多层扁球壳大挠度问题的近似解析解

Approximate Analytical Solution for the Large Deflection Problem of Multilayer Shal-low Spherical Shells

摘要: 本文基于Adomian分解法考虑在非均匀温度场内受横向均布载荷作用下的多层扁球壳大挠度问题。在边界固定夹紧、可移动夹紧情况下，分别给出了新近似解。在固定夹紧边界条件下，本文中得到的二次近似解析解对应的二次特征关系式与修正迭代法得到结果一致。在可移动夹紧边界条件下，通过误差分析说明了得到的Adomian近似解的收敛趋势。

Abstract: In this paper, based on the Adomian decomposition method, the large deflection problem of a mul-ti-layered oblate spherical shell subjected to a laterally uniform load in a non-uniform temperature field is considered. New approximate solutions are given in the case of fixed clamping and movable clamping of the boundary, respectively. The quadratic characteristic relation corresponding to the quadratic approximate analytical solution obtained in this paper under the fixed clamping bound-ary condition is consistent with the result obtained by the modified iteration method. Under the condition of movable clamping boundary, the convergence trend of the obtained Adomian approx-imate solution is illustrated by error analysis.

1. 引言

多层扁球壳在航空航天、机械工程、船舶制造等许多领域应用广泛，设计者可根据所想要获得的材料性能来制作多层扁球壳，所以多层扁球壳的研究具有实际应用意义，但考虑在非均匀温度场的影响下，受均布载荷作用的多层扁球壳大挠度问题的求解是一项具有挑战性的工作。罗祖道 [1] 等人讨论了有关双层金属扁壳的大挠度问题；李定坤 [2] 等人利用迭代法来研究双层金属扁壳的跳跃问题；朱永安 [3] 等人将修正迭代法应用于扁球壳的弯曲问题；赵伟东 [4] [5] 等人采用打靶法研究了扁球壳的非线性屈曲问题；王震鸣 [6] [7] 等人采用分解刚度法来研究各种载荷作用下多层扁壳的大挠度问题；叶开沅 [8] 等人利用修正迭代法求解了多层扁圆锥壳的大挠度方程；王新志 [9] 等人研究了多层扁球壳的非线性弯曲问题。

扁球壳的大挠度问题是非线性问题，对于求非线性问题的近似解，主要有严圣平，肖凡 [10] [11] 等人分别利用样条配点法和边界元法来求解大挠度方程；宋卫平 [12] 利用牛顿迭代法对扁球壳的非线性问题进行分析计算；刘人怀 [13] [14] 等人通过修正迭代法对扁球壳的大挠度问题进行研究。但Adomian分解法在求解扁球壳的非线性问题中应用很少，Adomian分解法 [15] [16] 对求解微分方程初边值问题非常实用，对于复杂的非线性微分方程，不需要先进行线性化。用Adomian多项式 [17] 的形式来代替非线性部分，由分解法得到的结果收敛较快，精度较高且便于计算机程序来实现，该方法广泛应用于各种方程，如二维Helmholtz方程 [18]、分数阶Volterra积分方程 [19] [20]、Fredholm积分方程 [21] [22]、Lane-Emden型初值问题 [23]。

本文在以上研究的基础上，基于Adomian分解法研究受非均匀温度场的影响下，多层扁球壳在固定夹紧和可移动夹紧边界条件下的大挠度问题，求出其近似解析解。具体内容为：在第2节回顾了大挠度问题的控制方程及边界条件；在第3节中利用Adomian分解法分析了不同边界条件下，多层扁球壳的大挠度问题；第4节讨论了多层扁球壳的特征关系式。

2. 该大挠度问题的控制方程及边界条件的回顾

多层扁球壳所满足的大挠度方程 [8] 为

$r\frac{{\text{d}}^{3}w}{\text{d}{r}^3}+\frac{{\text{d}}^{2}w}{\text{d}{r}^2}-{\lambda }^2\frac{\text{d}w}{\text{d}r}=\frac{q}{2D_{11}}{r}^2+\frac{T_{1}r}{D_{11}}\left(\theta +\frac{\text{d}w}{\text{d}r}\right)+\frac{D_{1t}}{D_{11}h}\frac{\text{d}\Delta t}{\text{d}r}+\frac{\left(D_{1t}-D_{2t}\right)}{D_{11}h}\Delta t,$ (1)

$r\frac{{\text{d}}^2\left(T_{1}r\right)}{\text{d}{r}^2}+\frac{\text{d}\left(T_{1}r\right)}{\text{d}r}-{\eta }^2\frac{\left(T_{1}r\right)}{r}=-l_1\frac{\text{d}w}{\text{d}r}\left(\theta +\frac{1}{2}\frac{\text{d}w}{\text{d}r}\right)+l_2{t}_0+l_{3}r\frac{\text{d}{t}_0}{\text{d}r},$ (2)

其中R为球壳半径，h为球壳厚度，w为中心挠度，q为横向均布载荷， $T_1$ 径向张力，C为多层壳体的抗拉刚度，D为多层壳体的弯曲刚度， $\Delta t=t_1-t_2$ 为温度改变量， $t_1,t_2$ 为上下表面温度，且

${\lambda }^2=\frac{D_{22}}{D_{11}},{\eta }^2=\frac{C_{22}}{C_{11}},t=t_0+\frac{z}{h}\Delta t,l_1=\frac{C_{11}{C}_{22}-C_{12}{C}_{12}}{C_{11}},$

$l_2=\frac{C_{1t}\left(C_{12}+C_{22}\right)-C_{2t}\left(C_{11}+C_{12}\right)}{C_{11}},l_3=\frac{C_{1t}{C}_{12}-C_{2t}{C}_{11}}{C_{11}}.$

