研究生《数值分析》复习题

1、 用改进的欧拉折线法求解 y n+1 = yn + hzn z n+1 = zn + h( yn+1 = yn + zn+1 2 2zn 1 yn (1 中点公式属于二阶显式龙格库塔方法，二阶精度，截断误 差 o(h3 ，具体参见课本 P287. (2 令 z = y ，则 y = z ，原方程等价于 y = z, y (0 = 0.1 z (0 = 0.2 h (zn + z n+1 2 2 2z 2 h 2zn n+1 = zn + ( + ( 2 1 yn 1 y n+1 z = 2z 2y + e2x , 用中点公式求解 K1 = zn 取 h = 0.1 解得 y1 = 0.09,

2、z1 = 0.8289. y = z z dz dy 2z 2 dz = dy 1y dy z= = A(y 12 dx 1 y= +1 Ax + B 代入初值得到 A = 1, B = 1，故精确解为 y = 1 则 y (0.1 = 0.09091, y (0.1 = 0.82645. 6、对初值问题 y = ax + b, x > 0 y (0 = 0 用改进欧拉公式计算. 1 (1 证明：数值解 yn = 2 ax2 n + bxn . (2 求数值解与准确解的误差并从误差阶角度分析原因. (1 h yn = yn1 + (axn + b + (axn+1 + b 2 由于 x0

3、 = 0，故 nh = xn ，则 yn = yn1 + ah xn + xn1 + bh 2 xn1 + xn2 xn + xn1 + ah + 2bh = yn2 + ah 2 2 = ··· = y0 + ah n 1 i=0 n 1 M1 = 2zn 2yn + e2xn h K2 = zn + M1 2 h h h M2 = 2(zn + M1 2(yn + K1 + e2(xn + 2 2 2 yn+1 = yn + hK2 zn+1 = zn + hM2 计算 x = 0.1 处： 1 ， x+1 K1 = z0 = 0.2 M1 = 2z0 2y0

4、 + e2x0 = 0.8 h K2 = z0 + M1 = 0.16 2 h h h M2 = 2(z0 + M1 2(y0 + K1 + e2(x0 + 2 = 1.0052 2 2 y1 = y0 + hK2 = 0.116 z1 = z0 + hM2 = 0.0995 计算 x = 0.2 处： K1 = z1 = 0.0995 M1 = 2z1 2y1 + e2x1 = 1.2544 h K2 = z1 + M1 = 0.0368 2 h h h M2 = 2(z1 + M1 2(y1 + K1 + e2(x1 + 2 = 1.5182 2 2 y2 = y1 + hK2 = 0.1197 z2 = z1 + hM2 = 0.0523 xi + xi+1 + bnh 2 THE END. sunxb10 Jan. 16th, 2015 1 = 0 + ah (i + h + bxn 2 i=0 ( n(n 1 n 2 = ah + + bxn 2 2 1 = ax2 + bxn 2 n 2 (2 准确解为 y = 1 2 ax + bx，数值解与准确解之间误差为零， 因为改进欧拉法具有二阶精度，截断误差是 o(h3 . 7、用龙格 -库塔方法中的中点公式计算常微方程初值问题 y 2y + 2y = e2x y (0 = 0.1, y