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Homework Help: Zero Homogenous Solutions with transient Particular Solutions- Physically Possible? & More Questions

  1. Mar 17, 2016 #1
    BIBO/stable systems
    NOTE: zero here does not mean the roots of the denominator in a transfer function

    TRUE/FALSE -Please provide feedback- some answers are based on ODE example listed below

    1/True) The Homogenous Solution is either zero or transient.; i.e. it can never be steady state
    2/True) The Homogenous Solution does not always solely represent the Transient Response. What's more, if the Homogenous Solution is zero then it does not represent any part of the Transient Response (and it should go without saying that it doesn't represent any part of the total response)
    -In general: Homogenous Solution, if non-zero, always and solely represents the Transient Response provided the Particular Solution is not transient

    3/True) The Particular Solution can either be zero, transient, or steady state
    4/True) The Particular Solution does not always represent the Steady State Response. That is, it can be transient and thus can represent part of the Transient Response. (and thus it wouldn't represent, at all, the Steady State Response)
    -In general: Particular Solution, if non-zero and non-transient, always represents the Steady State Response

    5) A transient Particular Solution with a zero-Homogenous Solution, with zero initial conditions. How?
    - Example below shows that this is strictly due to zero initial conditions.
    6) A transient Particular Solution with a zero-Homogenous Solution, given non-zero initial conditions. How?
    - I don't think this can ever happen. Can this ever be?
    7) A transient Particular Solution with a Homogenous Solution, given zero initial conditions. How?
    -I assume it's just a variation of the example below where the homogenous solution doesn't zero out
    8) A transient Particular Solution with a Homogenous Solution, due to non-zero initial conditions. How?
    -Again, just a variation of below where homogenous solution is non zero because of non zero initial conditions
    9) A steady state Particular Solution with a zero-Homogenous Solution, given zero initial conditions. How?
    Note: not talking about large passing of time where the transients have died out
    -This one is incredibly difficult for me to visualize;
    10) A steady state Particular Solution with a zero-Homogenous Solution, given zero initial conditions. How?
    -This one is also very difficult to visualize.

    11) A steady state Particular Solution with a Homogenous Solution, given zero initial conditions. How?
    -typical mechanical vibrations/ODE's/controls problem
    12) A steady state Particular Solution with a Homogenous Solution, given non-zero initial conditions. How?
    -Again typical mechanical vibrations/ODE's/controls problem

    13) Also, what real life examples has an input (forcing function?) that is exponentially decreasing with time or rather a particular solution that represents, or partially represents (if the homogenous solution is non zero) the Transient Response?


    Then visited Swarthmore College's Webpage: http://lpsa.swarthmore.edu/Transient/TransZIZStime.html
    What, physically, do the homogeneous and particular response represent. The particular response represents the response of the system after any initial transients have died out, but the homogeneous response doesn't really represent anything physical. The reason we use it is that it is mathematically correct and yields the right answer.

    Then I Emailed a Professor at Swarthmore about the website's statement:
    He gave me an example that resists the common taught rules of thumb in dynamics and controls courses:
    -> Homogenous Solution=Transient Response; Professor says this is not generally true
    -> Particular Solution=Steady State Response; Example says this isn't always true, though their website says it is

    He indicated that a zero-Homogenous Solution does not represent the Transient Response (example below), and non-zero-Particular Solutions can be transient and thus represent part (all, if zero homogenous solution) of the Transient Response

    Example Description: a first order OODe with a exponentially decreasing with time forcing function
    Resulted in: a transient Particular Solution with a zero-Homogoenous Solution, due to zero initial conditions
    and after apply initial conditions to total solution:
    y(t) = total solution = homogenous solution + transient solution = yh(t) + yp(t) = 0 + t*exp(-t)

    EXAMPLE BELOW WARNING LOTS OF MATH (skip to Bold Terms to see the same result as just above)
    ----------------------------------------------------------------------------Begin Math--------------------------------------------------------------------------

    dy/dt + y = exp(-t) Zero-Initial Conditions: y(0) = 0 & dy/dt(0) = 0

    To find the total solution I tried two techniques: 1) Integrating Factors 2) Summing Complementary (Homogenous) and Particular Solutions.

