Will the energy of a randomly driven harmonic oscillator grow to infinity or oscillate about a finite value?












9














The equation of motion for an undamped harmonic oscillator, with driver $f=f(t)$ is given by:
$$ddot{x}+x=f.$$
Let the initial conditions be given by:
$$x(0)=dot{x}(0)=0.$$
If $f=cos(t)$ then the solution is:
$$x(t)=frac{1}{2}tsin(t).$$
Hence, a resonance is setup and the energy of the oscillator will grow forever.
If $f=cos(omega t)$ where $omegane1$, the solution is:
$$x(t)=frac{2}{omega^2-1}sinleft(frac{omega-1}{2}tright)sinleft(frac{omega+1}{2}tright),$$
hence, the energy oscillates about some finite value. My question is, if $f$ were replaced with some continuous random driver where the frequency profile resmbled that of say gaussian white noise, would the energy of the oscillator grow forever or would it oscillate about some finite value?



Does anyone know of a simple function I could replace $f$ with to generate a continuous white noise driver?










share|cite|improve this question
























  • My gut says it definitely wouldn't grow to infinity. You might well get chaotic behavior, though.
    – Adrian Keister
    Dec 19 at 19:47






  • 3




    That's a very interesting question! Just some thoughts: multiplying the equation by $dot{x}$ we have the total energy of the system: $$dot{x} ddot{x}+dot{x}x = frac{1}{2} frac{d}{dt} dot{x}^2 + frac{1}{2} frac{d}{dt} x^2 = dot{E}.$$ Therefore, $dot{E} = f dot{x}$ and $$E = int_0^t f dot{x} dt.$$
    – rafa11111
    Dec 19 at 19:49










  • Nice one @rafa11111. That means in the long run, since $f$ is completely random, the integral is $0$. So the system will emit the same amount of energy it absorbs. On the short time scale however, the energy is not constant.
    – Andrei
    Dec 19 at 20:02










  • @Andrei I'm not sure... How can you ensure that the integral is $0$? I have the feeling that it should be $0$, but I don't know how can one show that.
    – rafa11111
    Dec 19 at 20:06










  • I'm wondering if your DE there can be re-cast into the form of a stochastic differential equation (SDE). That might be a fruitful area of inquiry.
    – Adrian Keister
    Dec 19 at 20:13
















9














The equation of motion for an undamped harmonic oscillator, with driver $f=f(t)$ is given by:
$$ddot{x}+x=f.$$
Let the initial conditions be given by:
$$x(0)=dot{x}(0)=0.$$
If $f=cos(t)$ then the solution is:
$$x(t)=frac{1}{2}tsin(t).$$
Hence, a resonance is setup and the energy of the oscillator will grow forever.
If $f=cos(omega t)$ where $omegane1$, the solution is:
$$x(t)=frac{2}{omega^2-1}sinleft(frac{omega-1}{2}tright)sinleft(frac{omega+1}{2}tright),$$
hence, the energy oscillates about some finite value. My question is, if $f$ were replaced with some continuous random driver where the frequency profile resmbled that of say gaussian white noise, would the energy of the oscillator grow forever or would it oscillate about some finite value?



Does anyone know of a simple function I could replace $f$ with to generate a continuous white noise driver?










share|cite|improve this question
























  • My gut says it definitely wouldn't grow to infinity. You might well get chaotic behavior, though.
    – Adrian Keister
    Dec 19 at 19:47






  • 3




    That's a very interesting question! Just some thoughts: multiplying the equation by $dot{x}$ we have the total energy of the system: $$dot{x} ddot{x}+dot{x}x = frac{1}{2} frac{d}{dt} dot{x}^2 + frac{1}{2} frac{d}{dt} x^2 = dot{E}.$$ Therefore, $dot{E} = f dot{x}$ and $$E = int_0^t f dot{x} dt.$$
    – rafa11111
    Dec 19 at 19:49










