[수치해석] 9. Euler's method

Euler's method

• 가장 단순한 미분법
• Initial value problem
  • $\frac{dy}{dt}=f(t,y),\ a\leq t\leq b, y(a)=\alpha$
  • [a,b] 범위를 n등분 : $t_i=a+ih,\ i=0,1,2,...,n$
  • $h=(b-a)/n=t_{i+1}-t_i$를 step size라고 정의
• euler's method 유도를 위한 taylor's theorem
  • $y(t_{i+1})=y(t_i)+(t_{i+1}-t_i)y'(t_i)+\frac{(t_{i+1}-t)^2}{2}y''(\zeta_i)\ (t_i\leq\zeta\leq t_{i+1})$
  • step size 공식에 의해 $y(t_{i+1})=y(t_i)+hf(t_i, y(t_i)) + \frac{h^2}{2}y''(\zeta_i)$

---

• Euler's method
  • $w_i\simeq y(t_i)$로 정으
  • $w_0 = \alpha$
  • $w_{i+1}=w_i+hf(t_i, w_i)$
• ex. h = 0.5, y(0)=0.5, $y'=y-t^2+1$
  • $w_0=y(0)=0.5$
  • $w_1=w_0+0.5(w_0-0^2+1)$
  • $w_2=w_1+0.5(w_1-(0.5)^2+1$
• $w_{j+1}=w_j+hf(t_j, w_j)$ : worward euler method
  • h가 커지면 발산할 위험이 있음
• $w_{j+1}\simeq w_j + hf(t_{j+1}, y_{j+1})$ : backward euler method
  • $w_{j+1}$을 풀어야 하나, stable한 장점이 있음

---

• Runge-Katta method (order 2)
  • $\int^{t_{j+1}}_{t_j}y'dt=\int^{t_{j+1}}_{t_j}f(t, y(t))dt\\\rarr y_{j+1}-y_j\simeq \frac{h}{2}[f(t_j, y_j) + f(t_{j+1}, y_{j+1})]$ : Implicit
  • Explicit : $y_{j+1}-y_j\simeq \frac{h}{2}[f(t_j, y_j) + f(t_{j+1}, \hat{y_{j+1})}]$
    • $y_{j+1}\simeq\hat{y_{j+1}}=y_j+hf(t_j, y_j)$
• Runge-Katta method (order 4)
  • Implicit : $y_{j+1}-y_j = \frac{h}{6}[f(t_j, y_j) + 4f(t_{j+1/2}, y_{j+1/2})+f(t_{j+1}, y_{j+1})]$
  • Explicit : $y_{j+1}-y_j = \frac{h}{6}[f(t_j, y_j) + 2f(t_{j+1/2}, \hat{y_{j+1/2}})+2f(t_{j+1/2}, \hat{\hat{y_{j+1/2}}})+f(t_{j+1}, \hat{y_{j+1}})]$
    • $\hat{y_{j+1/2}}=y_j+\frac{h}{2}f(t_j, y_j)$
    • $\hat{\hat{y_{j+1/2}}}=y_j+\frac{h}{2}f(t_j, \hat{y_j})$
    • $\hat{y_{j+1}}=y_j+hf(t_j+\hat{\hat{y_{j+1/2}}})$