Forward Euler Method: how to derive global error
$begingroup$
I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.
I'm given a system such that:
$$frac{du}{dt} = v ~ ~ & ~ ~ frac{dv}{dt} = -f(u)$$
with Hamiltonian $$H = frac{1}{2} left(frac{du}{dt}right)^2 + int f du.$$
I have to show that using the forward Euler method leads to a global error for $H$ that grows like $nh^2$ for step size $h$ and number of steps $n$.
I know that the global error can be calculated via $ epsilon = |U^n - U(T)|$ but I'm not sure how to apply it in this case.
Thanks for any help!
EDIT: So if I understand correct, I have to calculate $|H_{n+1}-H_n| $
numerical-methods
edited Dec 14 '18 at 12:23
LutzL
58k42054
asked Dec 14 '18 at 1:03
user3424575user3424575
124
$endgroup$
add a comment |
$begingroup$
I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.
I'm given a system such that:
$$frac{du}{dt} = v ~ ~ & ~ ~ frac{dv}{dt} = -f(u)$$
with Hamiltonian $$H = frac{1}{2} left(frac{du}{dt}right)^2 + int f du.$$
I have to show that using the forward Euler method leads to a global error for $H$ that grows like $nh^2$ for step size $h$ and number of steps $n$.
I know that the global error can be calculated via $ epsilon = |U^n - U(T)|$ but I'm not sure how to apply it in this case.
Thanks for any help!
EDIT: So if I understand correct, I have to calculate $|H_{n+1}-H_n| $
numerical-methods
edited Dec 14 '18 at 12:23
LutzL
58k42054
asked Dec 14 '18 at 1:03
user3424575user3424575
124
$endgroup$
add a comment |
$begingroup$
I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.
I'm given a system such that:
$$frac{du}{dt} = v ~ ~ & ~ ~ frac{dv}{dt} = -f(u)$$
with Hamiltonian $$H = frac{1}{2} left(frac{du}{dt}right)^2 + int f du.$$
I have to show that using the forward Euler method leads to a global error for $H$ that grows like $nh^2$ for step size $h$ and number of steps $n$.
I know that the global error can be calculated via $ epsilon = |U^n - U(T)|$ but I'm not sure how to apply it in this case.
Thanks for any help!
EDIT: So if I understand correct, I have to calculate $|H_{n+1}-H_n| $
numerical-methods
edited Dec 14 '18 at 12:23
LutzL
58k42054
asked Dec 14 '18 at 1:03
user3424575user3424575
124
$endgroup$
I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.
I'm given a system such that:
$$frac{du}{dt} = v ~ ~ & ~ ~ frac{dv}{dt} = -f(u)$$
with Hamiltonian $$H = frac{1}{2} left(frac{du}{dt}right)^2 + int f du.$$
I have to show that using the forward Euler method leads to a global error for $H$ that grows like $nh^2$ for step size $h$ and number of steps $n$.
I know that the global error can be calculated via $ epsilon = |U^n - U(T)|$ but I'm not sure how to apply it in this case.
Thanks for any help!
EDIT: So if I understand correct, I have to calculate $|H_{n+1}-H_n| $
numerical-methods
numerical-methods
edited Dec 14 '18 at 12:23
LutzL
58k42054
asked Dec 14 '18 at 1:03
user3424575user3424575
124
edited Dec 14 '18 at 12:23
LutzL
58k42054
asked Dec 14 '18 at 1:03
user3424575user3424575
124
edited Dec 14 '18 at 12:23
LutzL
58k42054
edited Dec 14 '18 at 12:23
LutzL
58k42054
edited Dec 14 '18 at 12:23
LutzL
58k42054
58k42054
asked Dec 14 '18 at 1:03
user3424575user3424575
124
asked Dec 14 '18 at 1:03
user3424575user3424575
124
asked Dec 14 '18 at 1:03
user3424575user3424575
124
124
add a comment |
add a comment |
1 Answer
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If you take the example $f(u)=u$, then you can interpret the Hamilton system in the complex plane as $$dot z=dot u+idot v = -i(u+iv)=-iz.$$
The Euler forward iteration thus produces elements
$$
z_{k+1}=z_k+h(-iz_k)=(1-ih)z_kimplies z_n=(1-ih)^nz_0.
$$
The Hamilton function is $H=frac12(v^2+u^2)=frac12|z|^2$, which gives on the Euler solution
$$
H_n=frac12|z_n|^2=frac12(1+h^2)^n|z_0|^2=e^{nh^2+O(nh^4)}H_0
$$
so that indeed $H_n-H_0=nh^2e^{frac12nh^2+O(nh^4)}H_0$.
Under more general conditions one gets
begin{align}
H_{k+1}&=frac12(v_k-hf(u_k))^2+F(u_k+hv_k) \
&=H_k-hv_kf(u_k)+frac12h^2f(u_k)^2 ~ + ~ f(u_k)(hv_k)+frac12f'(u_k)(hv_k)^2+O(h^3) \
&=H_k+frac12h^2bigl[f(u_k)^2+f'(u_k)v_k^2bigr]+O(h^3)
end{align}
Now you have to argue that the coefficient of the $h^2$ term remains bounded and the claim of the task follows.
