Solve general 2nd order ODE numerically with 2nd order time-differences











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I want to solve a 2nd order ODE of the following general form



$ddot{x} = f(dot{x}, x)$



where dot indicates a time-derivative. A simple numerical solution that is first-order accurate in time would be



$
begin{eqnarray}
x(t + Delta t) &=& x(t) + v(t)Delta t \
v(t + Delta t) &=& v(t) + f(v(t),x(t))Delta t \
end{eqnarray}
$



However, the accuracy of this solution is unsatisfactory. I would like a solution that is 2nd order accurate in time. Naively, I apply 2nd order central differences for 1st and 2nd order derivatives



$frac{x(t+Delta t) - 2 x(t) + x(t-Delta t)}{Delta t^2} = fbiggl(frac{x(t+Delta t) - x(t-Delta t)}{2Delta t}, x(t)biggr)$



In order to numerically solve this equation, I need to express future as a function of the past, that is, solve the above equation for $x(t+Delta t)$. However, for a general $f$ there might not be a way to solve the above equation explicitly.



Question: Is there a way to solve the above ODE numerically to 2nd order precision, if the only thing we are allowed to do with the RHS is finding its value for given inputs.



Note: For the initial conditions, $x$ and $v$ are known at $t=0$. If the proposed method requires more than 2 steps of memory, please indicate how to correctly initialize it.






differential-equations numerical-methods finite-differences







share|cite|improve this question




















  • 1




    For the second order formula for first derivative you have $$ dot{x} approx frac{x(t+Delta t)-x(t-Delta t)}{2Delta t} $$
    – rafa11111
    Nov 26 at 12:03










  • Thanks, my bad, will fix now
    – Aleksejs Fomins
    Nov 26 at 12:05






  • 1




    You're welcome. AFAIK, there are specific methods for solving movement equations, such as the Verlet integration or the Leapfrog integration. I'm not familiar with those methods, but I think you can have a good start looking at this, rather than trying to derive a scheme from scratch.
    – rafa11111
    Nov 26 at 12:11






  • 1




    Thanks a lot, I'll have a look. I used to know these things, but it was ages ago...
    – Aleksejs Fomins
    Nov 26 at 12:18






  • 1




    @rafa11111: Verlet integration is exactly this scheme for situations where $f$ does not depend on the first derivative $dot x$, in other words, where the force is conservative resp. the system Hamiltonian. The special properties of symplectic integration methods depend on this symplectic framework for the ODE system.
    – LutzL
    Nov 26 at 12:25

















up vote
1
down vote

favorite












I want to solve a 2nd order ODE of the following general form



$ddot{x} = f(dot{x}, x)$



where dot indicates a time-derivative. A simple numerical solution that is first-order accurate in time would be



$
begin{eqnarray}
x(t + Delta t) &=& x(t) + v(t)Delta t \
v(t + Delta t) &=& v(t) + f(v(t),x(t))Delta t \
end{eqnarray}
$



However, the accuracy of this solution is unsatisfactory. I would like a solution that is 2nd order accurate in time. Naively, I apply 2nd order central differences for 1st and 2nd order derivatives



$frac{x(t+Delta t) - 2 x(t) + x(t-Delta t)}{Delta t^2} = fbiggl(frac{x(t+Delta t) - x(t-Delta t)}{2Delta t}, x(t)biggr)$



In order to numerically solve this equation, I need to express future as a function of the past, that is, solve the above equation for $x(t+Delta t)$. However, for a general $f$ there might not be a way to solve the above equation explicitly.



Question: Is there a way to solve the above ODE numerically to 2nd order precision, if the only thing we are allowed to do with the RHS is finding its value for given inputs.



