Mixing
Clarification of assumptions made in deriving error of implicit midpoint rule
In my derivation for $y^prime = f(t,y)$, I begin by writing the method as an expression which should simplify to the error, by substitution of the exact solution
begin{equation} y(t_{n+1}) - y(t_n) - hfleft(t_n+frac{1}{2}h,,frac{1}{2}(y(t_{n+1}) + y(t_n))right)end{equation}
From here, it may be simplified in a straightforward way using Taylor expansions to the following
begin{equation}hy^prime(t_n) + frac{1}{2}h^2y^{primeprime}(t_n) + mathcal{O}(h^3) - hfleft(t_n+frac{1}{2}h,,yleft(t_{n}+frac{1}{2}hright)+ mathcal{O}(h^2)right)end{equation}
This is where I am unsure how to proceed. From this question, the top answer seems to suggest that it is a trivial matter to assume that we may assert $$f(t_n + 0.5h,y(t_n + 0.5h) + mathcal{O}(h^2)) = f(t_n + 0.5h,y(t_n + 0.5h)) + mathcal{O}(h^2) = y^prime(t_n+0.5h) + mathcal{O}(h^2)$$
but this does not seem obvious to me. While it is stipulated that $f$ is an analytic function, I don't see how a perturbation in the input would necessarily perturb the output on the order of $h^2$ in general.
If this is a good assumption to make, could someone please explain? If not, why? What is the correct way to proceed here?
differential-equations numerical-methods approximation-theory
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
add a comment |
In my derivation for $y^prime = f(t,y)$, I begin by writing the method as an expression which should simplify to the error, by substitution of the exact solution
begin{equation} y(t_{n+1}) - y(t_n) - hfleft(t_n+frac{1}{2}h,,frac{1}{2}(y(t_{n+1}) + y(t_n))right)end{equation}
From here, it may be simplified in a straightforward way using Taylor expansions to the following
begin{equation}hy^prime(t_n) + frac{1}{2}h^2y^{primeprime}(t_n) + mathcal{O}(h^3) - hfleft(t_n+frac{1}{2}h,,yleft(t_{n}+frac{1}{2}hright)+ mathcal{O}(h^2)right)end{equation}
This is where I am unsure how to proceed. From this question, the top answer seems to suggest that it is a trivial matter to assume that we may assert $$f(t_n + 0.5h,y(t_n + 0.5h) + mathcal{O}(h^2)) = f(t_n + 0.5h,y(t_n + 0.5h)) + mathcal{O}(h^2) = y^prime(t_n+0.5h) + mathcal{O}(h^2)$$
but this does not seem obvious to me. While it is stipulated that $f$ is an analytic function, I don't see how a perturbation in the input would necessarily perturb the output on the order of $h^2$ in general.
If this is a good assumption to make, could someone please explain? If not, why? What is the correct way to proceed here?
differential-equations numerical-methods approximation-theory
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
add a comment |
In my derivation for $y^prime = f(t,y)$, I begin by writing the method as an expression which should simplify to the error, by substitution of the exact solution
begin{equation} y(t_{n+1}) - y(t_n) - hfleft(t_n+frac{1}{2}h,,frac{1}{2}(y(t_{n+1}) + y(t_n))right)end{equation}
From here, it may be simplified in a straightforward way using Taylor expansions to the following
begin{equation}hy^prime(t_n) + frac{1}{2}h^2y^{primeprime}(t_n) + mathcal{O}(h^3) - hfleft(t_n+frac{1}{2}h,,yleft(t_{n}+frac{1}{2}hright)+ mathcal{O}(h^2)right)end{equation}
This is where I am unsure how to proceed. From this question, the top answer seems to suggest that it is a trivial matter to assume that we may assert $$f(t_n + 0.5h,y(t_n + 0.5h) + mathcal{O}(h^2)) = f(t_n + 0.5h,y(t_n + 0.5h)) + mathcal{O}(h^2) = y^prime(t_n+0.5h) + mathcal{O}(h^2)$$
but this does not seem obvious to me. While it is stipulated that $f$ is an analytic function, I don't see how a perturbation in the input would necessarily perturb the output on the order of $h^2$ in general.
If this is a good assumption to make, could someone please explain? If not, why? What is the correct way to proceed here?
differential-equations numerical-methods approximation-theory
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
In my derivation for $y^prime = f(t,y)$, I begin by writing the method as an expression which should simplify to the error, by substitution of the exact solution
begin{equation} y(t_{n+1}) - y(t_n) - hfleft(t_n+frac{1}{2}h,,frac{1}{2}(y(t_{n+1}) + y(t_n))right)end{equation}
From here, it may be simplified in a straightforward way using Taylor expansions to the following
begin{equation}hy^prime(t_n) + frac{1}{2}h^2y^{primeprime}(t_n) + mathcal{O}(h^3) - hfleft(t_n+frac{1}{2}h,,yleft(t_{n}+frac{1}{2}hright)+ mathcal{O}(h^2)right)end{equation}
This is where I am unsure how to proceed. From this question, the top answer seems to suggest that it is a trivial matter to assume that we may assert $$f(t_n + 0.5h,y(t_n + 0.5h) + mathcal{O}(h^2)) = f(t_n + 0.5h,y(t_n + 0.5h)) + mathcal{O}(h^2) = y^prime(t_n+0.5h) + mathcal{O}(h^2)$$
but this does not seem obvious to me. While it is stipulated that $f$ is an analytic function, I don't see how a perturbation in the input would necessarily perturb the output on the order of $h^2$ in general.