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you take the example $f(u)=u$, then you can interpret the Hamilton system in the complex plane as $$dot z=dot u+idot v = -i(u+iv)=-iz.$$
The Euler forward iteration thus produces elements
$$
z_{k+1}=z_k+h(-iz_k)=(1-ih)z_kimplies z_n=(1-ih)^nz_0.
$$
The Hamilton function is $H=frac12(v^2+u^2)=frac12|z|^2$, which gives on the Euler solution
$$
H_n=frac12|z_n|^2=frac12(1+h^2)^n|z_0|^2=e^{nh^2+O(nh^4)}H_0
$$
so that indeed $H_n-H_0=nh^2e^{frac12nh^2+O(nh^4)}H_0$.
Under more general conditions one gets
begin{align}
H_{k+1}&=frac12(v_k-hf(u_k))^2+F(u_k+hv_k) \
&=H_k-hv_kf(u_k)+frac12h^2f(u_k)^2 ~ + ~ f(u_k)(hv_k)+frac12f'(u_k)(hv_k)^2+O(h^3) \
&=H_k+frac12h^2bigl[f(u_k)^2+f'(u_k)v_k^2bigr]+O(h^3)
end{align}
Now you have to argue that the coefficient of the $h^2$ term remains bounded and the claim of the task follows.
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
$endgroup$
add a comment |
$begingroup$
If you take the example $f(u)=u$, then you can interpret the Hamilton system in the complex plane as $$dot z=dot u+idot v = -i(u+iv)=-iz.$$
The Euler forward iteration thus produces elements
$$
z_{k+1}=z_k+h(-iz_k)=(1-ih)z_kimplies z_n=(1-ih)^nz_0.
$$
The Hamilton function is $H=frac12(v^2+u^2)=frac12|z|^2$, which gives on the Euler solution
$$
H_n=frac12|z_n|^2=frac12(1+h^2)^n|z_0|^2=e^{nh^2+O(nh^4)}H_0
$$
so that indeed $H_n-H_0=nh^2e^{frac12nh^2+O(nh^4)}H_0$.
Under more general conditions one gets
begin{align}
H_{k+1}&=frac12(v_k-hf(u_k))^2+F(u_k+hv_k) \
&=H_k-hv_kf(u_k)+frac12h^2f(u_k)^2 ~ + ~ f(u_k)(hv_k)+frac12f'(u_k)(hv_k)^2+O(h^3) \
&=H_k+frac12h^2bigl[f(u_k)^2+f'(u_k)v_k^2bigr]+O(h^3)
end{align}
Now you have to argue that the coefficient of the $h^2$ term remains bounded and the claim of the task follows.
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
$endgroup$
add a comment |
$begingroup$
If you take the example $f(u)=u$, then you can interpret the Hamilton system in the complex plane as $$dot z=dot u+idot v = -i(u+iv)=-iz.$$
The Euler forward iteration thus produces elements
$$
z_{k+1}=z_k+h(-iz_k)=(1-ih)z_kimplies z_n=(1-ih)^nz_0.
$$
The Hamilton function is $H=frac12(v^2+u^2)=frac12|z|^2$, which gives on the Euler solution
$$
H_n=frac12|z_n|^2=frac12(1+h^2)^n|z_0|^2=e^{nh^2+O(nh^4)}H_0
$$
so that indeed $H_n-H_0=nh^2e^{frac12nh^2+O(nh^4)}H_0$.
Under more general conditions one gets
begin{align}
H_{k+1}&=frac12(v_k-hf(u_k))^2+F(u_k+hv_k) \
&=H_k-hv_kf(u_k)+frac12h^2f(u_k)^2 ~ + ~ f(u_k)(hv_k)+frac12f'(u_k)(hv_k)^2+O(h^3) \
&=H_k+frac12h^2bigl[f(u_k)^2+f'(u_k)v_k^2bigr]+O(h^3)
end{align}
Now you have to argue that the coefficient of the $h^2$ term remains bounded and the claim of the task follows.
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
$endgroup$
If you take the example $f(u)=u$, then you can interpret the Hamilton system in the complex plane as $$dot z=dot u+idot v = -i(u+iv)=-iz.$$
The Euler forward iteration thus produces elements
$$
z_{k+1}=z_k+h(-iz_k)=(1-ih)z_kimplies z_n=(1-ih)^nz_0.
$$
The Hamilton function is $H=frac12(v^2+u^2)=frac12|z|^2$, which gives on the Euler solution
$$
H_n=frac12|z_n|^2=frac12(1+h^2)^n|z_0|^2=e^{nh^2+O(nh^4)}H_0
$$
so that indeed $H_n-H_0=nh^2e^{frac12nh^2+O(nh^4)}H_0$.
Under more general conditions one gets
begin{align}
H_{k+1}&=frac12(v_k-hf(u_k))^2+F(u_k+hv_k) \
&=H_k-hv_kf(u_k)+frac12h^2f(u_k)^2 ~ + ~ f(u_k)(hv_k)+frac12f'(u_k)(hv_k)^2+O(h^3) \
&=H_k+frac12h^2bigl[f(u_k)^2+f'(u_k)v_k^2bigr]+O(h^3)
end{align}
Now you have to argue that the coefficient of the $h^2$ term remains bounded and the claim of the task follows.
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
answered Dec 14 '18 at 12:20
LutzLLutzL
58k42054
58k42054
add a comment |
add a comment |
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