Outline 第十 讲 变分法初步(二 北京大学物理学院 数学物理方法课程组 2007年春
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Outline 讲授要点 ③泛函的条件极值(续 举例 0微分方程的变分 微分方程定解问题 微分方程本征值问题
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Outline 讲授要点 ③泛函的条件极值(续 举例 ②微分方程的变分形式 微分方程定解问题 微分方程本征值问题 O Ravleigh-ritz 基本思路
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Outline 讲授要点 ③泛函的条件极值(续 举例 ②微分方程的变分形式 微分方程定解问题 微分方程本征值问题 ③ Rayleigh-Riz方法 基本思路 举例
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References 吴崇试,《数学物理方法》,第21章
Isoperimetric Problem (continued) Variational Formulism of Differential Equation (Approximation Method (Rayleigh-Ritz) References ÇÂÁ§5êÆÔn{6§121Ù ù&§5êÆÔn{6§§15.1 nÎ!X1Á§5êÆÔn{6§115Ù C. S. Wu 1lù C©{ÐÚ
References 吴崇试,《数学物理方法》,第21章 梁昆淼,《数学物理方法》,§15.1 学物理方
Isoperimetric Problem (continued) Variational Formulism of Differential Equation (Approximation Method (Rayleigh-Ritz) References ÇÂÁ§5êÆÔn{6§121Ù ù&§5êÆÔn{6§§15.1 nÎ!X1Á§5êÆÔn{6§115Ù C. S. Wu 1lù C©{ÐÚ
References 吴崇试,《数学物理方法》,第21章 梁昆淼,《数学物理方法》,§15.1 ¤胡嗣柱、倪光炯,《数学物理方法》,第15章
Isoperimetric Problem (continued) Variational Formulism of Differential Equation (Approximation Method (Rayleigh-Ritz) References ÇÂÁ§5êÆÔn{6§121Ù ù&§5êÆÔn{6§§15.1 nÎ!X1Á§5êÆÔn{6§115Ù C. S. Wu 1lù C©{ÐÚ
讲授要点 ③泛函的条件极值(续 举例 微分方程的变分形式 微分方程定解问题 微分方程本征值问题 O Ravleigh-Ritz方法 基本思路 举例
Isoperimetric Problem (continued) Variational Formulism of Differential Equation (Approximation Method (Rayleigh-Ritz) Examples ùÇ: 1 ¼^4(Y) Þ~ 2 ©§C©/ª ©§½)¯K ©§ ¯K 3 Rayleigh-Ritz{ Ä g´ Þ~ C. S. Wu 1lù C©{ÐÚ
例181求泛函=/xy/d在边界条件 (0)有界,y(1)=0 和约束条件xy2d=1下的极值曲线 解】采用上面描述的 lagrange乘子法,可以得 到必要条件
Isoperimetric Problem (continued) Variational Formulism of Differential Equation (Approximation Method (Rayleigh-Ritz) Examples ~18.1 ¦¼ I[y] = Z 1 0 xy02 dx 3>.^ y(0)k., y(1) = 0 Úå^ Z 1 0 xy2dx = 1 e4 =)>æ^þ¡£ãLagrange¦f{§± 7^ ∂ ∂y − d dx ∂ ∂y0 xy02 − λxy2 = 0 = d dx x dy dx + λxy = 0 C. S. Wu 1lù C©{ÐÚ
例181求泛函=/xy/d在边界条件 (0)有界,y(1)=0 和约束条件xy2d=1下的极值曲线 【解】采用上面描述的 Lagrange乘子法,可以得 到必要条件 0d0 d d 即 0 d
Isoperimetric Problem (continued) Variational Formulism of Differential Equation (Approximation Method (Rayleigh-Ritz) Examples ~18.1 ¦¼ I[y] = Z 1 0 xy02 dx 3>.^ y(0)k., y(1) = 0 Úå^ Z 1 0 xy2dx = 1 e4 =)>æ^þ¡£ãLagrange¦f{§± 7^ ∂ ∂y − d dx ∂ ∂y0 xy02 − λxy2 = 0 = d dx x dy dx + λxy = 0 C. S. Wu 1lù C©{ÐÚ
