Update runge_kutta.md

Author: djeada

notes/7_ordinary_differential_equations/runge_kutta.md

@@ -6,51 +6,51 @@ The Runge-Kutta method is part of a family of iterative methods, both implicit a
 
 The first-order ODE that is typically used in the Runge-Kutta methods takes the following form:
 
-$$ \frac{du}{dt} = f(t, u), $$
+$$ \frac{du}{dt} = f(t, u)$$
 
 In this equation, $u$ is the unknown function of time $t$ that we are aiming to solve for, and $f$ is a function of $t$ and $u$.
 
 The general formulation of the 4th order Runge-Kutta method is presented below:
 
-$$k_1 = \Delta t \cdot f(t, u),$$
+$$k_1 = \Delta t \cdot f(t, u)$$
 
-$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right),$$
+$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right)$$
 
-$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right),$$
+$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right)$$
 
-$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3),$$
+$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3)$$
 
-$$u(t + \Delta t) = u(t) + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4).$$
+$$u(t + \Delta t) = u(t) + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$$
 
 ### Derivation of Runge-Kutta Method
 
 The derivation of the Runge-Kutta methods, especially the 4th order Runge-Kutta method, begins with an initial condition and attempts to estimate the solution value after a small step $\Delta t$. This is achieved by taking weighted averages of increments at the beginning, middle, and end of the interval. The purpose is to imitate a Taylor series expansion, but without the necessity of computing higher derivatives.
 
 Starting with a Taylor series expansion, we get:
 
-$$u(t+\Delta t) = u(t) + \Delta t u'(t) + \frac{\Delta t^2}{2!} u''(t) + \frac{\Delta t^3}{3!} u'''(t) + \frac{\Delta t^4}{4!} u''''(t) + O(\Delta t^5).$$
+$$u(t+\Delta t) = u(t) + \Delta t u'(t) + \frac{\Delta t^2}{2!} u''(t) + \frac{\Delta t^3}{3!} u'''(t) + \frac{\Delta t^4}{4!} u''''(t) + O(\Delta t^5)$$
 
 The first derivative $u'(t)$ is already provided by the ODE:
 
-$$u'(t) = f(t, u).$$
+$$u'(t) = f(t, u)$$
 
 Higher-order derivatives can also be represented in terms of $f(t, u)$ and its derivatives:
 
-$$u''(t) = \frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial u} \frac{du}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial u} f,$$
+$$u''(t) = \frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial u} \frac{du}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial u} f$$
 
 The Runge-Kutta method is designed to mimic the above Taylor series expansion by taking a suitable weighted average of increments at the start, middle, and end of the interval. It avoids explicitly calculating higher derivatives of $f(t, u)$, instead approximating these increments by evaluating $f(t, u)$ at several points within the interval.
 
 The 4th order Runge-Kutta method is thus given by:
 
-$$k_1 = \Delta t \cdot f(t, u),$$
+$$k_1 = \Delta t \cdot f(t, u)$$
 
-$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right),$$
+$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right)$$
 
-$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right),$$
+$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right)$$
 
-$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3),$$
+$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3)$$
 
-$$u(t + \Delta t) = u(t) + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4),$$
+$$u(t + \Delta t) = u(t) + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$$
 
 In this method, the weights of $k_1, k_2, k_3, k_4$ (1/6, 1/3, 1/3, 1/6) are chosen such that the error term is $O(\Delta t^5)$, indicating that the local truncation error at each step is proportional to the fifth power of the step size, and the global truncation error (after $N$ steps) is proportional to the fourth power of the step size.
 
@@ -73,13 +73,13 @@ with the initial condition $u(0)=1$. We want to estimate the value of $u$ at $t
 
 First, we calculate the four increments:
 
-$$k_1 = \Delta t \cdot f(t, u) = 0.05 \cdot 1 = 0.05,$$
+$$k_1 = \Delta t \cdot f(t, u) = 0.05 \cdot 1 = 0.05$$
 
-$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right) = 0.05 \cdot (1 + 0.05/2) = 0.05125,$$
+$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right) = 0.05 \cdot (1 + 0.05/2) = 0.05125$$
 
-$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right) = 0.05 \cdot (1 + 0.05125/2) = 0.0515625,$$
+$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right) = 0.05 \cdot (1 + 0.05125/2) = 0.0515625$$
 
-$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3) = 0.05 \cdot (1 + 0.0515625) = 0.052578125.$$
+$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3) = 0.05 \cdot (1 + 0.0515625) = 0.052578125$$
 
 Then, we calculate the next value of $u$:
 
@@ -93,21 +93,21 @@ $$u(0.05) = 1.05104166667.$$
 
 Similarly, we first calculate the four increments:
 
-$$k_1 = \Delta t \cdot f(t, u) = 0.05 \cdot 1.05104166667 = 0.0525520833335,$$
+$$k_1 = \Delta t \cdot f(t, u) = 0.05 \cdot 1.05104166667 = 0.0525520833335$$
 
-$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right) = 0.05 \cdot (1.05104166667 + 0.0525520833335/2) = 0.0531899049680,$$
+$$k_2 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_1}{2}\right) = 0.05 \cdot (1.05104166667 + 0.0525520833335/2) = 0.0531899049680$$
 
-$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right) = 0.05 \cdot (1.05104166667 + 0.0531899049680/2) = 0.0533905595735,$$
+$$k_3 = \Delta t \cdot f\left(t + \frac{\Delta t}{2}, u + \frac{k_2}{2}\right) = 0.05 \cdot (1.05104166667 + 0.0531899049680/2) = 0.0533905595735$$
 
-$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3) = 0.05 \cdot (1.05104166667 + 0.0533905595735) = 0.0552220602058.$$
+$$k_4 = \Delta t \cdot f(t + \Delta t, u + k_3) = 0.05 \cdot (1.05104166667 + 0.0533905595735) = 0.0552220602058$$
 
 Then, we calculate the next value of $u$:
 
 $$u(0.1) = u(0.05) + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$$
 
 $$u(0.1) = 1.05104166667 + \frac{1}{6}(0.0525520833335 + 2 \cdot 0.0531899049680 + 2 \cdot 0.0533905595735 + 0.0552220602058)$$
 
-$$u(0.1) = 1.10517087727.$$
+$$u(0.1) = 1.10517087727$$
 
 ### Advantages