Mixing
Convolution theorem for a function that doesn't vanish at infinity
$begingroup$
I'm studying a paper of Rapp and Kassal (1968), in which a second order differential function that doesn't vanish at $pm infty$ is solved. I will put here the parts of the paper that matter:
$hspace{45pt}$
$hspace{45pt}$
I think that there's a typo at $(26)$: the "$lim_{tto+infty}$" shouldn't appear there.
The differential equation that I'm trying to solve is $(24)$ which has the intial condition $(25)$. It seems that the authors of the paper applied a Fourier transform to $(24)$ and used the convolution theorem to get the solution $(26)$.
Let $y(t)=tilde{y}(t)-tilde{y}_0$. Let also $h(t)=-frac{1}{mu}frac{gamma A''}{L}text{sech}^2left(frac{v_o t}{2L}right)$ and $omega=sqrt{frac{f}{mu}}$.
I will define the Fourier transform of a function $f(t)$ as:
$$F(xi)=mathscr{F}[f(t)]=frac{1}{sqrt{2pi}}int_{-infty}^{+infty}f(t)e^{-ixi t},dt$$
Applying the Fourier trasform to $(24)$, we would get:
$$frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}-xi^2Y(xi)+w^2Y(xi)=H(xi)Leftrightarrow$$
$$Leftrightarrow Y(xi)=-frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}G(xi)+ H(xi)G(xi) $$
where $G(xi)=frac{1}{-xi^2+omega^2}$.
And then, by using the convolution theorem we would have:
$tilde{y}(t)-tilde{y}_0=$
$-frac{1}{sqrt{2pi}}int_{-infty}^{infty}mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}g(t-s),ds+frac{1}{sqrt{2pi}}int_{-infty}^{infty}h(s)g(t-s),ds$
My main problem is how to deal with the $mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}$ term. It should only depend on $s$. But for that, I would need to remove the dependency on $t$, which I don't know if it is even possible (since the operand function isn't convergent). Is there any other way to achive $(26)$? My attempt seems not to be a good one.
[Note: I actually tried to find the procedure to get the relation $(26)$ from $[11]$ reference, but I wasn't successful.]
ordinary-differential-equations fourier-analysis convolution
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
$endgroup$
add a comment |
$begingroup$
I'm studying a paper of Rapp and Kassal (1968), in which a second order differential function that doesn't vanish at $pm infty$ is solved. I will put here the parts of the paper that matter:
$hspace{45pt}$
$hspace{45pt}$
I think that there's a typo at $(26)$: the "$lim_{tto+infty}$" shouldn't appear there.
The differential equation that I'm trying to solve is $(24)$ which has the intial condition $(25)$. It seems that the authors of the paper applied a Fourier transform to $(24)$ and used the convolution theorem to get the solution $(26)$.
Let $y(t)=tilde{y}(t)-tilde{y}_0$. Let also $h(t)=-frac{1}{mu}frac{gamma A''}{L}text{sech}^2left(frac{v_o t}{2L}right)$ and $omega=sqrt{frac{f}{mu}}$.
I will define the Fourier transform of a function $f(t)$ as:
$$F(xi)=mathscr{F}[f(t)]=frac{1}{sqrt{2pi}}int_{-infty}^{+infty}f(t)e^{-ixi t},dt$$
Applying the Fourier trasform to $(24)$, we would get:
$$frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}-xi^2Y(xi)+w^2Y(xi)=H(xi)Leftrightarrow$$
$$Leftrightarrow Y(xi)=-frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}G(xi)+ H(xi)G(xi) $$
where $G(xi)=frac{1}{-xi^2+omega^2}$.
And then, by using the convolution theorem we would have:
$tilde{y}(t)-tilde{y}_0=$
$-frac{1}{sqrt{2pi}}int_{-infty}^{infty}mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}g(t-s),ds+frac{1}{sqrt{2pi}}int_{-infty}^{infty}h(s)g(t-s),ds$
My main problem is how to deal with the $mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}$ term. It should only depend on $s$. But for that, I would need to remove the dependency on $t$, which I don't know if it is even possible (since the operand function isn't convergent). Is there any other way to achive $(26)$? My attempt seems not to be a good one.
