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Differential equation d^2f/dx^2-(3-2i)f=0

  1. Mar 12, 2013 #1
    the function obeys the differential equation d^2f/dx^2-(3-2i)f=0 , and satisfy the condition f(0)=1 and f(x)----->o ,for x-----> infinity , for f=0 calculate the value of f(∏)?

    Can Anybody give me any hints how to go about this problem???

    What I know is the following;

    D^2f/Dy^2=(3-2i)f
    => D^2=(3-2i)
    => D=[itex]\pm[/itex][itex]\sqrt{}(3-2i)[/itex]
    →f(=Ae^[itex]\sqrt{}(3-2i)[/itex]x + Be^-[itex]\sqrt{}(3-2i)[/itex]x

    Please Help
     
  2. jcsd
  3. Mar 12, 2013 #2

    Ray Vickson

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    The way I did it successfully was to write separate DEs for the real and imaginary parts of f; this will give you two coupled linear second-order ODEs, which you can solve in the usual way: assume f = exp(r*x) and find r, etc.
     
  4. Mar 13, 2013 #3
    OK, so I separate the differential equation:

    d2f/dx2 - 3f = 0 and d2/dx2 + 2i = 0
    roots are [itex]\pm[/itex][itex]\sqrt{}3[/itex] and [itex]\pm[/itex][itex]\sqrt{}2i^2[/itex]

    [itex]\Rightarrow[/itex] f(x)= Asin[itex]\sqrt{}3[/itex]x + Bcos[itex]\sqrt{}3[/itex]x and f(x)=Csin[itex]\sqrt{}2i^2[/itex] + Dcos[itex]\sqrt{}2i^2[/itex]

    applying first condition f(0)=1,

    f(x)=Asin[itex]\sqrt{}3[/itex]x + 1.cos[itex]\sqrt{}3[/itex]x and f(x)=Csin[itex]\sqrt{}2i^2[/itex] + 1.cos[itex]\sqrt{}2i^2[/itex]

    please give me hints on how to go about the second condition, so as to determine the remaining constants.
    thanks!!!
     
  5. Mar 13, 2013 #4

    Ray Vickson

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    No. You absolutely cannot do what you tried above. You have two *coupled* DEs, and you need to set up a 2x2 system to find the characteristic roots. So, if f = F + iG (F,G real) you have
    [tex]D^2(F + iG) =(3-2i)(F + iG) = (3F + 2G) + i(3G - 2F),[/tex]
    or
    [tex] F'' = 3F + 2G\\
    G'' = -2F + 3G[/tex]
    When you write ##F= ae^{rx}, \: G = be^{rx}## (same r for both) you get the equations
    [tex] r^2 a e^{rx} = 3a e^{rx} + 2b e^{rx} \\
    r^2 b e^{rx} = -2ae^{rx} + 3b e^{rx} [/tex]
    so
    [tex] (r^2 - 3)a - 2b = 0\\
    2a + (r^2 - 3) b = 0 [/tex]
    In order to have a nonzero solution we need the determinant of the coefficients = 0, so we need ##(r^2-3)^2 + 4 = 0, ## so ##r =\pm \sqrt{3 + 2 i}## or ##r = \pm \sqrt{3 - 2 i}.## These are, of course, what you wrote before. However, we are still not done: you need to find the real and imaginary parts, so you need to write ##\sqrt(2 + 3i) = u + iv## with real u,v, etc. That will give you functions like ##e^{ux} \sin(vx)## and ##e^{ux} \cos(vx).## Do that first, and if the solution is not then obvious come back for some more hints.

    Note: you could have done this the way you tried in your first post, but what stopped you from making more progress then was your failure to write the real and imaginary parts of ##\sqrt{3 - 2i}.##
     
  6. Mar 14, 2013 #5

    ehild

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    You gave the general solution of the differential equation. Now find A and B so as the conditions f(0)=1 and lim(f)=0 at infinity fulfil.

    To see the asymptotic behaviour of f, write sqrt(3-2i) in the form u+iv.


    ehild
     
  7. Mar 14, 2013 #6

    I am really confused about how to go about this problem,
    following are the points that I am confused with

    1) are The complex roots of the equation [itex]\pm[/itex][itex]\sqrt{}3-2i[/itex], or as you have mentioned using matrices [itex]\pm[/itex][itex]\sqrt{}3\pm2i[/itex]

    2) If I try to solve it using the complex root of the form a+ib ; a,b are real
    [itex]\Rightarrow[/itex] e(a+ib)x=eaxebix
    [itex]\Rightarrow[/itex] =eaxcos(bx) + ieaxsin(bx)

    since complex root come in conjugate pairs
    [itex]\Rightarrow[/itex] e(a-ib)x = eaxcos(bx) - ieaxsin(bx)

    after solving, we get the solution as
    f(x)=Aeaxcos(bx) + Beaxsin(bx); where A & B are constants.