If this is a good assumption to make, could someone please explain? If not, why? What is the correct way to proceed here?
differential-equations numerical-methods approximation-theory
differential-equations numerical-methods approximation-theory
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
edited Dec 31 '18 at 22:52
Kyle Poe
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
asked Dec 31 '18 at 22:13
Kyle PoeKyle Poe
83
83
add a comment |
add a comment |
1 Answer
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In general you assume the Lipschitz condition, from there you get $|f(t,y+O(h^2))-f(t,y)|=O(h^2)$.
For an order $p$ method to get the order $p$ local error you also need in general that $f$ is $p$ times continuously differentiable in all variables. Or at least $p-1$ times differentiable, with the highest derivative Lipschitz continuous.
You should probably also use the central difference quotient for the first terms,
$$
y(t+h)-y(h)=hy'(t+0.5h)+O(h^3),
$$
then there are no second order terms in the formula that have to be compensated for.
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
Thank you. This makes sense, I wasn't thinking about an argument based on magnitude of the Lipschitz constant.
– Kyle Poe
Dec 31 '18 at 23:00
add a comment |
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1 Answer
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1 Answer
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In general you assume the Lipschitz condition, from there you get $|f(t,y+O(h^2))-f(t,y)|=O(h^2)$.
For an order $p$ method to get the order $p$ local error you also need in general that $f$ is $p$ times continuously differentiable in all variables. Or at least $p-1$ times differentiable, with the highest derivative Lipschitz continuous.
You should probably also use the central difference quotient for the first terms,
$$
y(t+h)-y(h)=hy'(t+0.5h)+O(h^3),
$$
then there are no second order terms in the formula that have to be compensated for.
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
Thank you. This makes sense, I wasn't thinking about an argument based on magnitude of the Lipschitz constant.
– Kyle Poe
Dec 31 '18 at 23:00
add a comment |
In general you assume the Lipschitz condition, from there you get $|f(t,y+O(h^2))-f(t,y)|=O(h^2)$.
For an order $p$ method to get the order $p$ local error you also need in general that $f$ is $p$ times continuously differentiable in all variables. Or at least $p-1$ times differentiable, with the highest derivative Lipschitz continuous.
You should probably also use the central difference quotient for the first terms,
$$
y(t+h)-y(h)=hy'(t+0.5h)+O(h^3),
$$
then there are no second order terms in the formula that have to be compensated for.
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
Thank you. This makes sense, I wasn't thinking about an argument based on magnitude of the Lipschitz constant.
– Kyle Poe
Dec 31 '18 at 23:00
add a comment |
In general you assume the Lipschitz condition, from there you get $|f(t,y+O(h^2))-f(t,y)|=O(h^2)$.
For an order $p$ method to get the order $p$ local error you also need in general that $f$ is $p$ times continuously differentiable in all variables. Or at least $p-1$ times differentiable, with the highest derivative Lipschitz continuous.
You should probably also use the central difference quotient for the first terms,
$$
y(t+h)-y(h)=hy'(t+0.5h)+O(h^3),
$$
then there are no second order terms in the formula that have to be compensated for.
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
In general you assume the Lipschitz condition, from there you get $|f(t,y+O(h^2))-f(t,y)|=O(h^2)$.
For an order $p$ method to get the order $p$ local error you also need in general that $f$ is $p$ times continuously differentiable in all variables. Or at least $p-1$ times differentiable, with the highest derivative Lipschitz continuous.
You should probably also use the central difference quotient for the first terms,
$$
y(t+h)-y(h)=hy'(t+0.5h)+O(h^3),
$$
then there are no second order terms in the formula that have to be compensated for.
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
edited Dec 31 '18 at 23:20
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
answered Dec 31 '18 at 22:54
LutzLLutzL
56.6k42054
56.6k42054
Thank you. This makes sense, I wasn't thinking about an argument based on magnitude of the Lipschitz constant.
– Kyle Poe
Dec 31 '18 at 23:00
add a comment |
Thank you. This makes sense, I wasn't thinking about an argument based on magnitude of the Lipschitz constant.
– Kyle Poe
Dec 31 '18 at 23:00
Thank you. This makes sense, I wasn't thinking about an argument based on magnitude of the Lipschitz constant.
– Kyle Poe
Dec 31 '18 at 23:00
Thank you. This makes sense, I wasn't thinking about an argument based on magnitude of the Lipschitz constant.
– Kyle Poe
Dec 31 '18 at 23:00
add a comment |
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