How to Choose Coefficients in the Runge-Kutta Method
📂Numerical Analysis How to Choose Coefficients in the Runge-Kutta Method Overview Let’s consider the ordinary differential equation given as follows. y y y t t t ( ′ ) (^{\prime}) ( ′ ) t t t
y ′ = f ( t , y ) , t ≥ t 0 , y ( t 0 ) = y 0 , t n + 1 = t n + h
y^{\prime} = f(t,y),\quad t \ge t_{0},\quad y(t_{0}) = y_{0}, \quad t_{n+1} = t_{n} + h
y ′ = f ( t , y ) , t ≥ t 0 , y ( t 0 ) = y 0 , t n + 1 = t n + h
The explicit Runge-Kutta method is a way to approximate y n + 1 = y ( t n + 1 ) y_{n+1} = y(t_{n+1}) y n + 1 = y ( t n + 1 ) y n = y ( t n ) y_{n} = y(t_{n}) y n = y ( t n )
y n + 1 = y n + h ∑ j = 1 ν b j f ( t n + c j h , ξ j ) , n = 0 , 1 , 2 , ⋯
y_{n+1} = y_{n} + h\sum_{j=1}^{\nu} b_{j} f(t_{n} + c_{j}h, \xi_{j}),\quad n=0,1,2,\cdots
y n + 1 = y n + h j = 1 ∑ ν b j f ( t n + c j h , ξ j ) , n = 0 , 1 , 2 , ⋯
Here, each ξ j \xi_{j} ξ j
ξ 1 = y n ξ 2 = y n + h a 2 , 1 f ( t n , ξ 1 ) ξ 3 = y n + h a 3 , 1 f ( t n , ξ 1 ) + h a 3 , 2 f ( t n + c 2 h , ξ 2 ) ⋮ ξ ν = y n + h ∑ i = 1 ν − 1 a ν , i f ( t n + c i h , ξ i )
\begin{align*}
\xi_{1} &= y_{n} \\
\xi_{2} &= y_{n} + ha_{2,1}f(t_{n}, \xi_{1}) \\
\xi_{3} &= y_{n} + ha_{3,1}f(t_{n}, \xi_{1}) + ha_{3,2}f(t_{n} + c_{2}h, \xi_{2}) \\
\vdots & \\
\xi_{\nu} &= y_{n} + h\sum_{i=1}^{\nu-1}a_{\nu, i}f(t_{n} + c_{i}h, \xi_{i})
\end{align*}
ξ 1 ξ 2 ξ 3 ⋮ ξ ν = y n = y n + h a 2 , 1 f ( t n , ξ 1 ) = y n + h a 3 , 1 f ( t n , ξ 1 ) + h a 3 , 2 f ( t n + c 2 h , ξ 2 ) = y n + h i = 1 ∑ ν − 1 a ν , i f ( t n + c i h , ξ i )
In the RK method, the coefficients below are chosen to be used. We will explore how to select appropriate coefficients.
A = [ a j , i ] = [ a 2 , 1 a 3 , 1 a 3 , 2 ⋮ a ν , 1 a ν , 2 ⋯ a ν , ν − 1 ] and b = [ b 1 b 2 ⋮ b ν ] and c = [ c 1 c 2 ⋮ c ν ]
A = [a_{j,i}]
= \begin{bmatrix}
\\
a_{2,1} & \\
a_{3,1} & a_{3,2} \\
\vdots & \\
a_{\nu,1} & a_{\nu,2} & \cdots & a_{\nu,\nu-1} &
\end{bmatrix}
\quad \text{and} \quad
\mathbf{b} = \begin{bmatrix}
b_{1} \\ b_{2} \\ \vdots \\ b_{\nu}
\end{bmatrix}
\quad \text{and} \quad
\mathbf{c} = \begin{bmatrix}
c_{1} \\ c_{2} \\ \vdots \\ c_{\nu}
\end{bmatrix}
A = [ a j , i ] = a 2 , 1 a 3 , 1 ⋮ a ν , 1 a 3 , 2 a ν , 2 ⋯ a ν , ν − 1 and b = b 1 b 2 ⋮ b ν and c = c 1 c 2 ⋮ c ν
For the case where ν = 2 \nu = 2 ν = 2 First, let’s consider the simplest case where ν = 2 \nu = 2 ν = 2 ξ 1 = y n \xi_{1} = y_{n} ξ 1 = y n ξ 2 = y n + h a 2 , 1 f ( t n , y n ) \xi_{2} = y_{n} + ha_{2,1}f(t_{n}, y_{n}) ξ 2 = y n + h a 2 , 1 f ( t n , y n ) y n + 1 y_{n+1} y n + 1
