Opening the book…
Each step in Euler’s method requires one evaluation of f and each step in the improved Euler method requires two evaluations of f. We’ll skip methods for which three evaluations of f are used and proceed to the Runge-Kutta method, the most widely used method, for which four evaluations of f are used to solve the initial value problem
\[\label{eq:3.3.1} y'=f(x,y),\quad y(x_0)=y_0.\]
The Runge-Kutta method computes approximate values \(y_1\), \(y_2\), …, \(y_n\) of the solution of Equation \ref{eq:3.3.1} at \(x_0\), \(x_0+h\), …, \(x_0+nh\) as follows: Given \(y_i\), compute
\[\begin{align*} k_{1i}&=f(x_i,y_i),\\ k_{2i}&=f \left(x_i+{h\over2},y_i+{h\over2}k_{1i}\right),\\ k_{3i}&=f\left(x_i+{h\over2},y_i+{h\over2}k_{2i}\right),\\ k_{4i}&=f(x_i+h,y_i+hk_{3i}),\end{align*}\]
and
\[y_{i+1}=y_i+{h\over6}(k_{1i}+2k_{2i}+2k_{3i}+k_{4i}).\nonumber \]
Example 3.3.1 illustrates the computational procedure indicated in the Runge-Kutta method.
Use the Runge-Kutta method with \(h=0.1\) to find approximate values for the solution of the initial value problem
\[\label{eq:3.3.2} y'+2y=x^3e^{-2x},\quad y(0)=1,\]
at \(x=0.1,0.2\).
Solution
Again we rewrite Equation \ref{eq:3.3.2} as
\[y'=-2y+x^3e^{-2x},\quad y(0)=1, \nonumber\]
which is of the form Equation \ref{eq:3.3.1}, with
\[f(x,y)=-2y+x^3e^{-2x},\ x_0=0,\mbox{ and}\ y_0=1. \nonumber\]
The Runge-Kutta method yields
\[\begin{aligned} k_{10} & = f(x_0,y_0) = f(0,1)=-2,\\ k_{20} & = f(x_0+h/2,y_0+hk_{10}/2)=f(.05,1+(.05)(-2))\\ &= f(.05,.9)=-2(.9)+(.05)^3e^{-.1}=-1.799886895,\\ k_{30} & = f(x_0+h/2,y_0+hk_{20}/2)=f(.05,1+(.05)(-1.799886895))\\ &= f(.05,.910005655)=-2(.910005655)+(.05)^3e^{-.1}=-1.819898206,\\ k_{40} & = f(x_0+h,y_0+hk_{30})=f(.1,1+(.1)(-1.819898206))\\ &=f(.1,.818010179)=-2(.818010179)+(.1)^3e^{-.2}=-1.635201628,\\ y_1&=y_0+{h\over6}(k_{10}+2k_{20}+2k_{30}+k_{40}),\\ &=1+{.1\over6}(-2+2(-1.799886895)+2(-1.819898206) -1.635201628)=.818753803,\\[4pt] k_{11} & = f(x_1,y_1) = f(.1,.818753803)=-2(.818753803))+(.1)^3e^{-.2}=-1.636688875,\\ k_{21} & = f(x_1+h/2,y_1+hk_{11}/2)=f(.15,.818753803+(.05)(-1.636688875))\\ &= f(.15,.736919359)=-2(.736919359)+(.15)^3e^{-.3}=-1.471338457,\\ k_{31} & = f(x_1+h/2,y_1+hk_{21}/2)=f(.15,.818753803+(.05)(-1.471338457))\\ &= f(.15,.745186880)=-2(.745186880)+(.15)^3e^{-.3}=-1.487873498,\\ k_{41} & = f(x_1+h,y_1+hk_{31})=f(.2,.818753803+(.1)(-1.487873498))\\ &=f(.2,.669966453)=-2(.669966453)+(.2)^3e^{-.4}=-1.334570346,\\ y_2&=y_1+{h\over6}(k_{11}+2k_{21}+2k_{31}+k_{41}),\\ &=.818753803+{.1\over6}(-1.636688875+2(-1.471338457)+2(-1.487873498)-1.334570346) \\&=.670592417.\end{aligned}\]
The Runge-Kutta method is sufficiently accurate for most applications.
