Finding the flow of a vector field

[docnet]

Homework Statement

Find the flow of the vector field V(x, y) = (y, x)

Relevant Equations

V(x, y) = (y, x)

In part c, plotting the vector field shows the vector field is symmetric in x and y in the sets {x=y}.

in {x=y}, the variables can be interchanged and the solution becomes

x = x°e^t
y = y°e^tHowever, these solutions do not work for anywhere except {x=y} and don't satisfy dx/dt = y and dy/dt = x

yX+xY corresponds to the system of ODES

dx/dt = y
dy/dt = xWe eliminate dt from the above system to obtain the following solution to the system of ODES

(x^2)/2 - (y^2)/2=c

However I believe we cannot parametrize this equation to obtain the flow of the vector field.

What I did find from algebraic manipulations is

x = (x°e^t)/2 + (y°e^t)/2 + x°/2 - y°/2
y = (x°e^t)/2 + (y°e^t)/2 - x°/2 + y°/2

which satisfies the values t = 0, x = x°, y= y°

We tried to find the flow by t and s and the calculator tells us this is the wrong answer. Any ideas at all would be appreciated. thank you.

 

[Infrared]

If $dx/dt=y$ and $dy/dt=x$, then you can substitute one equation into the other to find $y=dx/dt=d^2y/dt^2$ and similarly $x=dx^2/dt^2.$ Then solve and use initial conditions.

 

[docnet]

If $dx/dt=y$ and $dy/dt=x$, then you can substitute one equation into the other to find $y=dx/dt=d^2y/dt^2$ and similarly $x=dx^2/dt^2.$ Then solve and use initial conditions.

Thank you for the advice. We tried this method and obtained the following solutions.

x=x°e^t + c te^t
y=y°e^t + c te^t

We found that the constant c can be set to x°-y° or y°-x° to satisfy dx/dt=y or dy/dt=x. Yet, neither choice satisfies the two initial conditions at the same time.

This is a strange situation because the equations say c = x°-y° and c = y°-x°.

 

[pasmith]

Thank you for the advice. We tried this method and obtained the following solutions.

x=x°e^t + c te^t
y=y°e^t + c te^t

The solution of \ddot x = x is <br /> x = A \cosh t + B \sinh t and its derivative is <br /> \dot x = A\sinh t + B \cosh t. Since \cosh(0) = 1 and \sinh(0) = 0 it is easy to impose the initial conditions.

We found that the constant c can be set to x°-y° or y°-x° to satisfy dx/dt=y or dy/dt=x. Yet, neither choice satisfies the two initial conditions at the same time.

This is a strange situation because the equations say c = x°-y° and c = y°-x°.

There are two dependent variables Therefore there are two constants of integration. They don't have to be equal.

 

[vela]

yX+xY corresponds to the system of ODES

dx/dt = y
dy/dt = x

We eliminate dt from the above system to obtain the following solution to the system of ODES

(x^2)/2 - (y^2)/2=c

However I believe we cannot parametrize this equation to obtain the flow of the vector field.

You might as well absorb the factor of 1/2 into the constant $c$ so $x^2-y^2=c$. Depending on the sign of $c$, you can parameterize using the hyperbolic functions. If $c>0$, you can write
\begin{align*}
x &= \sqrt{c} \cosh (t-t_0) \\
y &= \sqrt{c} \sinh (t-t_0)
\end{align*} If $c<0$, you'd need to swap $\cosh$ and $\sinh$ and stick a minus sign in the square root.

Note this is the same solution as the one @pasmith gives in post #4.

What I did find from algebraic manipulations is

x = (x°e^t)/2 + (y°e^t)/2 + x°/2 - y°/2
y = (x°e^t)/2 + (y°e^t)/2 - x°/2 + y°/2

which satisfies the values t = 0, x = x°, y= y°

You seem to be losing solutions along the way as there should be $e^{-t}$ terms as well. I'm guessing when you took a square root somewhere, you neglected the negative solution.

 

[docnet]

Thank you Pasmith and vela, you are right, this method solves the problem!

using the hyperbolic solution to second order odes, I found the solution

x= x cosh t + y sinh t
y= x sinh t + y cosh t

here is a graph of the flow when x = -1 and y = -2 parametrized by t.

vela, you are right, there are e^-t terms that I missed as well. I was tired after trying for many hours to solve this problem.