y n + 1 = y n + h [ b 1 f ( t n + c 1 h , ξ 1 ) + b 2 f ( t n + c 2 h , ξ 2 ) ] = y n + h [ b 1 f ( t n , y n ) + b 2 f ( t n + c 2 h , y n + h a 2 , 1 f ( t n , y n ) ) ]
\begin{align}
y_{n+1}
&= y_{n} + h \Big[ b_{1}f(t_{n} + c_{1}h, \xi_{1}) + b_{2}f(t_{n} + c_{2}h, \xi_{2}) \Big] \nonumber \\
&= y_{n} + h \Big[ b_{1}f(t_{n}, y_{n}) + b_{2}f(t_{n} + c_{2}h, y_{n} + ha_{2,1}f(t_{n}, y_{n}) )\Big]
\end{align}
y n + 1 = y n + h [ b 1 f ( t n + c 1 h , ξ 1 ) + b 2 f ( t n + c 2 h , ξ 2 ) ] = y n + h [ b 1 f ( t n , y n ) + b 2 f ( t n + c 2 h , y n + h a 2 , 1 f ( t n , y n )) ]
Here, c 1 = 0 c_{1} = 0 c 1 = 0 f f f smooth . Then, the Taylor expansion at the point ( t n , y n ) (t_{n}, y_{n}) ( t n , y n ) f ( t n + c 2 h , y n + h a 2 , 1 f ( t n , y n ) ) f(t_{n} + c_{2}h, y_{n} + ha_{2,1}f(t_{n}, y_{n}) ) f ( t n + c 2 h , y n + h a 2 , 1 f ( t n , y n ))
f ( t n + c 2 h , y n + h a 2 , 1 f ( t n , y n ) ) = f ( t n , y n ) + ( c 2 h , h a 2 , 1 f ( t n , y n ) ) T ∇ f ( t n , y n ) + O ( h 2 ) = f ( t n , y n ) + h [ c 2 ∂ f ( t n , y n ) ∂ t + a 2 , 1 ∂ f ( t n , y n ) ∂ y f ( t n , y n ) ] + O ( h 2 )
\begin{align*}
& f(t_{n} + c_{2}h, y_{n} + ha_{2,1}f(t_{n}, y_{n}) ) \\
&= f(t_{n}, y_{n}) + (c_{2}h, ha_{2,1}f(t_{n}, y_{n}))^{T} \nabla f(t_{n}, y_{n}) + \mathcal{O}(h^{2}) \\
&= f(t_{n}, y_{n}) + h\left[c_{2} \dfrac{\partial f(t_{n}, y_{n})}{\partial t} + a_{2,1}\dfrac{\partial f(t_{n}, y_{n})}{\partial y}f(t_{n}, y_{n}) \right] + \mathcal{O}(h^{2}) \\
\end{align*}
f ( t n + c 2 h , y n + h a 2 , 1 f ( t n , y n )) = f ( t n , y n ) + ( c 2 h , h a 2 , 1 f ( t n , y n ) ) T ∇ f ( t n , y n ) + O ( h 2 ) = f ( t n , y n ) + h [ c 2 ∂ t ∂ f ( t n , y n ) + a 2 , 1 ∂ y ∂ f ( t n , y n ) f ( t n , y n ) ] + O ( h 2 )
∇ f \nabla f ∇ f gradient of f f f ( 1 ) (1) ( 1 )
y n + 1 = y n + h [ b 1 f ( t n , y n ) + b 2 f ( t n , y n ) + h b 2 c 2 ∂ f ( t n , y n ) ∂ t + h b 2 a 2 , 1 ∂ f ( t n , y n ) ∂ y f ( t n , y n ) + O ( h 2 ) ] = y n + h ( b 1 + b 2 ) f ( t n , y n ) + h 2 b 2 [ c 2 ∂ f ( t n , y n ) ∂ t + a 2 , 1 ∂ f ( t n , y n ) ∂ y f ( t n , y n ) ] + O ( h 3 )
\begin{align*}
y_{n+1} &= y_{n} + h \bigg[ b_{1}f(t_{n}, y_{n}) + b_{2}f(t_{n}, y_{n}) \\
& \qquad + hb_{2}c_{2} \dfrac{\partial f(t_{n}, y_{n})}{\partial t} + hb_{2}a_{2,1}\dfrac{\partial f(t_{n}, y_{n})}{\partial y}f(t_{n}, y_{n}) + \mathcal{O}(h^{2}) \bigg] \\
