Second order non-linear difference equation solver.
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Is there an algorithm which I can use to find an approximate solution of the following second-order difference equation in order to run simulations of the solution :
$nu(1 - beta) f(u_{n+1})-g(u_{n+1}-u_{n}) + beta g (u_{n+2} - u_{n+1}) = 0 $
with $nu > 0$ and $0 < beta < 1$ and $u_0 = 0$ and $u_N = C > 0$ and $N in mathbb{N}^{*}$ and $f$ and $g$ two real, positive and increasing functions.
Thank you.
algorithms recurrence-relations
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
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add a comment |
$begingroup$
Is there an algorithm which I can use to find an approximate solution of the following second-order difference equation in order to run simulations of the solution :
$nu(1 - beta) f(u_{n+1})-g(u_{n+1}-u_{n}) + beta g (u_{n+2} - u_{n+1}) = 0 $
with $nu > 0$ and $0 < beta < 1$ and $u_0 = 0$ and $u_N = C > 0$ and $N in mathbb{N}^{*}$ and $f$ and $g$ two real, positive and increasing functions.
Thank you.
algorithms recurrence-relations
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
$endgroup$
add a comment |
$begingroup$
Is there an algorithm which I can use to find an approximate solution of the following second-order difference equation in order to run simulations of the solution :
$nu(1 - beta) f(u_{n+1})-g(u_{n+1}-u_{n}) + beta g (u_{n+2} - u_{n+1}) = 0 $
with $nu > 0$ and $0 < beta < 1$ and $u_0 = 0$ and $u_N = C > 0$ and $N in mathbb{N}^{*}$ and $f$ and $g$ two real, positive and increasing functions.
Thank you.
algorithms recurrence-relations
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
$endgroup$
Is there an algorithm which I can use to find an approximate solution of the following second-order difference equation in order to run simulations of the solution :
$nu(1 - beta) f(u_{n+1})-g(u_{n+1}-u_{n}) + beta g (u_{n+2} - u_{n+1}) = 0 $
with $nu > 0$ and $0 < beta < 1$ and $u_0 = 0$ and $u_N = C > 0$ and $N in mathbb{N}^{*}$ and $f$ and $g$ two real, positive and increasing functions.
Thank you.
algorithms recurrence-relations
algorithms recurrence-relations
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
asked Dec 13 '18 at 15:30
AlexC75AlexC75
216
216
add a comment |
add a comment |
1 Answer
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Rewrite as
$$
u_n = u_{n+1} - g^{-1}bigl[(1-beta)nu f(u_{n+1})+beta g(u_{n+2}-u_{n+1})bigr].
$$
If you fix $u_N=C$ (as given in the problem) and also $u_{N-1}$, then you can use the previous recurrence formula to solve backward for $x_{N-2},x_{N-3},dots,x_1,x_0$. Then the problem becomes to find an appropriate value for $x_{N-1}$ so that you end up with $x_0=0$.
This is what is called shooting method for the boundary value problem $x_0=0$, $x_N=C$.
In precise terms, through the above backward solution, you get a map $phi:mathbb Rtomathbb R$ which gives you $x_0=phi(x_{N-1})$. Your goal is now to solve $phi(x)=0$. This can be done in several ways, noting in particular that if $f,g,g^{-1}$ are differentiable, then $phi$ is also differentiable, so you may be able to use Newton's method, beside the
bisection method.
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
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Thank you very much, seems to work perfectly. Great method.
$endgroup$
– AlexC75
Dec 13 '18 at 17:42
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@AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you.
$endgroup$
– Federico
Dec 13 '18 at 17:48
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Rewrite as
$$
u_n = u_{n+1} - g^{-1}bigl[(1-beta)nu f(u_{n+1})+beta g(u_{n+2}-u_{n+1})bigr].
$$
If you fix $u_N=C$ (as given in the problem) and also $u_{N-1}$, then you can use the previous recurrence formula to solve backward for $x_{N-2},x_{N-3},dots,x_1,x_0$. Then the problem becomes to find an appropriate value for $x_{N-1}$ so that you end up with $x_0=0$.
This is what is called shooting method for the boundary value problem $x_0=0$, $x_N=C$.
In precise terms, through the above backward solution, you get a map $phi:mathbb Rtomathbb R$ which gives you $x_0=phi(x_{N-1})$. Your goal is now to solve $phi(x)=0$. This can be done in several ways, noting in particular that if $f,g,g^{-1}$ are differentiable, then $phi$ is also differentiable, so you may be able to use Newton's method, beside the
bisection method.
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
$endgroup$
$begingroup$
Thank you very much, seems to work perfectly. Great method.
