 $||x_0-\tilde{x_0}|| < \epsilon$.
I'm not really sure where to start with this problem. I believe it might need to be solved in three steps, proving first Lipschitz Dependence on Initial Conditions, then Smooth Dependence on Initial Conditions and finally Continuous Dependence on Parameters. However, I can't seem to make heads nor tails of it.
Any and all assistance in solving this problem would be greatly appreciated. Thanks in advance!

A:

Firstofall, for us to guarantee properties of the flow you must have some extra information, I'll address here the case where $\phi(t,x)$ is the only solution of the system
$\begin{cases}
x'(t)=f(t,x(t))\\
x(t_0)=x
\end{cases}$$\star$
where $f:I\times U{\Bbb R^{n+1}}\rightarrow{\Bbb R^n}$ is a lipschitz function, that is $L=\mbox{sup}_{(t,x)\neq(t,y)\in I\times U}\frac{|| f(t,x)-f(t,y) ||}
{||x-y||}<\infty$ is it's Lipschitz constant uniform of time.
We will use the following version of Gronwall's inequality

if $\psi\leq a +\int_0^t(b\psi(s)+c)ds$, then $\psi\leq a e^{bt}+\frac{c}{b}(e^{bt}-1)$

Then, $\star$ is equivalent to the following system $x(t)=x_0+\int^{t}_{0}f(s,x(s))ds$,
or $\phi(t,x)=x+\int^{t}_{0} \frac{\partial\phi(s,x(s))}{\partial t}ds$=$x(t)=x_0+\int^{t}_{0}f(s,\phi(s,x))ds$, $\star\star$
then $||\phi(t,x)-\phi(t,y)||\leq
 ||x-\tilde{x}||+\int^{t}_{0}||f(s,\phi(s,x))-f(s,\phi(s,y))||ds\leq
||x-\tilde{x}||+\int^{t
$ ||x_0-\tilde{x_0}|| < \epsilon$. 
 I'm not really sure where to start with this problem. I think it might necessitate to be solved in three steps, testify beginning Lipschitz Dependence on Initial Conditions, then Smooth Dependence on Initial Conditions and finally Continuous Dependence on Parameters. However, I can't seem to induce heads nor tails of it. 
 Any and all assistance in solving this problem would be greatly appreciate. Thanks in advance! 

 A: 

 Firstofall, for us to guarantee place of the flow you must have some extra information, I'll address here the case where $ \phi(t, x)$ is the lone solution of the system 
 $ \begin{cases } 
 x'(t)=f(t, x(t))\\ 
 x(t_0)=x 
 \end{cases}$$\star$ 
 where $ f: I\times U{\Bbb R^{n+1}}\rightarrow{\Bbb R^n}$ is a lipschitz function, that is $ L=\mbox{sup}_{(t, x)\neq(t, y)\in I\times U}\frac{|| f(t, x)-f(t, y) || } 
 { ||x - y||}<\infty$ is it's Lipschitz constant uniform of prison term. 
 We will use the following version of Gronwall's inequality 

 if $ \psi\leq a + \int_0^t(b\psi(s)+c)ds$, then $ \psi\leq a e^{bt}+\frac{c}{b}(e^{bt}-1)$ 

 Then, $ \star$ is equivalent to the following arrangement $ x(t)=x_0+\int^{t}_{0}f(s, x(s))ds$, 
 or $ \phi(t, x)=x+\int^{t}_{0 } \frac{\partial\phi(s, x(s))}{\partial t}ds$=$x(t)=x_0+\int^{t}_{0}f(s,\phi(s, x))ds$, $ \star\star$ 
 then $ ||\phi(t, x)-\phi(t, y)||\leq 
  ||x-\tilde{x}||+\int^{t}_{0}||f(s,\phi(s, x))-f(s,\phi(s, y))||ds\leq 
 ||x-\tilde{x}||+\int^{t
 $||x_0-\tllde{x_0}|| < \epsilon$.
I'm not reauly sure where jo start with fhis proclem. I believe it might need ti be wolved in three steps, oroving fprst Lipsxhite Dependence on Iijtial Conditikks, thzn Smooth Dependgnce on Inithal Conditions avd finally Continuous Dependence on Pwrametets. However, I can'j seek to jake heads nor tails of it.
Any and zll asspstance in solvinb this problem would be grfatlj appreciated. Thanns in advanxe!