&= y_{n} + h (b_{1} + b_{2})f(t_{n}, y_{n}) \\
& \qquad + h^{2}b_{2} \left[ c_{2} \dfrac{\partial f(t_{n}, y_{n})}{\partial t} + a_{2,1}\dfrac{\partial f(t_{n}, y_{n})}{\partial y}f(t_{n}, y_{n}) \right] + \mathcal{O}(h^{3}) \tag{2}
\end{align*}
y n + 1 = y n + h [ b 1 f ( t n , y n ) + b 2 f ( t n , y n ) + h b 2 c 2 ∂ t ∂ f ( t n , y n ) + h b 2 a 2 , 1 ∂ y ∂ f ( t n , y n ) f ( t n , y n ) + O ( h 2 ) ] = y n + h ( b 1 + b 2 ) f ( t n , y n ) + h 2 b 2 [ c 2 ∂ t ∂ f ( t n , y n ) + a 2 , 1 ∂ y ∂ f ( t n , y n ) f ( t n , y n ) ] + O ( h 3 ) ( 2 )
However, the total derivative of y ′ = f ( t , y ) y^{\prime} = f(t, y) y ′ = f ( t , y ) d ( y ′ ) = ∂ f ( t , y ) ∂ t d t + ∂ f ( t , y ) ∂ y d y d (y^{\prime}) = \dfrac{\partial f(t, y)}{\partial t}dt + \dfrac{\partial f(t, y)}{\partial y}dy d ( y ′ ) = ∂ t ∂ f ( t , y ) d t + ∂ y ∂ f ( t , y ) d y ′ = d d t ^{\prime} = \dfrac{d}{dt} ′ = d t d y ′ ′ y^{\prime \prime} y ′′
y ′ ′ = ∂ f ( t , y ) ∂ t + ∂ f ( t , y ) ∂ y d y d t = ∂ f ( t , y ) ∂ t + ∂ f ( t , y ) ∂ y f ( t , y )
y^{\prime \prime} = \dfrac{\partial f(t, y)}{\partial t} + \dfrac{\partial f(t, y)}{\partial y}\dfrac{dy}{dt} = \dfrac{\partial f(t, y)}{\partial t} + \dfrac{\partial f(t, y)}{\partial y}f(t, y)
y ′′ = ∂ t ∂ f ( t , y ) + ∂ y ∂ f ( t , y ) d t d y = ∂ t ∂ f ( t , y ) + ∂ y ∂ f ( t , y ) f ( t , y )
By the Taylor theorem , we obtain the following:
y n + 1 = y n + h y ′ + h 2 2 y ′ ′ + O ( h 3 ) = y n + h f ( t n , y n ) + h 2 2 ( ∂ f ( t , y ) ∂ t + ∂ f ( t , y ) ∂ y f ( t , y ) ) + O ( h 3 )
\begin{align*}
y_{n+1} &= y_{n} + hy^{\prime} + \dfrac{h^{2}}{2}y^{\prime \prime} + \mathcal{O}(h^{3}) \\
&= y_{n} + hf(t_{n},y_{n}) + \dfrac{h^{2}}{2}\left( \dfrac{\partial f(t, y)}{\partial t} + \dfrac{\partial f(t, y)}{\partial y}f(t, y) \right) + \mathcal{O}(h^{3}) \tag{3} \\
\end{align*}
y n + 1 = y n + h y ′ + 2 h 2 y ′′ + O ( h 3 ) = y n + h f ( t n , y n ) + 2 h 2 ( ∂ t ∂ f ( t , y ) + ∂ y ∂ f ( t , y ) f ( t , y ) ) + O ( h 3 ) ( 3 )
By comparing ( 2 ) (2) ( 2 ) ( 3 ) (3) ( 3 )
b 1 + b 2 = 1 , b 2 c 2 = 1 2 , a 2 , 1 = c 2
b_{1} + b_{2} = 1, \qquad b_{2}c_{2} = \dfrac{1}{2}, \qquad a_{2,1} = c_{2}
b 1 + b 2 = 1 , b 2 c 2 = 2 1 , a 2 , 1 = c 2
The following RK tables satisfy this:
c A b t 0 1 2 1 2 0 1 , 0 2 3 2 3 1 2 1 4 3 4 , 0 1 1 1 2 1 2 1 2 ,
