- #1
motherh
- 27
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Hi, I want to answer the following question:
x=x(t) is continuous on [0,T) and satisfies
1 ≤ x(t) ≤ C[itex]_{1}[/itex] + C[itex]_{2}[/itex]∫[itex]^{t}_{0}[/itex] x(s)(1+logx(s)) ds
for 0 ≤ t < T. Prove x(t) is bounded on [0,T].Using Gronwall's inequality I get to
x(t) ≤ C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex] ∫[itex]^{t}_{0}[/itex] (1+logx(s)) ds )
≤ C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex]t + C[itex]_{2}[/itex]∫[itex]^{t}_{0}[/itex] logx(s) ds )
Can I say that this is less than C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex]T + ∫[itex]^{T}_{0}[/itex] logx(s) ds ) ?
I'm not too sure where to proceed from here. Would it be helpful to use x(s) > logx(s)?
Any help is appreciated!
x=x(t) is continuous on [0,T) and satisfies
1 ≤ x(t) ≤ C[itex]_{1}[/itex] + C[itex]_{2}[/itex]∫[itex]^{t}_{0}[/itex] x(s)(1+logx(s)) ds
for 0 ≤ t < T. Prove x(t) is bounded on [0,T].Using Gronwall's inequality I get to
x(t) ≤ C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex] ∫[itex]^{t}_{0}[/itex] (1+logx(s)) ds )
≤ C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex]t + C[itex]_{2}[/itex]∫[itex]^{t}_{0}[/itex] logx(s) ds )
Can I say that this is less than C[itex]_{1}[/itex]exp( C[itex]_{2}[/itex]T + ∫[itex]^{T}_{0}[/itex] logx(s) ds ) ?
I'm not too sure where to proceed from here. Would it be helpful to use x(s) > logx(s)?
Any help is appreciated!
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