Compute the flow $Phi_t$ of $mathbb{X}(x,y)=(y,x)$












2












$begingroup$


Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
$mathbb{X}$



I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.



The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$



My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.



Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?



It is probably a problem with definitions, but any help would be greatly appreciated.






calculus ordinary-differential-equations derivatives







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
    $mathbb{X}$



    I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.



    The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$



    My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.



    Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?



    It is probably a problem with definitions, but any help would be greatly appreciated.






    calculus ordinary-differential-equations derivatives







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
      $mathbb{X}$



      I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.



      The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$



      My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.



      Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?



      It is probably a problem with definitions, but any help would be greatly appreciated.






      calculus ordinary-differential-equations derivatives







      share|cite|improve this question









      $endgroup$




      Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
      $mathbb{X}$



      I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.



      The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$



      My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.



      Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?



      It is probably a problem with definitions, but any help would be greatly appreciated.






      calculus ordinary-differential-equations derivatives




      calculus ordinary-differential-equations derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 27 '14 at 21:02









      Sam HoustonSam Houston

      1,2141132




      1,2141132






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          No, it is not a problem with definition.



          Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
          $$
          A=begin{bmatrix}
          0&1\
          1&0
          end{bmatrix}
          $$
          and $z=(x,y)^top$. Hence you need to calculate
          $$
          e^{tA}
          $$
          and see the result.



          Addition.



          There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
          $$
          e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
          $$
          Simply by noting that $A^2=I$ I find
          $$
          e^{tA}=begin{bmatrix}
          1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
          t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
          end{bmatrix}=begin{bmatrix}
          cosh t&sinh t\
          sinh t&cosh t
          end{bmatrix},
          $$
          where, as usual,
          $$
          cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
          $$
          Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            yes I remember reading that, thanks for the help, will give it another go
            $endgroup$
            – Sam Houston
            Dec 27 '14 at 21:15










          • $begingroup$
            Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
            $endgroup$
            – Artem
            Dec 27 '14 at 21:16






          • 1




            $begingroup$
            @Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
            $endgroup$
            – Artem
            Dec 27 '14 at 21:24






          • 1




            $begingroup$
            Did you hear about the matrix exponent?
            $endgroup$
            – Artem
            Dec 27 '14 at 22:37






          • 1




            $begingroup$
            See the addition. If you cannot get through it, you should start with the book I advised you.
            $endgroup$
            – Artem
            Dec 27 '14 at 23:14



















          0












          $begingroup$

          Another way to see this result without needing to use the exponential map is the following:



          For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            No, it is not a problem with definition.



            Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
            $$
            A=begin{bmatrix}
            0&1\
            1&0
            end{bmatrix}
            $$
            and $z=(x,y)^top$. Hence you need to calculate
            $$
            e^{tA}
            $$
            and see the result.



            Addition.



            There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
            $$
            e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
            $$
            Simply by noting that $A^2=I$ I find
            $$
            e^{tA}=begin{bmatrix}
            1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
            t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
            end{bmatrix}=begin{bmatrix}
            cosh t&sinh t\
            sinh t&cosh t
            end{bmatrix},
            $$
            where, as usual,
            $$
            cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
            $$
            Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              yes I remember reading that, thanks for the help, will give it another go
              $endgroup$
              – Sam Houston
              Dec 27 '14 at 21:15










            • $begingroup$
              Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:16






            • 1




              $begingroup$
              @Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:24






            • 1




              $begingroup$
              Did you hear about the matrix exponent?
              $endgroup$
              – Artem
              Dec 27 '14 at 22:37






            • 1




              $begingroup$
              See the addition. If you cannot get through it, you should start with the book I advised you.
              $endgroup$
              – Artem
              Dec 27 '14 at 23:14
















            4












            $begingroup$

            No, it is not a problem with definition.



            Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
            $$
            A=begin{bmatrix}
            0&1\
            1&0
            end{bmatrix}
            $$
            and $z=(x,y)^top$. Hence you need to calculate
            $$
            e^{tA}
            $$
            and see the result.



            Addition.



            There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
            $$
            e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
            $$
            Simply by noting that $A^2=I$ I find
            $$
            e^{tA}=begin{bmatrix}
            1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
            t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
            end{bmatrix}=begin{bmatrix}
            cosh t&sinh t\
            sinh t&cosh t
            end{bmatrix},
            $$
            where, as usual,
            $$
            cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
            $$
            Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              yes I remember reading that, thanks for the help, will give it another go
              $endgroup$
              – Sam Houston
              Dec 27 '14 at 21:15










            • $begingroup$
              Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:16






            • 1




              $begingroup$
              @Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:24






            • 1




              $begingroup$
              Did you hear about the matrix exponent?
              $endgroup$
              – Artem
              Dec 27 '14 at 22:37






            • 1




              $begingroup$
              See the addition. If you cannot get through it, you should start with the book I advised you.
              $endgroup$
              – Artem
              Dec 27 '14 at 23:14














            4












            4








            4





            $begingroup$

            No, it is not a problem with definition.



            Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
            $$
            A=begin{bmatrix}
            0&1\
            1&0
            end{bmatrix}
            $$
            and $z=(x,y)^top$. Hence you need to calculate
            $$
            e^{tA}
            $$
            and see the result.



            Addition.



            There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
            $$
            e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
            $$
            Simply by noting that $A^2=I$ I find
            $$
            e^{tA}=begin{bmatrix}
            1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
            t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
            end{bmatrix}=begin{bmatrix}
            cosh t&sinh t\
            sinh t&cosh t
            end{bmatrix},
            $$
            where, as usual,
            $$
            cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
            $$
            Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.






            share|cite|improve this answer











            $endgroup$



            No, it is not a problem with definition.



            Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
            $$
            A=begin{bmatrix}
            0&1\
            1&0
            end{bmatrix}
            $$
            and $z=(x,y)^top$. Hence you need to calculate
            $$
            e^{tA}
            $$
            and see the result.



            Addition.



            There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
            $$
            e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
            $$
            Simply by noting that $A^2=I$ I find
            $$
            e^{tA}=begin{bmatrix}
            1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
            t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
            end{bmatrix}=begin{bmatrix}
            cosh t&sinh t\
            sinh t&cosh t
            end{bmatrix},
            $$
            where, as usual,
            $$
            cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
            $$
            Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 27 '14 at 23:14

























            answered Dec 27 '14 at 21:13









            ArtemArtem

            11.4k32245




            11.4k32245












            • $begingroup$
              yes I remember reading that, thanks for the help, will give it another go
              $endgroup$
              – Sam Houston
              Dec 27 '14 at 21:15










            • $begingroup$
              Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:16






            • 1




              $begingroup$
              @Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:24






            • 1




              $begingroup$
              Did you hear about the matrix exponent?
              $endgroup$
              – Artem
              Dec 27 '14 at 22:37






            • 1




              $begingroup$
              See the addition. If you cannot get through it, you should start with the book I advised you.
              $endgroup$
              – Artem
              Dec 27 '14 at 23:14


















            • $begingroup$
              yes I remember reading that, thanks for the help, will give it another go
              $endgroup$
              – Sam Houston
              Dec 27 '14 at 21:15










            • $begingroup$
              Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:16






            • 1




              $begingroup$
              @Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
              $endgroup$
              – Artem
              Dec 27 '14 at 21:24






            • 1




              $begingroup$
              Did you hear about the matrix exponent?
              $endgroup$
              – Artem
              Dec 27 '14 at 22:37






            • 1




              $begingroup$
              See the addition. If you cannot get through it, you should start with the book I advised you.
              $endgroup$
              – Artem
              Dec 27 '14 at 23:14
















            $begingroup$
            yes I remember reading that, thanks for the help, will give it another go
            $endgroup$
            – Sam Houston
            Dec 27 '14 at 21:15




            $begingroup$
            yes I remember reading that, thanks for the help, will give it another go
            $endgroup$
            – Sam Houston
            Dec 27 '14 at 21:15












            $begingroup$
            Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
            $endgroup$
            – Artem
            Dec 27 '14 at 21:16




            $begingroup$
            Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
            $endgroup$
            – Artem
            Dec 27 '14 at 21:16




            1




            1




            $begingroup$
            @Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
            $endgroup$
            – Artem
            Dec 27 '14 at 21:24




            $begingroup$
            @Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
            $endgroup$
            – Artem
            Dec 27 '14 at 21:24




            1




            1




            $begingroup$
            Did you hear about the matrix exponent?
            $endgroup$
            – Artem
            Dec 27 '14 at 22:37




            $begingroup$
            Did you hear about the matrix exponent?
            $endgroup$
            – Artem
            Dec 27 '14 at 22:37




            1




            1




            $begingroup$
            See the addition. If you cannot get through it, you should start with the book I advised you.
            $endgroup$
            – Artem
            Dec 27 '14 at 23:14




            $begingroup$
            See the addition. If you cannot get through it, you should start with the book I advised you.
            $endgroup$
            – Artem
            Dec 27 '14 at 23:14











            0












            $begingroup$

            Another way to see this result without needing to use the exponential map is the following:



            For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Another way to see this result without needing to use the exponential map is the following:



              For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Another way to see this result without needing to use the exponential map is the following:



                For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.






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                $endgroup$



                Another way to see this result without needing to use the exponential map is the following:



                For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 '18 at 4:43









                Brandon Thomas Van OverBrandon Thomas Van Over

                3,83721127




                3,83721127






























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