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Runge-Kutta method, Matrix exponential
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9.0px (5.0) by clicking Small A among Small A, Big A on the top side of this document.Table of Contents
PH.Posting History
1.Differential equation with matrices (RLC circuit)
2.\( \exp(A_{ij}) \)
3.Evolution of variables
4.Numerical implementation
4.1.Euler's method (Up to 1st-order)
5.Let's try to satisfy/content higher orders.
5.1.Taylor expansion of $\Psi$
6.Let's match the coefficients up to $\textrm{O}(h^4)$.
6.1.Case $k=0$
6.2.Case $k=1$
6.3.Case $k=2$
6.4.Case $k=3$
7.Evolution of variables when it explicitely depends on itself.
8.Numerical implementation to satisfy/content higher orders.
8.1.Taylor expansion of $\Psi$
9.Let's match the coefficients up to $\textrm{O}(h^k)$.
9.1.Order of $h$. $k=0$.
9.2.Order of $h^2$. $k=1$.
9.3.Order of $h^3$. $k=2$.
9.4.Order of $h^4$. $k=3$.
9.5.Order of $h^5$. $k=4$.
10.Written in papers by color ballpoint pen
11.Written in papers 2.
12.Candidate set $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$
12.1.Let's consider $b_0 = \frac{1}{6}$, $b_1 = \frac{1}{3}$, $b_2 = \frac{1}{3}$, $b_3 = \frac{1}{6}$ and $c_0 = 0$, $c_1 = \frac{1}{2}$, $c_2 = \frac{1}{2}$, $c_3 = 1$.
12.2.Let's consider $b_0 = \frac{1}{8}$, $b_1 = \frac{3}{8}$, $b_2 = \frac{3}{8}$, $b_3 = \frac{1}{8}$ and $c_0 = 0$, $c_1 = \frac{1}{3}$, $c_2 = \frac{2}{3}$, $c_3 = 1$.
12.3.Let's consider $b_0 = \frac{16}{135}$, $b_1 = 0$, $b_2 = \frac{6656}{12825}$, $b_3 = \frac{28561}{56430}$, $b_4 = -\frac{9}{50}$, $b_5 = \frac{2}{55}$ and $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$.
12.4.Let's consider $b_0 = \frac{25}{216}$, $b_1 = 0$, $b_2 = \frac{1408}{2565}$, $b_3 = \frac{2197}{4104}$, $b_4 = -\frac{1}{5}$, $b_5 = 0$ and $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$.
13.Adaptive step size
Refs.References and Related Articles
T1.Differential equation with matrices (RLC circuit)
▼ Show/Hide
(1-1)
\[ \begin{align*}
\frac{d x_k}{d t} &= A_{ki} x_i + B_{ki} y_i \\
y_k &= C_{ki} x_i + D_{ki} y_i
\end{align*} \]
where dummy (repeated) indices means summation over all (e.g. \( A_{ki} x_i \equiv \sum_{i} A_{ki} x_i \) ).
▲ Hide
T2.\( \exp(A_{ij}) \)
▼ Show/Hide
Taylor series definition of matrix exponential
[01]
.Ref. [01] en.wikipedia.org :: Matrix exponential
(2-1)
\[ \exp(A_{ij}) = \sum_{k=0}^{\infty} \frac{1}{k!} A^k \]
If \( A_{ij} \) can be diagonalized as
(2-2)
\[ A_{ij} = U_{ik} \delta_{kl} D_{kl} U^{-1}_{lj} \]
, then
(2-3)
\[ \exp(A_{ij}) = \exp(U_{ik} \delta_{kl} D_{kl} U^{-1}_{lj})
= U_{ik} \exp(\delta_{kl} D_{kl}) U^{-1}_{lj} \]
where
(2-4)
\[ \exp(\delta_{kl} D_{kl}) = \delta_{kl} \exp(D_{kl}) . \]
Explicitely speaking,
(2-5)
\[ A = \begin{bmatrix}
a_1 & 0 & \cdots & 0 \\
0 & a_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_n
\end{bmatrix} , \]
then its exponential can be obtained by exponentiating each entry on the main diagonal:
(2-6)
\[ e^A = \begin{bmatrix}
e^{a_1} & 0 & \cdots & 0 \\
0 & e^{a_2} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & e^{a_n}
\end{bmatrix} . \]
▲ Hide
T3.Evolution of variables
▼ Show/Hide
Let an evolution of $\Psi$ from the initial value be specified as follows:
(3-1)
\[ \Psi \big( \{ x^{\mu} \} \big)
= \Psi \big( \{ x^{\mu} (\lambda) \} \big) . \]
The evolution equation is
(3-2)
\[ \frac{d \Psi \big( \{ x^{\mu} (\lambda) \} \big)}{d \lambda}
= f \Big( \{ x^{\mu} (\lambda) \} ; \lambda \Big) . \]
Then we want to find the final value $\Psi_f$ from the initial value $\Psi_i$ where
(3-3)
\[ \Psi_f \equiv \Psi \big( \{ x^{\mu} (\lambda_f) \} \big)
~~~ \text{and} ~~~
\Psi_i \equiv \Psi \big( \{ x^{\mu} (\lambda_i) \} \big) . \]
And let's think the case that the path $x^{\mu} (\lambda)$ is set to be fixed.
Then
(3-4)
\[ \begin{align*}
\Psi_f - \Psi_i & = \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} (\lambda) \} ; \lambda \Big) ~ d \lambda \\
&= \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} \} ; \lambda \Big) ~ d \lambda
\end{align*} \]
where $f()$ is explicitely expressed by $\{ x^{\mu} \}$, $\lambda$.