\begin{array}{c|c} \mathbf{c} & A \\ \hline & \mathbf{b}^{t} \end{array} \qquad
\begin{array}{c|c} 0 \\ \frac{1}{2} & \frac{1}{2} \\[0.5em] \hline & 0 & 1 \end{array}, \qquad
\begin{array}{c|c} 0 \\ \frac{2}{3} & \frac{2}{3} \\[0.5em] \hline \phantom{\dfrac{1}{2}} & \frac{1}{4} & \frac{3}{4} \end{array}, \qquad
\begin{array}{c|c} 0 \\ 1 & 1 \\[0.5em] \hline \phantom{\dfrac{1}{2}} & \frac{1}{2} & \frac{1}{2} \end{array}, \qquad
c A b t 0 2 1 2 1 0 1 , 0 3 2 2 1 3 2 4 1 4 3 , 0 1 2 1 1 2 1 2 1 ,
For the case where ν = 3 \nu = 3 ν = 3 First, by the Taylor theorem , we get the following:
y n + 1 = y n + h y ′ + h 2 2 y ′ ′ + h 3 6 y ′ ′ ′ + O ( h 4 ) = y n + h f n + h 2 2 ( ∂ f ∂ t + ∂ f ∂ y f ) + h 3 6 ( ∂ 2 f ∂ t 2 + ∂ 2 f ∂ y 2 f 2 + ∂ f ∂ t ∂ f ∂ y ) + O ( h 4 ) (4)
\begin{aligned}
y_{n+1} &= y_{n} + hy^{\prime} + \dfrac{h^{2}}{2}y^{\prime \prime} + \dfrac{h^{3}}{6}y^{\prime \prime \prime} + \mathcal{O}(h^{4}) \\
&= y_{n} + hf_{n} + \dfrac{h^{2}}{2}\left( \dfrac{\partial f}{\partial t} + \dfrac{\partial f}{\partial y}f \right) + \dfrac{h^{3}}{6}\left( \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{\partial^{2} f}{\partial y^{2}}f^{2} + \dfrac{\partial f}{\partial t}\dfrac{\partial f}{\partial y} \right) + \mathcal{O}(h^{4})
\tag{4}
\end{aligned}
y n + 1 = y n + h y ′ + 2 h 2 y ′′ + 6 h 3 y ′′′ + O ( h 4 ) = y n + h f n + 2 h 2 ( ∂ t ∂ f + ∂ y ∂ f f ) + 6 h 3 ( ∂ t 2 ∂ 2 f + ∂ y 2 ∂ 2 f f 2 + ∂ t ∂ f ∂ y ∂ f ) + O ( h 4 ) ( 4 )
And let us assume ∑ i = 1 j − 1 a j , i = c j \sum\limits_{i=1}^{j-1} a_{j,i} = c_{j} i = 1 ∑ j − 1 a j , i = c j ν = 3 \nu = 3 ν = 3
c 1 = 0 , c 2 = a 2 , 1 , c 3 = a 3 , 1 + a 3 , 2
c_{1} = 0,\qquad c_{2} = a_{2,1},\qquad c_{3} = a_{3,1} + a_{3,2}
c 1 = 0 , c 2 = a 2 , 1 , c 3 = a 3 , 1 + a 3 , 2
If we denote it as f n = f ( t n , y n ) f_{n} = f(t_{n}, y_{n}) f n = f ( t n , y n )
ξ 1 = y n , ξ 2 = y n + h c 2 f n , ξ 3 = y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n )
\xi_{1} = y_{n}, \quad \xi_{2} = y_{n} + hc_{2} f_{n}, \\[1em] \xi_{3} = y_{n} + ha_{3,1}f_{n} + ha_{3,2}f(t_{n} + c_{2}h, y_{n} + hc_{2} f_{n})
ξ 1 = y n , ξ 2 = y n + h c 2 f n , ξ 3 = y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n )
Therefore, y n + 1 y_{n+1} y n + 1
y n + 1 = y n + h b 1 f n + h b 2 f ( t n + c 2 h , ξ 2 ) + h b 3 f ( t n + c 3 h , ξ 3 ) = y n + h b 1 f n + h b 2 f ( t n + c 2 h , y n + h c 2 f n ) + h b 3 f ( t n + c 3 h , y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n ) )