[Note: I actually tried to find the procedure to get the relation $(26)$ from $[11]$ reference, but I wasn't successful.]
ordinary-differential-equations fourier-analysis convolution
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
$endgroup$
$begingroup$
Is this not just the variation of constants formula? And then the initial value is shifted towards $-infty$.
$endgroup$
– LutzL
Jan 18 at 18:00
add a comment |
$begingroup$
I'm studying a paper of Rapp and Kassal (1968), in which a second order differential function that doesn't vanish at $pm infty$ is solved. I will put here the parts of the paper that matter:
$hspace{45pt}$
$hspace{45pt}$
I think that there's a typo at $(26)$: the "$lim_{tto+infty}$" shouldn't appear there.
The differential equation that I'm trying to solve is $(24)$ which has the intial condition $(25)$. It seems that the authors of the paper applied a Fourier transform to $(24)$ and used the convolution theorem to get the solution $(26)$.
Let $y(t)=tilde{y}(t)-tilde{y}_0$. Let also $h(t)=-frac{1}{mu}frac{gamma A''}{L}text{sech}^2left(frac{v_o t}{2L}right)$ and $omega=sqrt{frac{f}{mu}}$.
I will define the Fourier transform of a function $f(t)$ as:
$$F(xi)=mathscr{F}[f(t)]=frac{1}{sqrt{2pi}}int_{-infty}^{+infty}f(t)e^{-ixi t},dt$$
Applying the Fourier trasform to $(24)$, we would get:
$$frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}-xi^2Y(xi)+w^2Y(xi)=H(xi)Leftrightarrow$$
$$Leftrightarrow Y(xi)=-frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}G(xi)+ H(xi)G(xi) $$
where $G(xi)=frac{1}{-xi^2+omega^2}$.
And then, by using the convolution theorem we would have:
$tilde{y}(t)-tilde{y}_0=$
$-frac{1}{sqrt{2pi}}int_{-infty}^{infty}mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}g(t-s),ds+frac{1}{sqrt{2pi}}int_{-infty}^{infty}h(s)g(t-s),ds$
My main problem is how to deal with the $mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}$ term. It should only depend on $s$. But for that, I would need to remove the dependency on $t$, which I don't know if it is even possible (since the operand function isn't convergent). Is there any other way to achive $(26)$? My attempt seems not to be a good one.
[Note: I actually tried to find the procedure to get the relation $(26)$ from $[11]$ reference, but I wasn't successful.]
ordinary-differential-equations fourier-analysis convolution
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
$endgroup$
I'm studying a paper of Rapp and Kassal (1968), in which a second order differential function that doesn't vanish at $pm infty$ is solved. I will put here the parts of the paper that matter:
$hspace{45pt}$
$hspace{45pt}$
I think that there's a typo at $(26)$: the "$lim_{tto+infty}$" shouldn't appear there.
The differential equation that I'm trying to solve is $(24)$ which has the intial condition $(25)$. It seems that the authors of the paper applied a Fourier transform to $(24)$ and used the convolution theorem to get the solution $(26)$.
Let $y(t)=tilde{y}(t)-tilde{y}_0$. Let also $h(t)=-frac{1}{mu}frac{gamma A''}{L}text{sech}^2left(frac{v_o t}{2L}right)$ and $omega=sqrt{frac{f}{mu}}$.
I will define the Fourier transform of a function $f(t)$ as:
$$F(xi)=mathscr{F}[f(t)]=frac{1}{sqrt{2pi}}int_{-infty}^{+infty}f(t)e^{-ixi t},dt$$
Applying the Fourier trasform to $(24)$, we would get:
$$frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}-xi^2Y(xi)+w^2Y(xi)=H(xi)Leftrightarrow$$
$$Leftrightarrow Y(xi)=-frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}G(xi)+ H(xi)G(xi) $$
where $G(xi)=frac{1}{-xi^2+omega^2}$.