▲ Hide
T4.Numerical implementation
▼ Show/Hide
T4.1.Euler's method (Up to 1st-order)
(4-1)
\[ \Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + \frac{\lambda_f - \lambda_i}{N} (n+1) \bigg) \big\} \Big)
- \Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + \frac{\lambda_f - \lambda_i}{N} n \bigg) \big\} \Big) \]
becomes, with defining $h \equiv \frac{\lambda_f - \lambda_i}{N}$,
(4-2)
\[ \begin{align*}
\Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + h (n+1) \bigg) \big\} \Big)
- \Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + h n \bigg) \big\} \Big)
&= \sum_{j=0}^{n-1} f \Big( \{ x^{\mu} (\lambda_i + h j) \} ; \lambda_i + h j \Big) \cdot h \\
&= \sum_{j=0}^{n-1} f \Big( \lambda_i + h j \Big) \cdot h .
\end{align*} \]
▲ Hide
T5.Let's try to satisfy/content higher orders.
▼ Show/Hide
We can candidate that the below equation is correct upto the fourth order $\mathrm{O}(h^4)$.
Difining $\lambda_n \equiv \lambda_i + h n$,
(5-1)
\[ \Psi \big( \{ x^{\mu} (\lambda_{n+1}) \} \big)
:=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_k b_k f_k , \]
where
(5-2)
\[ \begin{align*}
f_0 &= f \Big( \{ x^{\mu} (\lambda_n) \} ; \lambda_n \Big) \\
f_1 &= f \Big( \{ x^{\mu} (\lambda_n + c_1 \cdot h) \} ; \lambda_n + c_1 \cdot h \Big) \\
f_2 &= f \Big( \{ x^{\mu} (\lambda_n + c_2 \cdot h) \} ; \lambda_n + c_2 \cdot h \Big) \\
f_3 &= f \Big( \{ x^{\mu} (\lambda_n + c_3 \cdot h) \} ; \lambda_n + c_3 \cdot h \Big) .
\end{align*} \]
Generally $f$ can be written by
(5-3)
\[ f = f \Big( \{ x^{\mu} (\lambda_n + c \cdot h) \} ; \lambda_n + c \cdot h \Big) . \]
With index $p$,
(5-4)
\[ f_p = f \Big( \{ x^{\mu} (\lambda_n + c_p \cdot h) \} ; \lambda_n + c_p \cdot h \Big) . \]
T5.1.Taylor expansion of $\Psi$
(5-5)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \frac{d \Psi (\lambda_{n})}{d \lambda} \cdot h
+ \frac{1}{2!} \frac{d^2 \Psi (\lambda_{n})}{d \lambda^2} \cdot h^2
+ \frac{1}{3!} \frac{d^3 \Psi (\lambda_{n})}{d \lambda^3} \cdot h^3
+ \cdots \\
&= \sum_{k=0}^{\infty} \frac{1}{k!} \frac{d^k \Psi (\lambda)}{d \lambda^k} \bigg|_{\lambda = \lambda_n} \cdot h^k ~~ .
\end{align*} \]
Since $f_0 = \frac{d \Psi}{d \lambda}$,
(5-6)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ f_0 (\lambda_n) \cdot h
+ \frac{1}{2!} \frac{d f_0 (\lambda_n)}{d \lambda} \cdot h^2
+ \frac{1}{3!} \frac{d^2 f_0 (\lambda_n)}{d \lambda^2} \cdot h^3 + \cdots \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \sum_{k=0}^{\infty} \frac{1}{(k+1)!} \frac{d^k f_0 (\lambda_n)}{d \lambda^k} \cdot h^{k+1} ~~ .
\end{align*} \]
The taylor expansion of the Runge-Kutta method becomes
(5-7)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p b_p f_p (\lambda_n , h) \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_p (\lambda ; h)}{d h^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot h^{k} ,
\end{align*} \]
where
(5-8)
\[ \begin{align*}
\frac{d f_p \big( \lambda ; h \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}
&= \bigg[
\frac{d x^{\mu}}{d h}
\frac{\partial}{\partial x^{\mu}}
+ \frac{d \lambda}{d h}
\frac{\partial}{\partial \lambda}
\bigg] f_p \bigg|_{\lambda=\lambda_n , ~ h=0} \\
&=
c_p \cdot \bigg[
\frac{d x^{\mu}}{d \lambda}
\frac{\partial}{\partial x^{\mu}}
+
\frac{\partial}{\partial \lambda}
\bigg] f_p \bigg|_{\lambda=\lambda_n, ~ h=0} \\
&=
c_p \cdot
\frac{d f_p}{d \lambda} \bigg|_{\lambda=\lambda_n, ~ h=0}
\end{align*} \]
and
(5-9)
\[ \begin{align*}
\frac{d^k f_p \big( \lambda ; h \big)}{d h^k} \bigg|_{\lambda=\lambda_n, ~ h=0}
&= \bigg[
c_p \cdot
\frac{d}{d \lambda}
\bigg]^k f_p \big( \lambda ; h \big) \Bigg|_{\lambda=\lambda_n, ~ h=0} \\
&=
c_p^k \cdot \frac{d^k f_p}{d \lambda^k}
\big( \lambda ; h \big) \Bigg|_{\lambda=\lambda_n, ~ h=0} .
\end{align*} \]
Since \( \frac{d^k f_p}{d \lambda^k} \bigg|_{\lambda=\lambda_n, ~ h=0} = \frac{d^k f_0}{d \lambda^k} \bigg|_{\lambda=\lambda_n, ~ h=0} \) , therefore Eq.
(5-7)
(5-7)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p b_p f_p (\lambda_n , h) \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_p (\lambda ; h)}{d h^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot h^{k} ,
\end{align*} \]
(5-10)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_0 (\lambda ; h)}{d \lambda^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot c_p^k ~ h^{k+1} ,
\end{align*} \]
▲ Hide
T6.Let's match the coefficients up to $\textrm{O}(h^4)$.
▼ Show/Hide
Matching the coefficients of Eq.
(5-6)
(5-6)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ f_0 (\lambda_n) \cdot h
+ \frac{1}{2!} \frac{d f_0 (\lambda_n)}{d \lambda} \cdot h^2
+ \frac{1}{3!} \frac{d^2 f_0 (\lambda_n)}{d \lambda^2} \cdot h^3 + \cdots \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \sum_{k=0}^{\infty} \frac{1}{(k+1)!} \frac{d^k f_0 (\lambda_n)}{d \lambda^k} \cdot h^{k+1} ~~ .