Note: For the initial conditions, $x$ and $v$ are known at $t=0$. If the proposed method requires more than 2 steps of memory, please indicate how to correctly initialize it.






differential-equations numerical-methods finite-differences







share|cite|improve this question




















  • 1




    For the second order formula for first derivative you have $$ dot{x} approx frac{x(t+Delta t)-x(t-Delta t)}{2Delta t} $$
    – rafa11111
    Nov 26 at 12:03










  • Thanks, my bad, will fix now
    – Aleksejs Fomins
    Nov 26 at 12:05






  • 1




    You're welcome. AFAIK, there are specific methods for solving movement equations, such as the Verlet integration or the Leapfrog integration. I'm not familiar with those methods, but I think you can have a good start looking at this, rather than trying to derive a scheme from scratch.
    – rafa11111
    Nov 26 at 12:11






  • 1




    Thanks a lot, I'll have a look. I used to know these things, but it was ages ago...
    – Aleksejs Fomins
    Nov 26 at 12:18






  • 1




    @rafa11111: Verlet integration is exactly this scheme for situations where $f$ does not depend on the first derivative $dot x$, in other words, where the force is conservative resp. the system Hamiltonian. The special properties of symplectic integration methods depend on this symplectic framework for the ODE system.
    – LutzL
    Nov 26 at 12:25















up vote
1
down vote

favorite









up vote
1
down vote

favorite











I want to solve a 2nd order ODE of the following general form



$ddot{x} = f(dot{x}, x)$



where dot indicates a time-derivative. A simple numerical solution that is first-order accurate in time would be



$
begin{eqnarray}
x(t + Delta t) &=& x(t) + v(t)Delta t \
v(t + Delta t) &=& v(t) + f(v(t),x(t))Delta t \
end{eqnarray}
$



However, the accuracy of this solution is unsatisfactory. I would like a solution that is 2nd order accurate in time. Naively, I apply 2nd order central differences for 1st and 2nd order derivatives



$frac{x(t+Delta t) - 2 x(t) + x(t-Delta t)}{Delta t^2} = fbiggl(frac{x(t+Delta t) - x(t-Delta t)}{2Delta t}, x(t)biggr)$



In order to numerically solve this equation, I need to express future as a function of the past, that is, solve the above equation for $x(t+Delta t)$. However, for a general $f$ there might not be a way to solve the above equation explicitly.



Question: Is there a way to solve the above ODE numerically to 2nd order precision, if the only thing we are allowed to do with the RHS is finding its value for given inputs.



Note: For the initial conditions, $x$ and $v$ are known at $t=0$. If the proposed method requires more than 2 steps of memory, please indicate how to correctly initialize it.






differential-equations numerical-methods finite-differences







share|cite|improve this question















I want to solve a 2nd order ODE of the following general form



$ddot{x} = f(dot{x}, x)$



where dot indicates a time-derivative. A simple numerical solution that is first-order accurate in time would be



$
begin{eqnarray}
x(t + Delta t) &=& x(t) + v(t)Delta t \
v(t + Delta t) &=& v(t) + f(v(t),x(t))Delta t \
end{eqnarray}
$



However, the accuracy of this solution is unsatisfactory. I would like a solution that is 2nd order accurate in time. Naively, I apply 2nd order central differences for 1st and 2nd order derivatives



$frac{x(t+Delta t) - 2 x(t) + x(t-Delta t)}{Delta t^2} = fbiggl(frac{x(t+Delta t) - x(t-Delta t)}{2Delta t}, x(t)biggr)$



In order to numerically solve this equation, I need to express future as a function of the past, that is, solve the above equation for $x(t+Delta t)$. However, for a general $f$ there might not be a way to solve the above equation explicitly.



Question: Is there a way to solve the above ODE numerically to 2nd order precision, if the only thing we are allowed to do with the RHS is finding its value for given inputs.



Note: For the initial conditions, $x$ and $v$ are known at $t=0$. If the proposed method requires more than 2 steps of memory, please indicate how to correctly initialize it.






differential-equations numerical-methods finite-differences




differential-equations numerical-methods finite-differences






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 12:05

























asked Nov 26 at 11:22









Aleksejs Fomins

435211




435211








  • 1




    For the second order formula for first derivative you have $$ dot{x} approx frac{x(t+Delta t)-x(t-Delta t)}{2Delta t} $$
    – rafa11111
    Nov 26 at 12:03










  • Thanks, my bad, will fix now
    – Aleksejs Fomins
    Nov 26 at 12:05






  • 1




    You're welcome. AFAIK, there are specific methods for solving movement equations, such as the Verlet integration or the Leapfrog integration. I'm not familiar with those methods, but I think you can have a good start looking at this, rather than trying to derive a scheme from scratch.
    – rafa11111
    Nov 26 at 12:11






