Bounding a solution for an ODE
Given
$$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
I need to bound the solution to that ODE.
I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.
differential-equations
asked Nov 27 '18 at 15:55
galah92
25418
add a comment |
Given
$$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
I need to bound the solution to that ODE.
I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.
differential-equations
asked Nov 27 '18 at 15:55
galah92
25418
add a comment |
Given
$$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
I need to bound the solution to that ODE.
I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.
differential-equations
asked Nov 27 '18 at 15:55
galah92
25418
Given
$$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
I need to bound the solution to that ODE.
I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.
differential-equations
differential-equations
asked Nov 27 '18 at 15:55
galah92
25418
asked Nov 27 '18 at 15:55
galah92
25418
asked Nov 27 '18 at 15:55
galah92
25418
asked Nov 27 '18 at 15:55
galah92
25418
asked Nov 27 '18 at 15:55
galah92
25418
25418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
$$dot{x}(t)leq t$$ for all $t$, and integrating we have
$$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.
Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.
answered Nov 27 '18 at 16:06
Leo163
1,600512
Thanks. I can't detect the use of the IV though. Why is it needed?
– galah92
Nov 27 '18 at 16:19
You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
– Leo163
Nov 27 '18 at 16:22
Oh, of course..
– galah92
Nov 27 '18 at 16:23
add a comment |
Well, the unique solution is $x(t)=0$ for all $t$.
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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oldest
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Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
$$dot{x}(t)leq t$$ for all $t$, and integrating we have
$$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.
Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.
answered Nov 27 '18 at 16:06
Leo163
1,600512
Thanks. I can't detect the use of the IV though. Why is it needed?
– galah92
Nov 27 '18 at 16:19
You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
– Leo163
Nov 27 '18 at 16:22
Oh, of course..
– galah92
Nov 27 '18 at 16:23
add a comment |
Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
$$dot{x}(t)leq t$$ for all $t$, and integrating we have
$$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.
Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.
answered Nov 27 '18 at 16:06
Leo163
1,600512
Thanks. I can't detect the use of the IV though. Why is it needed?
– galah92
Nov 27 '18 at 16:19
You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
– Leo163
Nov 27 '18 at 16:22
Oh, of course..
– galah92
Nov 27 '18 at 16:23
add a comment |
Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
$$dot{x}(t)leq t$$ for all $t$, and integrating we have
$$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.
Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.
answered Nov 27 '18 at 16:06
Leo163
1,600512
Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
$$dot{x}(t)leq t$$ for all $t$, and integrating we have
$$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.
Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.
answered Nov 27 '18 at 16:06
Leo163
1,600512
edited Nov 27 '18 at 16:25
answered Nov 27 '18 at 16:06
Leo163
1,600512
answered Nov 27 '18 at 16:06
Leo163
1,600512
answered Nov 27 '18 at 16:06
Leo163
1,600512
1,600512
Thanks. I can't detect the use of the IV though. Why is it needed?
– galah92
Nov 27 '18 at 16:19
You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
– Leo163
Nov 27 '18 at 16:22
Oh, of course..
– galah92
Nov 27 '18 at 16:23
add a comment |
Thanks. I can't detect the use of the IV though. Why is it needed?
– galah92
Nov 27 '18 at 16:19
You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
– Leo163
Nov 27 '18 at 16:22
Oh, of course..
– galah92
Nov 27 '18 at 16:23
Thanks. I can't detect the use of the IV though. Why is it needed?
– galah92
Nov 27 '18 at 16:19
Thanks. I can't detect the use of the IV though. Why is it needed?
– galah92
Nov 27 '18 at 16:19
You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
– Leo163
Nov 27 '18 at 16:22
You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
– Leo163
Nov 27 '18 at 16:22
Oh, of course..
– galah92
Nov 27 '18 at 16:23
Oh, of course..
– galah92
Nov 27 '18 at 16:23
add a comment |
Well, the unique solution is $x(t)=0$ for all $t$.
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
add a comment |
Well, the unique solution is $x(t)=0$ for all $t$.
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
add a comment |
Well, the unique solution is $x(t)=0$ for all $t$.
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
Well, the unique solution is $x(t)=0$ for all $t$.
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
answered Nov 27 '18 at 16:01
Julián Aguirre
67.6k24094
67.6k24094
add a comment |
add a comment |
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