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Integrate delte Diract Function

  1. Feb 15, 2012 #1
    Hi there. I came to join this forum to seek help, because I can't solve the following homework question:

    02.15.2012-20.37.11.png

    The solution to this can be split in 3 right?

    1. the homogenous

    #1:

    Eq:=

    c*x' +k*x =0

    Characteristic:

    c*r+k=0 --> r=-c/k

    Homogenious solution:

    y(x) = A*exp((-(c/k)*x))

    initi. condit X(0) = 0 --> A=0 ?? Homogen sol = 0??

    2. part sol. 1: the left hand side is equal to fc sin (wt)
    3. part sol. 2: the diract delta function.



    But I just can't figure out how to integrate the dirac delta function. I thought of using impartial integration but im not sure.....

    please help!
     
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2

    vela

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    What did you have from #1 and #2 so far?
     
  4. Feb 15, 2012 #3
    #1:

    Eq:=

    c*x' +k*x =0

    Characteristic:

    c*r+k=0 --> r=-c/k

    Homogenious solution:

    y(x) = A*exp((-(c/k)*x))

    initi. condit X(0) = 0 --> A=0 ?? Homogen sol = 0??


    #2 incoming
     
    Last edited: Feb 15, 2012
  5. Feb 15, 2012 #4

    vela

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    You don't want to apply the initial condition yet. Also, your variables should be x and t, not y and x.

    The basic strategy is this:

    1. Find the solutions for 0≤t<t0 (region I) and for t>t0 (region II). This means finding the general solution to the differential equation
    $$c\frac{dx}{dt}+kx(t) = f_c\sin\omega t.$$ The delta function is equal to zero when 0≤t<t0 and for t>t0 so it doesn't appear on the right-hand side.

    2. Apply any boundary or initial conditions to determine whatever arbitrary constants you can.

    3. Match the two solutions at t=t0. You do this by integrating the differential equation from t-ε to t+ε and taking the limit ε→0+. When you do this, you'll find what the discontinuity in x(t) should be at t=t0.
     
  6. Feb 16, 2012 #5
    Thank you for your help. I tried number 1 (think its ok see images, sorry for my bad handwriting, people say I should become a doctor)

    at 2. Im stuck I have 2 unknowns and only 1 boundary conditions??


    1. Homogenious solution

    http://www.imambaks.nl/IMAGE6_homo.png [Broken]


    2. Solution for function = f sinx (note that in the last few pages y = x(t), my mistake)

    http://www.imambaks.nl/IMAGE1.png [Broken]
    http://www.imambaks.nl/IMAGE2.png [Broken]
    http://www.imambaks.nl/IMAGE3.png [Broken]
    http://www.imambaks.nl/IMAGE_4.png [Broken]

    The combined solution with a problem :( :

    http://www.imambaks.nl/IMAGE7_solution.png [Broken]

    According to maple the constant C1 is not there ? I am confused, I though that with every integration a constant C is introduced as if one would differentiate the constant dissapears...
     
    Last edited by a moderator: May 5, 2017
  7. Feb 16, 2012 #6

    vela

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    Looks pretty good, but you made some algebraic errors with the constants. After the integration by parts, you should end up with
    \begin{align*}
    e^{(k/c)t}x(t) &= \frac{f_c}{c}\int e^{(k/c)t} \sin\omega t\,dt \\
    &= \frac{f_c}{c} e^{(k/c)t} \frac{-\omega\cos\omega t + \frac{k}{c} \sin \omega t}{\frac{k^2}{c^2}+\omega^2} + C
    \end{align*} which is close to what you have. You also need to tack on the constant of integration here, not later. Then when you solve for x(t) and simplify, you end up with
    $$x(t) = f_c \frac{-c \omega \cos \omega t + k \sin\omega t}{k^2+\omega^2c^2}+C e^{-\frac{k}{c}t}$$ The first term is the particular solution. The last term is the homogeneous solution. When you use this technique, the homogeneous solution is automatically included.

    So you have a solution for each of the two regions:
    \begin{align*}
    x_1(t) &= f_c \frac{-c \omega \cos \omega t + k \sin\omega t}{k^2+\omega^2c^2}+C_1 e^{-\frac{k}{c}t} \\
    x_2(t) &= f_c \frac{-c \omega \cos \omega t + k \sin\omega t}{k^2+\omega^2c^2}+C_2 e^{-\frac{k}{c}t}
    \end{align*} where x1(t) is for 0≤t<t0 and x2(t) is for t>t0.
     
  8. Feb 16, 2012 #7
    Hmm yeah I understand that the constant was at that moment (forgot to put it in, so I add it later). But what I don't understand is:

    02.16.2012-19.58.38.png

    When you solve the system for its INITIAL boundary condition you get C = (f *omega *c)/ (k^2+c^2*omega^2)
     
    Last edited: Feb 16, 2012
  9. Feb 16, 2012 #8

    vela

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    When you bring over the factor of exp(kt/c) from the left-hand side, it becomes the homogeneous solution.
     
  10. Feb 16, 2012 #9
    Thank you! I am a little bit confused :D Thought that one could only get a homogeous solution with using the characteristic equation.

    I will now make an attempt to include the dirac delta function in this mess.

    Update: I'm clueless.

    I understand the following:

    - the diract function is infinitely small (in time) and infinitely large at x(t0), this is thus between t-eps and t+eps.
    - the integral of a delta dirac between 0 and infinite = 1.

    What I have so far:

    - A solution (homogenous) and the particular one (external force).

    What I Don't Understand:

    - How to involve the dirac function into this solution. What do I need to integrate? I have 2 solutions: 0<t<t0 and t>t0.
     
    Last edited: Feb 16, 2012
  11. Feb 16, 2012 #10

    Ray Vickson

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    You have [itex] dx/dt + kx = f_c \sin(\omega t) [/itex] for [itex] t < t_0 \text{ and } t > t_0,[/itex] so you need to figure out what happens as you go through t = t_0. The presence of the δ function on the right indicates that the derivative dx/dt has a spike at t_0, which essentially means that x(t) has a jump discontinuity at t = t_0. The standard tool for seeing what conditions hold is integration: integrate the DE from [itex] t_0 - \epsilon[/itex] to [itex] t_0 + \epsilon.[/itex] On the left we have [tex] O(\epsilon) + \int_{t_0 - \epsilon}^{t_0 + \epsilon} \frac{dx(t)}{dt} dt = O(\epsilon) + x(t_0 + \epsilon) - x(t_0 - \epsilon), [/tex] where [itex]O(\epsilon)[/itex] stands for terms of size linear in ε. On the right we have [itex] O(\epsilon) + f_p. [/itex] Taking [itex] \epsilon \rightarrow 0[/itex] gives
    [itex] x(t_0 +) - x(t_0 -) = f_p. [/itex]

    RGV
     
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