$endgroup$
– AlexC75
Dec 13 '18 at 17:42
$begingroup$
@AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you.
$endgroup$
– Federico
Dec 13 '18 at 17:48
add a comment |
$begingroup$
Rewrite as
$$
u_n = u_{n+1} - g^{-1}bigl[(1-beta)nu f(u_{n+1})+beta g(u_{n+2}-u_{n+1})bigr].
$$
If you fix $u_N=C$ (as given in the problem) and also $u_{N-1}$, then you can use the previous recurrence formula to solve backward for $x_{N-2},x_{N-3},dots,x_1,x_0$. Then the problem becomes to find an appropriate value for $x_{N-1}$ so that you end up with $x_0=0$.
This is what is called shooting method for the boundary value problem $x_0=0$, $x_N=C$.
In precise terms, through the above backward solution, you get a map $phi:mathbb Rtomathbb R$ which gives you $x_0=phi(x_{N-1})$. Your goal is now to solve $phi(x)=0$. This can be done in several ways, noting in particular that if $f,g,g^{-1}$ are differentiable, then $phi$ is also differentiable, so you may be able to use Newton's method, beside the
bisection method.
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
$endgroup$
$begingroup$
Thank you very much, seems to work perfectly. Great method.
$endgroup$
– AlexC75
Dec 13 '18 at 17:42
$begingroup$
@AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you.
$endgroup$
– Federico
Dec 13 '18 at 17:48
add a comment |
$begingroup$
Rewrite as
$$
u_n = u_{n+1} - g^{-1}bigl[(1-beta)nu f(u_{n+1})+beta g(u_{n+2}-u_{n+1})bigr].
$$
If you fix $u_N=C$ (as given in the problem) and also $u_{N-1}$, then you can use the previous recurrence formula to solve backward for $x_{N-2},x_{N-3},dots,x_1,x_0$. Then the problem becomes to find an appropriate value for $x_{N-1}$ so that you end up with $x_0=0$.
This is what is called shooting method for the boundary value problem $x_0=0$, $x_N=C$.
In precise terms, through the above backward solution, you get a map $phi:mathbb Rtomathbb R$ which gives you $x_0=phi(x_{N-1})$. Your goal is now to solve $phi(x)=0$. This can be done in several ways, noting in particular that if $f,g,g^{-1}$ are differentiable, then $phi$ is also differentiable, so you may be able to use Newton's method, beside the
bisection method.
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
$endgroup$
Rewrite as
$$
u_n = u_{n+1} - g^{-1}bigl[(1-beta)nu f(u_{n+1})+beta g(u_{n+2}-u_{n+1})bigr].
$$
If you fix $u_N=C$ (as given in the problem) and also $u_{N-1}$, then you can use the previous recurrence formula to solve backward for $x_{N-2},x_{N-3},dots,x_1,x_0$. Then the problem becomes to find an appropriate value for $x_{N-1}$ so that you end up with $x_0=0$.
This is what is called shooting method for the boundary value problem $x_0=0$, $x_N=C$.
In precise terms, through the above backward solution, you get a map $phi:mathbb Rtomathbb R$ which gives you $x_0=phi(x_{N-1})$. Your goal is now to solve $phi(x)=0$. This can be done in several ways, noting in particular that if $f,g,g^{-1}$ are differentiable, then $phi$ is also differentiable, so you may be able to use Newton's method, beside the
bisection method.
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
edited Dec 13 '18 at 15:52
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
answered Dec 13 '18 at 15:47
FedericoFederico
5,124514
5,124514
$begingroup$
Thank you very much, seems to work perfectly. Great method.
$endgroup$
– AlexC75
Dec 13 '18 at 17:42
$begingroup$
@AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you.
$endgroup$
– Federico
Dec 13 '18 at 17:48
add a comment |
$begingroup$
Thank you very much, seems to work perfectly. Great method.
$endgroup$
– AlexC75
Dec 13 '18 at 17:42
$begingroup$
@AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you.
$endgroup$
– Federico
Dec 13 '18 at 17:48
$begingroup$
Thank you very much, seems to work perfectly. Great method.
$endgroup$
– AlexC75
Dec 13 '18 at 17:42
$begingroup$
Thank you very much, seems to work perfectly. Great method.
$endgroup$
– AlexC75
Dec 13 '18 at 17:42
$begingroup$
@AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you.
$endgroup$
– Federico
Dec 13 '18 at 17:48
$begingroup$
@AlexC75 Of course the feasibility of this approach depends on how good the functions $f$ and $g$ are. I'm glad it works fine for you.
$endgroup$
– Federico
Dec 13 '18 at 17:48
add a comment |
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