\end{align*} \]
(5-10)
(5-10)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_0 (\lambda ; h)}{d \lambda^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot c_p^k ~ h^{k+1} ,
\end{align*} \]
(6-1)
\[ \frac{1}{(k+1)!} = \sum_p \frac{1}{k!} b_p c_p^k \]
Therefore
(6-2)
\[ \sum_p b_p c_p^k = \frac{1}{k+1} ~~~~~ \text{for} ~~ k=0, 1, 2, \cdots \]
▲ Hide
T7.Evolution of variables when it explicitely depends on itself.
▼ Show/Hide
Let's say that the evolution equation is given by
(7-1)
\[ \frac{d \Psi \big( \{ x^{\mu} (\lambda) \} \big)}{d \lambda}
= f \Big( \{ x^{\mu} (\lambda) \} ; \lambda ; \Psi (\lambda) \Big) . \]
Then we want to find the final value $\Psi_f$ from the initial value $\Psi_i$ where
(7-2)
\[ \Psi_f \equiv \Psi \big( \{ x^{\mu} (\lambda_f) \} \big)
~~~ \text{and} ~~~
\Psi_i \equiv \Psi \big( \{ x^{\mu} (\lambda_i) \} \big) . \]
And let's think the case that the path $x^{\mu} (\lambda)$ is set to be fixed.
Then
(7-3)
\[ \begin{align*}
\Psi_f - \Psi_i & = \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} (\lambda) \} ; \lambda, \Psi (\lambda) \Big) ~ d \lambda \\
&= \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} \} ; \lambda, \Psi \Big) ~ d \lambda
\end{align*} \]
where $f()$ is explicitely expressed by $\{ x^{\mu} \}$, $\lambda$, and $\Psi(\lambda)$.
▲ Hide
T8.Numerical implementation to satisfy/content higher orders.
▼ Show/Hide
We can candidate that the below equation is correct upto the $m$-th order $\mathrm{O}(h^{m})$.
Difining $\lambda_n \equiv \lambda_i + h n$,
(8-1)
\[ \Psi \big( \{ x^{\mu} (\lambda_{n+1}) \} \big)
:=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_k b_k f_k , \]
where
(8-2)
\[ \begin{align*}
f_0 &= f \Big( \{ x^{\mu} (\lambda_n) \} ; \lambda_n ; \Psi (\lambda_n) \Big) \\
f_1 &= f \Big( \{ x^{\mu} (\lambda_n + c_1 \cdot h) \} ; \lambda_n + c_1 \cdot h ; \Psi (\lambda_n) + h \cdot a_{10} f_0 \Big) \\
f_2 &= f \Big( \{ x^{\mu} (\lambda_n + c_2 \cdot h) \} ; \lambda_n + c_2 \cdot h ; \Psi (\lambda_n) + h \cdot \big( a_{20} f_0 + a_{21} f_1 \big) \Big) \\
f_3 &= f \Big( \{ x^{\mu} (\lambda_n + c_3 \cdot h) \} ; \lambda_n + c_3 \cdot h ; \Psi (\lambda_n) + h \cdot \big( a_{30} f_0 + a_{31} f_1 + a_{32} f_2 \big) \Big) .
\end{align*} \]
Generally $f$ can be written by
(8-3)
\[ f = f \Big( \{ x^{\mu} (\lambda_n + c \cdot h) \} ; \lambda_n + c \cdot h ; \Psi (\lambda_n) + h \cdot \sum_{m} \big( a_{m} f_{m} \big) \Big) . \]
With index $p$,
(8-4)
\[ f_p = f \Big( \{ x^{\mu} (\lambda_n + c_p \cdot h) \} ; \lambda_n + c_p \cdot h ; \Psi (\lambda_n) + h \cdot \sum_{m} \big( a_{pm} f_{m} \big) \Big) . \]
T8.1.Taylor expansion of $\Psi$
(8-5)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \frac{d \Psi (\lambda_{n})}{d \lambda} \cdot h
+ \frac{1}{2!} \frac{d^2 \Psi (\lambda_{n})}{d \lambda^2} \cdot h^2
+ \frac{1}{3!} \frac{d^3 \Psi (\lambda_{n})}{d \lambda^3} \cdot h^3
+ \cdots \\
&= \sum_{k=0}^{\infty} \frac{1}{k!} \frac{d^k \Psi (\lambda)}{d \lambda^k} \bigg|_{\lambda = \lambda_n} \cdot h^k ~~ .
\end{align*} \]
Since $f_0 = \frac{d \Psi}{d \lambda}$,
(8-6)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ f_0 (\lambda_n) \cdot h
+ \frac{1}{2!} \frac{d f_0 (\lambda_n)}{d \lambda} \cdot h^2
+ \frac{1}{3!} \frac{d^2 f_0 (\lambda_n)}{d \lambda^2} \cdot h^3 + \cdots \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \sum_{k=0}^{\infty} \frac{1}{(k+1)!} \frac{d^k f_0 (\lambda_n)}{d \lambda^k} \cdot h^{k+1} ~~ .
\end{align*} \]
The taylor expansion of the Runge-Kutta method becomes
(8-7)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p b_p f_p (\lambda_n ; h ; \Psi) \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_p (\lambda ; h ; \Psi)}{d h^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot h^{k} ,
\end{align*} \]
where
(8-8)
\[ \begin{align*}
\frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}
&=
\bigg[
\frac{d x^{\mu}}{d h}
\frac{\partial}{\partial x^{\mu}}
+ \frac{d \lambda}{d h}
\frac{\partial}{\partial \lambda}
+ \frac{d \Psi}{d h}
\frac{\partial}{\partial \Psi}
\bigg] f_p \bigg|_{\lambda=\lambda_n , ~ h=0} \\
&=
\bigg[
c_p \cdot
\frac{d x^{\mu}}{d \lambda}
\frac{\partial}{\partial x^{\mu}}
+ c_p \cdot
\frac{\partial}{\partial \lambda}
+ \sum_m \big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \big)
\frac{\partial}{\partial \Psi}
\bigg] f_p \bigg|_{\lambda=\lambda_n, ~ h=0} .
