Kenan kılıçaslan

  • Baca Hesabı
  • Sürtünme Kaybı
  • Diferansiyel Denklem
  • Denklem Çözümü
Hesap Modülleri Hesap Modülleri

Yüksek Mertebeden Diferansiyel Denklemlerin Runge-Kutta-Fehlberg Methodu İle Çözümü

Matematikte Runge – Kutta – Fehlberg yöntemi (veya Fehlberg yöntemi), diferansiyel denklemlerin sayısal çözümü için sayısal analizde bir algoritmadır. Alman matematikçi Erwin Fehlberg tarafından geliştirilmiştir ve Runge-Kutta yöntemlerine dayanmaktadır.
\(\small y''=f(t,y,y')\),     \(\small y(t_0)=y_0\),    \(\small y'(t_0)=y_0\)


Aşağıdaki gibi tanımlanan bir başlangıç değer problemini ele alalım.

\(\small y''-2y'+2y=e^{2t}\sin t\) denklemini Runge – Kutta – Fehlberg yöntemi ile çözelim.
Bu denklem \(\small y''=e^{2t}\sin t +2y'-2y\) 'dir.

\(\small u=y'\),
1.Denklem: \(\small f_1(t,y,u)=u\)
2.Denklem: \(\small f_2(t,y,u)=y''=f(t,y,u)\) iki denklem sistemi

Yukardaki örnek denkleme uygulayalım.
1.Denklem: \(\small f_1(t,y,u)=u\)
2.Denklem: \(\small f_2(t,y,u)=e^{2t}\sin t +2u-2y\) iki denklem sistemi


Her adım, aşağıdaki altı değerin kullanılmasını gerektirir:
\begin{equation} \small k_{1y}=hf_1(t_n,y_n,u_n), \end{equation}
\begin{equation} \small k_{1u}=hf_2(t_n,y_n,u_n), \end{equation}

\begin{equation} \small k_{2y}=hf_1(t_n+\displaystyle \frac{1}{4}h,y_n+\displaystyle \frac{1}{4}k_{1y},u_n+\displaystyle \frac{1}{4}k_{1u}) \end{equation}
\begin{equation} \small k_{2u}=hf_2(t_n+\displaystyle \frac{1}{4}h,y_n+\displaystyle \frac{1}{4}k_{1y},u_n+\displaystyle \frac{1}{4}k_{1u}) \end{equation}

\begin{equation} \small \begin{matrix} k_{3y}= &hf_1(t_n+\displaystyle \frac{3}{8}h, &y_n+\displaystyle \frac{3}{32}k_{1y}+\displaystyle \frac{9}{32}k_{2y}, \\ & & u_n+\displaystyle \frac{3}{32}k_{1u}+\displaystyle \frac{9}{32}k_{2u} ) \end{matrix} \end{equation}
\begin{equation} \small \begin{matrix} k_{3u}=& hf_2(t_n+\displaystyle \frac{3}{8}h, & y_n+\displaystyle \frac{3}{32}k_{1y}+\displaystyle \frac{9}{32}k_{2y}, \\ & & u_n+\displaystyle \frac{3}{32}k_{1u}+\displaystyle \frac{9}{32}k_{2u}) \end{matrix} \end{equation}

\begin{equation} \small \begin{matrix} k_{4y}=&hf_1(t_n+\displaystyle \frac{12}{13}h, &y_n+\displaystyle \frac{1932}{2197}k_{1y}-\displaystyle \frac{7200}{2197}k_{2y}+\displaystyle \frac{7296}{2197}k_{3y}, \\ & & u_n+\displaystyle \frac{1932}{2197}k_{1u}-\displaystyle \frac{7200}{2197}k_{2u}+\displaystyle \frac{7296}{2197}k_{3u}) \end{matrix} \end{equation}
\begin{equation} \small \begin{matrix} k_{4u}= & hf_2(t_n+\displaystyle \frac{12}{13}h, &y_n+\displaystyle \frac{1932}{2197}k_{1y}-\displaystyle \frac{7200}{2197}k_{2y}+\displaystyle \frac{7296}{2197}k_{3y}, \\ & & u_n+\displaystyle \frac{1932}{2197}k_{1u}-\displaystyle \frac{7200}{2197}k_{2u}+\displaystyle \frac{7296}{2197}k_{3u}) \end{matrix} \end{equation}