    1) Integrating Factors
    mu = exp(int(coefficient of y*dt)) = exp(int(1*dt)) = exp(t)
    multiply ODE by mu:
    mu*dy/dt + mu*y = mu*exp(-t) = exp(t)*exp(-t) = 1
    noticing product rule:
    d/dt*[exp(t)*y] = 1
    integrating both sides:
    y*exp(t) = t+C
    Initial Conditions:
    y(t) = exp(-t)*(t+C) => y(0) = 0 = exp(-0)*(0+C) => C = 0
    Total Solution by Integrating Factor Technique:
    y(t) = t*exp(-t)
    2) Sum of Complementary (Homogenous) Solution and Particular Solution
    Complementary (Homogenous) Solution: set right hand side of ODE equal to zero, and then can just do typical integration or can assume solution to be yh = A*exp(t)
    I chose to do the typical integration:

    dy/dt + y = 0 => dy/dt = -y => 1/y *dy = -dt
    log|y| +C1 = -t log here is natural logarithm
    exp(log|y|) = exp(-t-C1) => yh(t) = exp(-t)*exp(-C1); if you did this by assumption/guess then your A would be equal to the exponential of C1

    Particular Solution: referred to textbook for this one because if I assume a particular solution of the form of that similar to RHS of ODE yp(t) = B*exp(-t) then I duplicate the complementary/homogenous solution i.e.: dy/dt + y = -Aexp(-t) + A*exp(-t) = exp(-t) = A*[-exp(-t)+exp(-t)] = exp(-t) = A[0] = exp(-t) => cannot determine A
    Instead assuming: yp(t) = B*t*exp(-t); by reasoning above
    dy/dt + y = exp(-t) = Aexp(-t)-A*t*exp(-t)+A*t*exp(-t) = exp(-t) = A*exp(-t) = exp(-t) => A = 1 => yp(t) = t*exp(-t) due to assumed solution [different than just B*exp(-t)]

    Apply Initial Conditions to Total Solution:
    y(t) = yh(t) + yp(t) = exp(-t)*exp(-C1) + t*exp(-t) => y(0) = 0 = exp(-0)*exp(-C1) +0*exp(-0) => exp(-C1) = 0 => C1 = infinitely large number that drives exponent to zero
    y(t) = exp(-t)*exp(-infinity) + t*exp(-t) = exp(-t)*0 + t*exp(-t) = 0 + t*exp(-t); => yh(t) = 0 due to zero initial conditions

    Total Solution by Complementary (Homogenous) Solution and Particular Solution :
    y(t) = 0 + t*exp(-t); notice both methods have same result

    --------------------------------------------------------------------------------------End Math------------------------------------------------------------------

    You can have, at least mathematically, a Particular Solution that is transient and a Homogenous Solution that is non-zero and thus both solutions together represent the Transient Response. (Again there wouldn't be a Steady State Response here, and thus the Total Response is the Transient Response and the only response )

    Their website says something to the effect that Particular Solution represents the Steady State after any initial transients have died out. What about when the Particular Solution is transient. Is that statement general enough?

    Similar Post linked here-though the above post is much clearer:
    Last edited: Mar 17, 2016
  2. jcsd
  3. Mar 18, 2016 #2


    Staff: Mentor

    The initial-value problem that you're considering here is very unusual, with two initial conditions. A first-order initial-value problem (IVP) normally consists of a first order differential equation and one initial condition that specifies the value of the unknown function at a specific time.

    In your problem, there are two initial conditions: y(0) = 0 and y'(0) = 0. Your solution of y(t) = ##te^{-t}## satisfies thel diff. equation and the initial condition that y(0) = 0, but it does not satisfy the second initial condition, that y'(0) = 0.
  4. Mar 18, 2016 #3

    That is an error on my part.


    dy/dt + y = exp(-t) Zero-Initial Condition: y(0) = 0
  5. Mar 18, 2016 #4


    Staff: Mentor

    Your post was quite long, but I think this is the gist of it:
    With regard to the IVP problem of your example, y' + y = e-t, y(0) = 0, I wouldn't categorize the particular solution as transient. True, it decays pretty quickly, but there's nothing that says the steady-state portion of the solution has to continue on with non-zero values for long periods of time.
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