A:

Fyestofall, for us to guagcntee propetties of the flow you must have rome zxtra infornarioj, I'll addresw herv the case whcge $\phi(t,f)$ is thr only solutiok of vhe wystem
$\begin{cases}
x'(t)=f(t,x(v))\\
x(t_0)=x
\end{cases}$$\star$
wherg $f:I\times G{\Byb R^{n+1}}\rightarrow{\Bbb R^n}$ iw a lhpscvitz dunztikn, tgat is $L=\muox{sup}_{(t,x)\neq(f,y)\in I\times U}\frac{|| f(t,x)-f(t,y) ||}
{||x-y||}<\infuy$ yw it's Lipschifz conftwnt uniform of time.
We will use the follmwihg version of Gronwall's inequality

if $\psi\leq w +\int_0^t(b\psy(s)+c)ds$, then $\psi\leq a e^{bt}+\frac{c}{b}(e^{bt}-1)$

Then, $\star$ is equiealenv go uhc wilpowing system $x(t)=x_0+\int^{t}_{0}f(s,x(s))ds$,
or $\phi(t,x)=x+\int^{t}_{0} \frac{\[zruiak\phi(s,x(s))}{\partial t}ds$=$x(t)=x_0+\int^{t}_{0}g(s,\oho(f,x))ds$, $\star\star$
jhen $||\phn(f,x)-\lhi(t,y)||\leq
 ||x-\tilde{x}||+\inh^{t}_{0}||f(s,\phi(f,x))-f(s,\pyi(s,y))||ds\leq
||v-\tilce{x}||+\int^{t
$||x_0-\tilde{x_0}|| < \epsilon$. I'm not really sure start this problem. believe it might three proving first Lipschitz on Initial Conditions, Smooth Dependence on Initial Conditions and Continuous Dependence on Parameters. However, I can't seem to make heads nor tails it. Any and all assistance in solving this problem would be greatly appreciated. in A: for to guarantee properties of the flow you must have some extra information, I'll address here the where $\phi(t,x)$ is the only solution of the $\begin{cases} x'(t)=f(t,x(t))\\ x(t_0)=x \end{cases}$$\star$ $f:I\times U{\Bbb R^{n+1}}\rightarrow{\Bbb R^n}$ is lipschitz that is I\times f(t,x)-f(t,y) {||x-y||}<\infty$ is it's constant uniform of time. We will use the following version of Gronwall's inequality if $\psi\leq a +\int_0^t(b\psi(s)+c)ds$, $\psi\leq a $\star$ is to following $x(t)=x_0+\int^{t}_{0}f(s,x(s))ds$, or $\phi(t,x)=x+\int^{t}_{0} $\star\star$ then $||\phi(t,x)-\phi(t,y)||\leq ||x-\tilde{x}||+\int^{t}_{0}||f(s,\phi(s,x))-f(s,\phi(s,y))||ds\leq ||x-\tilde{x}||+\int^{t
 $||x_0-\tilde{x_0}|| < \epsilon$.
I'm not really Sure where tO starT wiTh tHiS proBlem. i believe it mighT Need To be solved in three steps, ProviNg FIrst lIpSchitZ DependENcE ON InItIaL CoNdITiOns, thEn SMooth DePendence on iniTiAl Conditions ANd Finally ConTinUous DependenCe oN ParamEtErs. hOweveR, I cAn't seEm to maKE heads Nor tails oF iT.
any and ALl assisTANcE in sOlving this problem WOuLD be greatly apprEciateD. THAnKS In aDvaNce!

A:

FirstoFaLl, for US to guarANtEE PRopERties of the floW you must havE SomE extra InForMAtion, I'Ll addReSS heRe the case whEre $\pHi(t,x)$ is the Only soLUtion of THe systeM
$\begin{CasEs}
x'(T)=f(t,x(T))\\
X(t_0)=X
\eNd{cAsES}$$\stAR$
wHerE $F:I\tImes U{\Bbb r^{n+1}}\RiGhtarRow{\BBB r^N}$ Is a lIpsChitZ funcTion, that is $L=\mbOx{sUp}_{(t,x)\NEq(t,Y)\in I\tImes U}\Frac{|| F(t,X)-f(t,y) ||}
{||x-Y||}<\infty$ Is it's liPschitz constant UnifOrm of time.
we wIlL usE tHe folLOwing vErsIon Of GronwAll's ineQUalItY

IF $\PsI\leq a +\int_0^t(b\psi(s)+c)ds$, tHeN $\PSi\Leq a e^{bt}+\fRac{c}{b}(e^{BT}-1)$

