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I have this exercise :

"Consider the Cauchy problem's :

$$ ^C D^{\alpha}y(t)=f(t,y(t),y'(t)), t\in [0,T] ....(1) $$ $$ y(0)=y_0, y'(0)=y_1 .... (2) $$

Where $f:[0,T]\times\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R} $is a function , $y_0,y_1\in \mathbb{R}$ and $\alpha \in ]1,2]$

give a result of existence and uniqueness for the problem by using the contraction principle of Banach "

Can someone help me to solve this exercise ?

Thank you.

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  • $\begingroup$ Can you include the explicit definition of ${}^CD^\alpha$? $\endgroup$ – timur Sep 19 '13 at 14:06
  • $\begingroup$ $\displaystyle^C D^{\alpha}h (t)= \frac{1}{\Gamma(n-\alpha)}\int_0^t (t-s)^{n-\alpha-1} h^{(n)}(s) ds ; n=[\alpha]+1$ $\endgroup$ – Vrouvrou Sep 19 '13 at 14:23
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We can assume $1<\alpha<2$ since the case $\alpha=2$ is the usual ODE case. So the Cauchy problem is $$ \gamma\int_0^t(t-s)^{1-\alpha}y''(s) \, \mathrm{d}s = f(t,y(t),y'(t)), \qquad\qquad(*) $$ with $y(0)=y_0$ and $y'(0)=y_1$, where $\gamma=1/\Gamma(2-\alpha)$. Note that the integral in the left hand side is well defined if for instance we assume that $y\in C^2([0,t])$. In order to mimic the proof for the ODE case, we need to write the problem in a form like $$ y'(T) = \int_0^T \big[\textrm{some function of }t\textrm{ and }f(t,y(t),y'(t))\big] \,\mathrm{d}t. $$ To do so, we multiply $(*)$ by $(T-t)^{\alpha-2}$ and integrate over $t$ from $0$ to $T$. $$ \gamma \int_0^T\int_0^t(T-t)^{\alpha-2}(t-s)^{1-\alpha}y''(s) \, \mathrm{d}s \,\mathrm{d}t = \int_0^T (T-t)^{\alpha-2} f(t,y(t),y'(t))\,\mathrm{d}t. \qquad\qquad(**) $$ Interchanging the order of integration in the left hand side, which is justified under the assumption $y\in C^2([0,T])$, we get $$ \begin{split} \int_0^T\int_0^t(T-t)^{\alpha-2}(t-s)^{1-\alpha}y''(s) \, \mathrm{d}s \,\mathrm{d}t &= \int_0^T\int_s^T(T-t)^{\alpha-2}(t-s)^{1-\alpha}y''(s) \,\mathrm{d}t\, \mathrm{d}s\\ &= \Gamma(\alpha-1)\Gamma(2-\alpha)\int_0^T y''(s)\, \mathrm{d}s \\ &=\Gamma(\alpha-1)\Gamma(2-\alpha)\big( y'(T)-y'(0) \big), \end{split} $$ where we have used the fact that $$ \int_0^1x^{a-1}(1-x)^{b-1}\, \mathrm{d}x = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, $$ for $a>0$ and $b>0$. To conclude, we have $$ y'(T) = y_1 + c\int_0^T (T-t)^{\alpha-2} f(t,y(t),y'(t))\,\mathrm{d}t, \qquad\qquad(\dagger) $$ where the integration is justified if $f$ is continuous and $y\in C^1([0,T])$. In the converse direction, if $y\in C^2([0,T])$ then $(\dagger)$ implies $(*)$.

Since now it is in a standard form and the question is a homework question, I leave the rest as an exercise.

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  • $\begingroup$ Thank you , but i don't know what to do now , help me please i have an exam about this tomorow $\endgroup$ – Vrouvrou Sep 20 '13 at 20:41
  • $\begingroup$ @Vrouvrou: Write please here how to prove existence for usual ODEs. Solve this one: $y'(t)=f(y(t))$ and $y(0)=y_0$. I don't think I can help you if you don't know the usual ODE case. $\endgroup$ – timur Sep 20 '13 at 20:48
  • $\begingroup$ we use th of peano or cauchy Lipchitz, we stydy the equivalent form of the problem $y(t)=y_0+\int_{0}^t f(s,y(s)) ds$ $\endgroup$ – Vrouvrou Sep 20 '13 at 21:00
  • $\begingroup$ please please help me $\endgroup$ – Vrouvrou Sep 20 '13 at 21:07
  • $\begingroup$ @Vrouvrou: Look at the equivalent form that you mentioned, and compare it with $(\dagger)$. Apply the same method to $(\dagger)$. $\endgroup$ – timur Sep 21 '13 at 16:32

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