Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ricatti equation

  1. Mar 3, 2005 #1
    Consider the following Riccatti equation:

    dy/dx= -y^2+ 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.
    Go to Top of Page

    for part 1 here is what i have substituting y=u'/u

    (dy/dx)= -(u'/u)^2+a(x)(u'/u)+b(x)

    multiply everything by u to get:

    u(dy/dx)=-u(u')^2+a(x)(u')+b(x)u

    So am i done here or is there more??? (I have a really lousy professor who
    just barely got his phD and cant really teach. therefore i am having a lot of diffuculty understanding the matrial. i feel like i am teaching myself ODE....so please help)
     
  2. jcsd
  3. Mar 3, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

  4. Mar 3, 2005 #3

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    No, wish to rid the equation of y and just have u's. First, just treat a(x) and b(x) as a and b with the understanding that they are functions of x ok. Then we have:

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

    Letting [itex]y=\frac{u^{'}}{u}[/itex] and substituting this into the ODE, we get:

    [tex]\frac{uu^{''}-(u^{'})^2}{u^2}=-\frac{(u^{'})^2}{u^2}+\frac{au^'}{u}+b [/tex]

    You understand this part right?

    Now, multiplying throghout by [itex]u^2[/itex] and noting that there is a [itex]-(u^{'})^2 [/itex] on both sides which cancel, we have:

    [tex]u(u{''})=au(u{'})+bu^2[/tex]

    Divide out the u and place in standard form:

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

    That's a good start and it's important to understand this before going further. I assume you'll look at mine, then attempt to go through the steps on your own on paper. Next, do a similiar substitution with the following more general form of the Riccati equation:

    [tex]y^{'}+Qy+Ry^2=P[/tex]

    Using the substitution:

    [tex]y=\frac{u^{'}}{Ru}[/tex]

    With Q,R, and P functions of X so when you're differentiating y remember to differentiate R as well.
     
  5. Mar 3, 2005 #4

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Oh and we're not done here: either you or me should/could write a final report on this with at least one plot. Would be better for you if you're taking this in school but if I don't see any follow-up postings by Sun. night, I'll do so (I'm patient).
     
  6. Mar 3, 2005 #5
    thanks for the help salty, it really pointed me in the right direction. After some calculus I was able to get u''-au'-bu=0 . however, I still cant figure out number 3, which I've written again below:

    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).

    I got u''-1=0, and then rewrote it as d^y/dx^2 =1 . From here though I'm stuck.

    Also, what is your opinion on number 4? is it because of the properties of a linear equation?

    Thanks
     
  7. Mar 3, 2005 #6

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Well, if b=1 then the second order eq. in u(x) becomes:

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

    Solving this the usual way, you know how to right, gets:
    [tex]u(x)=c_1e^x+c_2e^{-x}[/tex]

    Now, you can plug in u'(0)=u(0)=1 into this and its derivative, get two equations, two unknowns, find [itex]c_1[/itex] and [itex]c_2[/itex].

    Also, #4: non-linear equations are much harder to solve than linear equations.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook