Solving a distribution differential equation
$begingroup$
The exercise is to solve
$$x u'(x) = delta(x).$$
By using the definitions
$$begin{cases}(u'|varphi) = (-u|varphi') \ (fu|varphi) = (u|fvarphi) end{cases}$$
we get to solve
$$(-u|varphi + xvarphi') = varphi(0).$$
How do one proceed? Obviously we can see that $(-delta|varphi)$ is a solution, but what about other distributions?
ordinary-differential-equations distribution-theory weak-derivatives
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
$endgroup$
add a comment |
$begingroup$
The exercise is to solve
$$x u'(x) = delta(x).$$
By using the definitions
$$begin{cases}(u'|varphi) = (-u|varphi') \ (fu|varphi) = (u|fvarphi) end{cases}$$
we get to solve
$$(-u|varphi + xvarphi') = varphi(0).$$
How do one proceed? Obviously we can see that $(-delta|varphi)$ is a solution, but what about other distributions?
ordinary-differential-equations distribution-theory weak-derivatives
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
$endgroup$
$begingroup$
Is $(,cdot,|,cdot,)$ meant to be an inner product of functions?
$endgroup$
– Christoph
Jan 27 at 7:48
$begingroup$
@Christoph $(u|varphi)$ is a notation for the action $u[varphi] = u(varphi)$ of some test function $varphi$ under the distribution $u$.
$endgroup$
– Markus Klyver
Jan 27 at 12:21
$begingroup$
Do you know the general solutions to $xv(x) = delta(x)$?
$endgroup$
– md2perpe
Jan 28 at 19:20
$begingroup$
@md2perpe I'd guess some linear combination of Dirac deltas and a constant term.
$endgroup$
– Markus Klyver
Jan 28 at 19:56
add a comment |
$begingroup$
The exercise is to solve
$$x u'(x) = delta(x).$$
By using the definitions
$$begin{cases}(u'|varphi) = (-u|varphi') \ (fu|varphi) = (u|fvarphi) end{cases}$$
we get to solve
$$(-u|varphi + xvarphi') = varphi(0).$$
How do one proceed? Obviously we can see that $(-delta|varphi)$ is a solution, but what about other distributions?
ordinary-differential-equations distribution-theory weak-derivatives
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
$endgroup$
The exercise is to solve
$$x u'(x) = delta(x).$$
By using the definitions
$$begin{cases}(u'|varphi) = (-u|varphi') \ (fu|varphi) = (u|fvarphi) end{cases}$$
we get to solve
$$(-u|varphi + xvarphi') = varphi(0).$$
How do one proceed? Obviously we can see that $(-delta|varphi)$ is a solution, but what about other distributions?
ordinary-differential-equations distribution-theory weak-derivatives
ordinary-differential-equations distribution-theory weak-derivatives
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
asked Jan 25 at 20:46
Markus KlyverMarkus Klyver
407414
407414
$begingroup$
Is $(,cdot,|,cdot,)$ meant to be an inner product of functions?
$endgroup$
– Christoph
Jan 27 at 7:48
$begingroup$
@Christoph $(u|varphi)$ is a notation for the action $u[varphi] = u(varphi)$ of some test function $varphi$ under the distribution $u$.
$endgroup$
– Markus Klyver
Jan 27 at 12:21
$begingroup$
Do you know the general solutions to $xv(x) = delta(x)$?
$endgroup$
– md2perpe
Jan 28 at 19:20
$begingroup$
@md2perpe I'd guess some linear combination of Dirac deltas and a constant term.
$endgroup$
– Markus Klyver
Jan 28 at 19:56
add a comment |
$begingroup$
Is $(,cdot,|,cdot,)$ meant to be an inner product of functions?
$endgroup$
– Christoph
Jan 27 at 7:48
$begingroup$
@Christoph $(u|varphi)$ is a notation for the action $u[varphi] = u(varphi)$ of some test function $varphi$ under the distribution $u$.
$endgroup$
– Markus Klyver
Jan 27 at 12:21
$begingroup$
Do you know the general solutions to $xv(x) = delta(x)$?
$endgroup$
– md2perpe
Jan 28 at 19:20
$begingroup$
@md2perpe I'd guess some linear combination of Dirac deltas and a constant term.
$endgroup$
– Markus Klyver
Jan 28 at 19:56
$begingroup$
Is $(,cdot,|,cdot,)$ meant to be an inner product of functions?
$endgroup$
– Christoph
Jan 27 at 7:48
$begingroup$
Is $(,cdot,|,cdot,)$ meant to be an inner product of functions?
$endgroup$
– Christoph
Jan 27 at 7:48
$begingroup$
@Christoph $(u|varphi)$ is a notation for the action $u[varphi] = u(varphi)$ of some test function $varphi$ under the distribution $u$.
$endgroup$
– Markus Klyver
Jan 27 at 12:21
$begingroup$
@Christoph $(u|varphi)$ is a notation for the action $u[varphi] = u(varphi)$ of some test function $varphi$ under the distribution $u$.
$endgroup$
– Markus Klyver
Jan 27 at 12:21
$begingroup$
Do you know the general solutions to $xv(x) = delta(x)$?
$endgroup$
– md2perpe
Jan 28 at 19:20
$begingroup$
Do you know the general solutions to $xv(x) = delta(x)$?
$endgroup$
– md2perpe
Jan 28 at 19:20
$begingroup$
@md2perpe I'd guess some linear combination of Dirac deltas and a constant term.
$endgroup$
– Markus Klyver
Jan 28 at 19:56
$begingroup$
@md2perpe I'd guess some linear combination of Dirac deltas and a constant term.
$endgroup$
– Markus Klyver
Jan 28 at 19:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Setting $v=u'$ we get the equation $xv=delta$. The general solution to this is $v=-delta'+adelta,$ where $a$ is a constant. To get the general solution to $xu' = delta$ we therefore need to solve $u' = -delta'+adelta,$ which is solved by $u=-delta+aH+b,$ where also $b$ is a constant.
Thus, the general solution to $xu'=delta$ is $u=-delta+aH+b,$ where $a$ and $b$ are constants.
The solutions $v=-delta'+adelta$ to $xv=delta$ can be seen as a particular solution $v=-delta'$ and solutions to the homogeneous equation $xv=0$ which are $v=cdelta.$
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
$endgroup$
$begingroup$
Note there's a derivate acting on u.
$endgroup$
– Markus Klyver
Jan 28 at 23:21
$begingroup$
@MarkusKlyver. I know. I had missed it at one place. Except for this my solution was correct.
$endgroup$
– md2perpe
Jan 29 at 6:54
$begingroup$
Could you add a concluding line about the total solution?
$endgroup$
– Markus Klyver
Jan 29 at 7:28
$begingroup$
@MarkusKlyver. A concluding line has been added (the second paragraph). I also added "Setting $v=u'$ /.../".
$endgroup$
– md2perpe
Jan 29 at 9:15
add a comment |
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$begingroup$
Setting $v=u'$ we get the equation $xv=delta$. The general solution to this is $v=-delta'+adelta,$ where $a$ is a constant. To get the general solution to $xu' = delta$ we therefore need to solve $u' = -delta'+adelta,$ which is solved by $u=-delta+aH+b,$ where also $b$ is a constant.
Thus, the general solution to $xu'=delta$ is $u=-delta+aH+b,$ where $a$ and $b$ are constants.
The solutions $v=-delta'+adelta$ to $xv=delta$ can be seen as a particular solution $v=-delta'$ and solutions to the homogeneous equation $xv=0$ which are $v=cdelta.$
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
$endgroup$
$begingroup$
Note there's a derivate acting on u.
$endgroup$
– Markus Klyver
Jan 28 at 23:21
$begingroup$
@MarkusKlyver. I know. I had missed it at one place. Except for this my solution was correct.
$endgroup$
– md2perpe
Jan 29 at 6:54
$begingroup$
Could you add a concluding line about the total solution?
$endgroup$
– Markus Klyver
Jan 29 at 7:28
$begingroup$
@MarkusKlyver. A concluding line has been added (the second paragraph). I also added "Setting $v=u'$ /.../".
$endgroup$
– md2perpe
Jan 29 at 9:15
add a comment |
$begingroup$
Setting $v=u'$ we get the equation $xv=delta$. The general solution to this is $v=-delta'+adelta,$ where $a$ is a constant. To get the general solution to $xu' = delta$ we therefore need to solve $u' = -delta'+adelta,$ which is solved by $u=-delta+aH+b,$ where also $b$ is a constant.
Thus, the general solution to $xu'=delta$ is $u=-delta+aH+b,$ where $a$ and $b$ are constants.
The solutions $v=-delta'+adelta$ to $xv=delta$ can be seen as a particular solution $v=-delta'$ and solutions to the homogeneous equation $xv=0$ which are $v=cdelta.$
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
$endgroup$
$begingroup$
Note there's a derivate acting on u.
$endgroup$
– Markus Klyver
Jan 28 at 23:21
$begingroup$
@MarkusKlyver. I know. I had missed it at one place. Except for this my solution was correct.
$endgroup$
– md2perpe
Jan 29 at 6:54
$begingroup$
Could you add a concluding line about the total solution?
$endgroup$
– Markus Klyver
Jan 29 at 7:28
$begingroup$
@MarkusKlyver. A concluding line has been added (the second paragraph). I also added "Setting $v=u'$ /.../".
