( $y(t,2h)$ means approximation with $2h$)

I don't understand why the Richardson extrapolation is mentioned. For what do I have to use it? Can I not just calculate $y(t,2h)$ and $y(t,h)$ and see the error?

sublimnes9
2022-08-07
Answered

One method consists in computing the numerical solution for an arbitrary $h$ and then $2h$. The Richardson extrapolation gives an estimate of $e=\underset{t}{max}|y(t,2h)-y(t,h)|$ of the error. When the error is smaller than the tolerance, we keep the result and start from $2y(t,h)-y(t,h)$. If the error is larger we restart with $h/2$ until we reach the tolerance.

( $y(t,2h)$ means approximation with $2h$)

I don't understand why the Richardson extrapolation is mentioned. For what do I have to use it? Can I not just calculate $y(t,2h)$ and $y(t,h)$ and see the error?

( $y(t,2h)$ means approximation with $2h$)

I don't understand why the Richardson extrapolation is mentioned. For what do I have to use it? Can I not just calculate $y(t,2h)$ and $y(t,h)$ and see the error?

You can still ask an expert for help

raffatoaq

Answered 2022-08-08
Author has **22** answers

You can just check the error, and not bother with Richardson extrapolation, that's fine. But once you've paid the computational price of calculating the error, you might as well use Richardson extrapolation to blackuce the error further; it costs you essentially nothing.

asked 2022-08-12

How does the Velocity Verlet method differ from the standard Euler method? Why do we need to add Acceleration / 2 to calculate position?

asked 2022-08-14

${y}^{\u2033}=11-y$

$y(2)=1;{y}^{\prime}(2)=-4$

and asked to use Euler's method to find $y(2.2)$ for $h=0.1$

To find ${y}^{\prime}$ I simply took the integral of ${y}^{\u2033}$ to get:

${y}^{\prime}=11x-yx$

However, this does not satisfy the condition given above, that ${y}^{\prime}(2)=-4$. Is this not the correct way of obtaining the first order derivative?

$y(2)=1;{y}^{\prime}(2)=-4$

and asked to use Euler's method to find $y(2.2)$ for $h=0.1$

To find ${y}^{\prime}$ I simply took the integral of ${y}^{\u2033}$ to get:

${y}^{\prime}=11x-yx$

However, this does not satisfy the condition given above, that ${y}^{\prime}(2)=-4$. Is this not the correct way of obtaining the first order derivative?

asked 2022-07-16

${Y}^{\prime}=AY$

$A=\left[\begin{array}{ccc}1& 2& -1\\ \phantom{\rule{mediummathspace}{0ex}}1& 0& 1\\ \phantom{\rule{mediummathspace}{0ex}}4& -4& 5\end{array}\right]$

$Y(0)=\left(\begin{array}{c}-1\\ 0\\ 0\end{array}\right)$

I have to solve this system using Eulers method, but what's Eulers method, like I know how to approach this using the eigenvalue method, but the problem explicitly states that it has to be solved using that method, yet my textbook doesn't provide an algorithm for such a method.

$A=\left[\begin{array}{ccc}1& 2& -1\\ \phantom{\rule{mediummathspace}{0ex}}1& 0& 1\\ \phantom{\rule{mediummathspace}{0ex}}4& -4& 5\end{array}\right]$

$Y(0)=\left(\begin{array}{c}-1\\ 0\\ 0\end{array}\right)$

I have to solve this system using Eulers method, but what's Eulers method, like I know how to approach this using the eigenvalue method, but the problem explicitly states that it has to be solved using that method, yet my textbook doesn't provide an algorithm for such a method.

asked 2022-07-21

I'm asked to decide if I should solve the system

$\dot{y}=\left(\begin{array}{cc}-600& 400\\ 400& -600\end{array}\right)y,\phantom{\rule{1em}{0ex}}t\in [{t}_{0},{t}_{e}],\phantom{\rule{1em}{0ex}}y({t}_{0})={y}_{0}$

with either the explicit Euler method or the implicit Euler method.

Using the explicit Euler method I would get the updating scheme

${y}_{n+1}=\left(\begin{array}{cc}1-600h& 400h\\ 400h& 1-600h\end{array}\right){y}_{n}$

where the eigenvalues of the driving matrix is

${\lambda}_{1}=401-600h,$

${\lambda}_{2}=-399-600h.$

For the solution to be stable these need to be less than one which gives the conditions

$h\ge \frac{4}{6}=\frac{2}{3},$

$h\ge -\frac{4}{6}=-\frac{2}{3}.$

The last condition doesn't say anything but the first condition seems restrictive since I can't choose $h$ as small as possible.