\begin{align*}
y_{n+1}
&= y_{n} + hb_{1}f_{n} + hb_{2}f(t_{n} + c_{2}h, \xi_{2}) + hb_{3}f(t_{n} + c_{3}h, \xi_{3}) \\
&= y_{n} + hb_{1}f_{n} + hb_{2}f(t_{n} + c_{2}h, y_{n} + hc_{2} f_{n}) \\
& \qquad + hb_{3}f(t_{n} + c_{3}h, y_{n} + ha_{3,1}f_{n} + ha_{3,2}f(t_{n} + c_{2}h, y_{n} + hc_{2} f_{n})) \tag{5}
\end{align*}
y n + 1 = y n + h b 1 f n + h b 2 f ( t n + c 2 h , ξ 2 ) + h b 3 f ( t n + c 3 h , ξ 3 ) = y n + h b 1 f n + h b 2 f ( t n + c 2 h , y n + h c 2 f n ) + h b 3 f ( t n + c 3 h , y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n )) ( 5 )
The terms containing b 2 b_{2} b 2
f ( t n + c 2 h , y n + h c 2 f ( t n , y n ) ) = f n + ( h c 2 ) ∂ f ∂ t + ( h c 2 f n ) ∂ f ∂ y + ( c 2 h ) 2 2 ∂ 2 f ∂ t 2 + ( c 2 h f n ) 2 2 ∂ 2 f ∂ y 2 + O ( h 3 )
\begin{align*}
& f(t_{n} + c_{2}h, y_{n} + hc_{2}f(t_{n}, y_{n}) ) \\
&= f_{n} + (hc_{2}) \dfrac{\partial f}{\partial t} + (hc_{2}f_{n})\dfrac{\partial f}{\partial y} + \dfrac{(c_{2}h)^{2}}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{(c_{2}h f_{n})^{2}}{2} \dfrac{\partial^{2} f}{\partial y^{2}} + \mathcal{O}(h^{3}) \\
\end{align*}
f ( t n + c 2 h , y n + h c 2 f ( t n , y n )) = f n + ( h c 2 ) ∂ t ∂ f + ( h c 2 f n ) ∂ y ∂ f + 2 ( c 2 h ) 2 ∂ t 2 ∂ 2 f + 2 ( c 2 h f n ) 2 ∂ y 2 ∂ 2 f + O ( h 3 )
Substituting this into ( 5 ) (5) ( 5 )
y n + 1 = y n + h ( b 1 + b 2 ) f n + h 2 b 2 c 2 ∂ f ∂ t + h 2 b 2 c 2 ∂ f ∂ y f n + h 3 c 2 2 b 2 2 ∂ 2 f ∂ t 2 + h 3 c 2 2 b 2 2 ∂ 2 f ∂ y 2 f n 2 + h b 3 f ( t n + c 3 h , y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n ) ) + O ( h 4 ) (6)
\begin{aligned}
y_{n+1}
&= y_{n} + h(b_{1} + b_{2})f_{n} + h^{2}b_{2}c_{2} \dfrac{\partial f}{\partial t} + h^{2}b_{2}c_{2} \dfrac{\partial f}{\partial y}f_{n} \\
& \qquad + \dfrac{h^{3}c_{2}^{2}b_{2}}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{h^{3}c_{2}^{2}b_{2}}{2} \dfrac{\partial^{2} f}{\partial y^{2}}f_{n}^{2} \\
& \qquad + hb_{3}f(t_{n} + c_{3}h, y_{n} + ha_{3,1}f_{n} + ha_{3,2}f(t_{n} + c_{2}h, y_{n} + hc_{2} f_{n})) + \mathcal{O}(h^{4}) \tag{6}
\end{aligned}
y n + 1 = y n + h ( b 1 + b 2 ) f n + h 2 b 2 c 2 ∂ t ∂ f + h 2 b 2 c 2 ∂ y ∂ f f n + 2 h 3 c 2 2 b 2 ∂ t 2 ∂ 2 f + 2 h 3 c 2 2 b 2 ∂ y 2 ∂ 2 f f n 2 + h b 3 f ( t n + c 3 h , y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n )) + O ( h 4 ) ( 6 )
Now, by calculating the remaining terms containing b 3 b_{3} b 3
