ODE with trigonometric functions of solutions

sampahmel
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Dear all,

Homework Statement



Draw behavior around (0,0) of solutions to the following nonlinear system

<br /> \left(<br /> \begin{array}{c}<br /> x&#039;(t) \\<br /> y&#039;(t)<br /> \end{array}\right) =\left(<br /> \begin{array}{cc}<br /> cos {x(t)} + sin {x(t)} + {x(t)}^2 + {x(t)}^2{y(t)}^3 \\<br /> -x(t) + {y(t)}^2 + y(t) + sin {y(t)} <br /> \end{array}\right)\left(<br />

Homework Equations


My questions are:
1.) If (0,0) is not specified, do we automatically take (0,0) and draw behavior around it? If no, which point should be chosen?

For the next two questions, I know the rules for this forum are that I have to show my work, but I really don't have anything. All I can tell you is typically I just convert the LHSs of equations into matrix product. But here, because there is trigonometric functions of x(t) & y(t) and also squares of them, I feel matrix cannot help me in this case.

2.) How do you deal with the fact that x(t) and y(t) have cos & sin before it (ie. cos {x(t)} + sin {x(t)} ), instead of cos (t) * x(t)?

3.) How do you deal with the squares of x(t) & the third power of y(t) when converting it into a matrix product?

If you can't show them, at least tell me the name of the method to deal with these problems so I can google them up. But preferably, at least show me how you convert them into matrix form, please.

Thank you.

The Attempt at a Solution

 
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The graph of this system is a curve which we don't know.
But it can pass by any point of the xy plane.
Once you fixed a point, then the curve evolves following its derivatives.
The problem asks you to tell which direction the curve takes after being in (0,0).

How does one deal with sin and cos... just calculate them.

I just don't get when you talk of matrix product.
 
My matrix product refers to:

\left[x\acute{}(t) y\acute{}(t)\right] =

\left[1 2 3 4\right] \left[x(t) y(t)\right]

I don't know how to write the matrix in two rows, so the x and x derivative are on the 1st row in their respective matrix, while y and y derivative are on the 2nd row.

{1,2} are on the 1st row while {3,4} are on the 2nd row.But if there is sin (x(t)) and square of x(t), how do I use matrix in this case?
 
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Or otherwise how do you know how to draw the local behavior as I usually find the eigenvalues and eigenvector to draw phase portrait. But in this case I don't know how to convert them into matrix form.
 
Quinzio said:
The graph of this system is a curve which we don't know.
But it can pass by any point of the xy plane.
Once you fixed a point, then the curve evolves following its derivatives.
The problem asks you to tell which direction the curve takes after being in (0,0).

How does one deal with sin and cos... just calculate them.

I don't understand how do you "fixed a point".
And the sin and cos, calculate them using what?
 
Sam, do you mean?

<br /> \left(<br /> \begin{array}{c}<br /> x&#039;(t) \\<br /> y&#039;(t)<br /> \end{array}\right) =\left(<br /> \begin{array}{cc}<br /> 1 &amp; 2 \\<br /> 3 &amp; 4 <br /> \end{array}\right)\left(<br /> \begin{array}{c}<br /> x(t) \\<br /> y(t)<br /> \end{array}\right)<br />
 
hunt_mat said:
Sam, do you mean?

<br /> \left(<br /> \begin{array}{c}<br /> x&#039;(t) \\<br /> y&#039;(t)<br /> \end{array}\right) =\left(<br /> \begin{array}{cc}<br /> 1 &amp; 2 \\<br /> 3 &amp; 4 <br /> \end{array}\right)\left(<br /> \begin{array}{c}<br /> x(t) \\<br /> y(t)<br /> \end{array}\right)<br />

Exactly. Thank you.

This, I know how to solve.
But the example I have given where there are sin x(t) and x(t)^2, I don't. Can anyone please enlighten me?<br /> \left(<br /> \begin{array}{c}<br /> x&#039;(t) \\<br /> y&#039;(t)<br /> \end{array}\right) =\left(<br /> \begin{array}{cc}<br /> cos {x(t)} + sin {x(t)} + {x(t)}^2 + {x(t)}^2{y(t)}^3 \\<br /> -x(t) + {y(t)}^2 + y(t) + sin {y(t)} <br /> \end{array}\right)<br /> <br />
 
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I don't think that you can solve it exactly but what you can do in linearise and then that will give you some idea of how the system evolves.
 
hunt_mat said:
I don't think that you can solve it exactly but what you can do in linearise and then that will give you some idea of how the system evolves.

This is something new that I have never come across. How can I go about to linearize it?
 
  • #10
If it is something new then that usually means that it isn't required, what course is this part of?
 
  • #11
Higher Calculus.
By the way, do I take logs to linearize it?
 
  • #12
Are you sure you actually need to solve the ODE analytically? Not sure if that's even possible. Or is the assignment only to qualitatively figure out what the solution looks like assuming you start from the origin? In that case it would be more a case of figuring out which terms dominate initially and going from there.
 
  • #13
No, you use the Jacobian to linearise it in the form of:

<br /> \left(\begin{array}{c}<br /> x&#039;(t) \\<br /> y&#039;(t)<br /> \end{array}\right) =J\left(\begin{array}{c}<br /> x(t) \\<br /> y(t)<br /> \end{array}\right) +o(x(t),y(t))<br />
 
  • #14
Thanks a lot guys. I think I got it after googling the Jacobian suggested by hunt_mat.
I will derive wrt to each variable and evaluate at the point I am trying to linearize. Then I can finally use the matrix product that I am familiar with.

Thanks.
 
  • #15
sampahmel said:
This is something new that I have never come across. How can I go about to linearize it?

I could not get LaTeX to work properly, so here it is in plain text. Dropping the t arguments, your first RHS is cos x + sin x + x^2 + x^2 y^3. Expanding this to first-order in small x and y, we get simply x. The second RHS is -x + y^2 + y + sin y; to first order in small x and y this is -x + 2y. Therefore, the linearized system is
x' = x and y' = -x + 2y.

RGV
 
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