$endgroup$
– md2perpe
Jan 29 at 9:15
add a comment |
$begingroup$
Setting $v=u'$ we get the equation $xv=delta$. The general solution to this is $v=-delta'+adelta,$ where $a$ is a constant. To get the general solution to $xu' = delta$ we therefore need to solve $u' = -delta'+adelta,$ which is solved by $u=-delta+aH+b,$ where also $b$ is a constant.
Thus, the general solution to $xu'=delta$ is $u=-delta+aH+b,$ where $a$ and $b$ are constants.
The solutions $v=-delta'+adelta$ to $xv=delta$ can be seen as a particular solution $v=-delta'$ and solutions to the homogeneous equation $xv=0$ which are $v=cdelta.$
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
$endgroup$
Setting $v=u'$ we get the equation $xv=delta$. The general solution to this is $v=-delta'+adelta,$ where $a$ is a constant. To get the general solution to $xu' = delta$ we therefore need to solve $u' = -delta'+adelta,$ which is solved by $u=-delta+aH+b,$ where also $b$ is a constant.
Thus, the general solution to $xu'=delta$ is $u=-delta+aH+b,$ where $a$ and $b$ are constants.
The solutions $v=-delta'+adelta$ to $xv=delta$ can be seen as a particular solution $v=-delta'$ and solutions to the homogeneous equation $xv=0$ which are $v=cdelta.$
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
edited Jan 29 at 9:13
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
answered Jan 28 at 20:19
md2perpemd2perpe
8,19611028
8,19611028
$begingroup$
Note there's a derivate acting on u.
$endgroup$
– Markus Klyver
Jan 28 at 23:21
$begingroup$
@MarkusKlyver. I know. I had missed it at one place. Except for this my solution was correct.
$endgroup$
– md2perpe
Jan 29 at 6:54
$begingroup$
Could you add a concluding line about the total solution?
$endgroup$
– Markus Klyver
Jan 29 at 7:28
$begingroup$
@MarkusKlyver. A concluding line has been added (the second paragraph). I also added "Setting $v=u'$ /.../".
$endgroup$
– md2perpe
Jan 29 at 9:15
add a comment |
$begingroup$
Note there's a derivate acting on u.
$endgroup$
– Markus Klyver
Jan 28 at 23:21
$begingroup$
@MarkusKlyver. I know. I had missed it at one place. Except for this my solution was correct.
$endgroup$
– md2perpe
Jan 29 at 6:54
$begingroup$
Could you add a concluding line about the total solution?
$endgroup$
– Markus Klyver
Jan 29 at 7:28
$begingroup$
@MarkusKlyver. A concluding line has been added (the second paragraph). I also added "Setting $v=u'$ /.../".
$endgroup$
– md2perpe
Jan 29 at 9:15
$begingroup$
Note there's a derivate acting on u.
$endgroup$
– Markus Klyver
Jan 28 at 23:21
$begingroup$
Note there's a derivate acting on u.
$endgroup$
– Markus Klyver
Jan 28 at 23:21
$begingroup$
@MarkusKlyver. I know. I had missed it at one place. Except for this my solution was correct.
$endgroup$
– md2perpe
Jan 29 at 6:54
$begingroup$
@MarkusKlyver. I know. I had missed it at one place. Except for this my solution was correct.
$endgroup$
– md2perpe
Jan 29 at 6:54
$begingroup$
Could you add a concluding line about the total solution?
$endgroup$
– Markus Klyver
Jan 29 at 7:28
$begingroup$
Could you add a concluding line about the total solution?
$endgroup$
– Markus Klyver
Jan 29 at 7:28
$begingroup$
@MarkusKlyver. A concluding line has been added (the second paragraph). I also added "Setting $v=u'$ /.../".
$endgroup$
– md2perpe
Jan 29 at 9:15
$begingroup$
@MarkusKlyver. A concluding line has been added (the second paragraph). I also added "Setting $v=u'$ /.../".
$endgroup$
– md2perpe
Jan 29 at 9:15
add a comment |
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$begingroup$
Is $(,cdot,|,cdot,)$ meant to be an inner product of functions?
$endgroup$
– Christoph
Jan 27 at 7:48
$begingroup$
@Christoph $(u|varphi)$ is a notation for the action $u[varphi] = u(varphi)$ of some test function $varphi$ under the distribution $u$.
$endgroup$
– Markus Klyver
Jan 27 at 12:21
$begingroup$
Do you know the general solutions to $xv(x) = delta(x)$?
$endgroup$
– md2perpe
Jan 28 at 19:20
$begingroup$
@md2perpe I'd guess some linear combination of Dirac deltas and a constant term.
$endgroup$
– Markus Klyver
Jan 28 at 19:56