  • Nice one @rafa11111. That means in the long run, since $f$ is completely random, the integral is $0$. So the system will emit the same amount of energy it absorbs. On the short time scale however, the energy is not constant.
    – Andrei
    Dec 19 at 20:02










  • @Andrei I'm not sure... How can you ensure that the integral is $0$? I have the feeling that it should be $0$, but I don't know how can one show that.
    – rafa11111
    Dec 19 at 20:06










  • I'm wondering if your DE there can be re-cast into the form of a stochastic differential equation (SDE). That might be a fruitful area of inquiry.
    – Adrian Keister
    Dec 19 at 20:13














9












9








9


1





The equation of motion for an undamped harmonic oscillator, with driver $f=f(t)$ is given by:
$$ddot{x}+x=f.$$
Let the initial conditions be given by:
$$x(0)=dot{x}(0)=0.$$
If $f=cos(t)$ then the solution is:
$$x(t)=frac{1}{2}tsin(t).$$
Hence, a resonance is setup and the energy of the oscillator will grow forever.
If $f=cos(omega t)$ where $omegane1$, the solution is:
$$x(t)=frac{2}{omega^2-1}sinleft(frac{omega-1}{2}tright)sinleft(frac{omega+1}{2}tright),$$
hence, the energy oscillates about some finite value. My question is, if $f$ were replaced with some continuous random driver where the frequency profile resmbled that of say gaussian white noise, would the energy of the oscillator grow forever or would it oscillate about some finite value?



Does anyone know of a simple function I could replace $f$ with to generate a continuous white noise driver?










share|cite|improve this question















The equation of motion for an undamped harmonic oscillator, with driver $f=f(t)$ is given by:
$$ddot{x}+x=f.$$
Let the initial conditions be given by:
$$x(0)=dot{x}(0)=0.$$
If $f=cos(t)$ then the solution is:
$$x(t)=frac{1}{2}tsin(t).$$
Hence, a resonance is setup and the energy of the oscillator will grow forever.
If $f=cos(omega t)$ where $omegane1$, the solution is:
$$x(t)=frac{2}{omega^2-1}sinleft(frac{omega-1}{2}tright)sinleft(frac{omega+1}{2}tright),$$
hence, the energy oscillates about some finite value. My question is, if $f$ were replaced with some continuous random driver where the frequency profile resmbled that of say gaussian white noise, would the energy of the oscillator grow forever or would it oscillate about some finite value?



Does anyone know of a simple function I could replace $f$ with to generate a continuous white noise driver?







physics random noise






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share|cite|improve this question













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edited Dec 19 at 20:18

























asked Dec 19 at 19:38









Peanutlex

605




605












  • My gut says it definitely wouldn't grow to infinity. You might well get chaotic behavior, though.
    – Adrian Keister
    Dec 19 at 19:47






  • 3




    That's a very interesting question! Just some thoughts: multiplying the equation by $dot{x}$ we have the total energy of the system: $$dot{x} ddot{x}+dot{x}x = frac{1}{2} frac{d}{dt} dot{x}^2 + frac{1}{2} frac{d}{dt} x^2 = dot{E}.$$ Therefore, $dot{E} = f dot{x}$ and $$E = int_0^t f dot{x} dt.$$
    – rafa11111
    Dec 19 at 19:49










  • Nice one @rafa11111. That means in the long run, since $f$ is completely random, the integral is $0$. So the system will emit the same amount of energy it absorbs. On the short time scale however, the energy is not constant.
    – Andrei
    Dec 19 at 20:02










  • @Andrei I'm not sure... How can you ensure that the integral is $0$? I have the feeling that it should be $0$, but I don't know how can one show that.
    – rafa11111
    Dec 19 at 20:06










  • I'm wondering if your DE there can be re-cast into the form of a stochastic differential equation (SDE). That might be a fruitful area of inquiry.
    – Adrian Keister
    Dec 19 at 20:13


















