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Investigate uniqueness of the following differential equation
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This is Problem 1.13 from Teschl's book (ODEs and Dynamical Systems). The equation is as follows:
$$
x' =
begincases
-tsqrt, &text if x ge 0\
tsqrt, &text if x le 0.
endcases
$$
It asks to show that the initial value problem $x(0) = x_0$ has unique global solution for every $x_0 in mathbbR$. There is a hint saying that if $x(t)$ is a solution so is $-x(t)$ and $x(-t)$, so it suffices to consider $x_0 ge 0$ and $t ge 0$. For $x_0 > 0$, I managed to find the solution
$$
x(t) =
begincases
(sqrtx_0 - frac14t^2)^2, &text if |t| le 2x_0^1/4\
0, &text otherwise.
endcases
$$
It is easy to see that this is a solution by using the Mean Value Theorem when calculating $x'(t)$ for $t = pm 2x_0^1/4$, because $x$ is continuous at these points and $x'(theta) to 0$ as $theta to t$. For $x_0 = 0$ the obvious solution is $x(t) = 0$ for all $t$. The problem is that I do not know how to show these solutions are unique.
If $x$ is a is solution and $x_0 > 0$, then by continuity there is an open interval $(t_1, t_2)$ containing $0$ where $x(t)$ is positive, and so it makes sense to write
$$
-s = fracx'(s)sqrtx(s),
$$
for $s in (t_1, t_2)$, and to integrate both sides
beginalign
-frac12t^2 = int_0^t -s ds = int_0^t fracx'(s)sqrtx(s) ds = 2(sqrtx(t) - sqrtx(0)).
endalign
This yields $x(t) = (sqrtx_0 - frac14t^2)^2$ for $t in (t_1, t_2)$. Assuming that $(t_1, t_2)$ is a maximal interval where $x(t)$ is positive, and since $(t_1, t_2) subseteq (-2x_0^1/4, 2x_0^1/4)$, we see by continuity of $x$ that $x(t_1) = x(t_2) = 0$ and thus $(t_1, t_2) = (-2x_0^1/4, 2x_0^1/4)$. How do I show that $x(t) = 0$ when $t$ is not in this interval? I suspect the same technique in this last step can be used to show that $x(t) = 0$ when $x_0 = 0$.
real-analysis differential-equations
asked Jul 14 at 23:20
Doughnut Pump
439312
add a comment |Â
up vote
2
down vote
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This is Problem 1.13 from Teschl's book (ODEs and Dynamical Systems). The equation is as follows:
$$
x' =
begincases
-tsqrt, &text if x ge 0\
tsqrt, &text if x le 0.
endcases
$$
It asks to show that the initial value problem $x(0) = x_0$ has unique global solution for every $x_0 in mathbbR$. There is a hint saying that if $x(t)$ is a solution so is $-x(t)$ and $x(-t)$, so it suffices to consider $x_0 ge 0$ and $t ge 0$. For $x_0 > 0$, I managed to find the solution
$$
x(t) =
begincases
(sqrtx_0 - frac14t^2)^2, &text if |t| le 2x_0^1/4\
0, &text otherwise.
endcases
$$
It is easy to see that this is a solution by using the Mean Value Theorem when calculating $x'(t)$ for $t = pm 2x_0^1/4$, because $x$ is continuous at these points and $x'(theta) to 0$ as $theta to t$. For $x_0 = 0$ the obvious solution is $x(t) = 0$ for all $t$. The problem is that I do not know how to show these solutions are unique.
If $x$ is a is solution and $x_0 > 0$, then by continuity there is an open interval $(t_1, t_2)$ containing $0$ where $x(t)$ is positive, and so it makes sense to write
$$
-s = fracx'(s)sqrtx(s),
$$
for $s in (t_1, t_2)$, and to integrate both sides
beginalign
-frac12t^2 = int_0^t -s ds = int_0^t fracx'(s)sqrtx(s) ds = 2(sqrtx(t) - sqrtx(0)).
endalign
This yields $x(t) = (sqrtx_0 - frac14t^2)^2$ for $t in (t_1, t_2)$. Assuming that $(t_1, t_2)$ is a maximal interval where $x(t)$ is positive, and since $(t_1, t_2) subseteq (-2x_0^1/4, 2x_0^1/4)$, we see by continuity of $x$ that $x(t_1) = x(t_2) = 0$ and thus $(t_1, t_2) = (-2x_0^1/4, 2x_0^1/4)$. How do I show that $x(t) = 0$ when $t$ is not in this interval? I suspect the same technique in this last step can be used to show that $x(t) = 0$ when $x_0 = 0$.
real-analysis differential-equations
asked Jul 14 at 23:20
Doughnut Pump
439312
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is Problem 1.13 from Teschl's book (ODEs and Dynamical Systems). The equation is as follows:
$$
x' =
begincases
-tsqrt, &text if x ge 0\
tsqrt, &text if x le 0.
endcases
$$
It asks to show that the initial value problem $x(0) = x_0$ has unique global solution for every $x_0 in mathbbR$. There is a hint saying that if $x(t)$ is a solution so is $-x(t)$ and $x(-t)$, so it suffices to consider $x_0 ge 0$ and $t ge 0$. For $x_0 > 0$, I managed to find the solution
$$
x(t) =
begincases
(sqrtx_0 - frac14t^2)^2, &text if |t| le 2x_0^1/4\
0, &text otherwise.
endcases
$$
It is easy to see that this is a solution by using the Mean Value Theorem when calculating $x'(t)$ for $t = pm 2x_0^1/4$, because $x$ is continuous at these points and $x'(theta) to 0$ as $theta to t$. For $x_0 = 0$ the obvious solution is $x(t) = 0$ for all $t$. The problem is that I do not know how to show these solutions are unique.
If $x$ is a is solution and $x_0 > 0$, then by continuity there is an open interval $(t_1, t_2)$ containing $0$ where $x(t)$ is positive, and so it makes sense to write
$$
-s = fracx'(s)sqrtx(s),
$$
for $s in (t_1, t_2)$, and to integrate both sides
beginalign
-frac12t^2 = int_0^t -s ds = int_0^t fracx'(s)sqrtx(s) ds = 2(sqrtx(t) - sqrtx(0)).
endalign
This yields $x(t) = (sqrtx_0 - frac14t^2)^2$ for $t in (t_1, t_2)$. Assuming that $(t_1, t_2)$ is a maximal interval where $x(t)$ is positive, and since $(t_1, t_2) subseteq (-2x_0^1/4, 2x_0^1/4)$, we see by continuity of $x$ that $x(t_1) = x(t_2) = 0$ and thus $(t_1, t_2) = (-2x_0^1/4, 2x_0^1/4)$. How do I show that $x(t) = 0$ when $t$ is not in this interval? I suspect the same technique in this last step can be used to show that $x(t) = 0$ when $x_0 = 0$.
real-analysis differential-equations
asked Jul 14 at 23:20
Doughnut Pump
439312
This is Problem 1.13 from Teschl's book (ODEs and Dynamical Systems). The equation is as follows:
$$
x' =
begincases
-tsqrt, &text if x ge 0\
tsqrt, &text if x le 0.
endcases
$$
It asks to show that the initial value problem $x(0) = x_0$ has unique global solution for every $x_0 in mathbbR$. There is a hint saying that if $x(t)$ is a solution so is $-x(t)$ and $x(-t)$, so it suffices to consider $x_0 ge 0$ and $t ge 0$. For $x_0 > 0$, I managed to find the solution
$$
x(t) =
begincases
(sqrtx_0 - frac14t^2)^2, &text if |t| le 2x_0^1/4\
0, &text otherwise.
endcases
$$
It is easy to see that this is a solution by using the Mean Value Theorem when calculating $x'(t)$ for $t = pm 2x_0^1/4$, because $x$ is continuous at these points and $x'(theta) to 0$ as $theta to t$. For $x_0 = 0$ the obvious solution is $x(t) = 0$ for all $t$. The problem is that I do not know how to show these solutions are unique.
If $x$ is a is solution and $x_0 > 0$, then by continuity there is an open interval $(t_1, t_2)$ containing $0$ where $x(t)$ is positive, and so it makes sense to write
$$
-s = fracx'(s)sqrtx(s),
$$
for $s in (t_1, t_2)$, and to integrate both sides
beginalign
-frac12t^2 = int_0^t -s ds = int_0^t fracx'(s)sqrtx(s) ds = 2(sqrtx(t) - sqrtx(0)).
endalign
This yields $x(t) = (sqrtx_0 - frac14t^2)^2$ for $t in (t_1, t_2)$. Assuming that $(t_1, t_2)$ is a maximal interval where $x(t)$ is positive, and since $(t_1, t_2) subseteq (-2x_0^1/4, 2x_0^1/4)$, we see by continuity of $x$ that $x(t_1) = x(t_2) = 0$ and thus $(t_1, t_2) = (-2x_0^1/4, 2x_0^1/4)$. How do I show that $x(t) = 0$ when $t$ is not in this interval? I suspect the same technique in this last step can be used to show that $x(t) = 0$ when $x_0 = 0$.
