User:Egm6321.f12.team6.bang/report3/problem*3.10

=Report 3=

Problem Statement

 * a) Use IFM to Show that the solution of $$ \displaystyle \dot x(t) = ax(t) + bu(t) $$ is
 * $$ \displaystyle x(t) = [\exp \{a(t-t_0)\}] \,x(t_0) + \int^t_{t_0} [\exp \{a(t-\tau)\}] \,b \,u(\tau)d\tau $$
 * Identify the integrating factor, the homogeneous solution, and the particular solution.
 * b) Show tha the solution of the L1-ODE-VC $$ \displaystyle \dot x(t) = a(t)x(t)+b(t)u(t) $$ is
 * $$ \displaystyle x(t) = \left[\exp \int^t_{t_0} a(\tau)d\tau \right]x(t_0) + \int^t_{t_0} [\exp \int^t_\tau a(s)ds] \,b(\tau) \,u(\tau)d\tau $$

Given

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$$ \displaystyle \dot x(t) = ax(t) + bu(t) $$
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 * (10.1)
 * }
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$$ \displaystyle \dot x(t) = a(t)x(t)+b(t)u(t) $$
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 * (10.2)
 * }

Nomenclature

 * $$ \displaystyle \dot {x} = \frac {dx}{dt} $$
 * $$ \displaystyle \tau $$ → Dummy Variable 1
 * $$ \displaystyle s $$ → Dummy Variable 2
 * $$ \displaystyle e^x = \exp(x) $$
 * $$ \displaystyle M_x = \frac{\partial M(u,x)}{\partial x} $$
 * $$ \displaystyle N_u = \frac{\partial N(u,x)}{\partial u} $$

Part a)

 * With given Equation 10.1, we have to evaluate that the given equation is exact or not. First of all, we are proceeding with the first condition. The Equation 10.1 can be rewritten again.
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$$ \displaystyle \underbrace {1}_{N(u,x)} \cdot \dot x(t) - \underbrace {[ax(t)+bu(t)]}_{M(u,x)} = 0 $$
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 * (10.3)
 * }
 * Since we can write a equation in form of $$ \displaystyle N(u,x)x'-M(u,x) = 0 $$, this satisfies the first condition.
 * Now, we can proceed to the second condition. The second condition has have $$ \displaystyle N_u = M_x $$. Therefore, taking a partial derivative for both part will give the following result.
 * $$ \displaystyle N_x = 0; \;\;\;\;\; N_u = 0 $$
 * $$ \displaystyle M_u = -b; \;\; M_x = -a $$
 * This is clearly stating that $$ \displaystyle N_x \neq M_u $$. Hence, it fails the second condition Therefore, we can conclude that this is not a exact equation. This is indicating that the equation needs a integrating factor to solve it. Based on Euler's Integrating Factor Method, the equation for h is the following.
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$$ \displaystyle \frac {h_u}{h} = -\frac{1}{N}(N_u-M_x) $$
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 * (10.4)
 * }
 * $$ \displaystyle \frac {h_u}{h} = -\frac{1}{1}(0-(-a)) $$
 * $$ \displaystyle \frac {h_u}{h} = -a $$
 * Also remember that both x, and u are dependent variable for t.
 * $$ \displaystyle \int \frac {h_u}{h} = \int^t -a(s)ds $$
 * $$ \displaystyle e^h = -at $$
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$$ \displaystyle h = e^{-at} $$
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 * (10.5)
 * }
 * Now, we can apply integrating factor to Equation 10.3.
 * $$ \displaystyle e^{-at} [\dot x(t) - (ax(t)+bu(t)] = 0 $$
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$$ \displaystyle e^{-at}\dot x(t) - e^{-at}ax(t) = e^{-at}bu(t) $$
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 * (10.6)
 * }
 * However, there is one more information that is needed in order to solve this L1-ODE-CC.
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$$ \displaystyle e^{-at}a = [e^{-at}]' $$
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 * (10.7)
 * }
 * Then, combining Equation 10.6 to Equation 10.7 will give the following expression.
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$$ \displaystyle e^{-at}\dot x(t) - [e^{-at}]'x(t) = e^{-at}bu(t) $$
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 * (10.8)
 * }
 * Now we can simplify this equation using Product Rule for derivative. Then, we will have following expression.
 * $$ \displaystyle [e^{-at}x(t)]' = e^{-at}bu(t) $$
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$$ \displaystyle e^{-at}x(t) = \int^t_{t_0} e^{-a\tau}bu(\tau)d\tau + k $$
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 * (10.9)
 * }
 * In order to solve for constant k, we need to set the intial condition when $$ \displaystyle t_0 $$.
 * $$ \displaystyle e^{-at_0}x(t_0) = \int^{t_0}_{t_0} e^{-a\tau}bu(\tau)d\tau + k $$
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$$ \displaystyle k = e^{-at_0}x(t_0) $$ Combining Equation 10.9 and Equation 10.10 will give the complete expression.
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 * (10.10)
 * }
 * $$ \displaystyle e^{-at}x(t) = e^{-at_0}x(t_0) + \int^t_{t_0} e^{-a\tau}bu(\tau)d\tau $$
 * $$ \displaystyle x(t) = \frac{e^{-at_0}}{e^{-at}}x(t_0) + \int^t_{t_0} \frac{e^{-a\tau}}{e^{-at}}bu(\tau)d\tau $$
 * $$ \displaystyle x(t) = e^{-at_0+at}x(t_0) + \int^t_{t_0} e^{-a\tau+at}bu(\tau)d\tau $$
 * $$ \displaystyle x(t) = e^{a(t-t_0)}x(t_0) + \int^t_{t_0} e^{a(t-\tau)}bu(\tau)d\tau $$
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$$ \displaystyle x(t) = [\exp\{a(t-t_0)\}]x(t_0) + \int^t_{t_0} [\exp\{a(t-\tau)\}]bu(\tau)d\tau $$
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 * (10.11)
 * }
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Therefore, the homogeneous solution is $$ \displaystyle [\exp\{a(t-t_0)\}]x(t_0) $$, and the particular solution is $$ \displaystyle \int^t_{t_0} [\exp\{a(t-\tau)\}]bu(\tau)d\tau $$
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 * }

