Answer on Question #76730 – Math – Quantitative Methods
Question
Write and solve an example of Runge Kutta Fehlberg Method.
Solution
The rate of change of the temperature d T ( t ) d t \frac{dT(t)}{dt} d t d T ( t ) T ( t ) T(t) T ( t ) T a T_a T a
d T ( t ) d t = − r ( T ( t ) − T a ) , r − positive constant characteristic of the system \frac{dT(t)}{dt} = -r(T(t) - T_a), \quad r - \text{positive constant characteristic of the system} d t d T ( t ) = − r ( T ( t ) − T a ) , r − positive constant characteristic of the system
The analytical solution of this differential equation:
T ( t ) = T a + C ⋅ e − r t T(t) = T_a + C \cdot e^{-rt} T ( t ) = T a + C ⋅ e − r t
Let for t = 0 t = 0 t = 0 T ( 0 ) = 5 T(0) = 5 T ( 0 ) = 5 r = 0.3 r = 0.3 r = 0.3 T a = 35 T_a = 35 T a = 35
d T ( t ) d t = − 0.3 ( T ( t ) − 35 ) with IVP: T ( 0 ) = 5 \frac{dT(t)}{dt} = -0.3(T(t) - 35) \text{ with IVP: } T(0) = 5 d t d T ( t ) = − 0.3 ( T ( t ) − 35 ) with IVP: T ( 0 ) = 5
And exact solution:
T ( t ) = 35 − 30 ⋅ e − 0.3 ⋅ t T(t) = 35 - 30 \cdot e^{-0.3 \cdot t} T ( t ) = 35 − 30 ⋅ e − 0.3 ⋅ t
The Runge-Kutta-Fehlberg method is single-step method. This method has a procedure to determine whether the correct step size h h h
k 1 = h ⋅ f ( x k , y k ) k_1 = h \cdot f(x_k, y_k) k 1 = h ⋅ f ( x k , y k ) k 2 = h ⋅ f ( x k + 1 4 h , y k + 1 4 k 1 ) k_2 = h \cdot f\left(x_k + \frac{1}{4}h, y_k + \frac{1}{4}k_1\right) k 2 = h ⋅ f ( x k + 4 1 h , y k + 4 1 k 1 ) k 3 = h ⋅ f ( x k + 3 8 h , y k + 3 32 k 1 + 9 32 k 2 ) 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 3 = h ⋅ f ( x k + 8 3 h , y k + 32 3 k 1 + 32 9 k 2 ) k 4 = h ⋅ f ( x k + 12 13 h , y k + 1932 2197 k 1 − 7200 2197 k 2 + 7296 2197 k 3 ) 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 4 = h ⋅ f ( x k + 13 12 h , y k + 2197 1932 k 1 − 2197 7200 k 2 + 2197 7296 k 3 ) k 5 = h ⋅ f ( x k + h , y k + 439 216 k 1 − 8 k 2 + 3680 513 k 3 − 845 4104 k 4 ) 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 5 = h ⋅ f ( x k + h , y k + 216 439 k 1 − 8 k 2 + 513 3680 k 3 − 4104 845 k 4 ) k 6 = h ⋅ f ( x k + 1 2 h , y k − 8 27 k 1 + 2 k 2 − 3544 2565 k 3 + 1859 4104 k 4 − 11 40 k 5 ) 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) k 6 = h ⋅ f ( x k + 2 1 h , y k − 27 8 k 1 + 2 k 2 − 2565 3544 k 3 + 4104 1859 k 4 − 40 11 k 5 )
Then we calculate the approximation to the solution with the help of the method of the fourth order:
y k + 1 4 = y k + 25 216 k 1 + 1408 2565 k 3 + 2197 4101 k 4 − 1 5 k 5 , Error = O ( h 4 ) y_{k+1}^4 = y_k + \frac{25}{216}k_1 + \frac{1408}{2565}k_3 + \frac{2197}{4101}k_4 - \frac{1}{5}k_5, \quad \text{Error} = O(h^4) y k + 1 4 = y k + 216 25 k 1 + 2565 1408 k 3 + 4101 2197 k 4 − 5 1 k 5 , Error = O ( h 4 )
And the approximation to the solution with the help of the method of the 5th order:
