Mixing
Differential equation with delta dirac as the input
$begingroup$
I have this formula: $ C frac{dv(t)}{d(t)} + Gv(t) = K delta(t) $
and I have to calculate $v(0^{+})$ , given the $v(0^{-})$, without using the Laplace transformation. Could someone provide some help?
ordinary-differential-equations dirac-delta
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
$endgroup$
add a comment |
$begingroup$
I have this formula: $ C frac{dv(t)}{d(t)} + Gv(t) = K delta(t) $
and I have to calculate $v(0^{+})$ , given the $v(0^{-})$, without using the Laplace transformation. Could someone provide some help?
ordinary-differential-equations dirac-delta
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
$endgroup$
add a comment |
$begingroup$
I have this formula: $ C frac{dv(t)}{d(t)} + Gv(t) = K delta(t) $
and I have to calculate $v(0^{+})$ , given the $v(0^{-})$, without using the Laplace transformation. Could someone provide some help?
ordinary-differential-equations dirac-delta
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
$endgroup$
I have this formula: $ C frac{dv(t)}{d(t)} + Gv(t) = K delta(t) $
and I have to calculate $v(0^{+})$ , given the $v(0^{-})$, without using the Laplace transformation. Could someone provide some help?
ordinary-differential-equations dirac-delta
ordinary-differential-equations dirac-delta
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
asked Oct 23 '16 at 21:44
Alice1nw0Alice1nw0
285
285
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT:
For $t> 0$, it is easy to see that $v(t)=Ae^{-Gt/C}$.
where $A$ is a constant that will be found by enforcing the initial condition on $v(t)$.
Now, note that the presence of the Dirac Delta implies theta $v'(t)$ is discontinuous at $t=0$. Formally, we can write
$$lim_{epsilon to 0^+}int_{-epsilon}^{epsilon}left(Cv'(t)+Gv(t)right),dt=Klim_{epsilonto 0^+}int_{-epsilon}^{epsilon} delta(t) ,dt tag1
$$
The right-hand side of $(1)$ equals $K$.
Since $v$ is continuous, the contribution from integrating $v$ vanishes as $epsilon to 0$.
The left-hand side is then, $Cv(0^+)-Cv(0^-)=CA-Cv(0^-)$. Now, simply solve for $A$.
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Well, during our lesson, we had K =1 and our professor told that the solution was C*( v(0+) - v(0-) ) = 1... I am still having trouble following his train of thought.
$endgroup$
– Alice1nw0
Oct 24 '16 at 0:26
$begingroup$
The integral of the bare Dirac delta (with no K in front and no transformation of the argument) over an interval which has zero in its interior is 1. Now compare that with the given answer...
$endgroup$
– Ian
Oct 24 '16 at 15:20
$begingroup$
@ian For any $epsilon>0$, the functional $$langle 1_{[-epsilon,epsilon]}, delta rangle=1$$where $delta$ is the Dirac Delta distribution and $1$ is the indicator function. So, I don't understand the comment.
$endgroup$
– Mark Viola
Oct 24 '16 at 16:04
$begingroup$
@Dr.MV I should have pinged Alice1nw0, my apologies.
$endgroup$
– Ian
Oct 24 '16 at 16:57
$begingroup$
@Alice1nw0 My first comment here was directed at you.
$endgroup$
– Ian
Oct 24 '16 at 17:01
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
For $t> 0$, it is easy to see that $v(t)=Ae^{-Gt/C}$.
where $A$ is a constant that will be found by enforcing the initial condition on $v(t)$.
Now, note that the presence of the Dirac Delta implies theta $v'(t)$ is discontinuous at $t=0$. Formally, we can write
$$lim_{epsilon to 0^+}int_{-epsilon}^{epsilon}left(Cv'(t)+Gv(t)right),dt=Klim_{epsilonto 0^+}int_{-epsilon}^{epsilon} delta(t) ,dt tag1
$$
The right-hand side of $(1)$ equals $K$.
Since $v$ is continuous, the contribution from integrating $v$ vanishes as $epsilon to 0$.
The left-hand side is then, $Cv(0^+)-Cv(0^-)=CA-Cv(0^-)$. Now, simply solve for $A$.
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Well, during our lesson, we had K =1 and our professor told that the solution was C*( v(0+) - v(0-) ) = 1... I am still having trouble following his train of thought.
$endgroup$
– Alice1nw0
Oct 24 '16 at 0:26
$begingroup$
The integral of the bare Dirac delta (with no K in front and no transformation of the argument) over an interval which has zero in its interior is 1. Now compare that with the given answer...