And then, by using the convolution theorem we would have:
$tilde{y}(t)-tilde{y}_0=$
$-frac{1}{sqrt{2pi}}int_{-infty}^{infty}mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}g(t-s),ds+frac{1}{sqrt{2pi}}int_{-infty}^{infty}h(s)g(t-s),ds$
My main problem is how to deal with the $mathscr{F}^{-1}left{frac{1}{sqrt{2pi}}left[dot{y}(t)+ixi y(t)right]e^{-ixi t}biggrvert_{t=-infty}^{t=+infty}right}$ term. It should only depend on $s$. But for that, I would need to remove the dependency on $t$, which I don't know if it is even possible (since the operand function isn't convergent). Is there any other way to achive $(26)$? My attempt seems not to be a good one.
[Note: I actually tried to find the procedure to get the relation $(26)$ from $[11]$ reference, but I wasn't successful.]
ordinary-differential-equations fourier-analysis convolution
ordinary-differential-equations fourier-analysis convolution
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
edited Feb 7 at 16:08
Élio Pereira
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
asked Jan 18 at 17:50
Élio PereiraÉlio Pereira
454515
454515
$begingroup$
Is this not just the variation of constants formula? And then the initial value is shifted towards $-infty$.
$endgroup$
– LutzL
Jan 18 at 18:00
add a comment |
$begingroup$
Is this not just the variation of constants formula? And then the initial value is shifted towards $-infty$.
$endgroup$
– LutzL
Jan 18 at 18:00
$begingroup$
Is this not just the variation of constants formula? And then the initial value is shifted towards $-infty$.
$endgroup$
– LutzL
Jan 18 at 18:00
$begingroup$
Is this not just the variation of constants formula? And then the initial value is shifted towards $-infty$.
$endgroup$
– LutzL
Jan 18 at 18:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the function, related to the variation of constant method,
$$
g_t(s)=cos(ω(t-s))(y(s)-bar y)+frac{sin(ω(t-s))}{ω}y'(s).
$$
Then
$$
g_t'(s)=frac{sin(ω(t-s))}ω[ω^2(y(s)-bar y)+y''(s)]=frac{sin(ω(t-s))}{ωμ}f(s)
$$
Now $g_t(t)=y(t)-bar y$ so that by integration
$$
y(t)-bar y-g_t(a)=frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
or
$$
y(t)-bar y=cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)+frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
Now if $a$ and $t$ are both close to $-infty$, then the right side forcing function is very small, its contribution goes to zero. The integral from $-infty$ to $a$ is just as small, as the function $f(s)$ in the integrand is essentially a falling exponential, looking in direction $sto -infty$. This means that the bounded oscillatory behavior demanded by
$$
lim_{tto-infty}Bigl[y(t)-bar y -B_isin(ωt+delta_i)Bigr]=0
$$
has to come from the first terms,
$$
cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)approx B_isin(ωt+δ_i)
$$
which then gives the solution without limit
$$
y(t)-bar y=B_isin(ωt+δ_i)+frac{1}{ωμ}int_{-infty}^tsin(ω(t-s))f(s),ds
$$
I do not think that you should read the limits in the cited image as the standard function limits. Here this symbolism seems to stand for "behaves asymptotically like". As the last solution formula is exact for all $t$, it is also asymptotically true for $tto+infty$.
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
$endgroup$
$begingroup$
It seems that there's a problem with the signals when you take the derivatives. The derivative of $cosleft[omega(t-s)right]$ in order to $s$ is $+sinleft[omega(t-s)right]$, due to the minus signal inside the argument of the function. I'm not familiar with the variation of constant method. How did you get the function $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 20 at 21:44
$begingroup$
Yes, you are right. Thanks and corrected.
