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Contradiction in Fourier analysis of differential equation - what rules did I break?
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I'm interested in the Fourier Spectrum of the solution of the differential equation on the domain $t in [0, infty]$:
$$ fracdpdt = -beta p $$
Doing Fourier Analysis on this differential equation I get:
$$ (i omega + beta ) P = 0 $$
which is interesting because this would seem to indicate that there are only trivial solutions to this equation.
The time domain solution to this equation is:
$$ p(t) = p(0) e^-beta t quad textfor t geq 0$$
If I take the fourier transform of this I get:
$$ P (omega) = int_0^infty p(0) e^-beta t e^-iomega t dt = p(0) int_0^infty e^-(iomega + beta) t dt $$
$$ P(omega) = - fracp(0)iomega + beta e^-(iomega + beta) t big |^t=infty_t = 0 = fracp(0)iomega + beta $$
Now I plut this into my original equation:
$$ (i omega + beta) P = (i omega + beta) fracp(0)i omega + beta = p(0) neq 0 $$
Where did I go wrong?
differential-equations fourier-analysis
asked Jul 22 at 19:45
Mike Flynn
5611517
add a comment |Â
up vote
0
down vote
favorite
I'm interested in the Fourier Spectrum of the solution of the differential equation on the domain $t in [0, infty]$:
$$ fracdpdt = -beta p $$
Doing Fourier Analysis on this differential equation I get:
$$ (i omega + beta ) P = 0 $$
which is interesting because this would seem to indicate that there are only trivial solutions to this equation.
The time domain solution to this equation is:
$$ p(t) = p(0) e^-beta t quad textfor t geq 0$$
If I take the fourier transform of this I get:
$$ P (omega) = int_0^infty p(0) e^-beta t e^-iomega t dt = p(0) int_0^infty e^-(iomega + beta) t dt $$
$$ P(omega) = - fracp(0)iomega + beta e^-(iomega + beta) t big |^t=infty_t = 0 = fracp(0)iomega + beta $$
Now I plut this into my original equation:
$$ (i omega + beta) P = (i omega + beta) fracp(0)i omega + beta = p(0) neq 0 $$
Where did I go wrong?
differential-equations fourier-analysis
asked Jul 22 at 19:45
Mike Flynn
5611517
I have came across some similiar problems always by homogeous ODE's. The same problem occurs when you are using a Laplace transform. I am not sure but I would claim that integral transforms are somehow not able to solve homogeous ODE's ore you have to consider that $mathcalF_t0(omega)=c$.
– mrtaurho
Jul 22 at 19:59
hmmm, using the Laplace transform the answer actually pops out because $Lfracdpdt = sLp - p(0)$, which gives the equation $(s + beta)Lp = p(0)$ which is correct if $s = iomega$. Why didn't this work for Fourier?
– Mike Flynn
Jul 22 at 20:11
Oh, then my mind fooled me. Yes, you are right concerning this and I know where I did wrong on my problem. ^^
– mrtaurho
Jul 22 at 20:15
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm interested in the Fourier Spectrum of the solution of the differential equation on the domain $t in [0, infty]$:
$$ fracdpdt = -beta p $$
Doing Fourier Analysis on this differential equation I get:
$$ (i omega + beta ) P = 0 $$
which is interesting because this would seem to indicate that there are only trivial solutions to this equation.
The time domain solution to this equation is:
$$ p(t) = p(0) e^-beta t quad textfor t geq 0$$
If I take the fourier transform of this I get:
$$ P (omega) = int_0^infty p(0) e^-beta t e^-iomega t dt = p(0) int_0^infty e^-(iomega + beta) t dt $$
$$ P(omega) = - fracp(0)iomega + beta e^-(iomega + beta) t big |^t=infty_t = 0 = fracp(0)iomega + beta $$
Now I plut this into my original equation:
$$ (i omega + beta) P = (i omega + beta) fracp(0)i omega + beta = p(0) neq 0 $$
Where did I go wrong?
differential-equations fourier-analysis
asked Jul 22 at 19:45
Mike Flynn
5611517
I'm interested in the Fourier Spectrum of the solution of the differential equation on the domain $t in [0, infty]$:
$$ fracdpdt = -beta p $$
Doing Fourier Analysis on this differential equation I get:
$$ (i omega + beta ) P = 0 $$
which is interesting because this would seem to indicate that there are only trivial solutions to this equation.