Examples Illustrating The Error in the Runge-Kutta Method
Table 3.3.1 shows results of using the Runge-Kutta method with step sizes \(h=0.1\) and \(h=0.05\) to find approximate values of the solution of the initial value problem
\[y'+2y=x^3e^{-2x},\quad y(0)=1 \nonumber\]
at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\). For comparison, it also shows the corresponding approximate values obtained with the improved Euler method in Example 3.2.2, and the values of the exact solution
\[y={e^{-2x}\over4}(x^4+4).\nonumber \]
The results obtained by the Runge-Kutta method are clearly better than those obtained by the improved Euler method in fact; the results obtained by the Runge-Kutta method with \(h=0.1\) are better than those obtained by the improved Euler method with \(h=0.05\).
| Improved Euler | Runge-Kutta | ||||
|---|---|---|---|---|---|
| x | h=0.1 | h=0.05 | h=0.1 | h-0.05 | Exact |
| 0.0 | 1.000000000 | 1.000000000 | 1.000000000 | 1.000000000 | 1.000000000 |
| 0.1 | 0.820040937 | 0.819050572 | 0.818753803 | 0.818751370 | 0.818751221 |
| 0.2 | 0.672734445 | 0.671086455 | 0.670592417 | 0.670588418 | 0.670588174 |
| 0.3 | 0.552597643 | 0.550543878 | 0.549928221 | 0.549923281 | 0.549922980 |
| 0.4 | 0.455160637 | 0.452890616 | 0.452210430 | 0.452205001 | 0.452204669 |
| 0.5 | 0.376681251 | 0.374335747 | 0.373633492 | 0.373627899 | 0.373627557 |
| 0.6 | 0.313970920 | 0.311652239 | 0.310958768 | 0.310953242 | 0.310952904 |
| 0.7 | 0.264287611 | 0.262067624 | 0.261404568 | 0.261399270 | 0.261398947 |
| 0.8 | 0.225267702 | 0.223194281 | 0.222575989 | 0.222571024 | 0.222570721 |
| 0.9 | 0.194879501 | 0.192981757 | 0.192416882 | 0.192412317 | 0.192412038 |
| 1.0 | 0.171388070 | 0.169680673 | 0.169173489 | 0.169169356 | 0.169169104 |
Table 3.3.2 shows analogous results for the nonlinear initial value problem
\[y'=-2y^2+xy+x^2,\ y(0)=1. \nonumber\]
We applied the improved Euler method to this problem in Example 3.2.3.
| Improved Euler | Runge-Kutta | ||||
|---|---|---|---|---|---|
| x | h=0.1 | h=0.05 | h=0.1 | h-0.05 | "exact" |
| 0.0 | 1.000000000 | 1.000000000 | 1.000000000 | 1.000000000 | 1.000000000 |
| 0.1 | 0.840500000 | 0.838288371 | 0.837587192 | 0.837584759 | 0.837584494 |
| 0.2 | 0.733430846 | 0.730556677 | 0.729644487 | 0.729642155 | 0.729641890 |
| 0.3 | 0.661600806 | 0.658552190 | 0.657582449 | 0.657580598 | 0.657580377 |
| 0.4 | 0.615961841 | 0.612884493 | 0.611903380 | 0.611901969 | 0.611901791 |
| 0.5 | 0.591634742 | 0.588558952 | 0.587576716 | 0.587575635 | 0.587575491 |
| 0.6 | 0.586006935 | 0.582927224 | 0.581943210 | 0.581942342 | 0.581942225 |
| 0.7 | 0.597712120 | 0.594618012 | 0.593630403 | 0.593629627 | 0.593629526 |
| 0.8 | 0.626008824 | 0.622898279 | 0.621908378 | 0.621907553 | 0.621907458 |
| 0.9 | 0.670351225 | 0.667237617 | 0.666251988 | 0.666250942 | 0.666250842 |
| 1.0 | 0.730069610 | 0.726985837 | 0.726017378 | 0.726015908 | 0.726015790 |
This page titled 3.3: The Runge-Kutta Method was last modified on Sun, 27 Nov 2022 03:09:21 GMT and is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.
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