\end{align*} \]
Since
(8-9)
\[ \begin{align*}
\frac{d f_0}{d \lambda} \bigg|_{\lambda=\lambda_n}
&=
\bigg[
\frac{d x^{\mu}}{d \lambda}
\frac{\partial}{\partial x^{\mu}}
+ \frac{\partial}{\partial \lambda}
+ \frac{d \Psi}{d \lambda}
\frac{\partial}{\partial \Psi}
\bigg] f_0 \bigg|_{\lambda=\lambda_n} \\
&=
\bigg[
\frac{d x^{\mu}}{d \lambda}
\frac{\partial}{\partial x^{\mu}}
+ \frac{\partial}{\partial \lambda}
+ f_0
\frac{\partial}{\partial \Psi}
\bigg] f_0 \bigg|_{\lambda=\lambda_n} ~~ ,
\end{align*} \]
Eq.
(8-8)
(8-8)
\[ \begin{align*}
\frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}
&=
\bigg[
\frac{d x^{\mu}}{d h}
\frac{\partial}{\partial x^{\mu}}
+ \frac{d \lambda}{d h}
\frac{\partial}{\partial \lambda}
+ \frac{d \Psi}{d h}
\frac{\partial}{\partial \Psi}
\bigg] f_p \bigg|_{\lambda=\lambda_n , ~ h=0} \\
&=
\bigg[
c_p \cdot
\frac{d x^{\mu}}{d \lambda}
\frac{\partial}{\partial x^{\mu}}
+ c_p \cdot
\frac{\partial}{\partial \lambda}
+ \sum_m \big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \big)
\frac{\partial}{\partial \Psi}
\bigg] f_p \bigg|_{\lambda=\lambda_n, ~ h=0} .
\end{align*} \]
(8-10)
\[ \begin{align*}
\frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}
&=
c_p \cdot \bigg( \frac{f_0}{d \lambda} - f_0 \frac{\partial f_0}{\partial \Psi} \bigg)
+ \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} \\
&=
c_p \cdot \frac{f_0}{d \lambda}
+ \bigg[ \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) - c_p f_0 \bigg] \frac{\partial f_p}{\partial \Psi} \bigg|_{\lambda=\lambda_n , ~ h=0}
\end{align*} \]
▲ Hide
T9.Let's match the coefficients up to $\textrm{O}(h^k)$.
▼ Show/Hide
Matching the coefficients of Eq.
(8-6)
(8-6)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ f_0 (\lambda_n) \cdot h
+ \frac{1}{2!} \frac{d f_0 (\lambda_n)}{d \lambda} \cdot h^2
+ \frac{1}{3!} \frac{d^2 f_0 (\lambda_n)}{d \lambda^2} \cdot h^3 + \cdots \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ \sum_{k=0}^{\infty} \frac{1}{(k+1)!} \frac{d^k f_0 (\lambda_n)}{d \lambda^k} \cdot h^{k+1} ~~ .
\end{align*} \]
(8-7)
(8-7)
\[ \begin{align*}
\Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big)
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p b_p f_p (\lambda_n ; h ; \Psi) \\
&=
\Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big)
+ h \cdot \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_p (\lambda ; h ; \Psi)}{d h^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot h^{k} ,
\end{align*} \]
(8-10)
(8-10)
\[ \begin{align*}
\frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}
&=
c_p \cdot \bigg( \frac{f_0}{d \lambda} - f_0 \frac{\partial f_0}{\partial \Psi} \bigg)
+ \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} \\
&=
c_p \cdot \frac{f_0}{d \lambda}
+ \bigg[ \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) - c_p f_0 \bigg] \frac{\partial f_p}{\partial \Psi} \bigg|_{\lambda=\lambda_n , ~ h=0}
\end{align*} \]
(9-1)
\[ \sum_p b_p \frac{1}{k!} \frac{d^k f_p}{d h^k} \bigg|_{\lambda=\lambda_n , ~ h=0}
= \frac{1}{(k+1)!} \frac{d^k f_0}{d \lambda^k} \bigg|_{\lambda=\lambda_n} \]
(9-2)
\[ \sum_p b_p \frac{d^k f_p}{d h^k} \bigg|_{\lambda=\lambda_n , ~ h=0}
= \frac{1}{k+1} \frac{d^k f_0}{d \lambda^k} \bigg|_{\lambda=\lambda_n} \]
T9.1.Order of $h$. $k=0$.
(9-3)
\[ \sum_p b_p f_p \bigg|_{\lambda = \lambda_n, ~ h = 0} = f_0 \bigg|_{\lambda = \lambda_n} . \]
Since
(9-4)
\[ f_p \bigg|_{\lambda = \lambda_n, ~ h = 0} = f_0 \bigg|_{\lambda = \lambda_n} \]
, therefore
(9-5)
\[ \therefore \sum_p b_p = 1 . \]
T9.2.Order of $h^2$. $k=1$.
