary function to be determined according to the boundary condition.



          Condition $u(0,y)=y$ :



          $0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$



          The function $F$ is determined :
          $$F(X)=-frac{X^2}{2}$$
          We put it into the general solution where $X=u+frac{x^2}{2}$
          $$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
          Solving for $u$ :
          $$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$



          This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 19:03












          • $begingroup$
            In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 21:55














          1












          1








          1





          $begingroup$

          $$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
          First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
          $$u+frac{x^2}{2}=c_1$$
          Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :



          $c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$



          $(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$



          $$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
          General solution on the form of implicite equation :
          $$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
          $F$ is an arbitrary function to be determined according to the boundary condition.



          Condition $u(0,y)=y$ :



          $0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$



          The function $F$ is determined :
          $$F(X)=-frac{X^2}{2}$$
          We put it into the general solution where $X=u+frac{x^2}{2}$
          $$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
          Solving for $u$ :
          $$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$



          This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.






          share|cite|improve this answer











          $endgroup$



          $$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
          First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
          $$u+frac{x^2}{2}=c_1$$
          Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :



          $c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$



          $(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$



          $$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
          General solution on the form of implicite equation :
          $$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
          $F$ is an arbitrary function to be determined according to the boundary condition.



          Condition $u(0,y)=y$ :



          $0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$



          The function $F$ is determined :
          $$F(X)=-frac{X^2}{2}$$
          We put it into the general solution where $X=u+frac{x^2}{2}$
          $$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
          Solving for $u$ :
          $$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$



          This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 18:43

























          answered Dec 11 '18 at 18:34









          JJacquelinJJacquelin

          45.3k21856




          45.3k21856












          • $begingroup$
            Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 19:03












          • $begingroup$
            In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 21:55


















          • $begingroup$
            Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
            $endgroup$
            – Brad Scott
            Dec 11 '18 at 19:03












          • $begingroup$
            In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
            $endgroup$
            – JJacquelin
            Dec 11 '18 at 21:55
















          $begingroup$
          Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
          $endgroup$
          – Brad Scott
          Dec 11 '18 at 19:03






          $begingroup$
          Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
          $endgroup$
          – Brad Scott
          Dec 11 '18 at 19:03














          $begingroup$
          In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
          $endgroup$
          – JJacquelin
          Dec 11 '18 at 21:55




          $begingroup$
          In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
          $endgroup$
          – JJacquelin
          Dec 11 '18 at 21:55


















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