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Runge-Kutta method, Matrix exponential

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# Runge-Kutta method, Matrix exponential
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## PH
  • 2023-05-22 : First posting.
## TOC ## Differential equation with matrices (RLC circuit) \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). ## \exp(A_{ij}) Taylor series definition of matrix exponential . \exp(A_{ij}) = \sum_{k=0}^{\infty} \frac{1}{k!} A^k If A_{ij} can be diagonalized as A_{ij} = U_{ik} \delta_{kl} D_{kl} U^{-1}_{lj} , then \exp(A_{ij}) = \exp(U_{ik} \delta_{kl} D_{kl} U^{-1}_{lj}) = U_{ik} \exp(\delta_{kl} D_{kl}) U^{-1}_{lj} where \exp(\delta_{kl} D_{kl}) = \delta_{kl} \exp(D_{kl}) . Explicitely speaking, 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: 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} . ## Evolution of variables Let an evolution of $\Psi$ from the initial value be specified as follows: \Psi \big( \{ x^{\mu} \} \big) = \Psi \big( \{ x^{\mu} (\lambda) \} \big) . The evolution equation is \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 \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 \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 $\{ x^{\mu} \}$, $\lambda$. ## Numerical implementation ### Euler's method (Up to 1st-order) \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 $h \equiv \frac{\lambda_f - \lambda_i}{N}$, \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*} ## Let's try to satisfy/content higher orders. We can candidate that the below equation is correct upto the fourth order $\mathrm{O}(h^4)$. Difining $\lambda_n \equiv \lambda_i + h n$, \Psi \big( \{ x^{\mu} (\lambda_{n+1}) \} \big) := \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_k b_k f_k , where \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 f = f \Big( \{ x^{\mu} (\lambda_n + c \cdot h) \} ; \lambda_n + c \cdot h \Big) . With index $p$, f_p = f \Big( \{ x^{\mu} (\lambda_n + c_p \cdot h) \} ; \lambda_n + c_p \cdot h \Big) . ### Taylor expansion of $\Psi$ \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 $f_0 = \frac{d \Psi}{d \lambda}$, \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 \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 \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 \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. becomes \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*} ## Let's match the coefficients up to $\textrm{O}(h^4)$. Matching the coefficients of Eq. and Eq. , \frac{1}{(k+1)!} = \sum_p \frac{1}{k!} b_p c_p^k Therefore \sum_p b_p c_p^k = \frac{1}{k+1} ~~~~~ \text{for} ~~ k=0, 1, 2, \cdots ### Case $k=0$ \sum_p b_p = 1 . ### Case $k=1$ \sum_p b_p c_p = \frac{1}{2} . ### Case $k=2$ \sum_p b_p c_p^2 = \frac{1}{3} . ### Case $k=3$ \sum_p b_p c_p^3 = \frac{1}{4} . ## Evolution of variables when it explicitely depends on itself. Let's say that the evolution equation is given by \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 \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 \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 $\Psi(\lambda)$. ## Numerical implementation to satisfy/content higher orders. We can candidate that the below equation is correct upto the $m$-th order $\mathrm{O}(h^{m})$. Difining $\lambda_n \equiv \lambda_i + h n$, \Psi \big( \{ x^{\mu} (\lambda_{n+1}) \} \big) := \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_k b_k f_k , where \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 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 $p$, 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) . ### Taylor expansion of $\Psi$ \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 $f_0 = \frac{d \Psi}{d \lambda}$, \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 \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 \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 \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. becomes \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*} ## Let's match the coefficients up to $\textrm{O}(h^k)$. Matching the coefficients of Eq. and Eq. where $\frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}$ is calculated by Eq. , \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} \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} ### Order of $h$. $k=0$. \sum_p b_p f_p \bigg|_{\lambda = \lambda_n, ~ h = 0} = f_0 \bigg|_{\lambda = \lambda_n} . Since f_p \bigg|_{\lambda = \lambda_n, ~ h = 0} = f_0 \bigg|_{\lambda = \lambda_n} , therefore \therefore \sum_p b_p = 1 . ### Order of $h^2$. $k=1$. \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 \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} , \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) \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 \therefore \sum_p b_p c_p = \frac{1}{2} and \therefore \sum_{p, m} b_p a_{pm} = \frac{1}{2} This could mean that \therefore \sum_m a_{pm} = c_p . ### Order of $h^3$. $k=2$. \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. , \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, \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. , Eq. becomes \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 \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 \sum_p b_p c_p^2 = \frac{1}{3} 2 \sum_{p, m} b_p a_{pm} c_m = \frac{1}{3} \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*} ### Order of $h^4$. $k=3$. \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 \sum_p b_p c_p^3 = \frac{1}{4} 3 \sum_{p, q} b_p a_{pq} c_q^2 = \frac{1}{4} 6 \sum_{p, q} b_p c_p a_{pq} c_q = \frac{3}{4} 3 \sum_{p, q} b_p c_p^2 a_{pq} = \frac{1}{4} 6 \sum_{p, q, m} b_p a_{pq} a_{qm} c_m = \frac{1}{4} ### Order of $h^5$. $k=4$. \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 \sum_p b_p c_p^4 = \frac{1}{5} 4 \sum_{p, m} b_p a_{pm} c_m^3 = \frac{1}{5} 6 \sum_{p, m} b_p c_p a_{pm} c_m^2 = \frac{3}{5} 2 \sum_{p, m} b_p c_p^2 a_{pm} c_m = \frac{3}{5} ## Written in papers by color ballpoint pen
## Written in papers 2.
## Candidate set $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$ 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. , , , , , , , , , , , , , $\cdots$ ### Let's consider $b_0 = \frac{1}{6}$, $b_1 = \frac{1}{3}$, $b_2 = \frac{1}{3}$, $b_3 = \frac{1}{6}$ and $c_0 = 0$, $c_1 = \frac{1}{2}$, $c_2 = \frac{1}{2}$, $c_3 = 1$. \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}
### Let's consider $b_0 = \frac{1}{8}$, $b_1 = \frac{3}{8}$, $b_2 = \frac{3}{8}$, $b_3 = \frac{1}{8}$ and $c_0 = 0$, $c_1 = \frac{1}{3}$, $c_2 = \frac{2}{3}$, $c_3 = 1$. \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}
### 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 $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$. \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}
### 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 $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$. \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}
## Adaptive step size
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 .
## RRA
  1. en.wikipedia.org :: Matrix exponential
  2. en.wikipedia.org :: Runge–Kutta methods
  3. kipid's blog :: 선형 대수학 간단한 정리들 (Linear Algebra), 2014-03-18
  4. ghebook.blogspot.com :: 1계 선형 상미분 방정식의 해법 (Solution of the First Order Linear Ordinary Differential Equation), 2011-11-16
  5. Method of Lagrange multipliers, and 최적화, 라그랑지 승수법 (Optimization with the Method of Lagrange multipliers)
  6. Numerical Differentiation.pdf by D. Levy
  7. en.wikipedia.org :: Numerical differentiation
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