\begin{equation} \small \begin{array}{lll } k_{5y}= & hf_1(t_n+h, &y_n+\displaystyle \frac{439}{216}k_{1y}-8\displaystyle k_{2y}+\displaystyle \frac{3680}{513}k_{3y}-\displaystyle \frac{845}{4104}k_{4y}, \\ & & u_n+\displaystyle \frac{439}{216}k_{1u}-8\displaystyle k_{2u}+\displaystyle \frac{3680}{513}k_{3u}-\displaystyle \frac{845}{4104}k_{4u}) \end{array} \end{equation}
\begin{equation} \small \begin{array}{lll } k_{5u}= &hf_2(t_n+h, &y_n+\displaystyle \frac{439}{216}k_{1y}-8\displaystyle k_{2y}+\displaystyle \frac{3680}{513}k_{3y}-\displaystyle \frac{845}{4104}k_{4y}, \\ & & u_n+\displaystyle \frac{439}{216}k_{1u}-8\displaystyle k_{2u}+\displaystyle \frac{3680}{513}k_{3u}-\displaystyle \frac{845}{4104}k_{4u}) \end{array} \end{equation}

\begin{equation} \small \begin{array}{rrr } k_{6y}= & hf_1(t_n+\displaystyle \frac{1}{2}h, & y_n-\displaystyle \frac{8}{27}k_{1y}+2 k_{2y}-\displaystyle \frac{3544}{2565}k_{3y}+\displaystyle \frac{1859}{4104}k_{4y}\\ & &-\displaystyle \frac{11}{40}k_{5y}, \\ & & y_n-\displaystyle \frac{8}{27}k_{1u}+2 k_{2u}-\displaystyle \frac{3544}{2565}k_{3u}+\displaystyle \frac{1859}{4104}k_{4u}\\ & &-\displaystyle \frac{11}{40}k_{5u}) \end{array} \end{equation}
\begin{equation} \small \begin{array}{rrr } k_{6u}= & hf_2(t_n+\displaystyle \frac{1}{2}h, &y_n-\displaystyle \frac{8}{27}k_{1y}+2 k_{2y}-\displaystyle \frac{3544}{2565}k_{3y}+\displaystyle \frac{1859}{4104}k_{4y} \\ & & -\displaystyle \frac{11}{40}k_{5y}, \\ & & y_n-\displaystyle \frac{8}{27}k_{1u}+2 k_{2u}-\displaystyle \frac{3544}{2565}k_{3u}+\displaystyle \frac{1859}{4104}k_{4u}\\ & &-\displaystyle \frac{11}{40}k_{5u}) \end{array} \end{equation}

5. dereceden klasik Runge-Kutta Yöntemi
\begin{equation} \small y_{n+1}=y_{n}+ \displaystyle \frac{16}{135}k_{1y}+\displaystyle \frac{6656}{12825}k_{3y}+\displaystyle \frac{28561}{56430}k_{4y}-\displaystyle \frac{9}{50}k_{5y}+\displaystyle \frac{2}{55}k_{6y} \end{equation}
\begin{equation} \small u_{n+1}=u_{n}+ \displaystyle \frac{16}{135}k_{1u}+\displaystyle \frac{6656}{12825}k_{3u}+\displaystyle \frac{28561}{56430}k_{4u}-\displaystyle \frac{9}{50}k_{5u}+\displaystyle \frac{2}{55}k_{6u} \end{equation}

Burada \(h\) adım uzunluğudur.
beyaz_sayfa_en_alt_oval