f ( t n + c 3 h , y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n ) ) = f n + c 3 h ∂ f ∂ t + ( h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n ) ) ∂ f ∂ y + h 2 c 3 2 2 ∂ 2 f ∂ t 2 + ( h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n ) ) 2 2 ∂ 2 f ∂ y 2 + O ( h 3 ) = f n + c 3 h ∂ f ∂ t + h a 3 , 1 f n ∂ f ∂ y + h a 3 , 2 [ f n + h c 2 ∂ f ∂ t + O ( h 2 ) ] ∂ f ∂ y + h 2 c 3 2 2 ∂ 2 f ∂ t 2 + ( h a 3 , 1 f n + h a 3 , 2 f n ) 2 2 ∂ 2 f ∂ y 2 + O ( h 3 ) = f n + c 3 h ∂ f ∂ t + h c 3 f n ∂ f ∂ y + h 2 c 2 a 3 , 2 ∂ f ∂ t ∂ f ∂ y + h 2 c 3 2 2 ∂ 2 f ∂ t 2 + ( h c 3 f n ) 2 2 ∂ 2 f ∂ y 2 + O ( h 3 )
\begin{align*}
& f(t_{n} + c_{3}h, y_{n} + ha_{3,1}f_{n} + ha_{3,2}f(t_{n} + c_{2}h, y_{n} + hc_{2} f_{n})) \\
&= f_{n} + c_{3}h \dfrac{\partial f}{\partial t} + \left( ha_{3,1}f_{n} + ha_{3,2}f(t_{n} + c_{2}h, y_{n} + hc_{2} f_{n}) \right) \dfrac{\partial f}{\partial y} \\
& \qquad + \dfrac{h^{2}c_{3}^{2}}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{\left( ha_{3,1}f_{n} + ha_{3,2}f(t_{n} + c_{2}h, y_{n} + hc_{2} f_{n}) \right)^{2}}{2} \dfrac{\partial^{2} f}{\partial y^{2}} + \mathcal{O}(h^{3}) \\
&= f_{n} + c_{3}h \dfrac{\partial f}{\partial t} + ha_{3,1}f_{n}\dfrac{\partial f}{\partial y} + ha_{3,2} \left[ f_{n} + hc_{2}\dfrac{\partial f}{\partial t} + \mathcal{O}(h^{2}) \right] \dfrac{\partial f}{\partial y} \\
& \qquad + \dfrac{h^{2}c_{3}^{2}}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{(ha_{3,1}f_{n} + ha_{3,2}f_{n})^{2}}{2} \dfrac{\partial^{2} f}{\partial y^{2}} + \mathcal{O}(h^{3}) \\
&= f_{n} + c_{3}h\dfrac{\partial f}{\partial t} + hc_{3}f_{n}\dfrac{\partial f}{\partial y} + h^{2}c_{2}a_{3,2}\dfrac{\partial f}{\partial t} \dfrac{\partial f}{\partial y} \\
& \qquad + \dfrac{h^{2}c_{3}^{2}}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{(hc_{3}f_{n})^{2}}{2} \dfrac{\partial^{2} f}{\partial y^{2}} + \mathcal{O}(h^{3}) \\
\end{align*}
f ( t n + c 3 h , y n + h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n )) = f n + c 3 h ∂ t ∂ f + ( h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n ) ) ∂ y ∂ f + 2 h 2 c 3 2 ∂ t 2 ∂ 2 f + 2 ( h a 3 , 1 f n + h a 3 , 2 f ( t n + c 2 h , y n + h c 2 f n ) ) 2 ∂ y 2 ∂ 2 f + O ( h 3 ) = f n + c 3 h ∂ t ∂ f + h a 3 , 1 f n ∂ y ∂ f + h a 3 , 2 [ f n + h c 2 ∂ t ∂ f + O ( h 2 ) ] ∂ y ∂ f + 2 h 2 c 3 2 ∂ t 2 ∂ 2 f + 2 ( h a 3 , 1 f n + h a 3 , 2 f n ) 2 ∂ y 2 ∂ 2 f + O ( h 3 ) = f n + c 3 h ∂ t ∂ f + h c 3 f n ∂ y ∂ f + h 2 c 2 a 3 , 2 ∂ t ∂ f ∂ y ∂ f + 2 h 2 c 3 2 ∂ t 2 ∂ 2 f + 2 ( h c 3 f n ) 2 ∂ y 2 ∂ 2 f + O ( h 3 )