(9-6)
\[ \sum_p b_p \frac{d f_p}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}
= \frac{1}{2} \frac{d f_0}{d \lambda}\bigg|_{\lambda=\lambda_n} . \]
Since
(9-7)
\[ \frac{d f_p}{d h}
=
c_p \frac{\partial f_p}{\partial \lambda} + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} , \]
(9-8)
\[ \sum_p b_p \bigg( c_p \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n}
+ \sum_m \Big( a_{pm} f_0 \Big) \frac{\partial f_0}{\partial \Psi} \bigg|_{\lambda=\lambda_n}
=
\frac{1}{2} \bigg( \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n} + \frac{d \Psi}{d \lambda} \frac{\partial f_0}{\partial \Psi}\bigg|_{\lambda=\lambda_n} \bigg) \]
(9-9)
\[ \sum_p b_p \bigg( c_p \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n}
+ \sum_m \Big( a_{pm} f_0 \Big) \frac{\partial f_0}{\partial \Psi} \bigg|_{\lambda=\lambda_n}
=
\frac{1}{2} \bigg( \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n} + f_0 \frac{\partial f_0}{\partial \Psi}\bigg|_{\lambda=\lambda_n} \bigg) \]
Therefore
(9-10)
\[ \therefore
\sum_p b_p c_p
=
\frac{1}{2} \]
and
(9-11)
\[ \therefore
\sum_{p, m} b_p a_{pm}
=
\frac{1}{2} \]
This could mean that
(9-12)
\[ \therefore
\sum_m a_{pm}
=
c_p . \]
T9.3.Order of $h^3$. $k=2$.
(9-13)
\[ \sum_{p} b_p \frac{d^2 f_p}{d h^2} \bigg|_{\lambda=\lambda_n , ~ h=0}
=
\frac{1}{3} \frac{d^2 f_0}{d \lambda^2} \bigg|_{\lambda=\lambda_n} \]
From Eq.
(9-7)
(9-7)
\[ \frac{d f_p}{d h}
=
c_p \frac{\partial f_p}{\partial \lambda} + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} , \]
(9-14)
\[ \begin{align*}
\frac{d^2 f_p}{d h^2}
&=
\frac{d}{d h} \bigg[ c_p \frac{\partial f_p}{\partial \lambda} + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} \bigg] \\
&=
c_p \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \lambda} \bigg)
+ \sum_m \Big( 2 a_{pm} \frac{d f_m}{d h} + h \cdot a_{pm} \frac{d^2 f_m}{d h^2} \Big) \frac{\partial f_p}{\partial \Psi}
+ \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \Psi} \bigg) .
\end{align*} \]
And condition $\lambda=\lambda_n$, $h=0$ applies,
(9-15)
\[ \begin{align*}
\sum_p b_p
\Bigg[
c_p \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \lambda} \bigg)
+ \sum_m 2 a_{pm} \frac{d f_m}{d h} \frac{\partial f_p}{\partial \Psi}
+ \sum_m a_{pm} f_m \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \Psi} \bigg)
\Bigg]
&=
\frac{1}{3} \frac{d}{d \lambda}
\Bigg[
\frac{\partial f_0}{\partial \lambda} + f_0 \frac{\partial f_0}{\partial \Psi}
\Bigg] \\
&=
\frac{1}{3}
\Bigg[
\frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \lambda} \bigg) + \frac{d f_0}{d \lambda} \frac{\partial f_0}{\partial \Psi} + f_0 \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \Psi} \bigg)
\Bigg]
\end{align*} \]
From the result Eq.
(9-12)
(9-12)
\[ \therefore
\sum_m a_{pm}
=
c_p . \]
(9-7)
(9-7)
\[ \frac{d f_p}{d h}
=
c_p \frac{\partial f_p}{\partial \lambda} + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} , \]
(9-16)
\[ \begin{align*}
\frac{d f_p}{d h} \bigg|_{\lambda=\lambda_n, ~ h=0}
&=
c_p \frac{\partial f_0}{\partial \lambda} + \sum_m a_{pm} f_0 \frac{\partial f_0}{\partial \Psi} \bigg|_{\lambda=\lambda_n} \\
&=
c_p \bigg( \frac{\partial f_0}{\partial \lambda} + f_0 \frac{\partial f_0}{\partial \Psi} \bigg) \bigg|_{\lambda=\lambda_n} \\
&=
c_p \frac{d f_0}{d \lambda} \bigg|_{\lambda=\lambda_n} ~ .
\end{align*} \]
Then
(9-17)
\[ \begin{align*}
\sum_p b_p
\Bigg[
c_p^2 \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \lambda} \bigg)
+ \sum_m 2 a_{pm} c_m \frac{d f_0}{d \lambda} \frac{\partial f_0}{\partial \Psi}
+ \sum_m a_{pm} f_0 c_p \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \Psi} \bigg)
\Bigg]
&=
\frac{1}{3}
\Bigg[
\frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \lambda} \bigg) + \frac{d f_0}{d \lambda} \frac{\partial f_0}{\partial \Psi} + f_0 \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \Psi} \bigg)
\Bigg]
\end{align*} \]
Therefore
(9-18)
\[ \sum_p b_p c_p^2 = \frac{1}{3} \]
(9-19)
\[ 2 \sum_{p, m} b_p a_{pm} c_m = \frac{1}{3} \]
(9-20)
\[ \begin{align*}
&\sum_{p, m} b_p a_{pm} f_0 c_p = \frac{1}{3} f_0 \\
&\sum_p b_p c_p^2 = \frac{1}{3}
\end{align*} \]
T9.4.Order of $h^4$. $k=3$.
(9-21)
\[ \sum_p b_p \frac{d^3 f_p}{d h^3} \bigg|_{\lambda=\lambda_n , ~ h=0}
= \frac{1}{4} \frac{d^3 f_0}{d \lambda^3} \bigg|_{\lambda=\lambda_n} \]
Then the results are
(9-22)
\[ \sum_p b_p c_p^3 = \frac{1}{4} \]
(9-23)
\[ 3 \sum_{p, q} b_p a_{pq} c_q^2 = \frac{1}{4} \]
(9-24)
\[ 6 \sum_{p, q} b_p c_p a_{pq} c_q = \frac{3}{4} \]
(9-25)
\[ 3 \sum_{p, q} b_p c_p^2 a_{pq} = \frac{1}{4} \]
(9-26)
\[ 6 \sum_{p, q, m} b_p a_{pq} a_{qm} c_m = \frac{1}{4} \]
T9.5.Order of $h^5$. $k=4$.