$endgroup$
– LutzL
Jan 21 at 9:39
$begingroup$
I understood everything you did. I just have doubts about the variation constant method. How can one obtain $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 21 at 11:48
1
$begingroup$
This is some kind of Wronskian. You can also go the traditional route of the variation-of-constant method where $y(t)=c_1(t)cos(ωt)+c_2(t)sin(ωt)$ with $$ c_1'(t)cos(ωt)+c_2'(t)sin(ωt)=0,\-μωc_1'(t)sin(ωt)+μωc_2'(t)cos(ωt)=f(t),$$ so that $$μωc_1'(s)=-f(s)sin(ωs), \ μωc_2'(s)=f(s)cos(ωs).$$ Then the above formula follows from trigonometric identities in $$μωy(t)=-cos(ωt)int f(s)sin(ωs),ds+sin(ωt)int f(s)cos(ωs),ds\~\=d_1cos(ωt)+d_2sin(ωt)+int f(s)sin(ω(t-s)),ds.$$
$endgroup$
– LutzL
Jan 21 at 12:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider the function, related to the variation of constant method,
$$
g_t(s)=cos(ω(t-s))(y(s)-bar y)+frac{sin(ω(t-s))}{ω}y'(s).
$$
Then
$$
g_t'(s)=frac{sin(ω(t-s))}ω[ω^2(y(s)-bar y)+y''(s)]=frac{sin(ω(t-s))}{ωμ}f(s)
$$
Now $g_t(t)=y(t)-bar y$ so that by integration
$$
y(t)-bar y-g_t(a)=frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
or
$$
y(t)-bar y=cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)+frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
Now if $a$ and $t$ are both close to $-infty$, then the right side forcing function is very small, its contribution goes to zero. The integral from $-infty$ to $a$ is just as small, as the function $f(s)$ in the integrand is essentially a falling exponential, looking in direction $sto -infty$. This means that the bounded oscillatory behavior demanded by
$$
lim_{tto-infty}Bigl[y(t)-bar y -B_isin(ωt+delta_i)Bigr]=0
$$
has to come from the first terms,
$$
cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)approx B_isin(ωt+δ_i)
$$
which then gives the solution without limit
$$
y(t)-bar y=B_isin(ωt+δ_i)+frac{1}{ωμ}int_{-infty}^tsin(ω(t-s))f(s),ds
$$
I do not think that you should read the limits in the cited image as the standard function limits. Here this symbolism seems to stand for "behaves asymptotically like". As the last solution formula is exact for all $t$, it is also asymptotically true for $tto+infty$.
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
$endgroup$
$begingroup$
It seems that there's a problem with the signals when you take the derivatives. The derivative of $cosleft[omega(t-s)right]$ in order to $s$ is $+sinleft[omega(t-s)right]$, due to the minus signal inside the argument of the function. I'm not familiar with the variation of constant method. How did you get the function $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 20 at 21:44
$begingroup$
Yes, you are right. Thanks and corrected.
$endgroup$
– LutzL
Jan 21 at 9:39
$begingroup$
I understood everything you did. I just have doubts about the variation constant method. How can one obtain $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 21 at 11:48
1
$begingroup$
This is some kind of Wronskian. You can also go the traditional route of the variation-of-constant method where $y(t)=c_1(t)cos(ωt)+c_2(t)sin(ωt)$ with $$ c_1'(t)cos(ωt)+c_2'(t)sin(ωt)=0,\-μωc_1'(t)sin(ωt)+μωc_2'(t)cos(ωt)=f(t),$$ so that $$μωc_1'(s)=-f(s)sin(ωs), \ μωc_2'(s)=f(s)cos(ωs).$$ Then the above formula follows from trigonometric identities in $$μωy(t)=-cos(ωt)int f(s)sin(ωs),ds+sin(ωt)int f(s)cos(ωs),ds\~\=d_1cos(ωt)+d_2sin(ωt)+int f(s)sin(ω(t-s)),ds.$$
$endgroup$
– LutzL
Jan 21 at 12:06
add a comment |
$begingroup$
Consider the function, related to the variation of constant method,
$$
g_t(s)=cos(ω(t-s))(y(s)-bar y)+frac{sin(ω(t-s))}{ω}y'(s).
