User:Egm6321.f10.team2.zou/mtg14

Problem Description
I think there maybe some typo in note Mtg.14.

As indicated in the lecture, (2)14-1 and (3)14-1 are both solutions of (1)14-1 when a(t) is treated as constant and varying coefficient respectively. However, I find both (2)14-1 and (3)14-1 cannot be verified by (1)14-1. Then I solved (1)14-1 and I believe (2)14-1 should be written as:

$$ \displaystyle x(t)=\left[ \exp a(t-{{t}_{0}}) \right]x({{t}_{0}})+\exp a(t-{{t}_{N}})*\int_^{t}{\left[ \exp a({{t}_{N}}-\tau ) \right]}bu(\tau )d\tau $$, (Eq 1)
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where $$\displaystyle {{t}_{N}} $$ is an arbitrary constant.

I am 100% sure that Eq 1 is the right solution of (1)14-1. Team please check.

In Prof. Vu-Quoc's lecture the (2)14-1 is:
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$$ \displaystyle x(t)=\left[ \exp a(t-{{t}_{0}}) \right]x({{t}_{0}})+\int_^{t}{\left[ \exp a(t-\tau ) \right]}bu(\tau )d\tau $$ (Eq 2) You can see the difference between Eq 1 and Eq 2 is that Prof. Vu-Quoc probably substitutes $$\displaystyle {{t}_{N}} $$ as $$\displaystyle t $$ and I think this is inappropriate because if $$\displaystyle {{t}_{N}} $$ is replaced by a non-constant or some quantity can be taken as a non-constant, the meaning of Euler intigrating factor would change a lot.
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Team please veryfy Eq 2 to see whether it is the right solution of (1)14-1.

By the way, I think (3)14-1 should also be changed as:


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$$ \displaystyle x(t)=\left[ \exp \int_^{t}{a(s)ds} \right]x({{t}_{0}})+\left[ \exp \int_^{t}{a(s)ds} \right]*\int_^{t}{\left[ \exp \int_{\tau }^{a(s)ds} \right]}bu(\tau )d\tau $$, (Eq 3)
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where $$\displaystyle {{t}_{N}} $$ is an arbitrary constant. The difference between Eq 3 and (3)14-1 is similar to that between Eq 1 and (2)14-1.

I will post my solving process later.

My solving process: Deriving (3)14-1
Given:


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$$ \displaystyle \dot{x}(t)=a(t)x(t)+b(t)u(t)$$ 
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Then the Euler integrating factor is:


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$$ \displaystyle h(t)=\exp \int_ – ^{t}{-a(s)ds}$$, (Eq 4)
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or:


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$$ \displaystyle h(t)=\exp \int_^{t}{-a(s)ds}+{{k}_{1}}$$, (Eq 5)
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where $$\displaystyle {{t}_{N}}$$ is an constant corresponding to integrating constant $$\displaystyle {{k}_{1}}$$. Since we have proved $$\displaystyle {{k}_{1}}$$ is unneccesary thus arbitrary, $$\displaystyle {{t}_{N}}$$ is also arbitrary.

Then we derive,


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$$ \displaystyle \begin{align} x(t) & =\exp \int_ – ^{t}{a(s)ds*\int_ – ^{t}{\left[ \exp \int_ – ^{\tau }{-a(s)ds} \right]}}*bu(\tau )d\tau \\ & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*bu(\tau )d\tau +{{k}_{2}} \right\} \end{align}$$, (Eq 6)
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where $$\displaystyle {{t}_{N}}$$ and $$\displaystyle {{t}_{M}}$$ are arbitrary constants, while $$\displaystyle {{t}_{m}}$$ is also a constant corresponding to $$\displaystyle {{k}_{2}}$$.

Because we assume that $$\displaystyle x({{t}_{0}})$$ is known, we have,


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$$ \displaystyle x({{t}_{o}})=\exp \int_^{a(s)ds}*\left\{ \int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*bu(\tau )d\tau +{{k}_{2}} \right\}$$. (Eq 7)
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From Eq 7 we get $$\displaystyle {{k}_{2}}$$:


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$$ \displaystyle {{k}_{2}}=\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*bu(\tau )d\tau $$. (Eq 8)
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Substituting Eq 8 back to Eq 6, we have,


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$$ \displaystyle \begin{align} x(t) & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*bu(\tau )d\tau +\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*bu(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \left[ \int_^{t}{a(s)ds}-\exp \int_^{a(s)ds} \right]+\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*bu(\tau )d\tau -\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*bu(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp \int_{\tau }^{a(s)ds} \right]}*bu(\tau )d\tau \end{align}$$ (Eq 9)
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We can see that the difference between Eq 9 and (3)14-1 is that (3)14-1 takes $$\displaystyle {{t}_{N}}$$ as $$\displaystyle t$$. The reasons I don't agree with this are: Firstly, substituting $$\displaystyle t$$ into $$\displaystyle {{t}_{N}}$$ will eliminate the term $$\displaystyle \exp \int_^{t}{a(s)ds}$$; Secondly, this will change the meaning of Euler integrating factor since $$\displaystyle {{t}_{N}}$$ is part of it.

Final Conclusion
By Erman's help, I verified that Prof. Vu-Quoc's results are 100% correct without any doubt.

The critical mistake I made is that I failed to distinguish the derivation rule of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau }$$ between that of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau }$$. Note that,


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$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau =}f(t)$$, (Eq 10)
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$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau =}f(t,t)+\int_^{t}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 11)
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More generally,
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$$ \displaystyle \frac{d}{dt}\int_{b(t)}^{a(t)}{f(\tau ,t)d\tau =}\frac{da(t)}{dt}f\left( a(t),t \right)-\frac{db(t)}{dt}f\left( b(t),t \right)+\int_{b(t)}^{a(t)}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 12)
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Please see Variable Limits.

However, my results are correct also, but need one step further to modify. In effect, Eq 9 can be written as:


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$$ \displaystyle \begin{align} & x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp \int_{\tau }^{a(s)ds} \right]}*bu(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp (\int_{\tau }^{t}{a(s)ds+\int_{t}^{a(s)ds)}} \right]}*bu(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\exp \int_{t}^{a(s)ds*}\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*bu(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*bu(\tau )d\tau \end{align}$$ (Eq 13)
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We can see that I have got the same answer to that in note Mtg 14.

Acknowledgement
Special thanks to Erman Oztekin who taught me the correct derivation rule and Leibniz who invented this rule! And I also want to thank my undergraduate calculus textbooks and their authors!