(9-27)
\[ \sum_p b_p \frac{d^4 f_p}{d h^4} \bigg|_{\lambda=\lambda_n , ~ h=0}
= \frac{1}{5} \frac{d^4 f_0}{d \lambda^4} \bigg|_{\lambda=\lambda_n} \]
The results are
(9-28)
\[ \sum_p b_p c_p^4 = \frac{1}{5} \]
(9-29)
\[ 4 \sum_{p, m} b_p a_{pm} c_m^3 = \frac{1}{5} \]
(9-30)
\[ 6 \sum_{p, m} b_p c_p a_{pm} c_m^2 = \frac{3}{5} \]
(9-31)
\[ 2 \sum_{p, m} b_p c_p^2 a_{pm} c_m = \frac{3}{5} \]
▲ Hide
T12.Candidate set $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$
▼ Show/Hide
As it is not easy to solve the above simultaneous equations for $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$, let's just check the well-known coefficients set $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$ to satisfy the above simultaneous equations Eq.
(9-5)
(9-5)
\[ \therefore \sum_p b_p = 1 . \]
(9-10)
(9-10)
\[ \therefore
\sum_p b_p c_p
=
\frac{1}{2} \]
(9-12)
(9-12)
\[ \therefore
\sum_m a_{pm}
=
c_p . \]
(9-18)
(9-18)
\[ \sum_p b_p c_p^2 = \frac{1}{3} \]
(9-19)
(9-19)
\[ 2 \sum_{p, m} b_p a_{pm} c_m = \frac{1}{3} \]
(9-22)
(9-22)
\[ \sum_p b_p c_p^3 = \frac{1}{4} \]
(9-23)
(9-23)
\[ 3 \sum_{p, q} b_p a_{pq} c_q^2 = \frac{1}{4} \]
(9-24)
(9-24)
\[ 6 \sum_{p, q} b_p c_p a_{pq} c_q = \frac{3}{4} \]
(9-25)
(9-25)
\[ 3 \sum_{p, q} b_p c_p^2 a_{pq} = \frac{1}{4} \]
(9-26)
(9-26)
\[ 6 \sum_{p, q, m} b_p a_{pq} a_{qm} c_m = \frac{1}{4} \]
(9-28)
(9-28)
\[ \sum_p b_p c_p^4 = \frac{1}{5} \]
(9-29)
(9-29)
\[ 4 \sum_{p, m} b_p a_{pm} c_m^3 = \frac{1}{5} \]
(9-30)
(9-30)
\[ 6 \sum_{p, m} b_p c_p a_{pm} c_m^2 = \frac{3}{5} \]
(9-31)
(9-31)
\[ 2 \sum_{p, m} b_p c_p^2 a_{pm} c_m = \frac{3}{5} \]
T12.1.Let's consider $b_0 = \frac{1}{6}$, $b_1 = \frac{1}{3}$, $b_2 = \frac{1}{3}$, $b_3 = \frac{1}{6}$ and $c_0 = 0$, $c_1 = \frac{1}{2}$, $c_2 = \frac{1}{2}$, $c_3 = 1$.
(12-1)
\[ \begin{matrix}
0 \\
1/2 & 1/2 \\
1/2 & 0 & 1/2 \\
1 & 0 & 0 & 1 \\
& 1/6 & 1/3 & 1/3 & 1/6
\end{matrix} \]
$\sum_p b_p c_p^0$=0.9999999999999999 ($\sum_p b_p c_p^0-\frac{1}{1}$=-1.1102230246251565e-16)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.3333333333333333 ($\sum_p b_p c_p^2-\frac{1}{3}$=0)
$\sum_p b_p c_p^3$=0.25 ($\sum_p b_p c_p^3-\frac{1}{4}$=0)
$\sum_p b_p c_p^4$=0.20833333333333331 ($\sum_p b_p c_p^4-\frac{1}{5}$=0.008333333333333304)
$\sum_p b_p c_p^5$=0.1875 ($\sum_p b_p c_p^5-\frac{1}{6}$=0.020833333333333343)
$\sum_p b_p c_p^6$=0.17708333333333331 ($\sum_p b_p c_p^6-\frac{1}{7}$=0.034226190476190466)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.5 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.5 ($\sum_m a_{2m} - c_{2}$=0)
$\sum_m a_{3m}$=1 ($\sum_m a_{3m} - c_{3}$=0)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.3333333333333333 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=0)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.75 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=0)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.75 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.3333333333333333 ($\sum_p b_p c_p^2-\frac{1}{3}$=0)
$\sum_p b_p c_p^3$=0.25 ($\sum_p b_p c_p^3-\frac{1}{4}$=0)
$\sum_p b_p c_p^4$=0.20833333333333331 ($\sum_p b_p c_p^4-\frac{1}{5}$=0.008333333333333304)
$\sum_p b_p c_p^5$=0.1875 ($\sum_p b_p c_p^5-\frac{1}{6}$=0.020833333333333343)
$\sum_p b_p c_p^6$=0.17708333333333331 ($\sum_p b_p c_p^6-\frac{1}{7}$=0.034226190476190466)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.5 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.5 ($\sum_m a_{2m} - c_{2}$=0)
$\sum_m a_{3m}$=1 ($\sum_m a_{3m} - c_{3}$=0)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.3333333333333333 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=0)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.75 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=0)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.75 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
T12.2.Let's consider $b_0 = \frac{1}{8}$, $b_1 = \frac{3}{8}$, $b_2 = \frac{3}{8}$, $b_3 = \frac{1}{8}$ and $c_0 = 0$, $c_1 = \frac{1}{3}$, $c_2 = \frac{2}{3}$, $c_3 = 1$.
