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Differential Equation

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the second order equation x''+ ax' + bx = 0:
    (a) Convert the equation to a 2 x 2 system.
    (b) Compute the eigenvalues.
    (c) In circuit and spring problems, both constants are nonnegative. Assume that they
    are actually positive, and show that the eigenvalues have a negative real part, and conclude
    that the trivial solution is asymptotically stable.

    2. Relevant equations



    3. The attempt at a solution


    (a). Let x1=x, x2=x'

    x1'=x2, x2'=-ax'-b''


    {{0,1},{-b,-a}}=A



    (b) det(A-λI)=
    \begin{bmatrix}
    -λ & 1 \\
    -b & -a-λ\\

    \end{bmatrix}

    = x2+aλ+b

    →λ=-a/2+-√(a2-4b)/2

    -a/2 is real and negative... but I just don't feel like I'm going about getting the solution the correct way.

    (c) No idea...
     
    Last edited by a moderator: Feb 29, 2012
  2. jcsd
  3. Feb 29, 2012 #2

    Mark44

    Staff: Mentor

    Typo above. The 2nd equation should be x2' = bx1 - ax2.
    Another typo above. The equation you're solving is λ2 + aλ + b = 0.
    Other than the typos I pointed out, I think you're on the right track. Based on what you have here, and not knowing the relative sizes of a and b, I don't see any way of showing that the part under the radical is either positive or negative. Is there some information you didn't include?
     
  4. Feb 29, 2012 #3
    Unfortunately, no. I directly copied and pasted the directions i here. The only thing that makes sense to me beyond the given directions is that 4b>a^2 so we would get a definite Real and Complex part.
     
  5. Mar 11, 2012 #4
    Alright, I went back and worked out part (b) to this:

    det(A-λI)=λ2+aλ+b

    →λ=(1/2)*(-a±√(a2-4b))

    Now, there are three possible cases:

    √(a2-4b)=0
    √(a2-4b)>0
    √(a2-4b)<0

    Case 1: If √(a2-4b))=0 then trivially, λ= -a/2
    By assumption, a ε ℝ and a>0 → -a/2<0


    Case 2: If √(a2-4b)>0, then sub-cases arise:

    (i). Consider λ=(1/2)*(-a-√(a2-4b))
    Claim (1/2)*(-a±√(a2-4b))<0
    Proof Assume (1/2)*(-a±√(a2-4b))>0
    -a/2 > √(a2-4b)/2

    By assumption √(a2-4b)>0 and -a/2<0, so it follows
    -a/2 >/ √(a2-4b)/2

    Therefore a contradiction has been reached and it follows that:
    λ=(1/2)*(-a±√(a2-4b))<0


    (ii.) Consider λ=(1/2)*(-a+√(a2-4b))
    Claim: (1/2)*(-a+√(a2-4b))<0
    Proof: Assume (1/2)*(-a+√(a2-4b))>0

    √(a2-4b)/2 > a/2
    (a2-4b)/4 > a2/4

    By assumption b>0 → -4b<0 → a2-4b< a2

    Therefore a contradiction has been reached and it follows that:
    λ=(1/2)*(-a+√(a2-4b))<0



    Case 3: If √(a2-4b)<0, then
    λ=(1/2)*(-a±√(a2-4b))

    Since √(a2-4b)<0, we can re-write this as
    i√(4b-a2)

    So λ=(1/2)*(-a±i√(4b-a2))

    and we can break the solution into real and complex parts. Therefore,
    the real part of the λ is -a/2<0


    So this takes care of part 2; however, I'm not sure how to go about part (c), any help appreciated (Sorry about not using latex, I tried and it didnt go well..)
     
  6. Mar 11, 2012 #5

    Mark44

    Staff: Mentor

    You need to lose the square root symbols. The three cases are the possible values of the discriminant, the quantity inside the radical.

    You don't need to say all of this.
    If a2-4b = 0, then λ= -a/2.
    Since by assumption a > 0, then -a/2 < 0.
    .
    This -- √(a2-4b)<0 -- makes no sense. If the quantity under the radical is >= 0, its square root will be >= 0.

    If the quantity under the radical is < 0, then you have an imaginary number, which cannot be compared to zero using < or >.
    For the c part, I'm not exactly sure what is being asked, but the systems involved are physical systems. For a damped spring-mass system, could you have an undriven system in which the mass started oscillating in cycles of greater and greater amplitude? This is connected to whether the eigenvalues are real or complex.

    Similarly, for the inductor-capacitor system with resistance, could you have an undriven system in which the voltage oscillated higher and higher?
     
  7. Mar 11, 2012 #6
    Ok, this is where I am getting confused with you in part (c): In circuit and spring problems, both constants are nonnegative. Assume that they
    are actually positive, and show that the eigenvalues have a negative real part, and conclude that the trivial solution is asymptotically stable.

    I see your point about Case 1,2,3.

    As for part (c), we need to take the eigenvalues we found earlier, consider the different possibilities (i.e. case 1,2,3) and find the matrix exponential and show that it behaves in an asymptotically stable manner at the origin.

    Now, the possible eigenvalues would be: λ=(1/2)*(-a±√(a2-4b))

    Plug λ into A-λI to obtain a new matrix (call it Z). Row-reduce Z to obtain an eigenvector. Repeat this until you have two linearly independent eigenvectors. Use these eigenvectors to form a new matrix (call it T) such that T-1AT=J (Jordan Form)

    Break J into its diagonal(D) and nilpotent(N) parts such that J= D+N

    eAt=TeJtT-1, so
    TeDteNtT-1

    If this is done correctly I should get solutions that are asymptotically stable. Does this process seem correct? (I wanted your opinion before I work all that out)
     
  8. Mar 12, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    From the expressions you have for the eigenvalues, and for a, b >0, it is easy to see that both eigenvalues have negative real parts. Look at three cases: (1) a^2 - 4b > 0. Then both eigenvalues are real, and you need to show both are < 0. (2) a^2 - 4b = 0: easy. (3) a^2 - 4b < 0. Now the eigenvalues are complex, but what are their real parts?

    RGV
     
  9. Mar 12, 2012 #8
    Guys thank you for all the help! I was just sitting here and I finally saw the solution, I was just over thinking this problem big time.

    I determined the Jordan normal form for A, found the matrix exponential for it. The trivial solution for the matrix exponential displays asymptotically stable behavior. Confirmed with the instructor today. Again, thanks for all the help!
     
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