  • My gut says it definitely wouldn't grow to infinity. You might well get chaotic behavior, though.
    – Adrian Keister
    Dec 19 at 19:47






  • 3




    That's a very interesting question! Just some thoughts: multiplying the equation by $dot{x}$ we have the total energy of the system: $$dot{x} ddot{x}+dot{x}x = frac{1}{2} frac{d}{dt} dot{x}^2 + frac{1}{2} frac{d}{dt} x^2 = dot{E}.$$ Therefore, $dot{E} = f dot{x}$ and $$E = int_0^t f dot{x} dt.$$
    – rafa11111
    Dec 19 at 19:49










  • Nice one @rafa11111. That means in the long run, since $f$ is completely random, the integral is $0$. So the system will emit the same amount of energy it absorbs. On the short time scale however, the energy is not constant.
    – Andrei
    Dec 19 at 20:02










  • @Andrei I'm not sure... How can you ensure that the integral is $0$? I have the feeling that it should be $0$, but I don't know how can one show that.
    – rafa11111
    Dec 19 at 20:06










  • I'm wondering if your DE there can be re-cast into the form of a stochastic differential equation (SDE). That might be a fruitful area of inquiry.
    – Adrian Keister
    Dec 19 at 20:13
















My gut says it definitely wouldn't grow to infinity. You might well get chaotic behavior, though.
– Adrian Keister
Dec 19 at 19:47




My gut says it definitely wouldn't grow to infinity. You might well get chaotic behavior, though.
– Adrian Keister
Dec 19 at 19:47




3




3




That's a very interesting question! Just some thoughts: multiplying the equation by $dot{x}$ we have the total energy of the system: $$dot{x} ddot{x}+dot{x}x = frac{1}{2} frac{d}{dt} dot{x}^2 + frac{1}{2} frac{d}{dt} x^2 = dot{E}.$$ Therefore, $dot{E} = f dot{x}$ and $$E = int_0^t f dot{x} dt.$$
– rafa11111
Dec 19 at 19:49




That's a very interesting question! Just some thoughts: multiplying the equation by $dot{x}$ we have the total energy of the system: $$dot{x} ddot{x}+dot{x}x = frac{1}{2} frac{d}{dt} dot{x}^2 + frac{1}{2} frac{d}{dt} x^2 = dot{E}.$$ Therefore, $dot{E} = f dot{x}$ and $$E = int_0^t f dot{x} dt.$$
– rafa11111
Dec 19 at 19:49












Nice one @rafa11111. That means in the long run, since $f$ is completely random, the integral is $0$. So the system will emit the same amount of energy it absorbs. On the short time scale however, the energy is not constant.
– Andrei
Dec 19 at 20:02




Nice one @rafa11111. That means in the long run, since $f$ is completely random, the integral is $0$. So the system will emit the same amount of energy it absorbs. On the short time scale however, the energy is not constant.
– Andrei
Dec 19 at 20:02












@Andrei I'm not sure... How can you ensure that the integral is $0$? I have the feeling that it should be $0$, but I don't know how can one show that.
– rafa11111
Dec 19 at 20:06




@Andrei I'm not sure... How can you ensure that the integral is $0$? I have the feeling that it should be $0$, but I don't know how can one show that.
– rafa11111
Dec 19 at 20:06












I'm wondering if your DE there can be re-cast into the form of a stochastic differential equation (SDE). That might be a fruitful area of inquiry.
– Adrian Keister
Dec 19 at 20:13




I'm wondering if your DE there can be re-cast into the form of a stochastic differential equation (SDE). That might be a fruitful area of inquiry.
– Adrian Keister
Dec 19 at 20:13










2 Answers
2






active

oldest

votes


















8














If you have a Gaussian random force, the equation becomes a Langevin equation. In physicists notation, you would write
$$ ddot x + x = lambda xi(t) tag{1}$$
with $lambda$ the strength of the random force and (the bracket denotes the stochastic average)
$$langle xi(t) rangle = 0, quad text{and} quad langle xi(t) xi(t') rangle = delta(t-t'),. tag{2}$$
Note that in mathematics instead stochastic differential equations are more conventional.