$endgroup$
– Ian
Oct 24 '16 at 15:20
$begingroup$
@ian For any $epsilon>0$, the functional $$langle 1_{[-epsilon,epsilon]}, delta rangle=1$$where $delta$ is the Dirac Delta distribution and $1$ is the indicator function. So, I don't understand the comment.
$endgroup$
– Mark Viola
Oct 24 '16 at 16:04
$begingroup$
@Dr.MV I should have pinged Alice1nw0, my apologies.
$endgroup$
– Ian
Oct 24 '16 at 16:57
$begingroup$
@Alice1nw0 My first comment here was directed at you.
$endgroup$
– Ian
Oct 24 '16 at 17:01
add a comment |
$begingroup$
HINT:
For $t> 0$, it is easy to see that $v(t)=Ae^{-Gt/C}$.
where $A$ is a constant that will be found by enforcing the initial condition on $v(t)$.
Now, note that the presence of the Dirac Delta implies theta $v'(t)$ is discontinuous at $t=0$. Formally, we can write
$$lim_{epsilon to 0^+}int_{-epsilon}^{epsilon}left(Cv'(t)+Gv(t)right),dt=Klim_{epsilonto 0^+}int_{-epsilon}^{epsilon} delta(t) ,dt tag1
$$
The right-hand side of $(1)$ equals $K$.
Since $v$ is continuous, the contribution from integrating $v$ vanishes as $epsilon to 0$.
The left-hand side is then, $Cv(0^+)-Cv(0^-)=CA-Cv(0^-)$. Now, simply solve for $A$.
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
Well, during our lesson, we had K =1 and our professor told that the solution was C*( v(0+) - v(0-) ) = 1... I am still having trouble following his train of thought.
$endgroup$
– Alice1nw0
Oct 24 '16 at 0:26
$begingroup$
The integral of the bare Dirac delta (with no K in front and no transformation of the argument) over an interval which has zero in its interior is 1. Now compare that with the given answer...
$endgroup$
– Ian
Oct 24 '16 at 15:20
$begingroup$
@ian For any $epsilon>0$, the functional $$langle 1_{[-epsilon,epsilon]}, delta rangle=1$$where $delta$ is the Dirac Delta distribution and $1$ is the indicator function. So, I don't understand the comment.
$endgroup$
– Mark Viola
Oct 24 '16 at 16:04
$begingroup$
@Dr.MV I should have pinged Alice1nw0, my apologies.
$endgroup$
– Ian
Oct 24 '16 at 16:57
$begingroup$
@Alice1nw0 My first comment here was directed at you.
$endgroup$
– Ian
Oct 24 '16 at 17:01
add a comment |
$begingroup$
HINT:
For $t> 0$, it is easy to see that $v(t)=Ae^{-Gt/C}$.
where $A$ is a constant that will be found by enforcing the initial condition on $v(t)$.
Now, note that the presence of the Dirac Delta implies theta $v'(t)$ is discontinuous at $t=0$. Formally, we can write
$$lim_{epsilon to 0^+}int_{-epsilon}^{epsilon}left(Cv'(t)+Gv(t)right),dt=Klim_{epsilonto 0^+}int_{-epsilon}^{epsilon} delta(t) ,dt tag1
$$
The right-hand side of $(1)$ equals $K$.
Since $v$ is continuous, the contribution from integrating $v$ vanishes as $epsilon to 0$.
The left-hand side is then, $Cv(0^+)-Cv(0^-)=CA-Cv(0^-)$. Now, simply solve for $A$.
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
$endgroup$
HINT:
For $t> 0$, it is easy to see that $v(t)=Ae^{-Gt/C}$.
where $A$ is a constant that will be found by enforcing the initial condition on $v(t)$.
Now, note that the presence of the Dirac Delta implies theta $v'(t)$ is discontinuous at $t=0$. Formally, we can write
$$lim_{epsilon to 0^+}int_{-epsilon}^{epsilon}left(Cv'(t)+Gv(t)right),dt=Klim_{epsilonto 0^+}int_{-epsilon}^{epsilon} delta(t) ,dt tag1
$$
The right-hand side of $(1)$ equals $K$.
Since $v$ is continuous, the contribution from integrating $v$ vanishes as $epsilon to 0$.
The left-hand side is then, $Cv(0^+)-Cv(0^-)=CA-Cv(0^-)$. Now, simply solve for $A$.