The time domain solution to this equation is:
$$ p(t) = p(0) e^-beta t quad textfor t geq 0$$
If I take the fourier transform of this I get:
$$ P (omega) = int_0^infty p(0) e^-beta t e^-iomega t dt = p(0) int_0^infty e^-(iomega + beta) t dt $$
$$ P(omega) = - fracp(0)iomega + beta e^-(iomega + beta) t big |^t=infty_t = 0 = fracp(0)iomega + beta $$
Now I plut this into my original equation:
$$ (i omega + beta) P = (i omega + beta) fracp(0)i omega + beta = p(0) neq 0 $$
Where did I go wrong?
differential-equations fourier-analysis
asked Jul 22 at 19:45
Mike Flynn
5611517
edited Jul 22 at 19:51
asked Jul 22 at 19:45
Mike Flynn
5611517
asked Jul 22 at 19:45
Mike Flynn
5611517
asked Jul 22 at 19:45
Mike Flynn
5611517
5611517
I have came across some similiar problems always by homogeous ODE's. The same problem occurs when you are using a Laplace transform. I am not sure but I would claim that integral transforms are somehow not able to solve homogeous ODE's ore you have to consider that $mathcalF_t0(omega)=c$.
– mrtaurho
Jul 22 at 19:59
hmmm, using the Laplace transform the answer actually pops out because $Lfracdpdt = sLp - p(0)$, which gives the equation $(s + beta)Lp = p(0)$ which is correct if $s = iomega$. Why didn't this work for Fourier?
– Mike Flynn
Jul 22 at 20:11
Oh, then my mind fooled me. Yes, you are right concerning this and I know where I did wrong on my problem. ^^
– mrtaurho
Jul 22 at 20:15
add a comment |Â
I have came across some similiar problems always by homogeous ODE's. The same problem occurs when you are using a Laplace transform. I am not sure but I would claim that integral transforms are somehow not able to solve homogeous ODE's ore you have to consider that $mathcalF_t0(omega)=c$.
– mrtaurho
Jul 22 at 19:59
hmmm, using the Laplace transform the answer actually pops out because $Lfracdpdt = sLp - p(0)$, which gives the equation $(s + beta)Lp = p(0)$ which is correct if $s = iomega$. Why didn't this work for Fourier?
– Mike Flynn
Jul 22 at 20:11
Oh, then my mind fooled me. Yes, you are right concerning this and I know where I did wrong on my problem. ^^
– mrtaurho
Jul 22 at 20:15
I have came across some similiar problems always by homogeous ODE's. The same problem occurs when you are using a Laplace transform. I am not sure but I would claim that integral transforms are somehow not able to solve homogeous ODE's ore you have to consider that $mathcalF_t0(omega)=c$.
– mrtaurho
Jul 22 at 19:59
I have came across some similiar problems always by homogeous ODE's. The same problem occurs when you are using a Laplace transform. I am not sure but I would claim that integral transforms are somehow not able to solve homogeous ODE's ore you have to consider that $mathcalF_t0(omega)=c$.
– mrtaurho
Jul 22 at 19:59
hmmm, using the Laplace transform the answer actually pops out because $Lfracdpdt = sLp - p(0)$, which gives the equation $(s + beta)Lp = p(0)$ which is correct if $s = iomega$. Why didn't this work for Fourier?
– Mike Flynn
Jul 22 at 20:11
hmmm, using the Laplace transform the answer actually pops out because $Lfracdpdt = sLp - p(0)$, which gives the equation $(s + beta)Lp = p(0)$ which is correct if $s = iomega$. Why didn't this work for Fourier?
– Mike Flynn
Jul 22 at 20:11
Oh, then my mind fooled me. Yes, you are right concerning this and I know where I did wrong on my problem. ^^
– mrtaurho
Jul 22 at 20:15
Oh, then my mind fooled me. Yes, you are right concerning this and I know where I did wrong on my problem. ^^
– mrtaurho
Jul 22 at 20:15
add a comment |Â
1 Answer
1
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oldest
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up vote
1
down vote
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In the first case you take the Fourier transform over all of $mathbb R,$ but in the second case you take it only over the positive half of $mathbb R.$
If we limit $p(t)$ to be non-zero only for $t>0$ then the first Fourier transform becomes
$$
mathcalFLHS
= int_0^infty p'(t) , e^-iomega t , dt \
= [p(t) , e^-iomega t]_0^infty - int_0^infty p(t) , (-iomega) e^-iomega t , dt \
= -p(0) + iomega int_0^infty p(t) e^-iomega t , dt \
= -p(0) + iomega P(omega)
$$
and
$$
mathcalFRHS
= int_0^infty (-beta p(t)) , e^-iomega t , dt \
= -beta int_0^infty p(t) , e^-iomega t , dt \
= -beta P(omega)
$$
Thus,
$$(iomega+beta)P(omega) = p(0)$$
so
$$P(omega) = p(0)/(iomega+beta).$$
This is the same as you got in the second case.
answered Jul 22 at 20:13
md2perpe
5,84511022
1
could you elaborate? I see the conceptual difference but I don't see how it changes the answer.
