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Using the contraction mapping principle for the solution of an ODE


Questions about the Picard–Lindelöf theorem for an ODEExistence solution pendulum equation using the contraction principleUsing contraction mapping theorem to prove existence/uniqueness of solutions of Linear first order ODEsOn the extension of the solution to a nonlinear ODEStrong Solutions to Nonlinear ODE by Contraction MappingUnique solution of IVP on IUniqueness of solution of an ODE in an interval implies uniqueness of solution in a subintervalShow that a mapping is a contraction in $ mathbbR^2 $Proving system of equations has unique solution using contraction mapping principleSolve differential equation via successive approximations (contraction mapping principle)













0












$begingroup$


Suppose we want to solve a simple ODE $$dotx=2t(1-x)=f(t,x)qquad (1)$$ with initial condition $x_0=2$ using the contraction mapping principle.



I showed that $f$ is Lipschitz continuous over $[0,1]timesmathbb R$ in $t$ with Lipschitz constant $L=max_tin[0,1]2t$. By considering a normed vector space $(C[0,delta],||cdot||_C)$, then using the contraction map with $P:C[0,delta]to C[0,delta]$ given by $$(Pphi)(t)=x_0+int_0^tf(t,phi(tau));dtau$$ I found by direct integration that the solution of $(1)$ is $$x=1+exp(-t^2).$$



But so far we have shown that there exists a unique solution over the interval $[0,delta]$, right?



Is the next step to provide an argument that shows that our solution is unique over the entire set of real numbers? I am not sure how to make this argument. I thought that maybe by iteration and increasing $delta$ one can show that the solution is unique over a bigger interval. But this doesn't seem to me very rigorous.



I'd appreciate any help/hint. Thank you.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Hint: for large $T$, $P: C[-T,T] rightarrow C[-T,T]$ defined by, $$(Pphi)(t) = x_0 + int_0^t f(tau,phi(tau)),dtau,$$ is no longer a contraction mapping. However, for any such fixed $T$, there exists a $k$ such that $P^k$ is a contraction mapping. Try proving this and you will have your result.
    $endgroup$
    – forgottenarrow
    Mar 29 at 5:23







  • 1




    $begingroup$
    There is a result stating that if $fcolon(a,b)timesmathbbRtomathbbR$ is continuous and there are continuous functions $M_1,M_2colon(a,b)to[0,infty)$ such that $lvert f(t,x)rvertle M_1(t)+M_2(t)lvert xrvert$ for all $tin(a,b)$ and all $xinmathbbR$ then any nonextendible solution of $dotx=f(t,x)$ is defined on the whole of $(a,b)$. The proof uses Grönwall's inequality.
    $endgroup$
    – user539887
    Mar 29 at 6:28
















0












$begingroup$


Suppose we want to solve a simple ODE $$dotx=2t(1-x)=f(t,x)qquad (1)$$ with initial condition $x_0=2$ using the contraction mapping principle.



I showed that $f$ is Lipschitz continuous over $[0,1]timesmathbb R$ in $t$ with Lipschitz constant $L=max_tin[0,1]2t$. By considering a normed vector space $(C[0,delta],||cdot||_C)$, then using the contraction map with $P:C[0,delta]to C[0,delta]$ given by $$(Pphi)(t)=x_0+int_0^tf(t,phi(tau));dtau$$ I found by direct integration that the solution of $(1)$ is $$x=1+exp(-t^2).$$



But so far we have shown that there exists a unique solution over the interval $[0,delta]$, right?



Is the next step to provide an argument that shows that our solution is unique over the entire set of real numbers? I am not sure how to make this argument. I thought that maybe by iteration and increasing $delta$ one can show that the solution is unique over a bigger interval. But this doesn't seem to me very rigorous.



I'd appreciate any help/hint. Thank you.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Hint: for large $T$, $P: C[-T,T] rightarrow C[-T,T]$ defined by, $$(Pphi)(t) = x_0 + int_0^t f(tau,phi(tau)),dtau,$$ is no longer a contraction mapping. However, for any such fixed $T$, there exists a $k$ such that $P^k$ is a contraction mapping. Try proving this and you will have your result.
    $endgroup$
    – forgottenarrow
    Mar 29 at 5:23







  • 1




    $begingroup$
    There is a result stating that if $fcolon(a,b)timesmathbbRtomathbbR$ is continuous and there are continuous functions $M_1,M_2colon(a,b)to[0,infty)$ such that $lvert f(t,x)rvertle M_1(t)+M_2(t)lvert xrvert$ for all $tin(a,b)$ and all $xinmathbbR$ then any nonextendible solution of $dotx=f(t,x)$ is defined on the whole of $(a,b)$. The proof uses Grönwall's inequality.
    $endgroup$
    – user539887
    Mar 29 at 6:28














0












0








0





$begingroup$


Suppose we want to solve a simple ODE $$dotx=2t(1-x)=f(t,x)qquad (1)$$ with initial condition $x_0=2$ using the contraction mapping principle.



I showed that $f$ is Lipschitz continuous over $[0,1]timesmathbb R$ in $t$ with Lipschitz constant $L=max_tin[0,1]2t$. By considering a normed vector space $(C[0,delta],||cdot||_C)$, then using the contraction map with $P:C[0,delta]to C[0,delta]$ given by $$(Pphi)(t)=x_0+int_0^tf(t,phi(tau));dtau$$ I found by direct integration that the solution of $(1)$ is $$x=1+exp(-t^2).$$



But so far we have shown that there exists a unique solution over the interval $[0,delta]$, right?



