Question

Write down only one application of Runge Kutta Method to solve some real world problem

Answer 1

Answer on Question #75363 – Math – Quantitative Methods

Question

Write down only one application of Runge Kutta Method to solve some real world problem.

Solution

The rate of change of the temperature $dT(t)/dt$, is (by Newton's Law of Cooling) proportional to the difference between the temperature of the object $T(t)$ and the ambient temperature $Ta$. This means that:

$$\frac{dT(t)}{dt} = -r(T(t) - Ta), \quad r - \text{positive constant characteristic of the system}$$

The analytical solution of this differential equation:

$$T(t) = Ta + C \cdot e^{-rt}$$

Let for $t = 0$, $T(0) = 90$ and $r = 0.1$ and $Ta = 30$ then:

$$\frac{dT(t)}{dt} = -0.1(T(t) - 30) \text{ with IVP: } T(0) = 90$$

And exact solution:

$$T(t) = 30 + 60 \cdot e^{-0.1 \cdot t}$$

The Runge-Kutta-Fehlberg method is single-step method. This method has a procedure to determine whether the correct step size $h$ is used. Each step requires the use of the following six values:

$$k_1 = h \cdot f(x_k, y_k)$$ $$k_2 = h \cdot f\left(x_k + \frac{1}{4}h, y_k + \frac{1}{4}k_1\right)$$ $$k_3 = h \cdot f\left(x_k + \frac{3}{8}h, y_k + \frac{3}{32}k_1 + \frac{9}{32}k_2\right)$$ $$k_4 = h \cdot f\left(x_k + \frac{12}{13}h, y_k + \frac{1932}{2197}k_1 - \frac{7200}{2197}k_2 + \frac{7296}{2197}k_3\right)$$ $$k_5 = h \cdot f\left(x_k + h, y_k + \frac{439}{216}k_1 - 8k_2 + \frac{3680}{513}k_3 - \frac{845}{4104}k_4\right)$$ $$k_6 = h \cdot f\left(x_k + \frac{1}{2}h, y_k - \frac{8}{27}k_1 + 2k_2 - \frac{3544}{2565}k_3 + \frac{1859}{4104}k_4 - \frac{11}{40}k_5\right)$$

Then we calculate the approximation to the solution with the help of the method of the fourth order:

$$w_{k+1} = y_k + \frac{25}{216}k_1 + \frac{1408}{2565}k_3 + \frac{2197}{4101}k_4 - \frac{1}{5}k_5, \quad Error = O(h^4)$$

And the approximation to the solution with the help of the method of the 5th order:

$$y_{k+1} = y_k + \frac{16}{135}k_1 + \frac{6656}{12825}k_3 + \frac{28561}{56430}k_4 - \frac{9}{50}k_5 + \frac{2}{55}k_6, \quad Error = O(h^5)$$

At each step, two different approximations for the solution are made and compared. The optimal step size is $(\delta \cdot h)$.

$$R = \frac{1}{h} |w_{k+1} - y_{k+1}|, \quad \delta = \left(\frac{\varepsilon}{2R}\right)^{\frac{1}{4}} \varepsilon - specified\ accuracy$$

Problem solving and error analysis:

The classical fourth-order Runge-Kutta method for computations with a constant integration step requires the calculation of four coefficients:

$$k _ {3} = f \left(x _ {k} + \frac {1}{2} h, y _ {k} + \frac {1}{2} k _ {2}\right)$$ $$k _ {4} = f \left(x _ {k} + h, y _ {k} + h k _ {3}\right)$$

Then we calculate the approximation to the solution with the help of the method of the fourth order:

$$y _ {k + 1} = y _ {k} + \frac {h}{6} \left(k _ {1} + 2 k _ {2} + 2 k _ {3} + k _ {4}\right)$$

Problem solving and error analysis:

$$\begin{array}{r l} \mathrm {R u n g e 4 (y 0 , s t e p , D)} := & \left\{ \begin{array}{l l} \left(x _ {0} \leftarrow 0 y _ {0} \leftarrow y 0 h \leftarrow \mathrm {s t e p}\right) \\ \mathrm {f o r i \in 0 . 1 0 0} \\ \left[ \begin{array}{l} x _ {i + 1} \leftarrow x _ {i} + h \\ k 1 \leftarrow D \left(x _ {i}, y _ {i}\right) \\ k 2 \leftarrow D \left(x _ {i} + \frac {h}{2}, y _ {i} + \frac {h \cdot k 1}{2}\right) \\ k 3 \leftarrow D \left(x _ {i} + \frac {h}{2}, y _ {i} + \frac {h \cdot k 2}{2}\right) \\ k 4 \leftarrow D \left(x _ {i} + h, y _ {i} + h \cdot k 3\right) \\ y _ {i + 1} \leftarrow y _ {i} + h \left(\frac {k 1 + 2 \cdot k 2 + 2 \cdot k 3 + k 4}{6}\right) \\ \mathrm {r e t u r n a u g m e n t (x , y)} \end{array} \right. \end{array} \right. \end{array}$$

Error calculation.

T - exact value

AT - approximation

$$Z := \operatorname {R u n g e} 4 \left(y _ {0}, 0. 1 5, D\right)$$ $$i := 0 \dots \text {r o w s} (Z) - 1 \quad \text {t i m e} _ {i} := Z _ {i, 0} \quad T _ {i} := f T (\text {t i m e} _ {i}) \quad A T _ {i} := Z _ {i, 1} \quad | T - A T | = 0. 0 0 0 0 0 0 8 0 1 6$$

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