real-analysis differential-equations
asked Jul 14 at 23:20
Doughnut Pump
439312
edited Jul 15 at 12:50
asked Jul 14 at 23:20
Doughnut Pump
439312
asked Jul 14 at 23:20
Doughnut Pump
439312
asked Jul 14 at 23:20
Doughnut Pump
439312
439312
add a comment |Â
add a comment |Â
2 Answers
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Alright, I've done it. Suppose first $x_0 = 0$ and let $x(t)$ be a solution. Assume by contradiction that $x(t) > 0$ for some $t > 0$ and set
$$
d = infa < t: x(s) > 0 text for every s in (a, t) .
$$
By continuity $d$ is well defined, $0 le d < t$ and $x(d) = 0$. By the Mean Value Theorem, we see that
$$
x(t) = x(t) - x(d) = x'(theta)(t - d),
$$
for some $theta in (d, t)$, so
$$
x(t) = -thetasqrt(t - d) le 0
$$
which is a contradiction. If $x(t) < 0$ the proof should be analogous and we can do the same thing to finish what I wrote above for the case $x_0 > 0$.
answered Jul 15 at 0:14
Doughnut Pump
439312
Yes, this part is O.K. I would only define $d$ as the supremum of the set of those $a<t$ such that $x(a)=0$ (easier to show it is nonempty). Remember that you should also show that we have uniqueness at each point where the value of the solution is nonzero, and that any solution is global. Can you do that?
– user539887
Jul 15 at 7:34
@user539887 What do you mean by "uniqueness at each point where the value of the solution is nonzero"? I have shown it is unique for every choice of $x_0$, isn't that all there is to it?
– Doughnut Pump
Jul 15 at 9:28
You showed only that the solution cannot branch at a point where its value is $0$. At other points it cannot branch, either, but here one needs to use, for example, the local Lipschitz continuity at such a point. Or, alternatively, you can argue that your arguments in the last paragraph of your question show that the obtained formula is the only possible.
– user539887
Jul 15 at 10:52
add a comment |Â
up vote
0
down vote
To prove uniqueness, let's assume two solutions $x(t)$ and $y(t)$, $x(0)=y(0)=x_0>0$. Let $u=x-y$ with $u(0)=u_0=0$. We subtract the equations for $x$ and $y$ and obtain
$$
u'(t)=-tfracusqrtx+sqrty,
$$
which yields
$$
u(t)=u_0 expleft(-int_0^t fracssqrtx(s)+sqrty(s)dsright).
$$
Since $u_0=0$, then $uequiv0$.
answered Jul 16 at 4:34
pabodu
1415
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Alright, I've done it. Suppose first $x_0 = 0$ and let $x(t)$ be a solution. Assume by contradiction that $x(t) > 0$ for some $t > 0$ and set
$$
d = infa < t: x(s) > 0 text for every s in (a, t) .
$$
By continuity $d$ is well defined, $0 le d < t$ and $x(d) = 0$. By the Mean Value Theorem, we see that
$$
x(t) = x(t) - x(d) = x'(theta)(t - d),
$$
for some $theta in (d, t)$, so
$$
x(t) = -thetasqrt(t - d) le 0
$$
which is a contradiction. If $x(t) < 0$ the proof should be analogous and we can do the same thing to finish what I wrote above for the case $x_0 > 0$.
answered Jul 15 at 0:14
Doughnut Pump
439312
Yes, this part is O.K. I would only define $d$ as the supremum of the set of those $a<t$ such that $x(a)=0$ (easier to show it is nonempty). Remember that you should also show that we have uniqueness at each point where the value of the solution is nonzero, and that any solution is global. Can you do that?
– user539887
Jul 15 at 7:34
@user539887 What do you mean by "uniqueness at each point where the value of the solution is nonzero"? I have shown it is unique for every choice of $x_0$, isn't that all there is to it?
– Doughnut Pump
Jul 15 at 9:28
You showed only that the solution cannot branch at a point where its value is $0$. At other points it cannot branch, either, but here one needs to use, for example, the local Lipschitz continuity at such a point. Or, alternatively, you can argue that your arguments in the last paragraph of your question show that the obtained formula is the only possible.
