wendi1019gt

2022-08-14

Use the two second-order multi-step methods

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(3{f}_{i}-{f}_{i-1})$

and

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}({f}_{i+1}+{f}_{i})$

as a pblackictor-corrector method to compute an approximation to $y(0.3)$, with stepsize $h=0.1$, for the IVP;

${y}^{\prime}(t)=3ty,y(0)=-1.$

Use Euler’s method to start.

I do not understand how to use these methods to approximate $y(0.3)$. Moreover I am not sure how Euler's method fits into this question. Could someone clarify this question please?

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(3{f}_{i}-{f}_{i-1})$

and

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}({f}_{i+1}+{f}_{i})$

as a pblackictor-corrector method to compute an approximation to $y(0.3)$, with stepsize $h=0.1$, for the IVP;

${y}^{\prime}(t)=3ty,y(0)=-1.$

Use Euler’s method to start.

I do not understand how to use these methods to approximate $y(0.3)$. Moreover I am not sure how Euler's method fits into this question. Could someone clarify this question please?

Madilyn Dunn

Beginner2022-08-15Added 16 answers

The

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(3f({t}_{i},{\omega}_{i})-f({t}_{i-1},{\omega}_{i-1}))$

is an explicit two-step (using three points $i-1$,$i$ and $i+1$) method. In contrast, the second one

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(f({t}_{i+1},{\omega}_{i+1})+f({t}_{i},{\omega}_{i}))$

is an implicit one-step (using two points $i$ and $i+1$) method. It is implicit since you can not easily solve it for ${\omega}_{i+1}$, you have an equation for it

${\omega}_{i+1}-\frac{h}{2}f({t}_{i-1},{\omega}_{i+1})={\omega}_{i}+\frac{h}{2}f({t}_{i},{\omega}_{i}).$

Instead of solving that equation you're advised to use a pblackictor-corrector scheme, i.e. use ${\omega}_{i+1}$, obtained from the first scheme and plug it into $f({\omega}_{i+1})$ term of the second one. I'll rewrite it using tilde notation (${\stackrel{~}{\omega}}_{i+1}$ is a pblackicted value, while ${\omega}_{i+1}$ is a corrected one):

$\text{Pblackictor:}{\stackrel{~}{\omega}}_{i+1}={\omega}_{i}+\frac{h}{2}(3f({t}_{i},{\omega}_{i})-f({t}_{i+1},{\omega}_{i-1}))\phantom{\rule{0ex}{0ex}}\text{Corrector:}{\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(f({t}_{i+1},{\stackrel{~}{\omega}}_{i+1})+f({t}_{i},{\omega}_{i})).$

Now the evaluation is really straitforward, you just take ${\omega}_{i-1},{\omega}_{i}$, compute ${\stackrel{~}{\omega}}_{i+1}$ and finally ${\omega}_{i+1}$. The only problem is that you cannot start with that. The only initial value you have is ${\omega}_{0}=-1$ which is not sufficient to compute ${\omega}_{1}$ using pblackictor-corrector pair (it would need also ${\omega}_{-1}$). To overcome this problem you can use some other method to compute ${\omega}_{1}$, explicit Euler for example:

${\omega}_{1}={\omega}_{0}+hf({t}_{0},{\omega}_{0}).$

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(3f({t}_{i},{\omega}_{i})-f({t}_{i-1},{\omega}_{i-1}))$

is an explicit two-step (using three points $i-1$,$i$ and $i+1$) method. In contrast, the second one

${\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(f({t}_{i+1},{\omega}_{i+1})+f({t}_{i},{\omega}_{i}))$

is an implicit one-step (using two points $i$ and $i+1$) method. It is implicit since you can not easily solve it for ${\omega}_{i+1}$, you have an equation for it

${\omega}_{i+1}-\frac{h}{2}f({t}_{i-1},{\omega}_{i+1})={\omega}_{i}+\frac{h}{2}f({t}_{i},{\omega}_{i}).$

Instead of solving that equation you're advised to use a pblackictor-corrector scheme, i.e. use ${\omega}_{i+1}$, obtained from the first scheme and plug it into $f({\omega}_{i+1})$ term of the second one. I'll rewrite it using tilde notation (${\stackrel{~}{\omega}}_{i+1}$ is a pblackicted value, while ${\omega}_{i+1}$ is a corrected one):

$\text{Pblackictor:}{\stackrel{~}{\omega}}_{i+1}={\omega}_{i}+\frac{h}{2}(3f({t}_{i},{\omega}_{i})-f({t}_{i+1},{\omega}_{i-1}))\phantom{\rule{0ex}{0ex}}\text{Corrector:}{\omega}_{i+1}={\omega}_{i}+\frac{h}{2}(f({t}_{i+1},{\stackrel{~}{\omega}}_{i+1})+f({t}_{i},{\omega}_{i})).$

Now the evaluation is really straitforward, you just take ${\omega}_{i-1},{\omega}_{i}$, compute ${\stackrel{~}{\omega}}_{i+1}$ and finally ${\omega}_{i+1}$. The only problem is that you cannot start with that. The only initial value you have is ${\omega}_{0}=-1$ which is not sufficient to compute ${\omega}_{1}$ using pblackictor-corrector pair (it would need also ${\omega}_{-1}$). To overcome this problem you can use some other method to compute ${\omega}_{1}$, explicit Euler for example:

${\omega}_{1}={\omega}_{0}+hf({t}_{0},{\omega}_{0}).$

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