Let us assume that $x(0)=dot x(0)=0$.
We can solve (1) to obtain
$$ x(t) = lambda int_0^t!sin(t-t') xi(t'),dt',. tag{3}$$
This is a stochastic solution as it depends on the random function $xi(t)$. However, from (3) together with (2) we can calculate statistical predictions. For example the average position is given by
$$langle x(t) rangle =0,,$$
which is not unexpected (just compare it to a random walk). So on average the oscillator does diverge as it does not even move.



Of course, the more reasonable measure if the harmonic oscillator performs an unbounded oscillation is the variance. We obtain
$$langle x(t)^2 rangle = lambda^2 int_0^t int_0^t!sin(t-t') sin(t-t'') langlexi(t')xi(t'')rangle,dt'',dt' =lambda^2 int_0^t sin^2(t-t'),dt' = lambda^2 left(frac{t}2 - frac{sin(2t)}{4}right),. $$



From this we see that the typical amplitude of the oscillation, given by $sqrt{langle x(t)^2 rangle }$ behaves as
$$ sqrt{langle x(t)^2 rangle} sim lambda sqrt{frac{t}{2}}$$
for $ttoinfty$; i.e., the oscillation grows without bounds. However, the amplitude of the oscillation only grows as $sqrt{t}$ instead of proportional to $t$.






share|cite|improve this answer





















  • Thanks. How does equation (1) reduce to the Langevin equation: $$mddot{x}=-lambdadot{x}+eta(t)?$$ Also, what is the stochastic average and how it is calculated?
    – Peanutlex
    Dec 19 at 22:20






  • 1




    It is a Langevin equation not the Langevin equation. The stochastic average simply encodes the notion that $xi(t)$ is a random force with zero mean and "standard deviation" $lambda$.
    – Fabian
    Dec 19 at 22:28



















2















Does anyone know of a simple function I could replace f with to generate a continuous white noise driver?




I'm not sure if this approach will help you, but here it is. My approach to problems such as the one you presented is related to implementation of said problem and then derive some, at least numeric, result. Having this said, I would do as follows.



Consider a finite discrete Gaussian white noise which is stored in an array $G(nDelta t)$, with $n$ from $0$ to $N$. This can be interpolated using a polynomial $P(t)$ of order $N-1$ which is unique. This polynomial should be the solution you are looking for for $tin[0,NDelta t]$. From this you can compute your driver function using



$$f(t)=frac{d^2 P}{dt^2}(t) + P(t)$$






share|cite|improve this answer





















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    2 Answers
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    2 Answers
    2






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    active

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    8














    If you have a Gaussian random force, the equation becomes a Langevin equation. In physicists notation, you would write
    $$ ddot x + x = lambda xi(t) tag{1}$$
    with $lambda$ the strength of the random force and (the bracket denotes the stochastic average)
    $$langle xi(t) rangle = 0, quad text{and} quad langle xi(t) xi(t') rangle = delta(t-t'),. tag{2}$$
    Note that in mathematics instead stochastic differential equations are more conventional.



    Let us assume that $x(0)=dot x(0)=0$.
    We can solve (1) to obtain
    $$ x(t) = lambda int_0^t!sin(t-t') xi(t'),dt',. tag{3}$$
    This is a stochastic solution as it depends on the random function $xi(t)$. However, from (3) together with (2) we can calculate statistical predictions. For example the average position is given by
    $$langle x(t) rangle =0,,$$
    which is not unexpected (just compare it to a random walk). So on average the oscillator does diverge as it does not even move.



