Mixing
Truncation Error of a Predictor Corrector Method
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In a paper I am reading. A predictor corrector scheme is developed to solve equations of the form $$y^{prime}=y(y-1)g(y)$$ where $g(y)$ is a bounded and continuously differentiable function.
The scheme is given by:
$$y_{n+1}^{p}={y_{n}^c over y_{n}^c+(1-y_{n}^c)exp{(hg(y_{n}^c))}}$$
$$chi_{n+1}=chi_{n}exp{({h over 2}(g(y_{n}^c)+g(y_{n+1}^p)))}$$
$$y_{n+1}^c={1 over 1-chi_{n+1}}$$
where $y_{n}^c$ and $y_{n}^p$ are the corrector and predictor approximations.
I want to derive the truncation error for the predictor. $$y(x_n+h)-{y(x_n)over y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}}$$ I tried Taylor expansion of $y(x_n+h)$ then multiplied by $y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}$. How can I proceed after that, any hints?
differential-equations numerical-methods
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In a paper I am reading. A predictor corrector scheme is developed to solve equations of the form $$y^{prime}=y(y-1)g(y)$$ where $g(y)$ is a bounded and continuously differentiable function.
The scheme is given by:
$$y_{n+1}^{p}={y_{n}^c over y_{n}^c+(1-y_{n}^c)exp{(hg(y_{n}^c))}}$$
$$chi_{n+1}=chi_{n}exp{({h over 2}(g(y_{n}^c)+g(y_{n+1}^p)))}$$
$$y_{n+1}^c={1 over 1-chi_{n+1}}$$
where $y_{n}^c$ and $y_{n}^p$ are the corrector and predictor approximations.
I want to derive the truncation error for the predictor. $$y(x_n+h)-{y(x_n)over y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}}$$ I tried Taylor expansion of $y(x_n+h)$ then multiplied by $y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}$. How can I proceed after that, any hints?
differential-equations numerical-methods
asked 9 hours ago
Tres nom
1
New contributor
Tres nom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In a paper I am reading. A predictor corrector scheme is developed to solve equations of the form $$y^{prime}=y(y-1)g(y)$$ where $g(y)$ is a bounded and continuously differentiable function.
The scheme is given by:
$$y_{n+1}^{p}={y_{n}^c over y_{n}^c+(1-y_{n}^c)exp{(hg(y_{n}^c))}}$$
$$chi_{n+1}=chi_{n}exp{({h over 2}(g(y_{n}^c)+g(y_{n+1}^p)))}$$
$$y_{n+1}^c={1 over 1-chi_{n+1}}$$
where $y_{n}^c$ and $y_{n}^p$ are the corrector and predictor approximations.
I want to derive the truncation error for the predictor. $$y(x_n+h)-{y(x_n)over y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}}$$ I tried Taylor expansion of $y(x_n+h)$ then multiplied by $y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}$. How can I proceed after that, any hints?
differential-equations numerical-methods
asked 9 hours ago
Tres nom
1
New contributor
Tres nom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In a paper I am reading. A predictor corrector scheme is developed to solve equations of the form $$y^{prime}=y(y-1)g(y)$$ where $g(y)$ is a bounded and continuously differentiable function.
The scheme is given by:
$$y_{n+1}^{p}={y_{n}^c over y_{n}^c+(1-y_{n}^c)exp{(hg(y_{n}^c))}}$$
$$chi_{n+1}=chi_{n}exp{({h over 2}(g(y_{n}^c)+g(y_{n+1}^p)))}$$
$$y_{n+1}^c={1 over 1-chi_{n+1}}$$
where $y_{n}^c$ and $y_{n}^p$ are the corrector and predictor approximations.
I want to derive the truncation error for the predictor. $$y(x_n+h)-{y(x_n)over y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}}$$ I tried Taylor expansion of $y(x_n+h)$ then multiplied by $y(x_n)+(1-y(x_n))exp{(hg(y(x_n)))}$. How can I proceed after that, any hints?
differential-equations numerical-methods
differential-equations numerical-methods
asked 9 hours ago
Tres nom
1
New contributor
Tres nom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Tres nom
1
New contributor
Tres nom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 9 hours ago
Tres nom
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Tres nom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Tres nom
1
asked 9 hours ago
Tres nom
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1
New contributor
Tres nom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tres nom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
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