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1、第3章 波动方程初始问题的求解行波法 （达朗贝尔公式） （特征线积分法）Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.13.1 达朗贝尔公式（行波法）一维问题1 基本思想：先求出偏微分方
2、程的通解，然后用定解条件确定特解。 这一思想与常微分方程的解法是一样的。2 关键步骤：通过变量变换，将波动方程化为便于积分的齐次二阶 偏微分方程。通解法中有一种特殊的解法行波法, 即以自变量的 线性组合作变量代换，进行求解的一种方法，它对波动方 程类型的求解十分有效.数学物理方程数学物理方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2
3、.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.21.弦振动方程的初始问题-无界限的自由振动（1）物理解释：认为弦很长，考虑远离边界的某段弦在较短时间内的振 动，其中给定初始位移和速度，并且没有强迫外力作用。数学物理方程数学物理方程(3.1.1)Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides fo
4、r .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.3（2）一维波动方程的通解：作变换：任意函数数学物理方程数学物理方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. C
5、opyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.4（3）无界限弦自由振动的特解（考虑初始条件）：达朗贝尔公式数学物理方程数学物理方程仅对第一层变量进行积分(3.1.2)Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2
6、004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.5结论：达朗贝尔解表示沿x 轴正、反向传播的两列波速为a的波的叠加，故称为行波法。a. 只有初始位移时，代表以速度a 沿x 轴正向传播的波代表以速度a 沿x 轴负向传播的波（4）达朗贝尔公式的意义：数学物理方程数学物理方程b. 只有初始速度时：为 的积分原函数。Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with A
7、spose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.6示意图形（向右传播的波）：数学物理方程数学物理方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2
8、.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.7解：将初始条件代入达朗贝尔公式（5）达朗贝尔公式的应用：数学物理方程数学物理方程波动方程 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-201
9、1 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.8解：将初始条件代入达朗贝尔公式数学物理方程数学物理方程波动方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 A
10、spose Pty Ltd.9解：该问题不能直接运用达朗贝尔公式，但可按该思路进行。特征方程为：令：方程可化为 ：数学物理方程数学物理方程非波动方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty
11、Ltd.10所以，方程的通解为：带入初始条件：数学物理方程数学物理方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.11所以：因此：数学物理方程数学物理方程Evaluation only
12、.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.12解：该问题不能直接运用达朗贝尔原理，但可按该思路进行。特征方程为：令：方程可化为：数学物理方程数学物理方程非波动方程Evaluation only.Evaluatio
13、n only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.13所以，方程的通解为：带入初始条件：又因为：为有限值，所以：即：数学物理方程数学物理方程Evaluation only.Evaluation only. Created with Aspo
14、se.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.14所以：因此：数学物理方程数学物理方程Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Crea
15、ted with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.152 弦振动方程的初始问题-半无界弦的自由振动物理解释：认为弦很长，考虑弦线某端附近而远离另一端在较短时 间内的振动，其中给定初始位移和速度，没有强迫外力作 用，弦线一端被固定。数学物理方程数学物理方程(3.1.3)Evaluation only.Evaluation only. Created with Aspose.Slides f
16、or .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.16为了解决这个问题，可以将半无界弦延拓成无限长的弦， 由于边界条件 ，所以无限长弦的 必须为奇函数 ，其初始条件也必须为奇函数，即：数学物理方程数学物理方程(3.1.4)(3.1.5)Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyrig
