Mixing
First order ODE with Dirac delta funtcion
$begingroup$
I am looking for a direct method to solve this first order ODE with Dirac delta funtcion
$$frac{dU(t)}{dt}+k^2U(t)=frac{1}{sqrt{2pi}}delta(t)$$
with the initial condition $U(0)=frac{1}{sqrt{2pi}}$.
The solution to this problem is $$U(t)=frac{1}{sqrt{2pi}}e^{-k^2t}$$
My try
The integrating factor for this ode is
$$I=e^{int k^2 dt }=e^{ k^2 t }$$
then multiplying both sides of the differential equation by $,,e^{ k^2 t }$, we get
$$frac{d }{d t}left(e^{ k^2 t } U(t)right)=frac{1}{sqrt{2pi}}e^{ k^2 t } delta(t)$$
Integrating both sides, we have
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
From here on, I am lost. Any suggestions?
Back ground of this problem
The above ode we got after applying Fourier transform to the following PDE
$$u_{t}=u_{xx}+delta{(x)}delta{(t)}$$
with $u(x,0)=delta(x)$.
ordinary-differential-equations dirac-delta
asked Oct 17 '16 at 13:44
zhkzhk
484321
$endgroup$
|
show 4 more comments
$begingroup$
I am looking for a direct method to solve this first order ODE with Dirac delta funtcion
$$frac{dU(t)}{dt}+k^2U(t)=frac{1}{sqrt{2pi}}delta(t)$$
with the initial condition $U(0)=frac{1}{sqrt{2pi}}$.
The solution to this problem is $$U(t)=frac{1}{sqrt{2pi}}e^{-k^2t}$$
My try
The integrating factor for this ode is
$$I=e^{int k^2 dt }=e^{ k^2 t }$$
then multiplying both sides of the differential equation by $,,e^{ k^2 t }$, we get
$$frac{d }{d t}left(e^{ k^2 t } U(t)right)=frac{1}{sqrt{2pi}}e^{ k^2 t } delta(t)$$
Integrating both sides, we have
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
From here on, I am lost. Any suggestions?
Back ground of this problem
The above ode we got after applying Fourier transform to the following PDE
$$u_{t}=u_{xx}+delta{(x)}delta{(t)}$$
with $u(x,0)=delta(x)$.
ordinary-differential-equations dirac-delta
asked Oct 17 '16 at 13:44
zhkzhk
484321
$endgroup$
1
$begingroup$
No, a solution is $U(t) =frac{1}{sqrt{2pi}}e^{-k^2t} 1_{t > 0}$ (and $C e^{-k^2 t}$ are the solution to the homogeneous equation). Can you show us the steps for checking it is a solution ?
$endgroup$
– reuns
Oct 17 '16 at 13:51
$begingroup$
And $U(0) = 1/sqrt{2pi}$ is not correct (there is no solution that is continuous at $t= 0$)
$endgroup$
– reuns
Oct 17 '16 at 13:54
$begingroup$
@Ian or how to confuse the OP...
$endgroup$
– reuns
Oct 17 '16 at 14:06
$begingroup$
Anyway, you can dodge this difficulty by thinking physically a little bit: at the very beginning you have a unit mass at $x=0$, then you immediately add another unit mass at $x=0$ at the very beginning of the problem. Thus it is equivalent to solve $u_t=u_{xx},u(x,0)=2delta(x)$.
$endgroup$
– Ian
Oct 17 '16 at 14:07
$begingroup$
I have no idea what is going on! If I am missing something in the question or in the solution then please let me know?
$endgroup$
– zhk
Oct 17 '16 at 14:51
|
show 4 more comments
$begingroup$
I am looking for a direct method to solve this first order ODE with Dirac delta funtcion
$$frac{dU(t)}{dt}+k^2U(t)=frac{1}{sqrt{2pi}}delta(t)$$
with the initial condition $U(0)=frac{1}{sqrt{2pi}}$.
The solution to this problem is $$U(t)=frac{1}{sqrt{2pi}}e^{-k^2t}$$
My try
The integrating factor for this ode is
$$I=e^{int k^2 dt }=e^{ k^2 t }$$
then multiplying both sides of the differential equation by $,,e^{ k^2 t }$, we get
$$frac{d }{d t}left(e^{ k^2 t } U(t)right)=frac{1}{sqrt{2pi}}e^{ k^2 t } delta(t)$$
Integrating both sides, we have
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
From here on, I am lost. Any suggestions?
Back ground of this problem
The above ode we got after applying Fourier transform to the following PDE
$$u_{t}=u_{xx}+delta{(x)}delta{(t)}$$
with $u(x,0)=delta(x)$.
ordinary-differential-equations dirac-delta
asked Oct 17 '16 at 13:44
zhkzhk
484321
$endgroup$
I am looking for a direct method to solve this first order ODE with Dirac delta funtcion
$$frac{dU(t)}{dt}+k^2U(t)=frac{1}{sqrt{2pi}}delta(t)$$
with the initial condition $U(0)=frac{1}{sqrt{2pi}}$.
The solution to this problem is $$U(t)=frac{1}{sqrt{2pi}}e^{-k^2t}$$
My try
The integrating factor for this ode is
$$I=e^{int k^2 dt }=e^{ k^2 t }$$
then multiplying both sides of the differential equation by $,,e^{ k^2 t }$, we get
$$frac{d }{d t}left(e^{ k^2 t } U(t)right)=frac{1}{sqrt{2pi}}e^{ k^2 t } delta(t)$$
Integrating both sides, we have
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
From here on, I am lost. Any suggestions?
Back ground of this problem
The above ode we got after applying Fourier transform to the following PDE
$$u_{t}=u_{xx}+delta{(x)}delta{(t)}$$
with $u(x,0)=delta(x)$.
ordinary-differential-equations dirac-delta
ordinary-differential-equations dirac-delta
asked Oct 17 '16 at 13:44
zhkzhk
484321
asked Oct 17 '16 at 13:44
zhkzhk
484321
edited Oct 17 '16 at 17:11
zhk
asked Oct 17 '16 at 13:44
zhkzhk
484321
asked Oct 17 '16 at 13:44
zhkzhk
484321
asked Oct 17 '16 at 13:44
zhkzhk
484321
484321
1
$begingroup$
No, a solution is $U(t) =frac{1}{sqrt{2pi}}e^{-k^2t} 1_{t > 0}$ (and $C e^{-k^2 t}$ are the solution to the homogeneous equation). Can you show us the steps for checking it is a solution ?
$endgroup$
– reuns
Oct 17 '16 at 13:51
$begingroup$
And $U(0) = 1/sqrt{2pi}$ is not correct (there is no solution that is continuous at $t= 0$)
$endgroup$
– reuns
Oct 17 '16 at 13:54
$begingroup$
@Ian or how to confuse the OP...
$endgroup$
– reuns
Oct 17 '16 at 14:06
$begingroup$
Anyway, you can dodge this difficulty by thinking physically a little bit: at the very beginning you have a unit mass at $x=0$, then you immediately add another unit mass at $x=0$ at the very beginning of the problem. Thus it is equivalent to solve $u_t=u_{xx},u(x,0)=2delta(x)$.
$endgroup$
– Ian
Oct 17 '16 at 14:07
$begingroup$
I have no idea what is going on! If I am missing something in the question or in the solution then please let me know?
$endgroup$
– zhk
Oct 17 '16 at 14:51
|
show 4 more comments
1
$begingroup$
No, a solution is $U(t) =frac{1}{sqrt{2pi}}e^{-k^2t} 1_{t > 0}$ (and $C e^{-k^2 t}$ are the solution to the homogeneous equation). Can you show us the steps for checking it is a solution ?
$endgroup$
– reuns
Oct 17 '16 at 13:51
$begingroup$
And $U(0) = 1/sqrt{2pi}$ is not correct (there is no solution that is continuous at $t= 0$)
$endgroup$
– reuns
Oct 17 '16 at 13:54
$begingroup$
@Ian or how to confuse the OP...
$endgroup$
– reuns
Oct 17 '16 at 14:06
$begingroup$
Anyway, you can dodge this difficulty by thinking physically a little bit: at the very beginning you have a unit mass at $x=0$, then you immediately add another unit mass at $x=0$ at the very beginning of the problem. Thus it is equivalent to solve $u_t=u_{xx},u(x,0)=2delta(x)$.
$endgroup$
– Ian
Oct 17 '16 at 14:07
$begingroup$
I have no idea what is going on! If I am missing something in the question or in the solution then please let me know?
$endgroup$
– zhk
Oct 17 '16 at 14:51
1
1
$begingroup$
No, a solution is $U(t) =frac{1}{sqrt{2pi}}e^{-k^2t} 1_{t > 0}$ (and $C e^{-k^2 t}$ are the solution to the homogeneous equation). Can you show us the steps for checking it is a solution ?
$endgroup$
– reuns
Oct 17 '16 at 13:51
$begingroup$
No, a solution is $U(t) =frac{1}{sqrt{2pi}}e^{-k^2t} 1_{t > 0}$ (and $C e^{-k^2 t}$ are the solution to the homogeneous equation). Can you show us the steps for checking it is a solution ?
$endgroup$
– reuns
Oct 17 '16 at 13:51
$begingroup$
And $U(0) = 1/sqrt{2pi}$ is not correct (there is no solution that is continuous at $t= 0$)
$endgroup$
– reuns
Oct 17 '16 at 13:54
$begingroup$
And $U(0) = 1/sqrt{2pi}$ is not correct (there is no solution that is continuous at $t= 0$)
$endgroup$
– reuns
Oct 17 '16 at 13:54
$begingroup$
@Ian or how to confuse the OP...
$endgroup$
– reuns
Oct 17 '16 at 14:06
$begingroup$
@Ian or how to confuse the OP...
$endgroup$
– reuns
Oct 17 '16 at 14:06
$begingroup$
Anyway, you can dodge this difficulty by thinking physically a little bit: at the very beginning you have a unit mass at $x=0$, then you immediately add another unit mass at $x=0$ at the very beginning of the problem. Thus it is equivalent to solve $u_t=u_{xx},u(x,0)=2delta(x)$.
$endgroup$
– Ian
Oct 17 '16 at 14:07
$begingroup$
Anyway, you can dodge this difficulty by thinking physically a little bit: at the very beginning you have a unit mass at $x=0$, then you immediately add another unit mass at $x=0$ at the very beginning of the problem. Thus it is equivalent to solve $u_t=u_{xx},u(x,0)=2delta(x)$.
$endgroup$
– Ian
Oct 17 '16 at 14:07
$begingroup$
I have no idea what is going on! If I am missing something in the question or in the solution then please let me know?
$endgroup$
– zhk
Oct 17 '16 at 14:51
$begingroup$
I have no idea what is going on! If I am missing something in the question or in the solution then please let me know?
$endgroup$
– zhk
Oct 17 '16 at 14:51
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This maybe an answer to my question but needs your conformation/validation.
Continuing from where I left
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
$$Rightarrow U(t)=bigg[frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+Cbigg]e^{ -k^2 t }$$
Now recalling some properties of the Dirac delta function
- $$int_{-infty}^{infty}f(t)delta(t)dt=f(0)$$
- $$int_{a}^{b}f(t)delta(t)dt=
begin{cases}
f(0) & text{if } a<t<b\
0 & text{if } x=0
end{cases}
$$
Now since we are dealing with an IVP, where $tgeq0$, then adopting the above properties "we can have"
$$int{e^{ k^2 t } delta(t)}dt=e^{0}=1,$$
which gives
$$U(t)=bigg[frac{1}{sqrt{2pi}}+Cbigg]e^{ -k^2 t }$$
Finally utilizing the initial condition $U(0)=frac{1}{sqrt{2pi}}$, gives $C=0$, and hence
$$U(t)=frac{1}{sqrt{2pi}}e^{ -k^2 t }$$
Is this correct?
answered Oct 17 '16 at 17:30
zhkzhk
484321
$endgroup$
$begingroup$
It can't be correct for all times because there must be a jump at $t=0$. The usual way that one would interpret the solution to this ODE would be the result given by the Laplace transform, namely $(s+k^2)U(s)-frac{1}{sqrt{2 pi}}=frac{1}{sqrt{2 pi}}$ and so $U(s)=frac{2}{sqrt{2 pi}(s+k^2)}$. Thus going forward in time starting from $t=0$, you have $frac{2}{sqrt{2 pi}} e^{-k^2 t}$. Going backward in time you instead have $frac{1}{sqrt{2 pi}} e^{-k^2 t}$ (not that it makes any sense to send your original PDE back in time).
$endgroup$
– Ian
Oct 17 '16 at 17:34
$begingroup$
@Ian When I solve the ODE in Maple, it gives the solution $U(t)=bigg[frac{1}{sqrt{2pi}}H(t)+Cbigg]e^{-k^2t}$.
$endgroup$
– zhk
Oct 17 '16 at 17:49
$begingroup$
Which is what I wrote, with $C=frac{1}{sqrt{2 pi}}$.
$endgroup$
– Ian
Oct 17 '16 at 17:58
$begingroup$
@Ian It seems that $int{e^{ k^2 t } delta(t)}dt=H(t)$ but how ? If so, then how can we apply the initial condition to the Maple solution? I am thinking that $H(0)=1$ by definition then $C=0$ not the one you wrote.
$endgroup$
– zhk
Oct 18 '16 at 2:03
$begingroup$
$H (0) $ is not really defined. All you can really explain are the left and right limits. My interpretation of the problem is that your left limit should be $1/sqrt {2 pi} $ and your right limit should be $2/sqrt {2pi} $.
$endgroup$
– Ian
Oct 18 '16 at 10:20
add a comment |
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$begingroup$
This maybe an answer to my question but needs your conformation/validation.
Continuing from where I left
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
$$Rightarrow U(t)=bigg[frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+Cbigg]e^{ -k^2 t }$$
Now recalling some properties of the Dirac delta function
- $$int_{-infty}^{infty}f(t)delta(t)dt=f(0)$$
- $$int_{a}^{b}f(t)delta(t)dt=
begin{cases}
f(0) & text{if } a<t<b\
0 & text{if } x=0
end{cases}
$$
Now since we are dealing with an IVP, where $tgeq0$, then adopting the above properties "we can have"
$$int{e^{ k^2 t } delta(t)}dt=e^{0}=1,$$
which gives
$$U(t)=bigg[frac{1}{sqrt{2pi}}+Cbigg]e^{ -k^2 t }$$
Finally utilizing the initial condition $U(0)=frac{1}{sqrt{2pi}}$, gives $C=0$, and hence
$$U(t)=frac{1}{sqrt{2pi}}e^{ -k^2 t }$$
Is this correct?
answered Oct 17 '16 at 17:30
zhkzhk
484321
$endgroup$
$begingroup$
It can't be correct for all times because there must be a jump at $t=0$. The usual way that one would interpret the solution to this ODE would be the result given by the Laplace transform, namely $(s+k^2)U(s)-frac{1}{sqrt{2 pi}}=frac{1}{sqrt{2 pi}}$ and so $U(s)=frac{2}{sqrt{2 pi}(s+k^2)}$. Thus going forward in time starting from $t=0$, you have $frac{2}{sqrt{2 pi}} e^{-k^2 t}$. Going backward in time you instead have $frac{1}{sqrt{2 pi}} e^{-k^2 t}$ (not that it makes any sense to send your original PDE back in time).
$endgroup$
– Ian
Oct 17 '16 at 17:34
$begingroup$
@Ian When I solve the ODE in Maple, it gives the solution $U(t)=bigg[frac{1}{sqrt{2pi}}H(t)+Cbigg]e^{-k^2t}$.
$endgroup$
– zhk
Oct 17 '16 at 17:49
$begingroup$
Which is what I wrote, with $C=frac{1}{sqrt{2 pi}}$.
$endgroup$
– Ian
Oct 17 '16 at 17:58
$begingroup$
@Ian It seems that $int{e^{ k^2 t } delta(t)}dt=H(t)$ but how ? If so, then how can we apply the initial condition to the Maple solution? I am thinking that $H(0)=1$ by definition then $C=0$ not the one you wrote.
$endgroup$
– zhk
Oct 18 '16 at 2:03
$begingroup$
$H (0) $ is not really defined. All you can really explain are the left and right limits. My interpretation of the problem is that your left limit should be $1/sqrt {2 pi} $ and your right limit should be $2/sqrt {2pi} $.
$endgroup$
– Ian
Oct 18 '16 at 10:20
add a comment |
$begingroup$
This maybe an answer to my question but needs your conformation/validation.
Continuing from where I left
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
$$Rightarrow U(t)=bigg[frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+Cbigg]e^{ -k^2 t }$$
Now recalling some properties of the Dirac delta function
- $$int_{-infty}^{infty}f(t)delta(t)dt=f(0)$$
- $$int_{a}^{b}f(t)delta(t)dt=
begin{cases}
f(0) & text{if } a<t<b\
0 & text{if } x=0
end{cases}
$$
Now since we are dealing with an IVP, where $tgeq0$, then adopting the above properties "we can have"
$$int{e^{ k^2 t } delta(t)}dt=e^{0}=1,$$
which gives
$$U(t)=bigg[frac{1}{sqrt{2pi}}+Cbigg]e^{ -k^2 t }$$
Finally utilizing the initial condition $U(0)=frac{1}{sqrt{2pi}}$, gives $C=0$, and hence
$$U(t)=frac{1}{sqrt{2pi}}e^{ -k^2 t }$$
Is this correct?
answered Oct 17 '16 at 17:30
zhkzhk
484321
$endgroup$
$begingroup$
It can't be correct for all times because there must be a jump at $t=0$. The usual way that one would interpret the solution to this ODE would be the result given by the Laplace transform, namely $(s+k^2)U(s)-frac{1}{sqrt{2 pi}}=frac{1}{sqrt{2 pi}}$ and so $U(s)=frac{2}{sqrt{2 pi}(s+k^2)}$. Thus going forward in time starting from $t=0$, you have $frac{2}{sqrt{2 pi}} e^{-k^2 t}$. Going backward in time you instead have $frac{1}{sqrt{2 pi}} e^{-k^2 t}$ (not that it makes any sense to send your original PDE back in time).
$endgroup$
– Ian
Oct 17 '16 at 17:34
$begingroup$
@Ian When I solve the ODE in Maple, it gives the solution $U(t)=bigg[frac{1}{sqrt{2pi}}H(t)+Cbigg]e^{-k^2t}$.
$endgroup$
– zhk
Oct 17 '16 at 17:49
$begingroup$
Which is what I wrote, with $C=frac{1}{sqrt{2 pi}}$.
$endgroup$
– Ian
Oct 17 '16 at 17:58
$begingroup$
@Ian It seems that $int{e^{ k^2 t } delta(t)}dt=H(t)$ but how ? If so, then how can we apply the initial condition to the Maple solution? I am thinking that $H(0)=1$ by definition then $C=0$ not the one you wrote.
$endgroup$
– zhk
Oct 18 '16 at 2:03
$begingroup$
$H (0) $ is not really defined. All you can really explain are the left and right limits. My interpretation of the problem is that your left limit should be $1/sqrt {2 pi} $ and your right limit should be $2/sqrt {2pi} $.
$endgroup$
– Ian
Oct 18 '16 at 10:20
add a comment |
$begingroup$
This maybe an answer to my question but needs your conformation/validation.
Continuing from where I left
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
$$Rightarrow U(t)=bigg[frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+Cbigg]e^{ -k^2 t }$$
Now recalling some properties of the Dirac delta function
- $$int_{-infty}^{infty}f(t)delta(t)dt=f(0)$$
- $$int_{a}^{b}f(t)delta(t)dt=
begin{cases}
f(0) & text{if } a<t<b\
0 & text{if } x=0
end{cases}
$$
Now since we are dealing with an IVP, where $tgeq0$, then adopting the above properties "we can have"
$$int{e^{ k^2 t } delta(t)}dt=e^{0}=1,$$
which gives
$$U(t)=bigg[frac{1}{sqrt{2pi}}+Cbigg]e^{ -k^2 t }$$
Finally utilizing the initial condition $U(0)=frac{1}{sqrt{2pi}}$, gives $C=0$, and hence
$$U(t)=frac{1}{sqrt{2pi}}e^{ -k^2 t }$$
Is this correct?
answered Oct 17 '16 at 17:30
zhkzhk
484321
$endgroup$
This maybe an answer to my question but needs your conformation/validation.
Continuing from where I left
$$e^{ k^2 t } U(t)=frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+C$$
$$Rightarrow U(t)=bigg[frac{1}{sqrt{2pi}}int{e^{ k^2 t } delta(t)}dt+Cbigg]e^{ -k^2 t }$$
Now recalling some properties of the Dirac delta function
- $$int_{-infty}^{infty}f(t)delta(t)dt=f(0)$$
- $$int_{a}^{b}f(t)delta(t)dt=
begin{cases}
f(0) & text{if } a<t<b\
0 & text{if } x=0
end{cases}
$$
Now since we are dealing with an IVP, where $tgeq0$, then adopting the above properties "we can have"
$$int{e^{ k^2 t } delta(t)}dt=e^{0}=1,$$
which gives
$$U(t)=bigg[frac{1}{sqrt{2pi}}+Cbigg]e^{ -k^2 t }$$
Finally utilizing the initial condition $U(0)=frac{1}{sqrt{2pi}}$, gives $C=0$, and hence
$$U(t)=frac{1}{sqrt{2pi}}e^{ -k^2 t }$$
Is this correct?
answered Oct 17 '16 at 17:30
zhkzhk
484321
answered Oct 17 '16 at 17:30
zhkzhk
484321
answered Oct 17 '16 at 17:30
zhkzhk
484321
answered Oct 17 '16 at 17:30
zhkzhk
484321
484321
$begingroup$
It can't be correct for all times because there must be a jump at $t=0$. The usual way that one would interpret the solution to this ODE would be the result given by the Laplace transform, namely $(s+k^2)U(s)-frac{1}{sqrt{2 pi}}=frac{1}{sqrt{2 pi}}$ and so $U(s)=frac{2}{sqrt{2 pi}(s+k^2)}$. Thus going forward in time starting from $t=0$, you have $frac{2}{sqrt{2 pi}} e^{-k^2 t}$. Going backward in time you instead have $frac{1}{sqrt{2 pi}} e^{-k^2 t}$ (not that it makes any sense to send your original PDE back in time).
$endgroup$
– Ian
Oct 17 '16 at 17:34
$begingroup$
@Ian When I solve the ODE in Maple, it gives the solution $U(t)=bigg[frac{1}{sqrt{2pi}}H(t)+Cbigg]e^{-k^2t}$.
$endgroup$
– zhk
Oct 17 '16 at 17:49
$begingroup$
Which is what I wrote, with $C=frac{1}{sqrt{2 pi}}$.
$endgroup$
– Ian
Oct 17 '16 at 17:58
$begingroup$
@Ian It seems that $int{e^{ k^2 t } delta(t)}dt=H(t)$ but how ? If so, then how can we apply the initial condition to the Maple solution? I am thinking that $H(0)=1$ by definition then $C=0$ not the one you wrote.
$endgroup$
– zhk
Oct 18 '16 at 2:03
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$H (0) $ is not really defined. All you can really explain are the left and right limits. My interpretation of the problem is that your left limit should be $1/sqrt {2 pi} $ and your right limit should be $2/sqrt {2pi} $.
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– Ian
Oct 18 '16 at 10:20
add a comment |
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It can't be correct for all times because there must be a jump at $t=0$. The usual way that one would interpret the solution to this ODE would be the result given by the Laplace transform, namely $(s+k^2)U(s)-frac{1}{sqrt{2 pi}}=frac{1}{sqrt{2 pi}}$ and so $U(s)=frac{2}{sqrt{2 pi}(s+k^2)}$. Thus going forward in time starting from $t=0$, you have $frac{2}{sqrt{2 pi}} e^{-k^2 t}$. Going backward in time you instead have $frac{1}{sqrt{2 pi}} e^{-k^2 t}$ (not that it makes any sense to send your original PDE back in time).
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– Ian
Oct 17 '16 at 17:34
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@Ian When I solve the ODE in Maple, it gives the solution $U(t)=bigg[frac{1}{sqrt{2pi}}H(t)+Cbigg]e^{-k^2t}$.
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– zhk
Oct 17 '16 at 17:49
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Which is what I wrote, with $C=frac{1}{sqrt{2 pi}}$.
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– Ian
Oct 17 '16 at 17:58
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@Ian It seems that $int{e^{ k^2 t } delta(t)}dt=H(t)$ but how ? If so, then how can we apply the initial condition to the Maple solution? I am thinking that $H(0)=1$ by definition then $C=0$ not the one you wrote.
$endgroup$
– zhk
Oct 18 '16 at 2:03
$begingroup$
$H (0) $ is not really defined. All you can really explain are the left and right limits. My interpretation of the problem is that your left limit should be $1/sqrt {2 pi} $ and your right limit should be $2/sqrt {2pi} $.
