Mixing
Solution to ODE with Dirac Delta satisfies ODE
$begingroup$
I am working on a problem where I have the following ODE. $$mdot{v}+bv=delta_I(t)$$
where
$$delta_I(t)=begin{cases}0, & text{for}&tne0\ I, & text{for} &t=0end{cases}.$$
The solution $v(t)$ was derived using Laplace transforms, the ODE in the Laplace domain is (with $0$ initial conditions)$$(ms+b)V(s)=I$$ giving $$v(t)=frac{I}{m}e^{-bt/m}.$$
How does this solution satisfy the original ODE though? At $tne0$ everything is good,$$-frac{Ib}{m}e^{-bt/m}+frac{Ib}{m}e^{-bt/m}=0,$$ while at $t=0$, $$-frac{Ib}{m}+frac{Ib}{m}=I$$ the result seems to be saying $0=1$ which is obviously false. What am I missing here?
ordinary-differential-equations laplace-transform dirac-delta
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
$endgroup$
|
show 3 more comments
$begingroup$
I am working on a problem where I have the following ODE. $$mdot{v}+bv=delta_I(t)$$
where
$$delta_I(t)=begin{cases}0, & text{for}&tne0\ I, & text{for} &t=0end{cases}.$$
The solution $v(t)$ was derived using Laplace transforms, the ODE in the Laplace domain is (with $0$ initial conditions)$$(ms+b)V(s)=I$$ giving $$v(t)=frac{I}{m}e^{-bt/m}.$$
How does this solution satisfy the original ODE though? At $tne0$ everything is good,$$-frac{Ib}{m}e^{-bt/m}+frac{Ib}{m}e^{-bt/m}=0,$$ while at $t=0$, $$-frac{Ib}{m}+frac{Ib}{m}=I$$ the result seems to be saying $0=1$ which is obviously false. What am I missing here?
ordinary-differential-equations laplace-transform dirac-delta
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
$endgroup$
$begingroup$
You don't seem to have included the dirac Delta in solving the equation. You should solve the homogeneous eqn first and then find the inhomogeneous part using say variation of parameters or a Laplace or Fourier transform. m.wolframalpha.com/input/?i=y%27%2By%3Ddirac+delta
$endgroup$
– Tyberius
Jan 14 '18 at 2:11
$begingroup$
The dirac Delta is the RHS of the very first equation. The problem is using an impulse of magnitude $I$. The final solution was obtained using Laplace transforms.
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:18
$begingroup$
I meant that your solution didn't seem to have any term that would lead to a dirac Delta on the right hand side. Let me check and see if I can find a solution.
$endgroup$
– Tyberius
Jan 14 '18 at 2:22
$begingroup$
It is possible I made a mistake
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:23
$begingroup$
In your second displayed equation, the $I$ should be $Ie^{-t}.$
$endgroup$
– B. Goddard
Jan 14 '18 at 2:54
|
show 3 more comments
$begingroup$
I am working on a problem where I have the following ODE. $$mdot{v}+bv=delta_I(t)$$
where
$$delta_I(t)=begin{cases}0, & text{for}&tne0\ I, & text{for} &t=0end{cases}.$$
The solution $v(t)$ was derived using Laplace transforms, the ODE in the Laplace domain is (with $0$ initial conditions)$$(ms+b)V(s)=I$$ giving $$v(t)=frac{I}{m}e^{-bt/m}.$$
How does this solution satisfy the original ODE though? At $tne0$ everything is good,$$-frac{Ib}{m}e^{-bt/m}+frac{Ib}{m}e^{-bt/m}=0,$$ while at $t=0$, $$-frac{Ib}{m}+frac{Ib}{m}=I$$ the result seems to be saying $0=1$ which is obviously false. What am I missing here?
ordinary-differential-equations laplace-transform dirac-delta
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
$endgroup$
I am working on a problem where I have the following ODE. $$mdot{v}+bv=delta_I(t)$$
where
$$delta_I(t)=begin{cases}0, & text{for}&tne0\ I, & text{for} &t=0end{cases}.$$
The solution $v(t)$ was derived using Laplace transforms, the ODE in the Laplace domain is (with $0$ initial conditions)$$(ms+b)V(s)=I$$ giving $$v(t)=frac{I}{m}e^{-bt/m}.$$
How does this solution satisfy the original ODE though? At $tne0$ everything is good,$$-frac{Ib}{m}e^{-bt/m}+frac{Ib}{m}e^{-bt/m}=0,$$ while at $t=0$, $$-frac{Ib}{m}+frac{Ib}{m}=I$$ the result seems to be saying $0=1$ which is obviously false. What am I missing here?
