Some help with approximation

Question

lυἶცἶ0210 · Answer

Use the RK4 method to approx y(0.2), where y(x) is the solution to the IVP:

$\Large y''-2y+2y=e^t~ cost $, (y(0)=1 \), $y'(0)=2 $ 

Use h=0.2

So far I got to $\color{green}{y'=u}$ and $\color{red}{u'=2u-2y+e^t*cost} $ 

This might be a silly question but I'm new to this.. what do you do with the t variable?

amistre64 · Answer

rana kutta?  never played with it, i know its an averaging process but i got no idea how it works

dan815 · Answer

hello, its 4th order euler's method?

dan815 · Answer

all this rk stuff its all coming off the basic euler's approximation princilpes

anonymous · Answer

This is review for me, so let me follow along at my own pace :)
$$\begin{cases}y''-2y'+2y=e^t\cos t\y(0)=1\y'(0)=2\end{cases}$$
So far, you've set up the system of first order ODEs using $u=y'$, which gives
$$\begin{cases}y'=u\u'=2u-2y+e^t\cos t\end{cases}$$
with initial values $y(0)=1,u(0)=2$.

First, recall we're using the following recursions: for $n\ge0$,
$$\begin{cases}y_{n+1}=y_n+\dfrac{h}{6}f\left(k_1+2k_2+2k_3+k_4\right)\\
t_{n+1}=t_n+h\end{cases}$$
with $h=0.2$, and
$$\begin{cases}k_1=f(t_n,y_n)\
k_2=f\left(t_n+\dfrac{h}{2},y_n+\dfrac{h}{2}k_1\right)\
k_3=f\left(t_n+\dfrac{h}{2},y_n+\dfrac{h}{2}k_2\right)\
k_4=f(t_n+h,y_n+hk_3)\end{cases}$$
None of this is new I hope?

anonymous · Answer

The setup above is what you'd use for a first-order ODE, so here's what we have to deal with when numerically solving a second-order ODE:
$$\begin{cases}
\begin{cases}
k_1=f(t_n,y_n,u_n)\\
l_1=g(t_n,y_n,u_n)
\end{cases}\\
\begin{cases}
k_2=f\left(t_n+\dfrac{h}{2},y_n+\dfrac{h}{2}k_1,u_n+\dfrac{h}{2}l_1\right)\\
l_2=g\left(t_n+\dfrac{h}{2},y_n+\dfrac{h}{2}k_1,u_n+\dfrac{h}{2}l_1\right)
\end{cases}\\
\begin{cases}
k_3=f\left(t_n+\dfrac{h}{2},y_n+\dfrac{h}{2}k_2,u_n+\dfrac{h}{2}l_2\right)\\
l_3=g\left(t_n+\dfrac{h}{2},y_n+\dfrac{h}{2}k_2,u_n+\dfrac{h}{2}l_2\right)
\end{cases}\\
\begin{cases}
k_4=f(t_n+h,y_n+hk_3,u_n+hl_3)\\
l_4=g(t_n+h,y_n+hk_3,u_n+hl_3)
\end{cases}
\end{cases}$$
where
$$\begin{cases}y'=f(t,y,u)\
u'=g(t,y,u)\end{cases}$$
and the recursions are the same:
\begin{cases}
y_{n+1}=y_n+\dfrac{h}{6}f\left(k_1+2k_2+2k_3+k_4\right)\\
u_{n+1}=u_n+\dfrac{h}{6}g\left(k_1+2k_2+2k_3+k_4\right)\\
t_{n+1}=t_n+h
\end{cases}

anonymous · Answer

I'd start with a set of  tables. I'm currently writing them up in LaTeX and verifying the computations with Mathematica. Might take a moment...

lυἶცἶ0210 · Answer

I love the way you lay things out, thanks :) 

But since there is a t and no x, we replace x with t in the function?

anonymous · Answer

Well $t$ is your dependent variable here. There is no $x$.