$$
Then
$$
g_t'(s)=frac{sin(ω(t-s))}ω[ω^2(y(s)-bar y)+y''(s)]=frac{sin(ω(t-s))}{ωμ}f(s)
$$
Now $g_t(t)=y(t)-bar y$ so that by integration
$$
y(t)-bar y-g_t(a)=frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
or
$$
y(t)-bar y=cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)+frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
Now if $a$ and $t$ are both close to $-infty$, then the right side forcing function is very small, its contribution goes to zero. The integral from $-infty$ to $a$ is just as small, as the function $f(s)$ in the integrand is essentially a falling exponential, looking in direction $sto -infty$. This means that the bounded oscillatory behavior demanded by
$$
lim_{tto-infty}Bigl[y(t)-bar y -B_isin(ωt+delta_i)Bigr]=0
$$
has to come from the first terms,
$$
cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)approx B_isin(ωt+δ_i)
$$
which then gives the solution without limit
$$
y(t)-bar y=B_isin(ωt+δ_i)+frac{1}{ωμ}int_{-infty}^tsin(ω(t-s))f(s),ds
$$
I do not think that you should read the limits in the cited image as the standard function limits. Here this symbolism seems to stand for "behaves asymptotically like". As the last solution formula is exact for all $t$, it is also asymptotically true for $tto+infty$.
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
$endgroup$
$begingroup$
It seems that there's a problem with the signals when you take the derivatives. The derivative of $cosleft[omega(t-s)right]$ in order to $s$ is $+sinleft[omega(t-s)right]$, due to the minus signal inside the argument of the function. I'm not familiar with the variation of constant method. How did you get the function $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 20 at 21:44
$begingroup$
Yes, you are right. Thanks and corrected.
$endgroup$
– LutzL
Jan 21 at 9:39
$begingroup$
I understood everything you did. I just have doubts about the variation constant method. How can one obtain $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 21 at 11:48
1
$begingroup$
This is some kind of Wronskian. You can also go the traditional route of the variation-of-constant method where $y(t)=c_1(t)cos(ωt)+c_2(t)sin(ωt)$ with $$ c_1'(t)cos(ωt)+c_2'(t)sin(ωt)=0,\-μωc_1'(t)sin(ωt)+μωc_2'(t)cos(ωt)=f(t),$$ so that $$μωc_1'(s)=-f(s)sin(ωs), \ μωc_2'(s)=f(s)cos(ωs).$$ Then the above formula follows from trigonometric identities in $$μωy(t)=-cos(ωt)int f(s)sin(ωs),ds+sin(ωt)int f(s)cos(ωs),ds\~\=d_1cos(ωt)+d_2sin(ωt)+int f(s)sin(ω(t-s)),ds.$$
$endgroup$
– LutzL
Jan 21 at 12:06
add a comment |
$begingroup$
Consider the function, related to the variation of constant method,
$$
g_t(s)=cos(ω(t-s))(y(s)-bar y)+frac{sin(ω(t-s))}{ω}y'(s).
$$
Then
$$
g_t'(s)=frac{sin(ω(t-s))}ω[ω^2(y(s)-bar y)+y''(s)]=frac{sin(ω(t-s))}{ωμ}f(s)
$$
Now $g_t(t)=y(t)-bar y$ so that by integration
$$
y(t)-bar y-g_t(a)=frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
or
$$
y(t)-bar y=cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)+frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
Now if $a$ and $t$ are both close to $-infty$, then the right side forcing function is very small, its contribution goes to zero. The integral from $-infty$ to $a$ is just as small, as the function $f(s)$ in the integrand is essentially a falling exponential, looking in direction $sto -infty$. This means that the bounded oscillatory behavior demanded by
$$
lim_{tto-infty}Bigl[y(t)-bar y -B_isin(ωt+delta_i)Bigr]=0
$$
has to come from the first terms,
$$
cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)approx B_isin(ωt+δ_i)
$$
which then gives the solution without limit
$$
y(t)-bar y=B_isin(ωt+δ_i)+frac{1}{ωμ}int_{-infty}^tsin(ω(t-s))f(s),ds
$$
I do not think that you should read the limits in the cited image as the standard function limits. Here this symbolism seems to stand for "behaves asymptotically like". As the last solution formula is exact for all $t$, it is also asymptotically true for $tto+infty$.