ordinary-differential-equations laplace-transform dirac-delta
ordinary-differential-equations laplace-transform dirac-delta
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
edited Jan 14 '18 at 2:32
WnGatRC456
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
asked Jan 14 '18 at 1:20
WnGatRC456WnGatRC456
10811
10811
$begingroup$
You don't seem to have included the dirac Delta in solving the equation. You should solve the homogeneous eqn first and then find the inhomogeneous part using say variation of parameters or a Laplace or Fourier transform. m.wolframalpha.com/input/?i=y%27%2By%3Ddirac+delta
$endgroup$
– Tyberius
Jan 14 '18 at 2:11
$begingroup$
The dirac Delta is the RHS of the very first equation. The problem is using an impulse of magnitude $I$. The final solution was obtained using Laplace transforms.
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:18
$begingroup$
I meant that your solution didn't seem to have any term that would lead to a dirac Delta on the right hand side. Let me check and see if I can find a solution.
$endgroup$
– Tyberius
Jan 14 '18 at 2:22
$begingroup$
It is possible I made a mistake
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:23
$begingroup$
In your second displayed equation, the $I$ should be $Ie^{-t}.$
$endgroup$
– B. Goddard
Jan 14 '18 at 2:54
|
show 3 more comments
$begingroup$
You don't seem to have included the dirac Delta in solving the equation. You should solve the homogeneous eqn first and then find the inhomogeneous part using say variation of parameters or a Laplace or Fourier transform. m.wolframalpha.com/input/?i=y%27%2By%3Ddirac+delta
$endgroup$
– Tyberius
Jan 14 '18 at 2:11
$begingroup$
The dirac Delta is the RHS of the very first equation. The problem is using an impulse of magnitude $I$. The final solution was obtained using Laplace transforms.
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:18
$begingroup$
I meant that your solution didn't seem to have any term that would lead to a dirac Delta on the right hand side. Let me check and see if I can find a solution.
$endgroup$
– Tyberius
Jan 14 '18 at 2:22
$begingroup$
It is possible I made a mistake
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:23
$begingroup$
In your second displayed equation, the $I$ should be $Ie^{-t}.$
$endgroup$
– B. Goddard
Jan 14 '18 at 2:54
$begingroup$
You don't seem to have included the dirac Delta in solving the equation. You should solve the homogeneous eqn first and then find the inhomogeneous part using say variation of parameters or a Laplace or Fourier transform. m.wolframalpha.com/input/?i=y%27%2By%3Ddirac+delta
$endgroup$
– Tyberius
Jan 14 '18 at 2:11
$begingroup$
You don't seem to have included the dirac Delta in solving the equation. You should solve the homogeneous eqn first and then find the inhomogeneous part using say variation of parameters or a Laplace or Fourier transform. m.wolframalpha.com/input/?i=y%27%2By%3Ddirac+delta
$endgroup$
– Tyberius
Jan 14 '18 at 2:11
$begingroup$
The dirac Delta is the RHS of the very first equation. The problem is using an impulse of magnitude $I$. The final solution was obtained using Laplace transforms.
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:18
$begingroup$
The dirac Delta is the RHS of the very first equation. The problem is using an impulse of magnitude $I$. The final solution was obtained using Laplace transforms.
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:18
$begingroup$
I meant that your solution didn't seem to have any term that would lead to a dirac Delta on the right hand side. Let me check and see if I can find a solution.
$endgroup$
– Tyberius
Jan 14 '18 at 2:22
$begingroup$
I meant that your solution didn't seem to have any term that would lead to a dirac Delta on the right hand side. Let me check and see if I can find a solution.