y k + 1 5 = y k + 16 135 k 1 + 6656 12825 k 3 + 28561 56430 k 4 − 9 50 k 5 + 2 55 k 6 , E r r o r = O ( h 5 ) y_{k+1}^{5} = 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}) y k + 1 5 = y k + 135 16 k 1 + 12825 6656 k 3 + 56430 28561 k 4 − 50 9 k 5 + 55 2 k 6 , E rror = O ( h 5 )
At each step, two different approximations for the solution are made and compared. The optimal step size is ( δ ⋅ h ) (\delta \cdot h) ( δ ⋅ h )
ε n + 1 = 1 h ∣ y k + 1 4 − y k + 1 5 ∣ , δ = 0.84 ( ε ε n + 1 ) 1 4 , ε − s p e c i f i e d a c c u r a c y \varepsilon_{n+1} = \frac{1}{h} \left| y_{k+1}^{4} - y_{k+1}^{5} \right|, \quad \delta = 0.84 \left( \frac{\varepsilon}{\varepsilon_{n+1}} \right)^{\frac{1}{4}}, \quad \varepsilon - specified\ accuracy ε n + 1 = h 1 ∣ ∣ y k + 1 4 − y k + 1 5 ∣ ∣ , δ = 0.84 ( ε n + 1 ε ) 4 1 , ε − s p ec i f i e d a cc u r a cy
Problem solving and error analysis:
D(t,T) := -0.3·(T - 35) y 0 : = 5 y_0 := 5 y 0 := 5
D(t,T) - 1st order ODE with IVP
\text {R u n g e} 4 5 (\mathrm {y} 0, \mathrm {D}) := \left\{ \begin{array}{l} \left(z _ {0} \leftarrow 0 \quad y _ {0} \leftarrow y 0 \quad h \leftarrow 1 0 ^ {- 2}\right) \\ \text {f o r} \quad i \in 0.. 3 0 \\ \left| z _ {i + 1} \leftarrow z _ {i} + h \\ k 1 \leftarrow D \left(z _ {i}, y _ {i}\right) \\ k 2 \leftarrow D \left(z _ {i} + \frac {h}{4}, y _ {i} + \frac {h \cdot k 1}{4}\right) \\ k 3 \leftarrow D \left(z _ {i} + \frac {3 \cdot h}{8}, y _ {i} + \frac {h \cdot 3 \cdot k 1}{3 2} + \frac {h \cdot 9 \cdot k 2}{3 2}\right) \\ k 4 \leftarrow D \left(z _ {i} + \frac {1 2 \cdot h}{1 3}, y _ {i} + \frac {h \cdot 1 9 3 2 \cdot k 1}{2 1 9 7} - \frac {h \cdot 7 2 0 0 \cdot k 2}{2 1 9 7} + \frac {h \cdot 7 2 9 6 \cdot k 3}{2 1 9 7}\right) \\ k 5 \leftarrow D \left(z _ {i} + h, y _ {i} + \frac {h \cdot 4 3 9 \cdot k 1}{2 1 6} - h \cdot 8 \cdot k 2 + \frac {h \cdot 3 6 8 0 \cdot k 3}{5 1 3} - \frac {h \cdot 8 4 5 \cdot k 4}{4 1 0 4}\right) \\ k 6 \leftarrow D \left(z _ {i} + \frac {h}{2}, y _ {i} - \frac {h \cdot 8 \cdot k 1}{2 7} + h \cdot 2 \cdot k 2 - \frac {h \cdot 3 5 4 4 \cdot k 3}{2 5 6 5} + \frac {h \cdot 1 8 5 9 \cdot k 4}{4 1 0 4} - \frac {h \cdot 1 1 \cdot k 4}{4 0}\right) \\ w _ {i + 1} \leftarrow y _ {i} + h \cdot \left(\frac {2 5 \cdot k 1}{2 1 6} + \frac {1 4 0 8 \cdot k 3}{2 5 6 5} + \frac {2 1 9 7 \cdot k 4}{4 1 0 1} - \frac {k 5}{5}\right) \\ z _ {i + 1} \leftarrow y _ {i} + h \cdot \left(\frac {1 6 \cdot k 1}{1 3 5} + \frac {6 6 5 6 \cdot k 3}{1 2 8 2 5} + \frac {2 8 5 6 1 \cdot k 4}{5 6 4 3 0} - \frac {9 \cdot k 5}{5 0} + \frac {2 \cdot k 6}{5 5}\right) \\ y _ {i + 1} \leftarrow z _ {i + 1} \\ h \leftarrow \left(\frac {h \cdot 1 0 ^ {- 2}}{\left| z _ {i + 1} - w _ {i + 1} \right|}\right) ^ {\frac {1}{4}} \cdot h \cdot 0.84 \end{array}
Error calculation.
T - exact value
AT - approximation
∣ T − A T ∣ = 7.80215501191 E − 005 \left| T - AT \right| = 7.80215501191E-005 ∣ T − A T ∣ = 7.80215501191 E − 005
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#340153 on Dec 2023
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