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
edited Oct 24 '16 at 14:47
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
answered Oct 23 '16 at 21:50
Mark ViolaMark Viola
134k1278176
134k1278176
$begingroup$
Well, during our lesson, we had K =1 and our professor told that the solution was C*( v(0+) - v(0-) ) = 1... I am still having trouble following his train of thought.
$endgroup$
– Alice1nw0
Oct 24 '16 at 0:26
$begingroup$
The integral of the bare Dirac delta (with no K in front and no transformation of the argument) over an interval which has zero in its interior is 1. Now compare that with the given answer...
$endgroup$
– Ian
Oct 24 '16 at 15:20
$begingroup$
@ian For any $epsilon>0$, the functional $$langle 1_{[-epsilon,epsilon]}, delta rangle=1$$where $delta$ is the Dirac Delta distribution and $1$ is the indicator function. So, I don't understand the comment.
$endgroup$
– Mark Viola
Oct 24 '16 at 16:04
$begingroup$
@Dr.MV I should have pinged Alice1nw0, my apologies.
$endgroup$
– Ian
Oct 24 '16 at 16:57
$begingroup$
@Alice1nw0 My first comment here was directed at you.
$endgroup$
– Ian
Oct 24 '16 at 17:01
add a comment |
$begingroup$
Well, during our lesson, we had K =1 and our professor told that the solution was C*( v(0+) - v(0-) ) = 1... I am still having trouble following his train of thought.
$endgroup$
– Alice1nw0
Oct 24 '16 at 0:26
$begingroup$
The integral of the bare Dirac delta (with no K in front and no transformation of the argument) over an interval which has zero in its interior is 1. Now compare that with the given answer...
$endgroup$
– Ian
Oct 24 '16 at 15:20
$begingroup$
@ian For any $epsilon>0$, the functional $$langle 1_{[-epsilon,epsilon]}, delta rangle=1$$where $delta$ is the Dirac Delta distribution and $1$ is the indicator function. So, I don't understand the comment.
$endgroup$
– Mark Viola
Oct 24 '16 at 16:04
$begingroup$
@Dr.MV I should have pinged Alice1nw0, my apologies.
$endgroup$
– Ian
Oct 24 '16 at 16:57
$begingroup$
@Alice1nw0 My first comment here was directed at you.
$endgroup$
– Ian
Oct 24 '16 at 17:01
$begingroup$
Well, during our lesson, we had K =1 and our professor told that the solution was C*( v(0+) - v(0-) ) = 1... I am still having trouble following his train of thought.
$endgroup$
– Alice1nw0
Oct 24 '16 at 0:26
$begingroup$
Well, during our lesson, we had K =1 and our professor told that the solution was C*( v(0+) - v(0-) ) = 1... I am still having trouble following his train of thought.
$endgroup$
– Alice1nw0
Oct 24 '16 at 0:26
$begingroup$
The integral of the bare Dirac delta (with no K in front and no transformation of the argument) over an interval which has zero in its interior is 1. Now compare that with the given answer...
$endgroup$
– Ian
Oct 24 '16 at 15:20
$begingroup$
The integral of the bare Dirac delta (with no K in front and no transformation of the argument) over an interval which has zero in its interior is 1. Now compare that with the given answer...
$endgroup$
– Ian
Oct 24 '16 at 15:20
$begingroup$
@ian For any $epsilon>0$, the functional $$langle 1_{[-epsilon,epsilon]}, delta rangle=1$$where $delta$ is the Dirac Delta distribution and $1$ is the indicator function. So, I don't understand the comment.
$endgroup$
– Mark Viola
Oct 24 '16 at 16:04
$begingroup$
@ian For any $epsilon>0$, the functional $$langle 1_{[-epsilon,epsilon]}, delta rangle=1$$where $delta$ is the Dirac Delta distribution and $1$ is the indicator function. So, I don't understand the comment.
$endgroup$
– Mark Viola
Oct 24 '16 at 16:04
$begingroup$
@Dr.MV I should have pinged Alice1nw0, my apologies.
$endgroup$
– Ian
Oct 24 '16 at 16:57
$begingroup$
@Dr.MV I should have pinged Alice1nw0, my apologies.
$endgroup$
– Ian
Oct 24 '16 at 16:57
$begingroup$
@Alice1nw0 My first comment here was directed at you.
$endgroup$
– Ian
Oct 24 '16 at 17:01
$begingroup$
@Alice1nw0 My first comment here was directed at you.
$endgroup$
– Ian
Oct 24 '16 at 17:01
add a comment |
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