    Of course, the more reasonable measure if the harmonic oscillator performs an unbounded oscillation is the variance. We obtain
    $$langle x(t)^2 rangle = lambda^2 int_0^t int_0^t!sin(t-t') sin(t-t'') langlexi(t')xi(t'')rangle,dt'',dt' =lambda^2 int_0^t sin^2(t-t'),dt' = lambda^2 left(frac{t}2 - frac{sin(2t)}{4}right),. $$



    From this we see that the typical amplitude of the oscillation, given by $sqrt{langle x(t)^2 rangle }$ behaves as
    $$ sqrt{langle x(t)^2 rangle} sim lambda sqrt{frac{t}{2}}$$
    for $ttoinfty$; i.e., the oscillation grows without bounds. However, the amplitude of the oscillation only grows as $sqrt{t}$ instead of proportional to $t$.






    share|cite|improve this answer





















    • Thanks. How does equation (1) reduce to the Langevin equation: $$mddot{x}=-lambdadot{x}+eta(t)?$$ Also, what is the stochastic average and how it is calculated?
      – Peanutlex
      Dec 19 at 22:20






    • 1




      It is a Langevin equation not the Langevin equation. The stochastic average simply encodes the notion that $xi(t)$ is a random force with zero mean and "standard deviation" $lambda$.
      – Fabian
      Dec 19 at 22:28
















    8














    If you have a Gaussian random force, the equation becomes a Langevin equation. In physicists notation, you would write
    $$ ddot x + x = lambda xi(t) tag{1}$$
    with $lambda$ the strength of the random force and (the bracket denotes the stochastic average)
    $$langle xi(t) rangle = 0, quad text{and} quad langle xi(t) xi(t') rangle = delta(t-t'),. tag{2}$$
    Note that in mathematics instead stochastic differential equations are more conventional.



    Let us assume that $x(0)=dot x(0)=0$.
    We can solve (1) to obtain
    $$ x(t) = lambda int_0^t!sin(t-t') xi(t'),dt',. tag{3}$$
    This is a stochastic solution as it depends on the random function $xi(t)$. However, from (3) together with (2) we can calculate statistical predictions. For example the average position is given by
    $$langle x(t) rangle =0,,$$
    which is not unexpected (just compare it to a random walk). So on average the oscillator does diverge as it does not even move.



    Of course, the more reasonable measure if the harmonic oscillator performs an unbounded oscillation is the variance. We obtain
    $$langle x(t)^2 rangle = lambda^2 int_0^t int_0^t!sin(t-t') sin(t-t'') langlexi(t')xi(t'')rangle,dt'',dt' =lambda^2 int_0^t sin^2(t-t'),dt' = lambda^2 left(frac{t}2 - frac{sin(2t)}{4}right),. $$



    From this we see that the typical amplitude of the oscillation, given by $sqrt{langle x(t)^2 rangle }$ behaves as
    $$ sqrt{langle x(t)^2 rangle} sim lambda sqrt{frac{t}{2}}$$
    for $ttoinfty$; i.e., the oscillation grows without bounds. However, the amplitude of the oscillation only grows as $sqrt{t}$ instead of proportional to $t$.






    share|cite|improve this answer





















    • Thanks. How does equation (1) reduce to the Langevin equation: $$mddot{x}=-lambdadot{x}+eta(t)?$$ Also, what is the stochastic average and how it is calculated?
      – Peanutlex
      Dec 19 at 22:20






    • 1




      It is a Langevin equation not the Langevin equation. The stochastic average simply encodes the notion that $xi(t)$ is a random force with zero mean and "standard deviation" $lambda$.
      – Fabian
      Dec 19 at 22:28














    8












    8








    8






    If you have a Gaussian random force, the equation becomes a Langevin equation. In physicists notation, you would write
    $$ ddot x + x = lambda xi(t) tag{1}$$
    with $lambda$ the strength of the random force and (the bracket denotes the stochastic average)
    $$langle xi(t) rangle = 0, quad text{and} quad langle xi(t) xi(t') rangle = delta(t-t'),. tag{2}$$
    Note that in mathematics instead stochastic differential equations are more conventional.



