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Math Help - D.E involving the heaviside step function

  1. #1
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    D.E involving the heaviside step function

    Q: The differential equation y"+y=H(x)-H(x-a), where H is the heaviside step function, and a is a positive parameter, represents a simple harmonic oscillator subject to a constant force for a finite time. Solve the equation in three intervals of x and by applying appropriate matching conditions.


    I've used x<0, 0<x<a and x>a as my three time intervals.
    I managed to get:
    y"+y=0 for x<0
    y"+y=1 for 0<x<a
    y"+y=0 for x>a.
    Boundary conditions that I've come up with so far are y(0)=0 and y'(0)=1 and y'(a)=-1
    The solutions so far that I've been able to get are:
    y=0 for x<0
    y=1-cosx for 0<x<a
    but I'm stuck on the last time interval.
    For all three intervals I used y=Asinx+Bcosx as my complementary function, and added a particular integral where neccessary, but the answers that I've been given tell me that for x>a, y=cos(x-a)-cosx. I dont see where this comes from, please enlighten me.

    Edit: some1 mentioned using Laplace transforms for this but we havent covered that yet, so I'm thinking maybe there's another way to attack the question?
    Last edited by free_to_fly; October 26th 2009 at 12:37 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by free_to_fly View Post
    Q: The differential equation y"+y=H(x)-H(x-a), where H is the heaviside step function, and a is a positive parameter, represents a simple harmonic oscillator subject to a constant force for a finite time. Solve the equation in three intervals of x and by applying appropriate matching conditions.


    I've used x<0, 0<x<a and x>a as my three time intervals.
    I managed to get:
    y"+y=0 for x<0
    y"+y=1 for 0<x<a
    y"+y=0 for x>a.
    Boundary conditions that I've come up with so far are y(0)=0 and y'(0)=1 and y'(a)=-1
    The solutions so far that I've been able to get are:
    y=0 for x<0
    y=1-cosx for 0<x<a
    but I'm stuck on the last time interval.
    For all three intervals I used y=Asinx+Bcosx as my complementary function, and added a particular integral where neccessary, but the answers that I've been given tell me that for x>a, y=cos(x-a)-cosx. I dont see where this comes from, please enlighten me.

    Edit: some1 mentioned using Laplace transforms for this but we havent covered that yet, so I'm thinking maybe there's another way to attack the question?
    Define three regions:

    1. x\le 0
    2. 0\le x \le a
    3. a \le x

    Don't worry about the knots at x=0 abd x=0 being in two regions, we will meke the solution continuous there with solutions y_1(x),\ y_2(x) and y_3(x) in each of these.

    Initial consitions y_1(-1)=0,\ y'_1(-1)=0 and continuity conditions at the knots:

    y_2(0)=y_1(0),\ y_2'(0)=y_1'(0)
     <br />
y_3(a)=y_2(a),\ y_3'(a)=y_2'(a)<br />

    Now solve the IVP problem for y_1 in region 1, use that to set the initial conditions for an IVP for y_2 in region 2 etc.

    CB
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  3. #3
    MHF Contributor chisigma's Avatar
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    Remembering that if we set...

    Y(s)= \mathcal{L}\{y(t)\} (1)

    ... is...

     \mathcal{L}\{y^{''} (t)\}= s^{2}\cdot Y(s) - s\cdot y(0) - y^{'} (0) (2)

    ... the DE in terms of Laplace Transform is written as...

     s^{2}\cdot Y(s) - s\cdot y(0) - y^{'} (0) + Y(s) = \frac{1-e^{-as}}{s} (3)

    ... and from (3) we obtain...

    Y(s) = \frac{1-e^{-as}}{s\cdot (1+s^{2})} + \frac{s\cdot y(0)}{1+s^{2}} + \frac{y^{'}(0)}{1+s^{2}} (4)

    Now You can obtain y(t) performing the inverse LT of (4)...

    Kind regards

    \chi \sigma
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by chisigma View Post
    Remembering that if we set...

    Y(s)= \mathcal{L}\{y(t)\} (1)

    ... is...

     \mathcal{L}\{y^{''} (t)\}= s^{2}\cdot Y(s) - s\cdot y(0) - y^{'} (0) (2)

    ... the DE in terms of Laplace Transform is written as...

     s^{2}\cdot Y(s) - s\cdot y(0) - y^{'} (0) + Y(s) = \frac{1-e^{-as}}{s} (3)

    ... and from (3) we obtain...

    Y(s) = \frac{1-e^{-as}}{s\cdot (1+s^{2})} + \frac{s\cdot y(0)}{1+s^{2}} + \frac{y^{'}(0)}{1+s^{2}} (4)

    Now You can obtain y(t) performing the inverse LT of (4)...

    Kind regards

    \chi \sigma
    Does not answer the question since "Solve the equation in three intervals of x and by applying appropriate matching conditions" implies solving the equation in the three intervals and matching valuse and derivatives at the knots.

    CB
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  5. #5
    MHF Contributor chisigma's Avatar
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    The general solution of a linear second order DE is of the type...

    y(t) = y_{free} (t) + y_{forced} (t) (1)

    ... where y_{free} (t) is the general solution of the 'imcomplete' DE and y_{forced} (t) is a particular solution of the 'complete' DE. In our case is...

    Y(s) = Y_{free} (s) + Y_{forced} (s)

    Y_{free} (s) = \frac{y(0)\cdot s + y^{'} (0)}{1+s^{2}}

    Y_{forced} (s) = \frac{1-e^{-as}}{s\cdot (1+s^{2})} (2)

    Performing the inverse LT of the (2) we obtain...

    y_{free} (t) = y(0)\cdot \cos t + y^{'} (0)\cdot \sin t

    y_{forced} (t) = (1-\cos t)\cdot H(t) - \{1-\cos (t-a)\}\cdot H(t-a) (3)

    The main difference between y_{free} (t) and y_{forced} (t) is that the first is defined for -\infty < t < + \infty and the second is \ne 0 only for t>0. Once You know y(0) , y^{'} (0) and a the (3) supply to You the whole solution of the DE without great efforts and saving time... that's the main advantage of the use of LT ...

    Kind regards

    \chi \sigma
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