(12-2)
\[ \begin{matrix}
0 \\
1/3 & 1/3 \\
2/3 & -1/3 & 1 \\
1 & 1 & -1 & 1 \\
& 1/8 & 3/8 & 3/8 & 1/8
\end{matrix} \]
$\sum_p b_p c_p^0$=1 ($\sum_p b_p c_p^0-\frac{1}{1}$=0)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.3333333333333333 ($\sum_p b_p c_p^2-\frac{1}{3}$=0)
$\sum_p b_p c_p^3$=0.25 ($\sum_p b_p c_p^3-\frac{1}{4}$=0)
$\sum_p b_p c_p^4$=0.2037037037037037 ($\sum_p b_p c_p^4-\frac{1}{5}$=0.0037037037037036813)
$\sum_p b_p c_p^5$=0.17592592592592593 ($\sum_p b_p c_p^5-\frac{1}{6}$=0.009259259259259273)
$\sum_p b_p c_p^6$=0.15843621399176955 ($\sum_p b_p c_p^6-\frac{1}{7}$=0.015579071134626699)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.3333333333333333 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.6666666666666667 ($\sum_m a_{2m} - c_{2}$=1.1102230246251565e-16)
$\sum_m a_{3m}$=1 ($\sum_m a_{3m} - c_{3}$=0)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.33333333333333337 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=5.551115123125783e-17)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.75 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=0)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.75 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.3333333333333333 ($\sum_p b_p c_p^2-\frac{1}{3}$=0)
$\sum_p b_p c_p^3$=0.25 ($\sum_p b_p c_p^3-\frac{1}{4}$=0)
$\sum_p b_p c_p^4$=0.2037037037037037 ($\sum_p b_p c_p^4-\frac{1}{5}$=0.0037037037037036813)
$\sum_p b_p c_p^5$=0.17592592592592593 ($\sum_p b_p c_p^5-\frac{1}{6}$=0.009259259259259273)
$\sum_p b_p c_p^6$=0.15843621399176955 ($\sum_p b_p c_p^6-\frac{1}{7}$=0.015579071134626699)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.3333333333333333 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.6666666666666667 ($\sum_m a_{2m} - c_{2}$=1.1102230246251565e-16)
$\sum_m a_{3m}$=1 ($\sum_m a_{3m} - c_{3}$=0)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.33333333333333337 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=5.551115123125783e-17)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.75 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=0)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.75 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
T12.3.Let's consider $b_0 = \frac{16}{135}$, $b_1 = 0$, $b_2 = \frac{6656}{12825}$, $b_3 = \frac{28561}{56430}$, $b_4 = -\frac{9}{50}$, $b_5 = \frac{2}{55}$ and $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$.
(12-3)
\[ \begin{matrix}
0 \\
1/4 & 1/4 \\
3/8 & 3/32 & 9/32 \\
12/13 & 1932/2197 & -7200/2197 & 7296/2197 \\
1 & 439/216 & -8 & 3680/513 & -845/4104 \\
1/2 & -8/27 & 2 & -3544/2565 & 1859/4104 & -11/40 \\
& 16/135 & 0 & 6656/12825 & 28561/56430 & -9/50 & 2/55
\end{matrix} \]
$\sum_p b_p c_p^0$=1 ($\sum_p b_p c_p^0-\frac{1}{1}$=0)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.33333333333333337 ($\sum_p b_p c_p^2-\frac{1}{3}$=5.551115123125783e-17)
$\sum_p b_p c_p^3$=0.25000000000000006 ($\sum_p b_p c_p^3-\frac{1}{4}$=5.551115123125783e-17)
$\sum_p b_p c_p^4$=0.20000000000000007 ($\sum_p b_p c_p^4-\frac{1}{5}$=5.551115123125783e-17)
$\sum_p b_p c_p^5$=0.16418269230769236 ($\sum_p b_p c_p^5-\frac{1}{6}$=-0.0024839743589742946)
$\sum_p b_p c_p^6$=0.13511695636094687 ($\sum_p b_p c_p^6-\frac{1}{7}$=-0.007740186496195983)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.25 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.375 ($\sum_m a_{2m} - c_{2}$=0)
$\sum_m a_{3m}$=0.9230769230769229 ($\sum_m a_{3m} - c_{3}$=-2.220446049250313e-16)
$\sum_m a_{4m}$=0.9999999999999997 ($\sum_m a_{4m} - c_{4}$=-3.3306690738754696e-16)
$\sum_m a_{5m}$=0.4999999999999999 ($\sum_m a_{5m} - c_{5}$=-1.1102230246251565e-16)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.3333333333333333 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=0)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.7499999999999997 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=-3.3306690738754696e-16)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.7500000000000002 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5000000000000002)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.33333333333333337 ($\sum_p b_p c_p^2-\frac{1}{3}$=5.551115123125783e-17)
$\sum_p b_p c_p^3$=0.25000000000000006 ($\sum_p b_p c_p^3-\frac{1}{4}$=5.551115123125783e-17)
$\sum_p b_p c_p^4$=0.20000000000000007 ($\sum_p b_p c_p^4-\frac{1}{5}$=5.551115123125783e-17)
$\sum_p b_p c_p^5$=0.16418269230769236 ($\sum_p b_p c_p^5-\frac{1}{6}$=-0.0024839743589742946)
$\sum_p b_p c_p^6$=0.13511695636094687 ($\sum_p b_p c_p^6-\frac{1}{7}$=-0.007740186496195983)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.25 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.375 ($\sum_m a_{2m} - c_{2}$=0)
$\sum_m a_{3m}$=0.9230769230769229 ($\sum_m a_{3m} - c_{3}$=-2.220446049250313e-16)
$\sum_m a_{4m}$=0.9999999999999997 ($\sum_m a_{4m} - c_{4}$=-3.3306690738754696e-16)
$\sum_m a_{5m}$=0.4999999999999999 ($\sum_m a_{5m} - c_{5}$=-1.1102230246251565e-16)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.3333333333333333 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=0)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.7499999999999997 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=-3.3306690738754696e-16)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.7500000000000002 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5000000000000002)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
T12.4.Let's consider $b_0 = \frac{25}{216}$, $b_1 = 0$, $b_2 = \frac{1408}{2565}$, $b_3 = \frac{2197}{4104}$, $b_4 = -\frac{1}{5}$, $b_5 = 0$ and $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$.
