Solving Riccatti Equation - Help Needed

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In summary, Riccatti equation yields y(x) which approaches a limiting factor as x increases without bound.
  • #1
yoyo
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Having problem with this PLEASE HELP ME!

Consider the following Riccatti equation:

dy/dx= -y+ a(x)y + b(x) (Eq. 2)

Here a(x) and b(x) are arbitrary functions.

1. Set y(x)= u'(x)/ u(x) where u(x) is a function to be determined. Use (Eq. 2) to show that u(x) satisfies a linear differential equation of second order.

2. Set a=0 and b=1. Solve (Eq. 2) by separating variables using y(0)=0 as the initial condition.

3. Find the function u(x) corresponding to the case a=0, b=1 and y(0)=0. When solving the second order equation for u(x) assume that u'(0)= y(0) and u(0)= 1. Compare the solution of (Eq. 2) from Part 2 with u'(x)/u(x).

4. Eplain why it is nice to replace a non-linear first order differential equation with a linear second order one.
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  • #2
It looks very simple.I think you meant to write:
[tex]\frac{dy}{dx}=-y^{2}+a(x)y+b(x) [/tex] (2)...

How about posting some of your work??

Daniel.
 
  • #3
I guess it could also have been:

[tex]\frac{dy}{dx}=-y+a(x)y^{2}+b(x) [/tex]

but given the requirements of part 2, I think you are right, it should be:

[tex]\frac{dy}{dx}=-y^{2}+a(x)y+b(x) [/tex]

Yoyo - the problem you have been given is very explicit about what is required. If you show a little of what you have done and where you are having difficulty it will be easier to help.

J.
 
  • #4
Please, first thing first: The Riccati equation has [itex] y^2 [/itex] as Daniel pointed out.

Thus, let's assume then that yoyo made a typo and go with:

[tex]y^{'}=-y^2+a(x)y+b(x)[/tex]

Yoyo, can you report the answer to the first question, i.e., substitute [itex]y=\frac{u^{'}}{u}[/itex] in the equation above and some things cancel out, divide by u to clean up a bit, and then end up with a second-order ODE?

If this thing dies out and nobody responds for a few days, I'm gonna' follow-up with a full report and I don't care if nobody reads it either. Whatever. You know if you jump out of a plane, you have to write a Riccati equation on you sleeve else you won't know how to fall right.

Salty
 
  • #5
A summary

For:

[tex]y^{'}=-y^2+ay+b[/tex]

Letting [itex]y=\frac{u^{'}}{u}[/itex] and substituting this into the ODE, we get the converted version in terms of u(x):

[tex]u^{''}-au^{'}-bu=0 [/tex]

Letting a=0 and b=1 we can avoid converting to u(x) and separate variables:

[tex]\frac{dy}{dx}=1-y^2[/tex]

Separating variables gives:

[tex]\frac{dy}{1-y^2}=dx[/tex]

Factoring via partial-fraction decomposition, we get:

[tex](\frac{1}{2(1+y)}+\frac{1}{2(1-y)})dy=dx[/tex]

Integrate indefinitely, convert from logarithms to exponents, and keep up with the constant of integration, K, produces:

[tex]y(x)=\frac{Ke^{2x}-1}{1+Ke^{2x}}[/tex]

Substituting the initial conditions y(0), we find K=1 or you could have just performed a definite integration above, (you know, from y to [itex]y_0[/itex]), so that:

[tex]y(x)=\frac{e^{2x}-1}{1+e^{2x}}[/tex]


Starting with the equation for u(x), we have:

[tex]u^{''}-u=0[/tex]

or:

[tex]u(x)=c_1e^x+c_2e^{-x}[/tex]

Now, the initial conditions for u(x) are:

u(0)=1 and u'(0)=0. Substituting these, we get:

[tex]u(x)=\frac{1}{2}e^x+\frac{1}{2}e^{-x}[/tex]

Since [itex]y(x)=\frac{u^{'}(x)}{u(x)}[/itex], upon differentiating this expression, we get for y:

[tex]y(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}[/tex]

Multiplying top and bottom by [itex]e^x[/itex] yields:

[tex]y(x)=\frac{e^{2x}-1}{e^{2x}+1}[/tex]

which is the same answer as above.

A plot is attached. Notice that as x increases without bound, y approaches a limiting factor. Remember I stated that the Riccati equation is used to describe jumping from a plane? Note in free-fall, you reach a "terminal velocity". Isn't someone here working on that problem?
 

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1. What is the Riccatti equation?

The Riccatti equation is a type of nonlinear differential equation that is used to describe the behavior of certain physical systems. It is named after Italian mathematician Jacopo Riccati, who first studied it in the 18th century.

2. Why is it important to solve the Riccatti equation?

The Riccatti equation has many applications in physics, engineering, and other fields. It can be used to model the behavior of systems such as oscillators, control systems, and chemical reactions. Solving the equation allows us to predict the behavior of these systems and make informed decisions.

3. What are the methods for solving the Riccatti equation?

There are several methods for solving the Riccatti equation, including the substitution method, the linearization method, and the matrix method. The most commonly used method is the substitution method, which involves substituting a new variable to transform the equation into a linear form.

4. What are the challenges in solving the Riccatti equation?

The main challenge in solving the Riccatti equation is that it is a nonlinear equation, meaning that it does not follow the typical rules of linear equations. This makes it more difficult to find exact solutions and often requires the use of numerical methods. Another challenge is that the equation may have multiple solutions, making it important to determine the most relevant solution for a given problem.

5. What are the real-world applications of the Riccatti equation?

The Riccatti equation has many real-world applications, including in control systems, population dynamics, and finance. It is also used in the study of quantum mechanics and in the design of electronic circuits. Overall, the Riccatti equation is a powerful tool for understanding and predicting the behavior of dynamic systems in various fields.

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