Part b)

 * From previous equation, we have derived the Equation 10.1 in terms of x(t).
 * $$ \displaystyle \dot x(t) = ax(t) + bu(t) $$ becomes
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$$ \displaystyle x(t) = e^{a(t-t_0)}x(t_0) + \int^t_{t_0} e^{a(t-\tau)}bu(\tau)d\tau $$
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 * (10.12)
 * }
 * The difference between Equation 10.1 and Equation 10.2 is that now the coefficient, or constant terms both a and b are dependent variable respect to the independent variable t. In this case, we have to take an account of changes in variable a and b respect to the time. For the homogeneous solution, we can find this relationship.
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$$ \displaystyle a(t) = \int^t_{t_0} a(s)ds $$
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 * (10.13)
 * }
 * The homogeneous part of the answer becomes the following equation.
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$$ \displaystyle x_h(t) = e^{\int^t_{t_0} a(s)ds}x(t_0) $$
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 * (10.14)
 * }
 * The particular part is little bit more complicated. We have to take the integral twice for both a, and b. Also, the second dummy variable s is introduced to avoid overlapping usage of τ. Then we can find the following relationship.
 * $$ \displaystyle a(t) = \int^t_{t_0}\int^t_{\tau} a(s)ds \, d\tau $$
 * $$ \displaystyle b(t) = \int^t_{t_0} b(\tau)d\tau $$
 * Put this into the particular equation will give the following equation.
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$$ \displaystyle x_p(t) = \int^t_{t_0}e^{\int^t_{\tau} a(s)ds}b(\tau)u(\tau)d(\tau) $$
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 * (10.15)
 * }
 * Since $$ \displaystyle x(t) = x_h(t) + x_p(t) $$, we have the equation as following.
 * $$ \displaystyle x(t) = e^{\int^t_{t_0} a(s)ds}x(t_0) + \int^t_{t_0}e^{\int^t_{\tau} a(s)ds}b(\tau)u(\tau)d(\tau) $$
 * $$ \displaystyle $$
 * Rewriting this equation will give,
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$$ \displaystyle x(t) = \left[\exp \int^t_{t_0} a(\tau)d\tau \right]x(t_0) + \int^t_{t_0} [\exp \int^t_\tau a(s)ds] \,b(\tau) \,u(\tau)d\tau $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * (10.16)
 * }