  • 1




    Thanks a lot, I'll have a look. I used to know these things, but it was ages ago...
    – Aleksejs Fomins
    Nov 26 at 12:18






  • 1




    @rafa11111: Verlet integration is exactly this scheme for situations where $f$ does not depend on the first derivative $dot x$, in other words, where the force is conservative resp. the system Hamiltonian. The special properties of symplectic integration methods depend on this symplectic framework for the ODE system.
    – LutzL
    Nov 26 at 12:25
















  • 1




    For the second order formula for first derivative you have $$ dot{x} approx frac{x(t+Delta t)-x(t-Delta t)}{2Delta t} $$
    – rafa11111
    Nov 26 at 12:03










  • Thanks, my bad, will fix now
    – Aleksejs Fomins
    Nov 26 at 12:05






  • 1




    You're welcome. AFAIK, there are specific methods for solving movement equations, such as the Verlet integration or the Leapfrog integration. I'm not familiar with those methods, but I think you can have a good start looking at this, rather than trying to derive a scheme from scratch.
    – rafa11111
    Nov 26 at 12:11






  • 1




    Thanks a lot, I'll have a look. I used to know these things, but it was ages ago...
    – Aleksejs Fomins
    Nov 26 at 12:18






  • 1




    @rafa11111: Verlet integration is exactly this scheme for situations where $f$ does not depend on the first derivative $dot x$, in other words, where the force is conservative resp. the system Hamiltonian. The special properties of symplectic integration methods depend on this symplectic framework for the ODE system.
    – LutzL
    Nov 26 at 12:25










1




1




For the second order formula for first derivative you have $$ dot{x} approx frac{x(t+Delta t)-x(t-Delta t)}{2Delta t} $$
– rafa11111
Nov 26 at 12:03




For the second order formula for first derivative you have $$ dot{x} approx frac{x(t+Delta t)-x(t-Delta t)}{2Delta t} $$
– rafa11111
Nov 26 at 12:03












Thanks, my bad, will fix now
– Aleksejs Fomins
Nov 26 at 12:05




Thanks, my bad, will fix now
– Aleksejs Fomins
Nov 26 at 12:05




1




1




You're welcome. AFAIK, there are specific methods for solving movement equations, such as the Verlet integration or the Leapfrog integration. I'm not familiar with those methods, but I think you can have a good start looking at this, rather than trying to derive a scheme from scratch.
– rafa11111
Nov 26 at 12:11




You're welcome. AFAIK, there are specific methods for solving movement equations, such as the Verlet integration or the Leapfrog integration. I'm not familiar with those methods, but I think you can have a good start looking at this, rather than trying to derive a scheme from scratch.
– rafa11111
Nov 26 at 12:11




1




1




Thanks a lot, I'll have a look. I used to know these things, but it was ages ago...
– Aleksejs Fomins
Nov 26 at 12:18




Thanks a lot, I'll have a look. I used to know these things, but it was ages ago...
– Aleksejs Fomins
Nov 26 at 12:18




1




1




@rafa11111: Verlet integration is exactly this scheme for situations where $f$ does not depend on the first derivative $dot x$, in other words, where the force is conservative resp. the system Hamiltonian. The special properties of symplectic integration methods depend on this symplectic framework for the ODE system.
– LutzL
Nov 26 at 12:25






@rafa11111: Verlet integration is exactly this scheme for situations where $f$ does not depend on the first derivative $dot x$, in other words, where the force is conservative resp. the system Hamiltonian. The special properties of symplectic integration methods depend on this symplectic framework for the ODE system.
– LutzL
Nov 26 at 12:25












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










You could just iterate
$$
x^{[m+1]}(t+Δt)=2x(t)-x(t−Δt)+Δt^2fleft(frac{x^{[m]}(t+Δt)−x(t−Δt)}{2Δt},x(t)right)
$$
starting with a simple extrapolation $x^{[0]}(t+Δt)=2x(t)-x(t−Δt)$. Usually with the first or second iteration you should have reached sufficient accuracy.



For any twice differentiable functions you get $x(t+Δt)−2x(t)+x(t−Δt)=O(Δt^2)$, so that the error of the iteration is $O(Δt^2(LΔt/2)^k)$ for $k$ iterations, where $L$ is a Lipschitz constant of $f$ relative to its first argument. As the expected truncation error of the method is $O(Δt^4)$, $k=2$ iterations should be sufficient, all further iterations just reduce and regularize the error coefficient.