If I instead were to use the implicit Euler method I would get the updating scheme

${y}_{n+1}={\left(\begin{array}{cc}1-600h& 400h\\ 400h& 1-600h\end{array}\right)}^{-1}{y}_{n}.$

Now I can't solve for the eigenvalues of this system but I've heard the implicit Euler is unconditionally stable so it shouldn't matter.

So is the answer that I should choose implicit Euler because it is unconditionally stable or am I missing something? The order of consistency of both is 1 so that should not matter.

$\dot{y}=\left(\begin{array}{cc}-600& 400\\ 400& -600\end{array}\right)y,\phantom{\rule{1em}{0ex}}t\in [{t}_{0},{t}_{e}],\phantom{\rule{1em}{0ex}}y({t}_{0})={y}_{0}$

with either the explicit Euler method or the implicit Euler method.

Using the explicit Euler method I would get the updating scheme

${y}_{n+1}=\left(\begin{array}{cc}1-600h& 400h\\ 400h& 1-600h\end{array}\right){y}_{n}$

where the eigenvalues of the driving matrix is

${\lambda}_{1}=401-600h,$

${\lambda}_{2}=-399-600h.$

For the solution to be stable these need to be less than one which gives the conditions

$h\ge \frac{4}{6}=\frac{2}{3},$

$h\ge -\frac{4}{6}=-\frac{2}{3}.$

The last condition doesn't say anything but the first condition seems restrictive since I can't choose $h$ as small as possible.

If I instead were to use the implicit Euler method I would get the updating scheme

${y}_{n+1}={\left(\begin{array}{cc}1-600h& 400h\\ 400h& 1-600h\end{array}\right)}^{-1}{y}_{n}.$

Now I can't solve for the eigenvalues of this system but I've heard the implicit Euler is unconditionally stable so it shouldn't matter.

So is the answer that I should choose implicit Euler because it is unconditionally stable or am I missing something? The order of consistency of both is 1 so that should not matter.

asked 2022-08-13

I am attempting to compute an approximation of the solution with the forward Euler method in $[0,1]$ with step lengths ${h}_{1}=0.2$, ${h}_{2}=0.1$ given the initial value problem below

$\frac{dy}{dz}=\frac{1}{1+z}-y(z)\phantom{\rule{1em}{0ex}}y(0)=1$

I am not sure what to do when I am given two step sizes instead of one. I know how to compute it if it was given with a step size. Am I supposed to find out the approximation for two different step sizes? Or is there anything I am missing?

$\frac{dy}{dz}=\frac{1}{1+z}-y(z)\phantom{\rule{1em}{0ex}}y(0)=1$

I am not sure what to do when I am given two step sizes instead of one. I know how to compute it if it was given with a step size. Am I supposed to find out the approximation for two different step sizes? Or is there anything I am missing?

asked 2022-06-13

I am revising the modified euler method and would appreciate some help with this question:

The equation is

${y}^{\prime}=\frac{2}{x}y+{x}^{2}{e}^{x},y(1)=0$

Use modified euler method to calculate $y(1.1)$ taking $h=0.1$ (working to 4 decimal places)

Where do i start?

asked 2022-07-16

I need to solve the equation below with Euler's method:

${y}^{\u2033}+\pi y{e}^{x/3}(2{y}^{\prime}\mathrm{sin}(\pi x)+\pi y\mathrm{cos}(\pi x))=\frac{y}{9}$

for the initial conditions $y(0)=1$, ${y}^{\prime}(0)=-1/3$

So I know I need to turn the problem into a system of two first order differential equations.

Therefore ${u}_{1}={y}^{\prime}$ and ${u}_{2}={y}^{\u2033}$ I can now write the system as:

${u}_{1}={y}^{\prime}\phantom{\rule{0ex}{0ex}}{u}_{2}={\displaystyle \frac{y}{9}}-\pi y{e}^{x/3}(2{u}_{1}\mathrm{sin}(\pi x)-\pi y\mathrm{cos}(\pi x))$

How do I proceed from here?

${y}^{\u2033}+\pi y{e}^{x/3}(2{y}^{\prime}\mathrm{sin}(\pi x)+\pi y\mathrm{cos}(\pi x))=\frac{y}{9}$

for the initial conditions $y(0)=1$, ${y}^{\prime}(0)=-1/3$

So I know I need to turn the problem into a system of two first order differential equations.

Therefore ${u}_{1}={y}^{\prime}$ and ${u}_{2}={y}^{\u2033}$ I can now write the system as:

${u}_{1}={y}^{\prime}\phantom{\rule{0ex}{0ex}}{u}_{2}={\displaystyle \frac{y}{9}}-\pi y{e}^{x/3}(2{u}_{1}\mathrm{sin}(\pi x)-\pi y\mathrm{cos}(\pi x))$

How do I proceed from here?