(12-4)
\[ \begin{matrix}
0 \\
1/4 & 1/4 \\
3/8 & 3/32 & 9/32 \\
12/13 & 1932/2197 & -7200/2197 & 7296/2197 \\
1 & 439/216 & -8 & 3680/513 & -845/4104 \\
1/2 & -8/27 & 2 & -3544/2565 & 1859/4104 & -11/40 \\
& 25/216 & 0 & 1408/2565 & 2197/4104 & -1/5 & 0
\end{matrix} \]
$\sum_p b_p c_p^0$=1 ($\sum_p b_p c_p^0-\frac{1}{1}$=0)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.3333333333333334 ($\sum_p b_p c_p^2-\frac{1}{3}$=1.1102230246251565e-16)
$\sum_p b_p c_p^3$=0.25000000000000006 ($\sum_p b_p c_p^3-\frac{1}{4}$=5.551115123125783e-17)
$\sum_p b_p c_p^4$=0.1995192307692309 ($\sum_p b_p c_p^4-\frac{1}{5}$=-0.00048076923076911804)
$\sum_p b_p c_p^5$=0.1628374630177516 ($\sum_p b_p c_p^5-\frac{1}{6}$=-0.003829203648915064)
$\sum_p b_p c_p^6$=0.13269581922792456 ($\sum_p b_p c_p^6-\frac{1}{7}$=-0.010161323629218288)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.25 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.375 ($\sum_m a_{2m} - c_{2}$=0)
$\sum_m a_{3m}$=0.9230769230769229 ($\sum_m a_{3m} - c_{3}$=-2.220446049250313e-16)
$\sum_m a_{4m}$=0.9999999999999997 ($\sum_m a_{4m} - c_{4}$=-3.3306690738754696e-16)
$\sum_m a_{5m}$=0.4999999999999999 ($\sum_m a_{5m} - c_{5}$=-1.1102230246251565e-16)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.33333333333333354 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=2.220446049250313e-16)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.7500000000000002 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=2.220446049250313e-16)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.7500000000000004 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5000000000000004)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
$\sum_p b_p c_p^1$=0.5 ($\sum_p b_p c_p^1-\frac{1}{2}$=0)
$\sum_p b_p c_p^2$=0.3333333333333334 ($\sum_p b_p c_p^2-\frac{1}{3}$=1.1102230246251565e-16)
$\sum_p b_p c_p^3$=0.25000000000000006 ($\sum_p b_p c_p^3-\frac{1}{4}$=5.551115123125783e-17)
$\sum_p b_p c_p^4$=0.1995192307692309 ($\sum_p b_p c_p^4-\frac{1}{5}$=-0.00048076923076911804)
$\sum_p b_p c_p^5$=0.1628374630177516 ($\sum_p b_p c_p^5-\frac{1}{6}$=-0.003829203648915064)
$\sum_p b_p c_p^6$=0.13269581922792456 ($\sum_p b_p c_p^6-\frac{1}{7}$=-0.010161323629218288)
$\sum_m a_{0m}$=0 ($\sum_m a_{0m} - c_{0}$=0)
$\sum_m a_{1m}$=0.25 ($\sum_m a_{1m} - c_{1}$=0)
$\sum_m a_{2m}$=0.375 ($\sum_m a_{2m} - c_{2}$=0)
$\sum_m a_{3m}$=0.9230769230769229 ($\sum_m a_{3m} - c_{3}$=-2.220446049250313e-16)
$\sum_m a_{4m}$=0.9999999999999997 ($\sum_m a_{4m} - c_{4}$=-3.3306690738754696e-16)
$\sum_m a_{5m}$=0.4999999999999999 ($\sum_m a_{5m} - c_{5}$=-1.1102230246251565e-16)
$2 \sum_{p, m} b_p a_{pm} c_m$=0.33333333333333354 ($2 \sum_{p, m} b_p a_{pm} c_m - \frac{1}{3}$=2.220446049250313e-16)
$3 \sum_{p, q} b_p a_{pq} c_q^2$=0.25 ($3 \sum_{p, q} b_p a_{pq} c_q^2 - \frac{1}{4}$=0)
$6 \sum_{p, q} b_p c_p a_{pq} c_q$=0.7500000000000002 ($6 \sum_{p, q} b_p c_p a_{pq} c_q - \frac{3}{4}$=2.220446049250313e-16)
$3 \sum_{p, q} b_p c_p^2 a_{pq}$=0.7500000000000004 ($3 \sum_{p, q} b_p c_p^2 a_{pq} - \frac{1}{4}$=0.5000000000000004)
$6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q$=0.25 ($6 \sum_{p, m, q} b_p a_{pm} a_{mq} c_q - \frac{1}{4}$=0)
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T13.Adaptive step size
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During the integration, the step size is adapted such that the estimated error stays below a user-defined threshold: If the error is too high, a step is repeated with a lower step size; if the error is much smaller, the step size is increased to save time. This results in an (almost) optimal step size, which saves computation time. Moreover, the user does not have to spend time on finding an appropriate step size
[02]
.
Ref. [02] en.wikipedia.org :: Runge–Kutta methods
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TRefs.References and Related Articles
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- Ref. [01] en.wikipedia.org :: Matrix exponential
- Ref. [02] en.wikipedia.org :: Runge–Kutta methods
- Ref. [03] kipid's blog :: 선형 대수학 간단한 정리들 (Linear Algebra), 2014-03-18
- Ref. [04] ghebook.blogspot.com :: 1계 선형 상미분 방정식의 해법 (Solution of the First Order Linear Ordinary Differential Equation), 2011-11-16
- Ref. [05] Method of Lagrange multipliers, and 최적화, 라그랑지 승수법 (Optimization with the Method of Lagrange multipliers)
- Ref. [06] Numerical Differentiation.pdf by D. Levy
- Ref. [07] en.wikipedia.org :: Numerical differentiation
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