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
$endgroup$
Consider the function, related to the variation of constant method,
$$
g_t(s)=cos(ω(t-s))(y(s)-bar y)+frac{sin(ω(t-s))}{ω}y'(s).
$$
Then
$$
g_t'(s)=frac{sin(ω(t-s))}ω[ω^2(y(s)-bar y)+y''(s)]=frac{sin(ω(t-s))}{ωμ}f(s)
$$
Now $g_t(t)=y(t)-bar y$ so that by integration
$$
y(t)-bar y-g_t(a)=frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
or
$$
y(t)-bar y=cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)+frac{1}{ωμ}int_a^tsin(ω(t-s))f(s),ds
$$
Now if $a$ and $t$ are both close to $-infty$, then the right side forcing function is very small, its contribution goes to zero. The integral from $-infty$ to $a$ is just as small, as the function $f(s)$ in the integrand is essentially a falling exponential, looking in direction $sto -infty$. This means that the bounded oscillatory behavior demanded by
$$
lim_{tto-infty}Bigl[y(t)-bar y -B_isin(ωt+delta_i)Bigr]=0
$$
has to come from the first terms,
$$
cos(ω(t-a))(y(a)-bar y)+frac{sin(ω(t-a))}{ω}y'(a)approx B_isin(ωt+δ_i)
$$
which then gives the solution without limit
$$
y(t)-bar y=B_isin(ωt+δ_i)+frac{1}{ωμ}int_{-infty}^tsin(ω(t-s))f(s),ds
$$
I do not think that you should read the limits in the cited image as the standard function limits. Here this symbolism seems to stand for "behaves asymptotically like". As the last solution formula is exact for all $t$, it is also asymptotically true for $tto+infty$.
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
edited Jan 21 at 9:39
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
answered Jan 19 at 10:32
LutzLLutzL
59.1k42056
59.1k42056
$begingroup$
It seems that there's a problem with the signals when you take the derivatives. The derivative of $cosleft[omega(t-s)right]$ in order to $s$ is $+sinleft[omega(t-s)right]$, due to the minus signal inside the argument of the function. I'm not familiar with the variation of constant method. How did you get the function $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 20 at 21:44
$begingroup$
Yes, you are right. Thanks and corrected.
$endgroup$
– LutzL
Jan 21 at 9:39
$begingroup$
I understood everything you did. I just have doubts about the variation constant method. How can one obtain $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 21 at 11:48
1
$begingroup$
This is some kind of Wronskian. You can also go the traditional route of the variation-of-constant method where $y(t)=c_1(t)cos(ωt)+c_2(t)sin(ωt)$ with $$ c_1'(t)cos(ωt)+c_2'(t)sin(ωt)=0,\-μωc_1'(t)sin(ωt)+μωc_2'(t)cos(ωt)=f(t),$$ so that $$μωc_1'(s)=-f(s)sin(ωs), \ μωc_2'(s)=f(s)cos(ωs).$$ Then the above formula follows from trigonometric identities in $$μωy(t)=-cos(ωt)int f(s)sin(ωs),ds+sin(ωt)int f(s)cos(ωs),ds\~\=d_1cos(ωt)+d_2sin(ωt)+int f(s)sin(ω(t-s)),ds.$$
$endgroup$
– LutzL
Jan 21 at 12:06
add a comment |
$begingroup$
It seems that there's a problem with the signals when you take the derivatives. The derivative of $cosleft[omega(t-s)right]$ in order to $s$ is $+sinleft[omega(t-s)right]$, due to the minus signal inside the argument of the function. I'm not familiar with the variation of constant method. How did you get the function $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 20 at 21:44
$begingroup$
Yes, you are right. Thanks and corrected.