    3) If the above general solution is correct, how do I write [itex]\sqrt{}3-2i[/itex] in terms of u+vi

    Please help
     
  8. Mar 14, 2013 #7

    Ray Vickson

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    to get z = √(3 - 2i) there are two methods:
    (1) Directly: set ##z = u+iv,## then require that ##(u+iv)^2 = 3 - 2i.## This gives two equations for u and v when you expand out the left-hand-side.
    (2) Polar form: write ##3 - 2i = r e^{i \theta}.## Finding ##r, \theta## is a standard problem that I will not say any more about. Once you know ##r, \theta## we have
    ##z = \pm \, r^{1/2} e^{i \theta/2} = \pm [r^{1/2} \cos(\theta/2) + i r^{1/2} \sin(\theta/2)].##
     
  9. Mar 15, 2013 #8
    Converting z = √(3 - 2i) into polar form I get
    z =[itex]\pm[/itex] r/2 eiθ/2 = [itex]\pm[/itex][131/4e163i/2]
    [itex]\Rightarrow[/itex] z=-2+05i or 2-0.5i

    Now , should I use these z as my new complex roots and use it in my general solution?

    Please help, I am having a real hard time figuring the out the solution of this problem.
     
    Last edited: Mar 15, 2013
  10. Mar 15, 2013 #9

    HallsofIvy

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    You had, back in your first post, the correct general solution:
    [tex]f(x)= Ae^{\sqrt{3- 2i}x}+ Be^{-\sqrt{3- i}x}[/tex]
    You are told that f(0)= 1 so f(0)= A+ B= 1. You are also told that the function value goes to 0 as x goes to infinity which means that you cannot have an exponential with positive real part in the exponent. That tells you that A= 0.

    I have no idea what "for f=0 calculate the value of f(∏)" means! It's not hard to "calculate the value of f(∏) but I don't understand "for f= 0".
     
  11. Mar 15, 2013 #10
    that's a typo, It's calculate the value of f(∏)?

    I can't understand how to go about the second condition, f(x)→0 as x→∞
    lim e^-x as x→∞ is zero [itex]\Rightarrow[/itex] B=0
    lim e^x as x→∞ is infinity

    Substituting B=0 in condition 1 , we get A=1
    therefore,
    f(x)= 1.e√(3-2i)x+ 0.e-√(3- 2i)x
    hence, the general solution is
    [itex]\Rightarrow[/itex] f(x) = e√(3- 2i)x


    Now can I replace the √(3-2i) term with it's polar form??

    Please tell me if what I have done is correct.
     
    Last edited: Mar 15, 2013
  12. Mar 15, 2013 #11

    ehild

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    Write √(3-2i) in the form u+iv.

    ehild
     
  13. Mar 15, 2013 #12
    the general solution is
    f(x)= e√(3-2i)x

    Converting z = √(3 - 2i) into polar form I get
    z =± r1/2 eiθ/2 = ±[131/4 e163i/2]
    ⇒ z=-2+0.5i or 2-0.5i


    the general solution becomes
    ⇒ f(x)= e(-2+0.5i)x + e(2-0.5i)x


    Is this correct??
     
  14. Mar 15, 2013 #13

    SammyS

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    (2 - 0.5i)2 = 3.75 - 2 i

    So, no that's not correct.
     
  15. Mar 16, 2013 #14
    could you please explain how to go about, so as to get (2 - 0.5i)2

    also, wrt the above quote, my general equation now become

    f(x)= e3.75-2i
     
  16. Mar 16, 2013 #15

    SammyS

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    That's not correct either.

    What I said meant that since (2 - 0.5i)2 = 3.75 - 2i , then (2 - 0.5i)2 ≠ 3 + 2i,

    so that it can't be true that (2 - 0.5i) is the same as (3 - 2i)1/2 .
     
  17. Mar 16, 2013 #16

    ehild

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    √(3-2i)=u+iv.


    The general solution is

    f(x)=Ae(u-iv)x+Be-(u+iv)x.


    3-2i=(u+iv)2=u2-v2+2iuv

    Comparing real and imaginary parts, you get a system of equations for u and v.

    3=u2-v2
    2uv=-2, so v=-1/u

    Substituting for v into the first equation: 3=u2-1/u2 so u2 obeys the quadratic equation
    u4-3u2-1.

    The solution can not be negative so you are left with a single solution for u2, and u=√(u2)

    As u is positive the first term exponentially grows as x tends to infinity. Therefore A=0.

    Your solution which tends to zero at infinity is f(x)=Be-(u-iv)x. f(0)=1, so B=1.

    Find the numerical value of both u and v.

    ehild
     
  18. Mar 16, 2013 #17
    Thank you guys for all your help,

    writing √(3-2i) interms of u+vi, we get

    (1.81-0.55i), which I had previously rounded up into (2-0.5i)

    Hence after applying the conditions, the general solution is

    f(x)=e-(1.81-0.55i)x
    which is approximately
    f(x)≈e-(2-0.5i)x
    [itex]\Rightarrow[/itex] f(x)≈e-2xe(0.5i)x
    [itex]\Rightarrow[/itex] f(x)≈e-2x (cos(0.5)x+isin(0.5)x)


    Hence,
    f(∏)≈e-2∏(-1+0)
    [itex]\Rightarrow[/itex] f(∏)≈ -e-2∏


    thanks once again, I learnt alot of things from this problem..
     
  19. Mar 16, 2013 #18

    ehild

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    it is the solution which corresponds to the initial conditions.

    cos(0.5∏)=0 and sin(0.5∏)=1, so your f(∏)≈-e-2∏ is not correct.

    ehild
     
  20. Mar 16, 2013 #19
    oops,
    f(∏)≈e-2∏ (0+i) ≈ ie-2∏
     
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