$endgroup$
– Tyberius
Jan 14 '18 at 2:22
$begingroup$
It is possible I made a mistake
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:23
$begingroup$
It is possible I made a mistake
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:23
$begingroup$
In your second displayed equation, the $I$ should be $Ie^{-t}.$
$endgroup$
– B. Goddard
Jan 14 '18 at 2:54
$begingroup$
In your second displayed equation, the $I$ should be $Ie^{-t}.$
$endgroup$
– B. Goddard
Jan 14 '18 at 2:54
|
show 3 more comments
1 Answer
1
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$begingroup$
So I start by find the solution to the homogeneous eqn: $$my'+by=0 to y_h=Ce^{-bt/m}$$
Then, we can determine the inhomogeneous part using a Fourier transform of the original eqn: $$my'+by=delta(t) to iomega m tilde{y}+btilde{y}=1 to tilde{y}=frac{1}{m}frac{1}{b/m+iomega}$$
From a table of Fourier transforms $$frac{1}{b/m+iomega} to theta(t)e^{-bt/m}$$ where $theta(t)$ is the heaviside step function.
So, $f_i= theta(t)e^{-bt/m}/m$, which when combined with the homogeneous solution will give the general solution for this differential equation.
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
$endgroup$
$begingroup$
Thanks, I've seen this method before for similar problems. Can you say what I'm doing wrong with the Laplace transform?
$endgroup$
– WnGatRC456
Jan 14 '18 at 3:21
2
$begingroup$
@NWernerC456 The Laplace transform is defined for t greater than or equal to zero. For this range of t, $theta(t)=1$, which is the result you got. To include t less than zero, you need to replace the implicit 1 in front of your solution with $theta(t)$, which will give you the same inhomogeneous part as in my answer.
$endgroup$
– Tyberius
Jan 14 '18 at 4:04
add a comment |
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$begingroup$
So I start by find the solution to the homogeneous eqn: $$my'+by=0 to y_h=Ce^{-bt/m}$$
Then, we can determine the inhomogeneous part using a Fourier transform of the original eqn: $$my'+by=delta(t) to iomega m tilde{y}+btilde{y}=1 to tilde{y}=frac{1}{m}frac{1}{b/m+iomega}$$
From a table of Fourier transforms $$frac{1}{b/m+iomega} to theta(t)e^{-bt/m}$$ where $theta(t)$ is the heaviside step function.
So, $f_i= theta(t)e^{-bt/m}/m$, which when combined with the homogeneous solution will give the general solution for this differential equation.
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
$endgroup$
$begingroup$
Thanks, I've seen this method before for similar problems. Can you say what I'm doing wrong with the Laplace transform?
$endgroup$
– WnGatRC456
Jan 14 '18 at 3:21
2
$begingroup$
@NWernerC456 The Laplace transform is defined for t greater than or equal to zero. For this range of t, $theta(t)=1$, which is the result you got. To include t less than zero, you need to replace the implicit 1 in front of your solution with $theta(t)$, which will give you the same inhomogeneous part as in my answer.
$endgroup$
– Tyberius
Jan 14 '18 at 4:04
add a comment |
$begingroup$
So I start by find the solution to the homogeneous eqn: $$my'+by=0 to y_h=Ce^{-bt/m}$$
Then, we can determine the inhomogeneous part using a Fourier transform of the original eqn: $$my'+by=delta(t) to iomega m tilde{y}+btilde{y}=1 to tilde{y}=frac{1}{m}frac{1}{b/m+iomega}$$
From a table of Fourier transforms $$frac{1}{b/m+iomega} to theta(t)e^{-bt/m}$$ where $theta(t)$ is the heaviside step function.
So, $f_i= theta(t)e^{-bt/m}/m$, which when combined with the homogeneous solution will give the general solution for this differential equation.
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
$endgroup$
$begingroup$
Thanks, I've seen this method before for similar problems. Can you say what I'm doing wrong with the Laplace transform?
$endgroup$
– WnGatRC456
Jan 14 '18 at 3:21
2
$begingroup$
@NWernerC456 The Laplace transform is defined for t greater than or equal to zero. For this range of t, $theta(t)=1$, which is the result you got. To include t less than zero, you need to replace the implicit 1 in front of your solution with $theta(t)$, which will give you the same inhomogeneous part as in my answer.
$endgroup$
– Tyberius
Jan 14 '18 at 4:04
add a comment |
$begingroup$
So I start by find the solution to the homogeneous eqn: $$my'+by=0 to y_h=Ce^{-bt/m}$$
Then, we can determine the inhomogeneous part using a Fourier transform of the original eqn: $$my'+by=delta(t) to iomega m tilde{y}+btilde{y}=1 to tilde{y}=frac{1}{m}frac{1}{b/m+iomega}$$
From a table of Fourier transforms $$frac{1}{b/m+iomega} to theta(t)e^{-bt/m}$$ where $theta(t)$ is the heaviside step function.