Is the next step to provide an argument that shows that our solution is unique over the entire set of real numbers? I am not sure how to make this argument. I thought that maybe by iteration and increasing $delta$ one can show that the solution is unique over a bigger interval. But this doesn't seem to me very rigorous.



I'd appreciate any help/hint. Thank you.










share|cite|improve this question









$endgroup$




Suppose we want to solve a simple ODE $$dotx=2t(1-x)=f(t,x)qquad (1)$$ with initial condition $x_0=2$ using the contraction mapping principle.



I showed that $f$ is Lipschitz continuous over $[0,1]timesmathbb R$ in $t$ with Lipschitz constant $L=max_tin[0,1]2t$. By considering a normed vector space $(C[0,delta],||cdot||_C)$, then using the contraction map with $P:C[0,delta]to C[0,delta]$ given by $$(Pphi)(t)=x_0+int_0^tf(t,phi(tau));dtau$$ I found by direct integration that the solution of $(1)$ is $$x=1+exp(-t^2).$$



But so far we have shown that there exists a unique solution over the interval $[0,delta]$, right?



Is the next step to provide an argument that shows that our solution is unique over the entire set of real numbers? I am not sure how to make this argument. I thought that maybe by iteration and increasing $delta$ one can show that the solution is unique over a bigger interval. But this doesn't seem to me very rigorous.



I'd appreciate any help/hint. Thank you.







ordinary-differential-equations nonlinear-system contraction-operator






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 29 at 2:36









johnny09johnny09

705624




705624







  • 1




    $begingroup$
    Hint: for large $T$, $P: C[-T,T] rightarrow C[-T,T]$ defined by, $$(Pphi)(t) = x_0 + int_0^t f(tau,phi(tau)),dtau,$$ is no longer a contraction mapping. However, for any such fixed $T$, there exists a $k$ such that $P^k$ is a contraction mapping. Try proving this and you will have your result.
    $endgroup$
    – forgottenarrow
    Mar 29 at 5:23







  • 1




    $begingroup$
    There is a result stating that if $fcolon(a,b)timesmathbbRtomathbbR$ is continuous and there are continuous functions $M_1,M_2colon(a,b)to[0,infty)$ such that $lvert f(t,x)rvertle M_1(t)+M_2(t)lvert xrvert$ for all $tin(a,b)$ and all $xinmathbbR$ then any nonextendible solution of $dotx=f(t,x)$ is defined on the whole of $(a,b)$. The proof uses Grönwall's inequality.
    $endgroup$
    – user539887
    Mar 29 at 6:28













  • 1




    $begingroup$
    Hint: for large $T$, $P: C[-T,T] rightarrow C[-T,T]$ defined by, $$(Pphi)(t) = x_0 + int_0^t f(tau,phi(tau)),dtau,$$ is no longer a contraction mapping. However, for any such fixed $T$, there exists a $k$ such that $P^k$ is a contraction mapping. Try proving this and you will have your result.
    $endgroup$
    – forgottenarrow
    Mar 29 at 5:23







  • 1




    $begingroup$
    There is a result stating that if $fcolon(a,b)timesmathbbRtomathbbR$ is continuous and there are continuous functions $M_1,M_2colon(a,b)to[0,infty)$ such that $lvert f(t,x)rvertle M_1(t)+M_2(t)lvert xrvert$ for all $tin(a,b)$ and all $xinmathbbR$ then any nonextendible solution of $dotx=f(t,x)$ is defined on the whole of $(a,b)$. The proof uses Grönwall's inequality.
    $endgroup$
    – user539887
    Mar 29 at 6:28








1




1




$begingroup$
Hint: for large $T$, $P: C[-T,T] rightarrow C[-T,T]$ defined by, $$(Pphi)(t) = x_0 + int_0^t f(tau,phi(tau)),dtau,$$ is no longer a contraction mapping. However, for any such fixed $T$, there exists a $k$ such that $P^k$ is a contraction mapping. Try proving this and you will have your result.
$endgroup$
– forgottenarrow
Mar 29 at 5:23





$begingroup$
Hint: for large $T$, $P: C[-T,T] rightarrow C[-T,T]$ defined by, $$(Pphi)(t) = x_0 + int_0^t f(tau,phi(tau)),dtau,$$ is no longer a contraction mapping. However, for any such fixed $T$, there exists a $k$ such that $P^k$ is a contraction mapping. Try proving this and you will have your result.
$endgroup$
– forgottenarrow
Mar 29 at 5:23





1




1




$begingroup$
There is a result stating that if $fcolon(a,b)timesmathbbRtomathbbR$ is continuous and there are continuous functions $M_1,M_2colon(a,b)to[0,infty)$ such that $lvert f(t,x)rvertle M_1(t)+M_2(t)lvert xrvert$ for all $tin(a,b)$ and all $xinmathbbR$ then any nonextendible solution of $dotx=f(t,x)$ is defined on the whole of $(a,b)$. The proof uses Grönwall's inequality.
$endgroup$
– user539887
Mar 29 at 6:28





$begingroup$
There is a result stating that if $fcolon(a,b)timesmathbbRtomathbbR$ is continuous and there are continuous functions $M_1,M_2colon(a,b)to[0,infty)$ such that $lvert f(t,x)rvertle M_1(t)+M_2(t)lvert xrvert$ for all $tin(a,b)$ and all $xinmathbbR$ then any nonextendible solution of $dotx=f(t,x)$ is defined on the whole of $(a,b)$. The proof uses Grönwall's inequality.
$endgroup$
– user539887
Mar 29 at 6:28











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