    Let us assume that $x(0)=dot x(0)=0$.
    We can solve (1) to obtain
    $$ x(t) = lambda int_0^t!sin(t-t') xi(t'),dt',. tag{3}$$
    This is a stochastic solution as it depends on the random function $xi(t)$. However, from (3) together with (2) we can calculate statistical predictions. For example the average position is given by
    $$langle x(t) rangle =0,,$$
    which is not unexpected (just compare it to a random walk). So on average the oscillator does diverge as it does not even move.



    Of course, the more reasonable measure if the harmonic oscillator performs an unbounded oscillation is the variance. We obtain
    $$langle x(t)^2 rangle = lambda^2 int_0^t int_0^t!sin(t-t') sin(t-t'') langlexi(t')xi(t'')rangle,dt'',dt' =lambda^2 int_0^t sin^2(t-t'),dt' = lambda^2 left(frac{t}2 - frac{sin(2t)}{4}right),. $$



    From this we see that the typical amplitude of the oscillation, given by $sqrt{langle x(t)^2 rangle }$ behaves as
    $$ sqrt{langle x(t)^2 rangle} sim lambda sqrt{frac{t}{2}}$$
    for $ttoinfty$; i.e., the oscillation grows without bounds. However, the amplitude of the oscillation only grows as $sqrt{t}$ instead of proportional to $t$.






    share|cite|improve this answer












    If you have a Gaussian random force, the equation becomes a Langevin equation. In physicists notation, you would write
    $$ ddot x + x = lambda xi(t) tag{1}$$
    with $lambda$ the strength of the random force and (the bracket denotes the stochastic average)
    $$langle xi(t) rangle = 0, quad text{and} quad langle xi(t) xi(t') rangle = delta(t-t'),. tag{2}$$
    Note that in mathematics instead stochastic differential equations are more conventional.



    Let us assume that $x(0)=dot x(0)=0$.
    We can solve (1) to obtain
    $$ x(t) = lambda int_0^t!sin(t-t') xi(t'),dt',. tag{3}$$
    This is a stochastic solution as it depends on the random function $xi(t)$. However, from (3) together with (2) we can calculate statistical predictions. For example the average position is given by
    $$langle x(t) rangle =0,,$$
    which is not unexpected (just compare it to a random walk). So on average the oscillator does diverge as it does not even move.



    Of course, the more reasonable measure if the harmonic oscillator performs an unbounded oscillation is the variance. We obtain
    $$langle x(t)^2 rangle = lambda^2 int_0^t int_0^t!sin(t-t') sin(t-t'') langlexi(t')xi(t'')rangle,dt'',dt' =lambda^2 int_0^t sin^2(t-t'),dt' = lambda^2 left(frac{t}2 - frac{sin(2t)}{4}right),. $$



    From this we see that the typical amplitude of the oscillation, given by $sqrt{langle x(t)^2 rangle }$ behaves as
    $$ sqrt{langle x(t)^2 rangle} sim lambda sqrt{frac{t}{2}}$$
    for $ttoinfty$; i.e., the oscillation grows without bounds. However, the amplitude of the oscillation only grows as $sqrt{t}$ instead of proportional to $t$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 19 at 21:30









    Fabian

    19.4k3674




    19.4k3674












    • Thanks. How does equation (1) reduce to the Langevin equation: $$mddot{x}=-lambdadot{x}+eta(t)?$$ Also, what is the stochastic average and how it is calculated?
      – Peanutlex
      Dec 19 at 22:20






    • 1




      It is a Langevin equation not the Langevin equation. The stochastic average simply encodes the notion that $xi(t)$ is a random force with zero mean and "standard deviation" $lambda$.
      – Fabian
      Dec 19 at 22:28


















    • Thanks. How does equation (1) reduce to the Langevin equation: $$mddot{x}=-lambdadot{x}+eta(t)?$$ Also, what is the stochastic average and how it is calculated?
      – Peanutlex
      Dec 19 at 22:20