Also, slight typo: The recursion for $u$ should be $u_{n+1}=u_n+\dfrac{h}{6}g\left(\color{red}{l_1+2l_2+2l_3+l_4}\right)$, not $k$.

anonymous · Answer

Oh another typo, there is not $f$ or $g$, I misread my notes. The recursions should be
$$\begin{cases}
y_{n+1}=y_n+\dfrac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\
u_{n+1}=u_n+\dfrac{h}{6}\left(l_1+2l_2+2l_3+l_4\right)\\
t_{n+1}=t_n+h
\end{cases}$$

anonymous · Answer

Here's what I'm getting for $n=1$.
$$\begin{matrix}
\begin{array}{c|c|c|c}
n&t_n&y_n&u_n\
\hline
0&0&1&2\
1&\color{lightgray}{0.2}&1.46399&2.65945
\end{array}
\\
\begin{array}{c|c|c|c}
k_1&k_2&k_3&k_4\
\hline
2&2.3&2.32996&2.65992
\end{array}
\\
\begin{array}{c|c|c|c}
l_1&l_2&l_3&l_4\
\hline
3&3.29965&3.29958&3.5849
\end{array}
\end{matrix}$$
(The grayed out part isn't needed for this step, but will be for the next one.)

anonymous · Answer

Here's the breakdown for the calculations. Quick disclaimer: I'm using approximations of the results in the tables.

The first set of things you need to compute is $k_1$ and $l_1$, using $t_0=0$, $y_0=1$, and $u_0=2$.
$$\begin{cases}
k_1=f(0,1,2)=2\\
l_1=g(0,1,2)=2(2)-2(1)+e^0\cos0=3
\end{cases}$$
which are due to $f(t,y,u)=u$ and $g(t,y,u)=2u-2y+e^t\cos t$.

Next, $k_2$ and $l_2$.
$$\begin{cases}
k_2=f\left(0+\dfrac{0.2}{2},1+\dfrac{0.2}{2}(2),2+\dfrac{0.2}{2}(3)\right)=f(0.1,1.2,2.3)=2.3\\
l_2=g\left(0+\dfrac{0.2}{2},1+\dfrac{0.2}{2}(2),2+\dfrac{0.2}{2}(3)\right)=g(0.1,1.2,2.3)\approx3.3
\end{cases}$$
Next, $k_3$ and $l_3$.
$$\begin{cases}
k_3\approx f\left(0+\dfrac{0.2}{2},1+\dfrac{0.2}{2}(2.3),2+\dfrac{0.2}{2}(3.3)\right)\approx f(0.1,1.23,2.33)\approx 2.3\\
l_3\approx g\left(0+\dfrac{0.2}{2},1+\dfrac{0.2}{2}(2.3),2+\dfrac{0.2}{2}(3.3)\right)\approx g(0.1,1.23,2.33)\approx 3.3
\end{cases}$$
Finally, $k_4$ and $l_4$.
$$\begin{cases}
k_4\approx f(0+0.2,1+0.2(2.3),2+0.2(3.3))\approx f(0.2,1.46,2.66)\approx 2.7\\
l_4\approx g(0+0.2,1+0.2(2.3),2+0.2(3.3))\approx g(0.2,1.46,2.66)\approx 3.6
\end{cases}$$
We're not quite done yet, though, because now we have to compute $y_1$ and $u_1$, which are
$$\begin{cases}
y_1\approx 1+\dfrac{0.2}{6}\left(2+2(2.3)+2(2.3)+2.7\right)\approx 1.5\\
u_1\approx 2+\dfrac{0.2}{6}\left(3+2(3.3)+2(3.3)+3.6\right)\approx 2.7
\end{cases}$$

anonymous · Answer

For $n\ge1$, rinse and repeat, replacing the appropriate $t_0,y_0,u_0$ with the newly computed $t_1,y_1,u_1$, and compute the new $k$s and $l$s. Since the step size is $h=0.2$, and you want to approximate the solution at $t=2$, you're looking at $n=10$ rounds of these computations.

I highly recommend writing a script to do all the computations for you.

lυἶცἶ0210 · Answer

I see where I messed up now. And yea, I had to do this on Excel with h=0.2 and 0.1.. it can be a pain xD

But now I see what to do.. thanks for all your time Siths :)

anonymous · Answer

You're welcome!

anonymous · Answer

Oh, to answer your initial question: you replace each successive $t$ by the step size increment. So for the first round of calculations, $t_0=0$, and the next would be $t_1=t_0+h=0+0.2=0.2$, and the next would be $t_2=t_1+h=0.2+0.2=0.4$, and so on.