You get convergence to the solution closest to the initial estimate by some contraction argument like the Banach fixed-point theorem as long as for the combined Lipschitz constant of the iteration $LΔt/2<1$.




For faster or more robust convergence, you would need to implement a Newton-like method.






share|cite|improve this answer























  • I would appreciate an elaboration of the statement that the initial extrapolation is not significantly damaging to the accuracy of the method (proof or citation). Also, a link to a more robust newton-like method would be appreciated.
    – Aleksejs Fomins
    Nov 26 at 11:42










  • Added some semi-quantitative estimates.
    – LutzL
    Nov 26 at 11:55










  • Ok, I think I get the convergence rate. One last thing I am worried about is whether it converges to the right thing. In principle, I can imagine that a nonlinear equation can have multiple solutions for $x(t+1)$. Is there an intuition why it should not converge to some spurious root of the equation?
    – Aleksejs Fomins
    Nov 26 at 12:03










  • Also, ignore my question about what a newton method is. I was being stupid. It is just a better way of finding root of an equation
    – Aleksejs Fomins
    Nov 26 at 12:04










  • I corrected an error, the Lipschitz constant of the iteration is $Δt^2cdot Lcdot 1/(2Δt)=LΔt/2$, thus 2 iterations are necessary to reach error order 4, 3 or more give better results (with rapidly diminishing improvements).
    – LutzL
    Nov 26 at 12:15











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










You could just iterate
$$
x^{[m+1]}(t+Δt)=2x(t)-x(t−Δt)+Δt^2fleft(frac{x^{[m]}(t+Δt)−x(t−Δt)}{2Δt},x(t)right)
$$
starting with a simple extrapolation $x^{[0]}(t+Δt)=2x(t)-x(t−Δt)$. Usually with the first or second iteration you should have reached sufficient accuracy.



For any twice differentiable functions you get $x(t+Δt)−2x(t)+x(t−Δt)=O(Δt^2)$, so that the error of the iteration is $O(Δt^2(LΔt/2)^k)$ for $k$ iterations, where $L$ is a Lipschitz constant of $f$ relative to its first argument. As the expected truncation error of the method is $O(Δt^4)$, $k=2$ iterations should be sufficient, all further iterations just reduce and regularize the error coefficient.



You get convergence to the solution closest to the initial estimate by some contraction argument like the Banach fixed-point theorem as long as for the combined Lipschitz constant of the iteration $LΔt/2<1$.




For faster or more robust convergence, you would need to implement a Newton-like method.






share|cite|improve this answer























  • I would appreciate an elaboration of the statement that the initial extrapolation is not significantly damaging to the accuracy of the method (proof or citation). Also, a link to a more robust newton-like method would be appreciated.
    – Aleksejs Fomins
    Nov 26 at 11:42










  • Added some semi-quantitative estimates.
    – LutzL
    Nov 26 at 11:55










  • Ok, I think I get the convergence rate. One last thing I am worried about is whether it converges to the right thing. In principle, I can imagine that a nonlinear equation can have multiple solutions for $x(t+1)$. Is there an intuition why it should not converge to some spurious root of the equation?
    – Aleksejs Fomins
    Nov 26 at 12:03










  • Also, ignore my question about what a newton method is. I was being stupid. It is just a better way of finding root of an equation
    – Aleksejs Fomins
    Nov 26 at 12:04










  • I corrected an error, the Lipschitz constant of the iteration is $Δt^2cdot Lcdot 1/(2Δt)=LΔt/2$, thus 2 iterations are necessary to reach error order 4, 3 or more give better results (with rapidly diminishing improvements).
    – LutzL
    Nov 26 at 12:15















up vote
0
down vote



accepted










You could just iterate
$$
x^{[m+1]}(t+Δt)=2x(t)-x(t−Δt)+Δt^2fleft(frac{x^{[m]}(t+Δt)−x(t−Δt)}{2Δt},x(t)right)
$$
starting with a simple extrapolation $x^{[0]}(t+Δt)=2x(t)-x(t−Δt)$. Usually with the first or second iteration you should have reached sufficient accuracy.



