Finding the flow of a vector field

In summary, the vector field is symmetric in x and y in the sets {x=y}. However, the solutions provided do not satisfy the initial conditions.
  • #1
docnet
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Homework Statement
Find the flow of the vector field V(x, y) = (y, x)
Relevant Equations
V(x, y) = (y, x)
Screen Shot 2020-10-24 at 2.34.40 AM.png


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

Screen Shot 2020-10-24 at 2.36.02 AM.png

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.
 
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  • #2
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.
 
  • #3
Infrared said:
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°.
 
  • #4
docnet said:
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 [itex]\ddot x = x[/itex] is [tex]
x = A \cosh t + B \sinh t[/tex] and its derivative is [tex]
\dot x = A\sinh t + B \cosh t.[/tex] Since [itex]\cosh(0) = 1[/itex] and [itex]\sinh(0) = 0[/itex] 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.
 
  • #5
docnet said:
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.
 
  • #6
Screen Shot 2020-10-24 at 3.39.23 PM.png


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.

Screen Shot 2020-10-24 at 3.39.05 PM.png


Screen Shot 2020-10-24 at 3.39.09 PM.png


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.
 
Last edited:

What is a vector field?

A vector field is a mathematical concept that describes the behavior of vectors at every point in a given space. It is represented by arrows or lines that show the direction and magnitude of the vectors at each point.

What is the flow of a vector field?

The flow of a vector field refers to the movement or change of vectors at each point in the field. It describes how the vectors are affected by the forces or conditions within the field.

How do you find the flow of a vector field?

To find the flow of a vector field, you can use mathematical equations and techniques such as vector calculus, differential equations, and computer simulations. These methods help to calculate the changes in the vectors at each point in the field.

What is the significance of finding the flow of a vector field?

Finding the flow of a vector field is important in many scientific fields, such as physics, engineering, and meteorology. It helps to understand and predict the behavior and movement of physical systems, such as fluids, electric fields, and weather patterns.

Can the flow of a vector field change over time?

Yes, the flow of a vector field can change over time due to various factors such as external forces, changes in the field's conditions, or interactions with other fields. This dynamic nature makes it a useful tool for studying and analyzing complex systems.

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