Find the streamline equation of a velocity field












0












$begingroup$


Assume the velocity field:
$$
v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
$$
$$
v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
$$



The streamline equation is the solution of the differential equation:
$$
ds=frac{dx}{v_x}=frac{dy}{v_y}
$$
$$
iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
$$
$$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
$$



indefinite integration and adjusting of the constant gives:
$$
left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
$$
$$
iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
$$



Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?






ordinary-differential-equations fluid-dynamics







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Assume the velocity field:
    $$
    v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
    $$
    $$
    v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
    $$



    The streamline equation is the solution of the differential equation:
    $$
    ds=frac{dx}{v_x}=frac{dy}{v_y}
    $$
    $$
    iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
    $$
    $$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
    $$



    indefinite integration and adjusting of the constant gives:
    $$
    left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
    $$
    $$
    iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
    $$



    Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?






    ordinary-differential-equations fluid-dynamics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Assume the velocity field:
      $$
      v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
      $$
      $$
      v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
      $$



      The streamline equation is the solution of the differential equation:
      $$
      ds=frac{dx}{v_x}=frac{dy}{v_y}
      $$
      $$
      iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
      $$
      $$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
      $$



      indefinite integration and adjusting of the constant gives:
      $$
      left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
      $$
      $$
      iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
      $$



      Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?






      ordinary-differential-equations fluid-dynamics







      share|cite|improve this question











      $endgroup$




      Assume the velocity field:
      $$
      v_x=-Ue^{-at}cos left( frac{pi x}{L} right)sin left( frac{pi y}{L} right)
      $$
      $$
      v_y=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)
      $$



      The streamline equation is the solution of the differential equation:
      $$
      ds=frac{dx}{v_x}=frac{dy}{v_y}
      $$
      $$
      iff -Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right) dx=-Ue^{-at}sin left( frac{pi x}{L} right)cos left( frac{pi y}{L} right)dy
      $$
      $$ Rightarrow tan left( frac{pi x}{L} right)dx=tan left( frac{pi y}{L} right)dy
      $$



      indefinite integration and adjusting of the constant gives:
      $$
      left| cos left( frac{pi x}{L} right)right|=c cdot left|cos left( frac{pi y}{L} right)right|
      $$
      $$
      iff cos^2 left( frac{pi x}{L} right)+lambda cdot cos^2 left( frac{pi y}{L} right)=0
      $$



      Given that we are given no initial point $(x_0,y_0)$ to determine $c$, how can a final result be obtained?






      ordinary-differential-equations fluid-dynamics




      ordinary-differential-equations fluid-dynamics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 11:30







      Jevaut

















      asked Dec 11 '18 at 18:13









      JevautJevaut

      1,151112




      1,151112






















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          $begingroup$

          You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes



          begin{equation}
          int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
          end{equation}



          which gives



          begin{equation}
          - frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
          end{equation}



          or



          begin{equation}
          lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
          end{equation}



          Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is



          begin{equation}
          f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
          end{equation}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
            $endgroup$
            – Jevaut
            Dec 12 '18 at 11:29










          • $begingroup$
            You are right, I'll edit the answer with squares to be similar to your derivation.
            $endgroup$
            – Sia
            Dec 12 '18 at 20:22











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          $begingroup$

          You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes



          begin{equation}
          int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
          end{equation}



          which gives



          begin{equation}
          - frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
          end{equation}



          or



          begin{equation}
          lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
          end{equation}



          Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is



          begin{equation}
          f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
          end{equation}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
            $endgroup$
            – Jevaut
            Dec 12 '18 at 11:29










          • $begingroup$
            You are right, I'll edit the answer with squares to be similar to your derivation.
            $endgroup$
            – Sia
            Dec 12 '18 at 20:22
















          2












          $begingroup$

          You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes



          begin{equation}
          int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
          end{equation}



          which gives



          begin{equation}
          - frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
          end{equation}



          or



          begin{equation}
          lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
          end{equation}



          Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is



          begin{equation}
          f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
          end{equation}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
            $endgroup$
            – Jevaut
            Dec 12 '18 at 11:29










          • $begingroup$
            You are right, I'll edit the answer with squares to be similar to your derivation.
            $endgroup$
            – Sia
            Dec 12 '18 at 20:22














          2












          2








          2





          $begingroup$

          You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes



          begin{equation}
          int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
          end{equation}



          which gives



          begin{equation}
          - frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
          end{equation}



          or



          begin{equation}
          lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
          end{equation}



          Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is



          begin{equation}
          f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
          end{equation}






          share|cite|improve this answer











          $endgroup$



          You should assume a point on streamline is given which then will be treated a paramter $(x_0,y_0)$ in your equation. Take definite integral from a given point $(x_0,y_0)$ where the streamline passes through, to some arbitrary point $(x,y)$. The integral equation becomes



          begin{equation}
          int_{x_0}^x tan left( frac{pi x}{L} right) mathrm{d} x = int_{y_0}^y tan left( frac{pi y}{L} right) mathrm{d} y
          end{equation}



          which gives



          begin{equation}
          - frac{L}{pi} left[ ln left| cos( frac{pi x}{L} ) right| right]_{x_0}^x = - frac{L}{pi} left[ ln left| cos( frac{pi y}{L} ) right| right]_{y_0}^y
          end{equation}



          or



          begin{equation}
          lnleft| frac{cos( frac{pi x}{L} )}{cos( frac{pi x_0}{L} )} right| = lnleft| frac{cos( frac{pi y}{L} )}{cos( frac{pi y_0}{L} )} right|
          end{equation}



          Thus the streamline curve $f(x,y) = 0$ that passes through a given point $(x_0,y_0)$ is



          begin{equation}
          f(x,y) = left( frac{ cos( frac{pi x}{L} )}{ cos( frac{pi y}{L} )} right)^2 - underbrace{ left( frac{cos( frac{pi x_0}{L} ) }{cos( frac{pi y_0}{L} )}right)^2}_{c^2} = 0.
          end{equation}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 1:47

























          answered Dec 11 '18 at 21:35









          SiaSia

          1064




          1064












          • $begingroup$
            Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
            $endgroup$
            – Jevaut
            Dec 12 '18 at 11:29










          • $begingroup$
            You are right, I'll edit the answer with squares to be similar to your derivation.
            $endgroup$
            – Sia
            Dec 12 '18 at 20:22


















          • $begingroup$
            Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
            $endgroup$
            – Jevaut
            Dec 12 '18 at 11:29










          • $begingroup$
            You are right, I'll edit the answer with squares to be similar to your derivation.
            $endgroup$
            – Sia
            Dec 12 '18 at 20:22
















          $begingroup$
          Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
          $endgroup$
          – Jevaut
          Dec 12 '18 at 11:29




          $begingroup$
          Isn't it more appropriate to square the absolutes (just to get rid of the cases)?
          $endgroup$
          – Jevaut
          Dec 12 '18 at 11:29












          $begingroup$
          You are right, I'll edit the answer with squares to be similar to your derivation.
          $endgroup$
          – Sia
          Dec 12 '18 at 20:22




          $begingroup$
          You are right, I'll edit the answer with squares to be similar to your derivation.
          $endgroup$
          – Sia
          Dec 12 '18 at 20:22


















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