For any twice differentiable functions you get $x(t+Δt)−2x(t)+x(t−Δt)=O(Δt^2)$, so that the error of the iteration is $O(Δt^2(LΔt/2)^k)$ for $k$ iterations, where $L$ is a Lipschitz constant of $f$ relative to its first argument. As the expected truncation error of the method is $O(Δt^4)$, $k=2$ iterations should be sufficient, all further iterations just reduce and regularize the error coefficient.



You get convergence to the solution closest to the initial estimate by some contraction argument like the Banach fixed-point theorem as long as for the combined Lipschitz constant of the iteration $LΔt/2<1$.




For faster or more robust convergence, you would need to implement a Newton-like method.






share|cite|improve this answer























  • I would appreciate an elaboration of the statement that the initial extrapolation is not significantly damaging to the accuracy of the method (proof or citation). Also, a link to a more robust newton-like method would be appreciated.
    – Aleksejs Fomins
    Nov 26 at 11:42










  • Added some semi-quantitative estimates.
    – LutzL
    Nov 26 at 11:55










  • Ok, I think I get the convergence rate. One last thing I am worried about is whether it converges to the right thing. In principle, I can imagine that a nonlinear equation can have multiple solutions for $x(t+1)$. Is there an intuition why it should not converge to some spurious root of the equation?
    – Aleksejs Fomins
    Nov 26 at 12:03










  • Also, ignore my question about what a newton method is. I was being stupid. It is just a better way of finding root of an equation
    – Aleksejs Fomins
    Nov 26 at 12:04










  • I corrected an error, the Lipschitz constant of the iteration is $Δt^2cdot Lcdot 1/(2Δt)=LΔt/2$, thus 2 iterations are necessary to reach error order 4, 3 or more give better results (with rapidly diminishing improvements).
    – LutzL
    Nov 26 at 12:15













up vote
0
down vote



accepted







up vote
0
down vote



accepted






You could just iterate
$$
x^{[m+1]}(t+Δt)=2x(t)-x(t−Δt)+Δt^2fleft(frac{x^{[m]}(t+Δt)−x(t−Δt)}{2Δt},x(t)right)
$$
starting with a simple extrapolation $x^{[0]}(t+Δt)=2x(t)-x(t−Δt)$. Usually with the first or second iteration you should have reached sufficient accuracy.



For any twice differentiable functions you get $x(t+Δt)−2x(t)+x(t−Δt)=O(Δt^2)$, so that the error of the iteration is $O(Δt^2(LΔt/2)^k)$ for $k$ iterations, where $L$ is a Lipschitz constant of $f$ relative to its first argument. As the expected truncation error of the method is $O(Δt^4)$, $k=2$ iterations should be sufficient, all further iterations just reduce and regularize the error coefficient.



You get convergence to the solution closest to the initial estimate by some contraction argument like the Banach fixed-point theorem as long as for the combined Lipschitz constant of the iteration $LΔt/2<1$.




For faster or more robust convergence, you would need to implement a Newton-like method.






share|cite|improve this answer














You could just iterate
$$
x^{[m+1]}(t+Δt)=2x(t)-x(t−Δt)+Δt^2fleft(frac{x^{[m]}(t+Δt)−x(t−Δt)}{2Δt},x(t)right)
$$
starting with a simple extrapolation $x^{[0]}(t+Δt)=2x(t)-x(t−Δt)$. Usually with the first or second iteration you should have reached sufficient accuracy.



For any twice differentiable functions you get $x(t+Δt)−2x(t)+x(t−Δt)=O(Δt^2)$, so that the error of the iteration is $O(Δt^2(LΔt/2)^k)$ for $k$ iterations, where $L$ is a Lipschitz constant of $f$ relative to its first argument. As the expected truncation error of the method is $O(Δt^4)$, $k=2$ iterations should be sufficient, all further iterations just reduce and regularize the error coefficient.



You get convergence to the solution closest to the initial estimate by some contraction argument like the Banach fixed-point theorem as long as for the combined Lipschitz constant of the iteration $LΔt/2<1$.