$endgroup$
– Ian
Oct 18 '16 at 10:20
$begingroup$
It can't be correct for all times because there must be a jump at $t=0$. The usual way that one would interpret the solution to this ODE would be the result given by the Laplace transform, namely $(s+k^2)U(s)-frac{1}{sqrt{2 pi}}=frac{1}{sqrt{2 pi}}$ and so $U(s)=frac{2}{sqrt{2 pi}(s+k^2)}$. Thus going forward in time starting from $t=0$, you have $frac{2}{sqrt{2 pi}} e^{-k^2 t}$. Going backward in time you instead have $frac{1}{sqrt{2 pi}} e^{-k^2 t}$ (not that it makes any sense to send your original PDE back in time).
$endgroup$
– Ian
Oct 17 '16 at 17:34
$begingroup$
It can't be correct for all times because there must be a jump at $t=0$. The usual way that one would interpret the solution to this ODE would be the result given by the Laplace transform, namely $(s+k^2)U(s)-frac{1}{sqrt{2 pi}}=frac{1}{sqrt{2 pi}}$ and so $U(s)=frac{2}{sqrt{2 pi}(s+k^2)}$. Thus going forward in time starting from $t=0$, you have $frac{2}{sqrt{2 pi}} e^{-k^2 t}$. Going backward in time you instead have $frac{1}{sqrt{2 pi}} e^{-k^2 t}$ (not that it makes any sense to send your original PDE back in time).
$endgroup$
– Ian
Oct 17 '16 at 17:34
$begingroup$
@Ian When I solve the ODE in Maple, it gives the solution $U(t)=bigg[frac{1}{sqrt{2pi}}H(t)+Cbigg]e^{-k^2t}$.
$endgroup$
– zhk
Oct 17 '16 at 17:49
$begingroup$
@Ian When I solve the ODE in Maple, it gives the solution $U(t)=bigg[frac{1}{sqrt{2pi}}H(t)+Cbigg]e^{-k^2t}$.
$endgroup$
– zhk
Oct 17 '16 at 17:49
$begingroup$
Which is what I wrote, with $C=frac{1}{sqrt{2 pi}}$.
$endgroup$
– Ian
Oct 17 '16 at 17:58
$begingroup$
Which is what I wrote, with $C=frac{1}{sqrt{2 pi}}$.
$endgroup$
– Ian
Oct 17 '16 at 17:58
$begingroup$
@Ian It seems that $int{e^{ k^2 t } delta(t)}dt=H(t)$ but how ? If so, then how can we apply the initial condition to the Maple solution? I am thinking that $H(0)=1$ by definition then $C=0$ not the one you wrote.
$endgroup$
– zhk
Oct 18 '16 at 2:03
$begingroup$
@Ian It seems that $int{e^{ k^2 t } delta(t)}dt=H(t)$ but how ? If so, then how can we apply the initial condition to the Maple solution? I am thinking that $H(0)=1$ by definition then $C=0$ not the one you wrote.
$endgroup$
– zhk
Oct 18 '16 at 2:03
$begingroup$
$H (0) $ is not really defined. All you can really explain are the left and right limits. My interpretation of the problem is that your left limit should be $1/sqrt {2 pi} $ and your right limit should be $2/sqrt {2pi} $.
$endgroup$
– Ian
Oct 18 '16 at 10:20
$begingroup$
$H (0) $ is not really defined. All you can really explain are the left and right limits. My interpretation of the problem is that your left limit should be $1/sqrt {2 pi} $ and your right limit should be $2/sqrt {2pi} $.
$endgroup$
– Ian
Oct 18 '16 at 10:20
add a comment |
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No, a solution is $U(t) =frac{1}{sqrt{2pi}}e^{-k^2t} 1_{t > 0}$ (and $C e^{-k^2 t}$ are the solution to the homogeneous equation). Can you show us the steps for checking it is a solution ?
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– reuns
Oct 17 '16 at 13:51
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And $U(0) = 1/sqrt{2pi}$ is not correct (there is no solution that is continuous at $t= 0$)
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– reuns
Oct 17 '16 at 13:54
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@Ian or how to confuse the OP...
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– reuns
Oct 17 '16 at 14:06
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Anyway, you can dodge this difficulty by thinking physically a little bit: at the very beginning you have a unit mass at $x=0$, then you immediately add another unit mass at $x=0$ at the very beginning of the problem. Thus it is equivalent to solve $u_t=u_{xx},u(x,0)=2delta(x)$.
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– Ian
Oct 17 '16 at 14:07
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I have no idea what is going on! If I am missing something in the question or in the solution then please let me know?
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– zhk
Oct 17 '16 at 14:51