– user539887
Jul 15 at 10:52
add a comment |Â
up vote
0
down vote
Alright, I've done it. Suppose first $x_0 = 0$ and let $x(t)$ be a solution. Assume by contradiction that $x(t) > 0$ for some $t > 0$ and set
$$
d = infa < t: x(s) > 0 text for every s in (a, t) .
$$
By continuity $d$ is well defined, $0 le d < t$ and $x(d) = 0$. By the Mean Value Theorem, we see that
$$
x(t) = x(t) - x(d) = x'(theta)(t - d),
$$
for some $theta in (d, t)$, so
$$
x(t) = -thetasqrt(t - d) le 0
$$
which is a contradiction. If $x(t) < 0$ the proof should be analogous and we can do the same thing to finish what I wrote above for the case $x_0 > 0$.
answered Jul 15 at 0:14
Doughnut Pump
439312
Yes, this part is O.K. I would only define $d$ as the supremum of the set of those $a<t$ such that $x(a)=0$ (easier to show it is nonempty). Remember that you should also show that we have uniqueness at each point where the value of the solution is nonzero, and that any solution is global. Can you do that?
– user539887
Jul 15 at 7:34
@user539887 What do you mean by "uniqueness at each point where the value of the solution is nonzero"? I have shown it is unique for every choice of $x_0$, isn't that all there is to it?
– Doughnut Pump
Jul 15 at 9:28
You showed only that the solution cannot branch at a point where its value is $0$. At other points it cannot branch, either, but here one needs to use, for example, the local Lipschitz continuity at such a point. Or, alternatively, you can argue that your arguments in the last paragraph of your question show that the obtained formula is the only possible.
– user539887
Jul 15 at 10:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Alright, I've done it. Suppose first $x_0 = 0$ and let $x(t)$ be a solution. Assume by contradiction that $x(t) > 0$ for some $t > 0$ and set
$$
d = infa < t: x(s) > 0 text for every s in (a, t) .
$$
By continuity $d$ is well defined, $0 le d < t$ and $x(d) = 0$. By the Mean Value Theorem, we see that
$$
x(t) = x(t) - x(d) = x'(theta)(t - d),
$$
for some $theta in (d, t)$, so
$$
x(t) = -thetasqrt(t - d) le 0
$$
which is a contradiction. If $x(t) < 0$ the proof should be analogous and we can do the same thing to finish what I wrote above for the case $x_0 > 0$.
answered Jul 15 at 0:14
Doughnut Pump
439312
Alright, I've done it. Suppose first $x_0 = 0$ and let $x(t)$ be a solution. Assume by contradiction that $x(t) > 0$ for some $t > 0$ and set
$$
d = infa < t: x(s) > 0 text for every s in (a, t) .
$$
By continuity $d$ is well defined, $0 le d < t$ and $x(d) = 0$. By the Mean Value Theorem, we see that
$$
x(t) = x(t) - x(d) = x'(theta)(t - d),
$$
for some $theta in (d, t)$, so
$$
x(t) = -thetasqrt(t - d) le 0
$$
which is a contradiction. If $x(t) < 0$ the proof should be analogous and we can do the same thing to finish what I wrote above for the case $x_0 > 0$.
answered Jul 15 at 0:14
Doughnut Pump
439312
answered Jul 15 at 0:14
Doughnut Pump
439312
answered Jul 15 at 0:14
Doughnut Pump
439312
answered Jul 15 at 0:14
Doughnut Pump
439312
439312
Yes, this part is O.K. I would only define $d$ as the supremum of the set of those $a<t$ such that $x(a)=0$ (easier to show it is nonempty). Remember that you should also show that we have uniqueness at each point where the value of the solution is nonzero, and that any solution is global. Can you do that?
– user539887
Jul 15 at 7:34
@user539887 What do you mean by "uniqueness at each point where the value of the solution is nonzero"? I have shown it is unique for every choice of $x_0$, isn't that all there is to it?
– Doughnut Pump
Jul 15 at 9:28
You showed only that the solution cannot branch at a point where its value is $0$. At other points it cannot branch, either, but here one needs to use, for example, the local Lipschitz continuity at such a point. Or, alternatively, you can argue that your arguments in the last paragraph of your question show that the obtained formula is the only possible.
– user539887
Jul 15 at 10:52
add a comment |Â
Yes, this part is O.K. I would only define $d$ as the supremum of the set of those $a<t$ such that $x(a)=0$ (easier to show it is nonempty). Remember that you should also show that we have uniqueness at each point where the value of the solution is nonzero, and that any solution is global. Can you do that?
– user539887
Jul 15 at 7:34
@user539887 What do you mean by "uniqueness at each point where the value of the solution is nonzero"? I have shown it is unique for every choice of $x_0$, isn't that all there is to it?