Substituting and organizing this into ( 6 ) (6) ( 6 )
y n + 1 = y n + h ( b 1 + b 2 ) f n + h 2 b 2 c 2 ∂ f ∂ t + h 2 b 2 c 2 ∂ f ∂ y f n + h 3 c 2 2 b 2 2 ∂ 2 f ∂ t 2 + h 3 c 2 2 b 2 2 ∂ 2 f ∂ y 2 f n 2 + h b 3 [ f n + c 3 h ∂ f ∂ t + h c 3 f n ∂ f ∂ y + h 2 c 2 a 3 , 2 ∂ f ∂ t ∂ f ∂ y + h 2 c 3 2 2 ∂ 2 f ∂ t 2 + ( h c 3 f n ) 2 2 ∂ 2 f ∂ y 2 ] + O ( h 4 ) = y n + h ( b 1 + b 2 + b 3 ) f n + h 2 ( b 2 c 2 + b 3 c 3 ) ∂ f ∂ t + h 2 ( b 2 c 2 + b 3 c 3 ) ∂ f ∂ y f n + h 3 ( b 2 c 2 2 + b 3 c 3 2 ) 2 ∂ 2 f ∂ t 2 + h 3 b 3 c 2 a 3 , 2 ∂ f ∂ t ∂ f ∂ y + h 3 ( b 2 c 2 2 + b 3 c 3 2 ) 2 f n 2 ∂ 2 f ∂ y 2 + O ( h 4 )
\begin{align*}
y_{n+1}
&= y_{n} + h(b_{1} + b_{2})f_{n} + h^{2}b_{2}c_{2} \dfrac{\partial f}{\partial t} + h^{2}b_{2}c_{2} \dfrac{\partial f}{\partial y}f_{n} \\
& \qquad + \dfrac{h^{3}c_{2}^{2}b_{2}}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{h^{3}c_{2}^{2}b_{2}}{2} \dfrac{\partial^{2} f}{\partial y^{2}}f_{n}^{2} \\
& \qquad + hb_{3} \left[ f_{n} + c_{3}h\dfrac{\partial f}{\partial t} + hc_{3}f_{n}\dfrac{\partial f}{\partial y} + h^{2}c_{2}a_{3,2}\dfrac{\partial f}{\partial t} \dfrac{\partial f}{\partial y} \right. \\
& \qquad + \left. \dfrac{h^{2}c_{3}^{2}}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + \dfrac{(hc_{3}f_{n})^{2}}{2} \dfrac{\partial^{2} f}{\partial y^{2}} \right] + \mathcal{O}(h^{4}) \\
&= y_{n} + h(b_{1} + b_{2} + b_{3})f_{n} + h^{2}(b_{2}c_{2} + b_{3}c_{3}) \dfrac{\partial f}{\partial t} + h^{2}(b_{2}c_{2} + b_{3}c_{3}) \dfrac{\partial f}{\partial y}f_{n} \\
& \qquad + \dfrac{h^{3}(b_{2}c_{2}^{2} + b_{3}c_{3}^{2})}{2} \dfrac{\partial^{2} f}{\partial t^{2}} + h^{3}b_{3}c_{2}a_{3,2}\dfrac{\partial f}{\partial t} \dfrac{\partial f}{\partial y} + \dfrac{h^{3}(b_{2}c_{2}^{2} + b_{3}c_{3}^{2})}{2} f_{n}^{2}\dfrac{\partial^{2} f}{\partial y^{2}} + \mathcal{O}(h^{4})
\end{align*}
y n + 1 = y n + h ( b 1 + b 2 ) f n + h 2 b 2 c 2 ∂ t ∂ f + h 2 b 2 c 2 ∂ y ∂ f f n + 2 h 3 c 2 2 b 2 ∂ t 2 ∂ 2 f + 2 h 3 c 2 2 b 2 ∂ y 2 ∂ 2 f f n 2 + h b 3 [ f n + c 3 h ∂ t ∂ f + h c 3 f n ∂ y ∂ f + h 2 c 2 a 3 , 2 ∂ t ∂ f ∂ y ∂ f + 2 h 2 c 3 2 ∂ t 2 ∂ 2 f + 2 ( h c 3 f n ) 2 ∂ y 2 ∂ 2 f ] + O ( h 4 ) = y n + h ( b 1 + b 2 + b 3 ) f n + h 2 ( b 2 c 2 + b 3 c 3 ) ∂ t ∂ f + h 2 ( b 2 c 2 + b 3 c 3 ) ∂ y ∂ f f n + 2 h 3 ( b 2 c 2 2 + b 3 c 3 2 ) ∂ t 2 ∂ 2 f + h 3 b 3 c 2 a 3 , 2 ∂ t ∂ f ∂ y ∂ f + 2 h 3 ( b 2 c 2 2 + b 3 c 3 2 ) f n 2 ∂ y 2 ∂ 2 f + O ( h 4 )