For faster or more robust convergence, you would need to implement a Newton-like method.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 12:13

























answered Nov 26 at 11:34









LutzL

54.9k42053




54.9k42053












  • I would appreciate an elaboration of the statement that the initial extrapolation is not significantly damaging to the accuracy of the method (proof or citation). Also, a link to a more robust newton-like method would be appreciated.
    – Aleksejs Fomins
    Nov 26 at 11:42










  • Added some semi-quantitative estimates.
    – LutzL
    Nov 26 at 11:55










  • Ok, I think I get the convergence rate. One last thing I am worried about is whether it converges to the right thing. In principle, I can imagine that a nonlinear equation can have multiple solutions for $x(t+1)$. Is there an intuition why it should not converge to some spurious root of the equation?
    – Aleksejs Fomins
    Nov 26 at 12:03










  • Also, ignore my question about what a newton method is. I was being stupid. It is just a better way of finding root of an equation
    – Aleksejs Fomins
    Nov 26 at 12:04










  • I corrected an error, the Lipschitz constant of the iteration is $Δt^2cdot Lcdot 1/(2Δt)=LΔt/2$, thus 2 iterations are necessary to reach error order 4, 3 or more give better results (with rapidly diminishing improvements).
    – LutzL
    Nov 26 at 12:15


















  • I would appreciate an elaboration of the statement that the initial extrapolation is not significantly damaging to the accuracy of the method (proof or citation). Also, a link to a more robust newton-like method would be appreciated.
    – Aleksejs Fomins
    Nov 26 at 11:42










  • Added some semi-quantitative estimates.
    – LutzL
    Nov 26 at 11:55










  • Ok, I think I get the convergence rate. One last thing I am worried about is whether it converges to the right thing. In principle, I can imagine that a nonlinear equation can have multiple solutions for $x(t+1)$. Is there an intuition why it should not converge to some spurious root of the equation?
    – Aleksejs Fomins
    Nov 26 at 12:03










  • Also, ignore my question about what a newton method is. I was being stupid. It is just a better way of finding root of an equation
    – Aleksejs Fomins
    Nov 26 at 12:04










  • I corrected an error, the Lipschitz constant of the iteration is $Δt^2cdot Lcdot 1/(2Δt)=LΔt/2$, thus 2 iterations are necessary to reach error order 4, 3 or more give better results (with rapidly diminishing improvements).
    – LutzL
    Nov 26 at 12:15
















I would appreciate an elaboration of the statement that the initial extrapolation is not significantly damaging to the accuracy of the method (proof or citation). Also, a link to a more robust newton-like method would be appreciated.
– Aleksejs Fomins
Nov 26 at 11:42




I would appreciate an elaboration of the statement that the initial extrapolation is not significantly damaging to the accuracy of the method (proof or citation). Also, a link to a more robust newton-like method would be appreciated.
– Aleksejs Fomins
Nov 26 at 11:42












Added some semi-quantitative estimates.
– LutzL
Nov 26 at 11:55




Added some semi-quantitative estimates.
– LutzL
Nov 26 at 11:55












Ok, I think I get the convergence rate. One last thing I am worried about is whether it converges to the right thing. In principle, I can imagine that a nonlinear equation can have multiple solutions for $x(t+1)$. Is there an intuition why it should not converge to some spurious root of the equation?
– Aleksejs Fomins
Nov 26 at 12:03




Ok, I think I get the convergence rate. One last thing I am worried about is whether it converges to the right thing. In principle, I can imagine that a nonlinear equation can have multiple solutions for $x(t+1)$. Is there an intuition why it should not converge to some spurious root of the equation?
– Aleksejs Fomins
Nov 26 at 12:03












Also, ignore my question about what a newton method is. I was being stupid. It is just a better way of finding root of an equation
– Aleksejs Fomins
Nov 26 at 12:04




Also, ignore my question about what a newton method is. I was being stupid. It is just a better way of finding root of an equation
– Aleksejs Fomins
Nov 26 at 12:04












I corrected an error, the Lipschitz constant of the iteration is $Δt^2cdot Lcdot 1/(2Δt)=LΔt/2$, thus 2 iterations are necessary to reach error order 4, 3 or more give better results (with rapidly diminishing improvements).
– LutzL
Nov 26 at 12:15




I corrected an error, the Lipschitz constant of the iteration is $Δt^2cdot Lcdot 1/(2Δt)=LΔt/2$, thus 2 iterations are necessary to reach error order 4, 3 or more give better results (with rapidly diminishing improvements).
– LutzL
Nov 26 at 12:15


















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