– Doughnut Pump
Jul 15 at 9:28
You showed only that the solution cannot branch at a point where its value is $0$. At other points it cannot branch, either, but here one needs to use, for example, the local Lipschitz continuity at such a point. Or, alternatively, you can argue that your arguments in the last paragraph of your question show that the obtained formula is the only possible.
– user539887
Jul 15 at 10:52
Yes, this part is O.K. I would only define $d$ as the supremum of the set of those $a<t$ such that $x(a)=0$ (easier to show it is nonempty). Remember that you should also show that we have uniqueness at each point where the value of the solution is nonzero, and that any solution is global. Can you do that?
– user539887
Jul 15 at 7:34
Yes, this part is O.K. I would only define $d$ as the supremum of the set of those $a<t$ such that $x(a)=0$ (easier to show it is nonempty). Remember that you should also show that we have uniqueness at each point where the value of the solution is nonzero, and that any solution is global. Can you do that?
– user539887
Jul 15 at 7:34
@user539887 What do you mean by "uniqueness at each point where the value of the solution is nonzero"? I have shown it is unique for every choice of $x_0$, isn't that all there is to it?
– Doughnut Pump
Jul 15 at 9:28
@user539887 What do you mean by "uniqueness at each point where the value of the solution is nonzero"? I have shown it is unique for every choice of $x_0$, isn't that all there is to it?
– Doughnut Pump
Jul 15 at 9:28
You showed only that the solution cannot branch at a point where its value is $0$. At other points it cannot branch, either, but here one needs to use, for example, the local Lipschitz continuity at such a point. Or, alternatively, you can argue that your arguments in the last paragraph of your question show that the obtained formula is the only possible.
– user539887
Jul 15 at 10:52
You showed only that the solution cannot branch at a point where its value is $0$. At other points it cannot branch, either, but here one needs to use, for example, the local Lipschitz continuity at such a point. Or, alternatively, you can argue that your arguments in the last paragraph of your question show that the obtained formula is the only possible.
– user539887
Jul 15 at 10:52
add a comment |Â
up vote
0
down vote
To prove uniqueness, let's assume two solutions $x(t)$ and $y(t)$, $x(0)=y(0)=x_0>0$. Let $u=x-y$ with $u(0)=u_0=0$. We subtract the equations for $x$ and $y$ and obtain
$$
u'(t)=-tfracusqrtx+sqrty,
$$
which yields
$$
u(t)=u_0 expleft(-int_0^t fracssqrtx(s)+sqrty(s)dsright).
$$
Since $u_0=0$, then $uequiv0$.
answered Jul 16 at 4:34
pabodu
1415
add a comment |Â
up vote
0
down vote
To prove uniqueness, let's assume two solutions $x(t)$ and $y(t)$, $x(0)=y(0)=x_0>0$. Let $u=x-y$ with $u(0)=u_0=0$. We subtract the equations for $x$ and $y$ and obtain
$$
u'(t)=-tfracusqrtx+sqrty,
$$
which yields
$$
u(t)=u_0 expleft(-int_0^t fracssqrtx(s)+sqrty(s)dsright).
$$
Since $u_0=0$, then $uequiv0$.
answered Jul 16 at 4:34
pabodu
1415
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To prove uniqueness, let's assume two solutions $x(t)$ and $y(t)$, $x(0)=y(0)=x_0>0$. Let $u=x-y$ with $u(0)=u_0=0$. We subtract the equations for $x$ and $y$ and obtain
$$
u'(t)=-tfracusqrtx+sqrty,
$$
which yields
$$
u(t)=u_0 expleft(-int_0^t fracssqrtx(s)+sqrty(s)dsright).
$$
Since $u_0=0$, then $uequiv0$.
answered Jul 16 at 4:34
pabodu
1415
To prove uniqueness, let's assume two solutions $x(t)$ and $y(t)$, $x(0)=y(0)=x_0>0$. Let $u=x-y$ with $u(0)=u_0=0$. We subtract the equations for $x$ and $y$ and obtain
$$
u'(t)=-tfracusqrtx+sqrty,
$$
which yields
$$
u(t)=u_0 expleft(-int_0^t fracssqrtx(s)+sqrty(s)dsright).
$$
Since $u_0=0$, then $uequiv0$.
answered Jul 16 at 4:34
pabodu
1415
answered Jul 16 at 4:34
pabodu
1415
answered Jul 16 at 4:34
pabodu
1415
answered Jul 16 at 4:34
pabodu
1415
1415
add a comment |Â
add a comment |Â
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