– Mike Flynn
Jul 22 at 20:19
The normal Fourier integrals, without the restriction $tin[0,infty]$, looks like $int_-infty^inftyf(t)e^ixcdot ymathrmdx$ and from this one the correspondance tables are computed. Therefore, if you use a different transform, in this case $int_0^inftyf(t)e^ixcdot ymathrmdx$ you will get a different solution I guess
– mrtaurho
Jul 22 at 20:23
@MikeFlynn. I have now added some calculations.
– md2perpe
Jul 22 at 20:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In the first case you take the Fourier transform over all of $mathbb R,$ but in the second case you take it only over the positive half of $mathbb R.$
If we limit $p(t)$ to be non-zero only for $t>0$ then the first Fourier transform becomes
$$
mathcalFLHS
= int_0^infty p'(t) , e^-iomega t , dt \
= [p(t) , e^-iomega t]_0^infty - int_0^infty p(t) , (-iomega) e^-iomega t , dt \
= -p(0) + iomega int_0^infty p(t) e^-iomega t , dt \
= -p(0) + iomega P(omega)
$$
and
$$
mathcalFRHS
= int_0^infty (-beta p(t)) , e^-iomega t , dt \
= -beta int_0^infty p(t) , e^-iomega t , dt \
= -beta P(omega)
$$
Thus,
$$(iomega+beta)P(omega) = p(0)$$
so
$$P(omega) = p(0)/(iomega+beta).$$
This is the same as you got in the second case.
answered Jul 22 at 20:13
md2perpe
5,84511022
1
could you elaborate? I see the conceptual difference but I don't see how it changes the answer.
– Mike Flynn
Jul 22 at 20:19
The normal Fourier integrals, without the restriction $tin[0,infty]$, looks like $int_-infty^inftyf(t)e^ixcdot ymathrmdx$ and from this one the correspondance tables are computed. Therefore, if you use a different transform, in this case $int_0^inftyf(t)e^ixcdot ymathrmdx$ you will get a different solution I guess
– mrtaurho
Jul 22 at 20:23
@MikeFlynn. I have now added some calculations.
– md2perpe
Jul 22 at 20:25
add a comment |Â
up vote
1
down vote
accepted
In the first case you take the Fourier transform over all of $mathbb R,$ but in the second case you take it only over the positive half of $mathbb R.$
If we limit $p(t)$ to be non-zero only for $t>0$ then the first Fourier transform becomes
$$
mathcalFLHS
= int_0^infty p'(t) , e^-iomega t , dt \
= [p(t) , e^-iomega t]_0^infty - int_0^infty p(t) , (-iomega) e^-iomega t , dt \
= -p(0) + iomega int_0^infty p(t) e^-iomega t , dt \
= -p(0) + iomega P(omega)
$$
and
$$
mathcalFRHS
= int_0^infty (-beta p(t)) , e^-iomega t , dt \
= -beta int_0^infty p(t) , e^-iomega t , dt \
= -beta P(omega)
$$
Thus,
$$(iomega+beta)P(omega) = p(0)$$
so
$$P(omega) = p(0)/(iomega+beta).$$
This is the same as you got in the second case.
answered Jul 22 at 20:13
md2perpe
5,84511022
1
could you elaborate? I see the conceptual difference but I don't see how it changes the answer.
– Mike Flynn
Jul 22 at 20:19
The normal Fourier integrals, without the restriction $tin[0,infty]$, looks like $int_-infty^inftyf(t)e^ixcdot ymathrmdx$ and from this one the correspondance tables are computed. Therefore, if you use a different transform, in this case $int_0^inftyf(t)e^ixcdot ymathrmdx$ you will get a different solution I guess
– mrtaurho
Jul 22 at 20:23
@MikeFlynn. I have now added some calculations.
– md2perpe
Jul 22 at 20:25
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In the first case you take the Fourier transform over all of $mathbb R,$ but in the second case you take it only over the positive half of $mathbb R.$
If we limit $p(t)$ to be non-zero only for $t>0$ then the first Fourier transform becomes
$$
mathcalFLHS
= int_0^infty p'(t) , e^-iomega t , dt \
= [p(t) , e^-iomega t]_0^infty - int_0^infty p(t) , (-iomega) e^-iomega t , dt \
= -p(0) + iomega int_0^infty p(t) e^-iomega t , dt \
= -p(0) + iomega P(omega)
$$
and
$$
mathcalFRHS
= int_0^infty (-beta p(t)) , e^-iomega t , dt \
= -beta int_0^infty p(t) , e^-iomega t , dt \
= -beta P(omega)
$$
Thus,
$$(iomega+beta)P(omega) = p(0)$$
so
$$P(omega) = p(0)/(iomega+beta).$$
This is the same as you got in the second case.