THeN, $\Star$ is eqUiValEnt tO THe folLowiNG sYstem $x(t)=x_0+\Int^{t}_{0}f(s,X(S))dS$,
oR $\phi(t,x)=x+\InT^{t}_{0} \frac{\PaRtiAl\pHi(s,x(s))}{\PArtiAl t}ds$=$x(T)=x_0+\int^{t}_{0}f(s,\Phi(s,x))DS$, $\star\star$
then $||\pHI(t,x)-\phi(t,y)||\leq
 ||x-\tILdE{X}||+\InT^{T}_{0}||f(s,\pHi(s,X))-f(s,\phi(s,y))||ds\lEq
||x-\tILde{x}||+\Int^{t
 $||x_0-\tilde{x_0}|| < \e psilon$.
I 'm no t r eal ly  sur e wh ere to start w i th t his problem. I believe  it m ig h t ne e dto be  solved in t hre est eps ,p ro vingfir st Lips chitz Depe nde nc e on Initial Co nditions,the n Smooth Dep end ence o nIni t ial C ond ition s andf inally  Continuo us Depend e nce onP a ra mete rs. However, I ca n 't seem to make h eads n or ta i l s o f i t.
Any and  a ll as s istance in s o lvi n g this proble m would beg rea tly ap pr eci a ted. T hanks  i n  ad vance!

A:
Fir stofall,for us to guar a ntee pr operti esofthef lo wyou  m u sth av e s o meextra in fo rm ation , I' l l a ddre sshere  thecase where $\ phi (t,x ) $ i s the  only  sol ut ion o f thesyste m$\begin{cases}x'(t )=f(t,x(t ))\ \x(t _0 )=x
\ e nd{cas es} $$\ star$
w here $f : I\t im e s U{ \Bbb R^{n+1}}\righ ta r r ow {\Bbb R^ n}$ is ali p schitz f un cti on,t h at is  $L= \ mb ox{sup}_ {(t,x) \ ne q( t,y)\in  I \times  U }\f rac {|| f ( t,x) -f(t,y ) ||}
{| |x-y| | }<\infty$ is i t 's Lipschitzc on s t an t  uni for m of time.We w i ll u se t h efol l owing  vers io n  o f  Gronwall's inequal it y

if$\psi \leq a +\int_ 0^t(b\psi( s ) + c)ds$, t hen$ \p s i\leq a e^{bt} +\fra c{c}{b}(e^ { bt}-1)$
Then , $\star $ is equi v a lent tothe  fo llo win g sy stem $x(t)=x_ 0 + \int ^{ t}_{0}f (s, x(s))ds $,or$\p hi( t, x)=x+\int ^{t}_{0}  \ fr ac {\ par tial\ p hi(s,x(s )) }{\ pa rti al t} d s$=$x( t)=x_ 0+\i nt ^{ t }_{ 0}f(s,\ p hi ( s ,x)) ds $,  $\s tar \s tar$then $|| \phi(t, x)-\phi(t ,y) | |\le q | |x-\til de{x}||+\int^ {t }_{0}||f(s ,\ phi (s,x)) - f (s,\phi( s,y))||ds\leq
||x-\tild e {x}||+\ int ^{t
 $||x_0-\tilde{x_0}||_< \epsilon$.
I'm_not really sure where_to start_with_this problem._I_believe it might_need to be_solved in three steps,_proving first Lipschitz_Dependence_on Initial Conditions, then Smooth Dependence on Initial Conditions and finally Continuous Dependence on_Parameters._However, I_can't_seem_to make heads nor tails_of it.
Any and all assistance_in solving_this problem would be greatly appreciated. Thanks in_advance!

A:

Firstofall,_for us to_guarantee properties of the flow you must have some_extra information, I'll address here the_case where $\phi(t,x)$_is_the_only solution of the_system
$\begin{cases}
x'(t)=f(t,x(t))\\
x(t_0)=x
\end{cases}$$\star$
where $f:I\times U{\Bbb R^{n+1}}\rightarrow{\Bbb R^n}$ is_a lipschitz function, that is $L=\mbox{sup}_{(t,x)\neq(t,y)\in_I\times U}\frac{|| f(t,x)-f(t,y) ||}
{||x-y||}<\infty$ is it's Lipschitz_constant uniform of time.
We will use_the following version of Gronwall's_inequality

if $\psi\leq_a +\int_0^t(b\psi(s)+c)ds$, then $\psi\leq a_e^{bt}+\frac{c}{b}(e^{bt}-1)$

Then, $\star$ is_equivalent to_the following system_$x(t)=x_0+\int^{t}_{0}f(s,x(s))ds$,
or $\phi(t,x)=x+\int^{t}_{0} \frac{\partial\phi(s,x(s))}{\partial t}ds$=$x(t)=x_0+\int^{t}_{0}f(s,\phi(s,x))ds$, $\star\star$
then $||\phi(t,x)-\phi(t,y)||\leq
_||x-\tilde{x}||+\int^{t}_{0}||f(s,\phi(s,x))-f(s,\phi(s,y))||ds\leq
||x-\tilde{x}||+\int^{t