$endgroup$
– LutzL
Jan 21 at 9:39
$begingroup$
I understood everything you did. I just have doubts about the variation constant method. How can one obtain $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 21 at 11:48
1
$begingroup$
This is some kind of Wronskian. You can also go the traditional route of the variation-of-constant method where $y(t)=c_1(t)cos(ωt)+c_2(t)sin(ωt)$ with $$ c_1'(t)cos(ωt)+c_2'(t)sin(ωt)=0,\-μωc_1'(t)sin(ωt)+μωc_2'(t)cos(ωt)=f(t),$$ so that $$μωc_1'(s)=-f(s)sin(ωs), \ μωc_2'(s)=f(s)cos(ωs).$$ Then the above formula follows from trigonometric identities in $$μωy(t)=-cos(ωt)int f(s)sin(ωs),ds+sin(ωt)int f(s)cos(ωs),ds\~\=d_1cos(ωt)+d_2sin(ωt)+int f(s)sin(ω(t-s)),ds.$$
$endgroup$
– LutzL
Jan 21 at 12:06
$begingroup$
It seems that there's a problem with the signals when you take the derivatives. The derivative of $cosleft[omega(t-s)right]$ in order to $s$ is $+sinleft[omega(t-s)right]$, due to the minus signal inside the argument of the function. I'm not familiar with the variation of constant method. How did you get the function $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 20 at 21:44
$begingroup$
It seems that there's a problem with the signals when you take the derivatives. The derivative of $cosleft[omega(t-s)right]$ in order to $s$ is $+sinleft[omega(t-s)right]$, due to the minus signal inside the argument of the function. I'm not familiar with the variation of constant method. How did you get the function $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 20 at 21:44
$begingroup$
Yes, you are right. Thanks and corrected.
$endgroup$
– LutzL
Jan 21 at 9:39
$begingroup$
Yes, you are right. Thanks and corrected.
$endgroup$
– LutzL
Jan 21 at 9:39
$begingroup$
I understood everything you did. I just have doubts about the variation constant method. How can one obtain $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 21 at 11:48
$begingroup$
I understood everything you did. I just have doubts about the variation constant method. How can one obtain $g_t(s)$?
$endgroup$
– Élio Pereira
Jan 21 at 11:48
1
1
$begingroup$
This is some kind of Wronskian. You can also go the traditional route of the variation-of-constant method where $y(t)=c_1(t)cos(ωt)+c_2(t)sin(ωt)$ with $$ c_1'(t)cos(ωt)+c_2'(t)sin(ωt)=0,\-μωc_1'(t)sin(ωt)+μωc_2'(t)cos(ωt)=f(t),$$ so that $$μωc_1'(s)=-f(s)sin(ωs), \ μωc_2'(s)=f(s)cos(ωs).$$ Then the above formula follows from trigonometric identities in $$μωy(t)=-cos(ωt)int f(s)sin(ωs),ds+sin(ωt)int f(s)cos(ωs),ds\~\=d_1cos(ωt)+d_2sin(ωt)+int f(s)sin(ω(t-s)),ds.$$
$endgroup$
– LutzL
Jan 21 at 12:06
$begingroup$
This is some kind of Wronskian. You can also go the traditional route of the variation-of-constant method where $y(t)=c_1(t)cos(ωt)+c_2(t)sin(ωt)$ with $$ c_1'(t)cos(ωt)+c_2'(t)sin(ωt)=0,\-μωc_1'(t)sin(ωt)+μωc_2'(t)cos(ωt)=f(t),$$ so that $$μωc_1'(s)=-f(s)sin(ωs), \ μωc_2'(s)=f(s)cos(ωs).$$ Then the above formula follows from trigonometric identities in $$μωy(t)=-cos(ωt)int f(s)sin(ωs),ds+sin(ωt)int f(s)cos(ωs),ds\~\=d_1cos(ωt)+d_2sin(ωt)+int f(s)sin(ω(t-s)),ds.$$
$endgroup$
– LutzL
Jan 21 at 12:06
add a comment |
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$begingroup$
Is this not just the variation of constants formula? And then the initial value is shifted towards $-infty$.
$endgroup$
– LutzL
Jan 18 at 18:00