By comparing the coefficients of the above expression and ( 4 ) (4) ( 4 )
b 1 + b 2 + b 3 = 1 , b 2 c 2 + b 3 c 3 = 1 2 , b 2 c 2 2 + b 3 c 3 2 = 1 3 , b 3 c 2 a 3 , 2 = 1 6
b_{1} + b_{2} + b_{3} = 1, \quad b_{2}c_{2} + b_{3}c_{3} = \dfrac{1}{2}, \quad b_{2}c_{2}^{2} + b_{3}c_{3}^{2} = \dfrac{1}{3}, \quad b_{3}c_{2}a_{3,2} = \dfrac{1}{6}
b 1 + b 2 + b 3 = 1 , b 2 c 2 + b 3 c 3 = 2 1 , b 2 c 2 2 + b 3 c 3 2 = 3 1 , b 3 c 2 a 3 , 2 = 6 1
The following RK tables satisfy this:
0 1 2 1 2 1 − 1 2 1 2 1 6 2 3 1 6 ( classical RK method )
\begin{array}{c|c} 0 \\ \frac{1}{2} & \frac{1}{2} \\[0.5em] 1 & -1 & 2 \\[0.5em] \hline \phantom{\dfrac{1}{2}} & \frac{1}{6} & \frac{2}{3} & \frac{1}{6} \end{array} \quad (\text{classical RK method})
0 2 1 1 2 1 2 1 − 1 6 1 2 3 2 6 1 ( classical RK method ) 0 2 3 2 3 2 3 0 2 3 1 2 1 4 3 8 3 8 ( Nystrom scheme )
\begin{array}{c|c} 0 \\ \frac{2}{3} & \frac{2}{3} \\[0.5em] \frac{2}{3} & 0 & \frac{2}{3} \\[0.5em] \hline \phantom{\dfrac{1}{2}} & \frac{1}{4} & \frac{3}{8} & \frac{3}{8} \end{array} \quad (\text{Nystrom scheme})
0 3 2 3 2 2 1 3 2 0 4 1 3 2 8 3 8 3 ( Nystrom scheme )
For the case where ν = 4 \nu = 4 ν = 4 Using the same method, we can find the conditions for the coefficients. According to the rules observed above, we can infer the following:
b 1 + b 2 + b 3 + b 4 = 1 , b 2 c 2 + b 3 c 3 + b 4 c 4 = 1 2 , b 2 c 2 2 + b 3 c 3 2 + b 4 c 4 2 = 1 3 , b 2 c 2 3 + b 3 c 3 3 + b 4 c 4 3 = 1 4
b_{1} + b_{2} + b_{3} + b_{4} = 1, \quad b_{2}c_{2} + b_{3}c_{3} + b_{4}c_{4} = \dfrac{1}{2}, \\[1em]
b_{2}c_{2}^{2} + b_{3}c_{3}^{2} + b_{4}c_{4}^{2} = \dfrac{1}{3}, \quad b_{2}c_{2}^{3} + b_{3}c_{3}^{3} + b_{4}c_{4}^{3} = \dfrac{1}{4}
b 1 + b 2 + b 3 + b 4 = 1 , b 2 c 2 + b 3 c 3 + b 4 c 4 = 2 1 , b 2 c 2 2 + b 3 c 3 2 + b 4 c 4 2 = 3 1 , b 2 c 2 3 + b 3 c 3 3 + b 4 c 4 3 = 4 1
The most famous coefficients in RK4 are as follows. The table below actually satisfies the above conditions.
0 1 2 1 2 1 2 0 1 2 1 0 0 1 0 0 1 6 1 3 1 3 1 6
\begin{array}{c|cccc}
0 & \\[0.5em]
\frac{1}{2} & \frac{1}{2} \\[0.5em]
\frac{1}{2} & 0 & \frac{1}{2} \\[0.5em]
1 & 0 & 0 & 1 \\[0.5em]
\hline {\displaystyle {\color{white}\dfrac{0}{0}}} & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} & \frac{1}{6}
\end{array}
0 2 1 2 1 1 0 0 2 1 0 0 6 1 2 1 0 3 1 1 3 1 6 1
See Also 