    • 1




      It is a Langevin equation not the Langevin equation. The stochastic average simply encodes the notion that $xi(t)$ is a random force with zero mean and "standard deviation" $lambda$.
      – Fabian
      Dec 19 at 22:28
















    Thanks. How does equation (1) reduce to the Langevin equation: $$mddot{x}=-lambdadot{x}+eta(t)?$$ Also, what is the stochastic average and how it is calculated?
    – Peanutlex
    Dec 19 at 22:20




    Thanks. How does equation (1) reduce to the Langevin equation: $$mddot{x}=-lambdadot{x}+eta(t)?$$ Also, what is the stochastic average and how it is calculated?
    – Peanutlex
    Dec 19 at 22:20




    1




    1




    It is a Langevin equation not the Langevin equation. The stochastic average simply encodes the notion that $xi(t)$ is a random force with zero mean and "standard deviation" $lambda$.
    – Fabian
    Dec 19 at 22:28




    It is a Langevin equation not the Langevin equation. The stochastic average simply encodes the notion that $xi(t)$ is a random force with zero mean and "standard deviation" $lambda$.
    – Fabian
    Dec 19 at 22:28











    2















    Does anyone know of a simple function I could replace f with to generate a continuous white noise driver?




    I'm not sure if this approach will help you, but here it is. My approach to problems such as the one you presented is related to implementation of said problem and then derive some, at least numeric, result. Having this said, I would do as follows.



    Consider a finite discrete Gaussian white noise which is stored in an array $G(nDelta t)$, with $n$ from $0$ to $N$. This can be interpolated using a polynomial $P(t)$ of order $N-1$ which is unique. This polynomial should be the solution you are looking for for $tin[0,NDelta t]$. From this you can compute your driver function using



    $$f(t)=frac{d^2 P}{dt^2}(t) + P(t)$$






    share|cite|improve this answer


























      2















      Does anyone know of a simple function I could replace f with to generate a continuous white noise driver?




      I'm not sure if this approach will help you, but here it is. My approach to problems such as the one you presented is related to implementation of said problem and then derive some, at least numeric, result. Having this said, I would do as follows.



      Consider a finite discrete Gaussian white noise which is stored in an array $G(nDelta t)$, with $n$ from $0$ to $N$. This can be interpolated using a polynomial $P(t)$ of order $N-1$ which is unique. This polynomial should be the solution you are looking for for $tin[0,NDelta t]$. From this you can compute your driver function using



      $$f(t)=frac{d^2 P}{dt^2}(t) + P(t)$$






      share|cite|improve this answer
























        2












        2








        2







        Does anyone know of a simple function I could replace f with to generate a continuous white noise driver?




        I'm not sure if this approach will help you, but here it is. My approach to problems such as the one you presented is related to implementation of said problem and then derive some, at least numeric, result. Having this said, I would do as follows.



        Consider a finite discrete Gaussian white noise which is stored in an array $G(nDelta t)$, with $n$ from $0$ to $N$. This can be interpolated using a polynomial $P(t)$ of order $N-1$ which is unique. This polynomial should be the solution you are looking for for $tin[0,NDelta t]$. From this you can compute your driver function using



        $$f(t)=frac{d^2 P}{dt^2}(t) + P(t)$$






        share|cite|improve this answer













        Does anyone know of a simple function I could replace f with to generate a continuous white noise driver?




        I'm not sure if this approach will help you, but here it is. My approach to problems such as the one you presented is related to implementation of said problem and then derive some, at least numeric, result. Having this said, I would do as follows.



        Consider a finite discrete Gaussian white noise which is stored in an array $G(nDelta t)$, with $n$ from $0$ to $N$. This can be interpolated using a polynomial $P(t)$ of order $N-1$ which is unique. This polynomial should be the solution you are looking for for $tin[0,NDelta t]$. From this you can compute your driver function using



        $$f(t)=frac{d^2 P}{dt^2}(t) + P(t)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 at 20:04









        Victor Palea

        301312




        301312






























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