So, $f_i= theta(t)e^{-bt/m}/m$, which when combined with the homogeneous solution will give the general solution for this differential equation.
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
$endgroup$
So I start by find the solution to the homogeneous eqn: $$my'+by=0 to y_h=Ce^{-bt/m}$$
Then, we can determine the inhomogeneous part using a Fourier transform of the original eqn: $$my'+by=delta(t) to iomega m tilde{y}+btilde{y}=1 to tilde{y}=frac{1}{m}frac{1}{b/m+iomega}$$
From a table of Fourier transforms $$frac{1}{b/m+iomega} to theta(t)e^{-bt/m}$$ where $theta(t)$ is the heaviside step function.
So, $f_i= theta(t)e^{-bt/m}/m$, which when combined with the homogeneous solution will give the general solution for this differential equation.
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
edited Jan 14 '18 at 6:25
Rohan
27.8k42444
27.8k42444
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
answered Jan 14 '18 at 3:18
TyberiusTyberius
7342724
7342724
$begingroup$
Thanks, I've seen this method before for similar problems. Can you say what I'm doing wrong with the Laplace transform?
$endgroup$
– WnGatRC456
Jan 14 '18 at 3:21
2
$begingroup$
@NWernerC456 The Laplace transform is defined for t greater than or equal to zero. For this range of t, $theta(t)=1$, which is the result you got. To include t less than zero, you need to replace the implicit 1 in front of your solution with $theta(t)$, which will give you the same inhomogeneous part as in my answer.
$endgroup$
– Tyberius
Jan 14 '18 at 4:04
add a comment |
$begingroup$
Thanks, I've seen this method before for similar problems. Can you say what I'm doing wrong with the Laplace transform?
$endgroup$
– WnGatRC456
Jan 14 '18 at 3:21
2
$begingroup$
@NWernerC456 The Laplace transform is defined for t greater than or equal to zero. For this range of t, $theta(t)=1$, which is the result you got. To include t less than zero, you need to replace the implicit 1 in front of your solution with $theta(t)$, which will give you the same inhomogeneous part as in my answer.
$endgroup$
– Tyberius
Jan 14 '18 at 4:04
$begingroup$
Thanks, I've seen this method before for similar problems. Can you say what I'm doing wrong with the Laplace transform?
$endgroup$
– WnGatRC456
Jan 14 '18 at 3:21
$begingroup$
Thanks, I've seen this method before for similar problems. Can you say what I'm doing wrong with the Laplace transform?
$endgroup$
– WnGatRC456
Jan 14 '18 at 3:21
2
2
$begingroup$
@NWernerC456 The Laplace transform is defined for t greater than or equal to zero. For this range of t, $theta(t)=1$, which is the result you got. To include t less than zero, you need to replace the implicit 1 in front of your solution with $theta(t)$, which will give you the same inhomogeneous part as in my answer.
$endgroup$
– Tyberius
Jan 14 '18 at 4:04
$begingroup$
@NWernerC456 The Laplace transform is defined for t greater than or equal to zero. For this range of t, $theta(t)=1$, which is the result you got. To include t less than zero, you need to replace the implicit 1 in front of your solution with $theta(t)$, which will give you the same inhomogeneous part as in my answer.
$endgroup$
– Tyberius
Jan 14 '18 at 4:04
add a comment |
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$begingroup$
You don't seem to have included the dirac Delta in solving the equation. You should solve the homogeneous eqn first and then find the inhomogeneous part using say variation of parameters or a Laplace or Fourier transform. m.wolframalpha.com/input/?i=y%27%2By%3Ddirac+delta
$endgroup$
– Tyberius
Jan 14 '18 at 2:11
$begingroup$
The dirac Delta is the RHS of the very first equation. The problem is using an impulse of magnitude $I$. The final solution was obtained using Laplace transforms.
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:18
$begingroup$
I meant that your solution didn't seem to have any term that would lead to a dirac Delta on the right hand side. Let me check and see if I can find a solution.
$endgroup$
– Tyberius
Jan 14 '18 at 2:22
$begingroup$
It is possible I made a mistake
$endgroup$
– WnGatRC456
Jan 14 '18 at 2:23
$begingroup$
In your second displayed equation, the $I$ should be $Ie^{-t}.$
$endgroup$
– B. Goddard
Jan 14 '18 at 2:54