answered Jul 22 at 20:13
md2perpe
5,84511022
In the first case you take the Fourier transform over all of $mathbb R,$ but in the second case you take it only over the positive half of $mathbb R.$
If we limit $p(t)$ to be non-zero only for $t>0$ then the first Fourier transform becomes
$$
mathcalFLHS
= int_0^infty p'(t) , e^-iomega t , dt \
= [p(t) , e^-iomega t]_0^infty - int_0^infty p(t) , (-iomega) e^-iomega t , dt \
= -p(0) + iomega int_0^infty p(t) e^-iomega t , dt \
= -p(0) + iomega P(omega)
$$
and
$$
mathcalFRHS
= int_0^infty (-beta p(t)) , e^-iomega t , dt \
= -beta int_0^infty p(t) , e^-iomega t , dt \
= -beta P(omega)
$$
Thus,
$$(iomega+beta)P(omega) = p(0)$$
so
$$P(omega) = p(0)/(iomega+beta).$$
This is the same as you got in the second case.
answered Jul 22 at 20:13
md2perpe
5,84511022
edited Jul 22 at 20:25
answered Jul 22 at 20:13
md2perpe
5,84511022
answered Jul 22 at 20:13
md2perpe
5,84511022
answered Jul 22 at 20:13
md2perpe
5,84511022
5,84511022
1
could you elaborate? I see the conceptual difference but I don't see how it changes the answer.
– Mike Flynn
Jul 22 at 20:19
The normal Fourier integrals, without the restriction $tin[0,infty]$, looks like $int_-infty^inftyf(t)e^ixcdot ymathrmdx$ and from this one the correspondance tables are computed. Therefore, if you use a different transform, in this case $int_0^inftyf(t)e^ixcdot ymathrmdx$ you will get a different solution I guess
– mrtaurho
Jul 22 at 20:23
@MikeFlynn. I have now added some calculations.
– md2perpe
Jul 22 at 20:25
add a comment |Â
1
could you elaborate? I see the conceptual difference but I don't see how it changes the answer.
– Mike Flynn
Jul 22 at 20:19
The normal Fourier integrals, without the restriction $tin[0,infty]$, looks like $int_-infty^inftyf(t)e^ixcdot ymathrmdx$ and from this one the correspondance tables are computed. Therefore, if you use a different transform, in this case $int_0^inftyf(t)e^ixcdot ymathrmdx$ you will get a different solution I guess
– mrtaurho
Jul 22 at 20:23
@MikeFlynn. I have now added some calculations.
– md2perpe
Jul 22 at 20:25
1
1
could you elaborate? I see the conceptual difference but I don't see how it changes the answer.
– Mike Flynn
Jul 22 at 20:19
could you elaborate? I see the conceptual difference but I don't see how it changes the answer.
– Mike Flynn
Jul 22 at 20:19
The normal Fourier integrals, without the restriction $tin[0,infty]$, looks like $int_-infty^inftyf(t)e^ixcdot ymathrmdx$ and from this one the correspondance tables are computed. Therefore, if you use a different transform, in this case $int_0^inftyf(t)e^ixcdot ymathrmdx$ you will get a different solution I guess
– mrtaurho
Jul 22 at 20:23
The normal Fourier integrals, without the restriction $tin[0,infty]$, looks like $int_-infty^inftyf(t)e^ixcdot ymathrmdx$ and from this one the correspondance tables are computed. Therefore, if you use a different transform, in this case $int_0^inftyf(t)e^ixcdot ymathrmdx$ you will get a different solution I guess
– mrtaurho
Jul 22 at 20:23
@MikeFlynn. I have now added some calculations.
– md2perpe
Jul 22 at 20:25
@MikeFlynn. I have now added some calculations.
– md2perpe
Jul 22 at 20:25
add a comment |Â
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I have came across some similiar problems always by homogeous ODE's. The same problem occurs when you are using a Laplace transform. I am not sure but I would claim that integral transforms are somehow not able to solve homogeous ODE's ore you have to consider that $mathcalF_t0(omega)=c$.
– mrtaurho
Jul 22 at 19:59
hmmm, using the Laplace transform the answer actually pops out because $Lfracdpdt = sLp - p(0)$, which gives the equation $(s + beta)Lp = p(0)$ which is correct if $s = iomega$. Why didn't this work for Fourier?
– Mike Flynn
Jul 22 at 20:11
Oh, then my mind fooled me. Yes, you are right concerning this and I know where I did wrong on my problem. ^^
– mrtaurho
Jul 22 at 20:15