Mixing
Show $|x(t)|$ is strictly decreasing for $t>0$ for the initial value problem $dot{x} = -x^{3}, x(0)=1 $
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Solving the inital value problem gives me $x(t)=frac{1}{(2t+1)^{1/2}}$ which is clearly striclty decreasing as $(2t+1)^{1/2}$ is strictly increasing. However is it possible to just say it's strictly decreasing as the derivative $dot{x} = -x^{3}$ is strictly decreasing for postive x, if not how would I solve this without having to explicitly solve the inital value problem.
real-analysis ordinary-differential-equations
edited Jan 20 at 16:24
Did
248k23225463
asked Jan 20 at 15:47
RogerRoger
847
$endgroup$
add a comment |
$begingroup$
Solving the inital value problem gives me $x(t)=frac{1}{(2t+1)^{1/2}}$ which is clearly striclty decreasing as $(2t+1)^{1/2}$ is strictly increasing. However is it possible to just say it's strictly decreasing as the derivative $dot{x} = -x^{3}$ is strictly decreasing for postive x, if not how would I solve this without having to explicitly solve the inital value problem.
real-analysis ordinary-differential-equations
edited Jan 20 at 16:24
Did
248k23225463
asked Jan 20 at 15:47
RogerRoger
847
$endgroup$
add a comment |
$begingroup$
Solving the inital value problem gives me $x(t)=frac{1}{(2t+1)^{1/2}}$ which is clearly striclty decreasing as $(2t+1)^{1/2}$ is strictly increasing. However is it possible to just say it's strictly decreasing as the derivative $dot{x} = -x^{3}$ is strictly decreasing for postive x, if not how would I solve this without having to explicitly solve the inital value problem.
real-analysis ordinary-differential-equations
edited Jan 20 at 16:24
Did
248k23225463
asked Jan 20 at 15:47
RogerRoger
847
$endgroup$
Solving the inital value problem gives me $x(t)=frac{1}{(2t+1)^{1/2}}$ which is clearly striclty decreasing as $(2t+1)^{1/2}$ is strictly increasing. However is it possible to just say it's strictly decreasing as the derivative $dot{x} = -x^{3}$ is strictly decreasing for postive x, if not how would I solve this without having to explicitly solve the inital value problem.
real-analysis ordinary-differential-equations
real-analysis ordinary-differential-equations
edited Jan 20 at 16:24
Did
248k23225463
asked Jan 20 at 15:47
RogerRoger
847
edited Jan 20 at 16:24
Did
248k23225463
asked Jan 20 at 15:47
RogerRoger
847
edited Jan 20 at 16:24
Did
248k23225463
edited Jan 20 at 16:24
Did
248k23225463
edited Jan 20 at 16:24
Did
248k23225463
248k23225463
asked Jan 20 at 15:47
RogerRoger
847
asked Jan 20 at 15:47
RogerRoger
847
asked Jan 20 at 15:47
RogerRoger
847
847
add a comment |
add a comment |
1 Answer
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Yes, you show that exactly like that. As $x(t)=0$ is a solution, and the right side is locally Lipschitz, the uniqueness theorem implies that the zero solution can not be crossed, all solutions have a constant sign.
Now the initial value $x(0)=1$ fixes the sign for the solution of the given IVP as positive, and the formula for the right side gives then the sign of the derivative as negative.
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Yes, you show that exactly like that. As $x(t)=0$ is a solution, and the right side is locally Lipschitz, the uniqueness theorem implies that the zero solution can not be crossed, all solutions have a constant sign.
Now the initial value $x(0)=1$ fixes the sign for the solution of the given IVP as positive, and the formula for the right side gives then the sign of the derivative as negative.
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
Yes, you show that exactly like that. As $x(t)=0$ is a solution, and the right side is locally Lipschitz, the uniqueness theorem implies that the zero solution can not be crossed, all solutions have a constant sign.
Now the initial value $x(0)=1$ fixes the sign for the solution of the given IVP as positive, and the formula for the right side gives then the sign of the derivative as negative.
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
Yes, you show that exactly like that. As $x(t)=0$ is a solution, and the right side is locally Lipschitz, the uniqueness theorem implies that the zero solution can not be crossed, all solutions have a constant sign.
Now the initial value $x(0)=1$ fixes the sign for the solution of the given IVP as positive, and the formula for the right side gives then the sign of the derivative as negative.
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
$endgroup$
Yes, you show that exactly like that. As $x(t)=0$ is a solution, and the right side is locally Lipschitz, the uniqueness theorem implies that the zero solution can not be crossed, all solutions have a constant sign.
Now the initial value $x(0)=1$ fixes the sign for the solution of the given IVP as positive, and the formula for the right side gives then the sign of the derivative as negative.
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
answered Jan 20 at 16:04
LutzLLutzL
59.3k42057
59.3k42057
add a comment |
add a comment |
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