考虑以下边界条件 [8]：

固定夹紧边界条件

$\{\begin{array}{l}r=0:\frac{\text{d}w}{\text{d}r}=0,T_1<+\infty ,\\ r=R:w=0,\frac{\text{d}w}{\text{d}r}=0,\frac{\text{d}\left(T_{1}r\right)}{\text{d}r}-\mu \frac{\left(T_{1}r\right)}{r}=-\left(C_{2t}-\mu C_{1t}\right)t_0,\end{array}$ (3)

其中 $\mu =\frac{C_{12}}{C_{11}}$。

可移动夹紧边界条件

$\{\begin{array}{l}r=0:\frac{\text{d}w}{\text{d}r}=0,T_1<+\infty ,\\ r=R:w=0,\frac{\text{d}w}{\text{d}r}=0,T_1=0.\end{array}$ (4)

为了将方程进一步简化，便于计算，取 $t_0,\Delta t$ 均为常量 [9]，令

$D_{1t}=D_{2t}=D_t,D_{11}=D_{22}=D,C_{1t}=C_{2t}=C_t,C_{11}=C_{22}=C.$

通过无量纲变换 [9]：

$\rho =\frac{r}{R},y=\sqrt{\frac{C\left(1-{\mu }^2\right)}{D}}w,\frac{\text{d}y}{\text{d}\rho }=\phi ,S=-\frac{R^2{T}_1\rho }{D},K=2H\sqrt{\frac{C\left(1-{\mu }^2\right)}{D}},P=\frac{1}{2}\sqrt{\frac{C\left(1-{\mu }^2\right)}{D^3}}{R}^{4}q,$

方程(1)~(2)变成为

$\rho \frac{{\text{d}}^2\phi }{\text{d}{\rho }^2}+\frac{\text{d}\phi }{\text{d}\rho }-\frac{\phi }{\rho }=P{\rho }^2-S\left(K\rho +\phi \right),$ (5)

$\rho \frac{{\text{d}}^{2}S}{\text{d}{\rho }^2}+\frac{\text{d}S}{\text{d}\rho }-\frac{S}{\rho }=\phi \left(K\rho +\frac{1}{2}\phi \right),$ (6)

对于固定夹紧的无量纲边界条件为：

$\{\begin{array}{l}\rho =0:\phi =0,S=0,\\ \rho =1:y=0,\phi =0,\frac{\text{d}S}{\text{d}\rho }-\mu \frac{S}{\rho }=\alpha ,\end{array}$ (7)

其中 $\alpha =\frac{R^2}{D}\left(1-\mu \right)C_t{t}_0$。

对于可移动夹紧的无量纲边界条件为

$\{\begin{array}{l}\rho =0:\phi =0,S=0,\\ \rho =1:y=0,\phi =0,S=0.\end{array}$ (8)

3. Adomian分解法的应用

3.1. 在固定夹紧边界条件下多层扁球壳的大挠度问题

3.1.1. 确定逆算子

为了确定逆算子 [24] [25]，先对方程(5)两边分别进行从1到 $\rho$ 的积分，再进行从0到 $\rho$ 的积分，如下所示：

${\displaystyle {\int }_0^{\rho }{\displaystyle {\int }_1^{\rho }\left(\rho \frac{{\text{d}}^2\phi }{\text{d}{\rho }^2}+\frac{\text{d}\phi }{\text{d}\rho }-\frac{\phi }{\rho }\right)\text{d}\rho \text{d}\rho }}={\displaystyle {\int }_0^{\rho }{\displaystyle {\int }_1^{\rho }\left(P{\rho }^2-S\left(K\rho +\phi \right)\right)\text{d}\rho \text{d}\rho }},$

考虑边界条件(7)，我们得到

$\begin{array}{l}\phi \left(\rho \right)=-\rho {\displaystyle {\int }_0^1\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }\left(P{\rho }^2-S\left(\rho \right)K\rho -S\left(\rho \right)\phi \left(\rho \right)\right)\text{d}\rho \text{d}\rho }}\\ +\frac{1}{\rho }{\displaystyle {\int }_0^{\rho }\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }\left(P{\rho }^2-S\left(\rho \right)K\rho -S\left(\rho \right)\phi \left(\rho \right)\right)\text{d}\rho \text{d}\rho }},\end{array}$

由此逆算子 $L_1^{-1}$ 被选取为如下：

$L_1^{-1}=-\rho {\displaystyle {\int }_0^1\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }}+\frac{1}{\rho }{\displaystyle {\int }_0^{\rho }\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }}.$ (9)

同样考虑方程(6)和边界条件(7)，逆算子 $L_2^{-1}$ 被确定为如下：

$L_2^{-1}=\frac{1+\mu }{1-\mu }\rho {\displaystyle {\int }_0^1\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }}+\frac{1}{\rho }{\displaystyle {\int }_0^{\rho }\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }}.$ (10)

3.1.2. 求解大挠度方程

将逆算子 $L_1^{-1},L_2^{-1}$ 分别作用于方程(5)和(6)两边，得到

$\phi \left(\rho \right)=L_1{}^{-1}\left(P{\rho }^2-S\left(\rho \right)K\rho -S\left(\rho \right)\phi \left(\rho \right)\right),$ (11)

$S\left(\rho \right)=\frac{\rho \alpha }{1-\mu }+L_2^{-1}\left(\phi \left(\rho \right)K\rho +\frac{1}{2}{\left(\phi \left(\rho \right)\right)}^2\right),$ (12)

根据Adomian分解法，将未知函数 $\phi \left(\rho \right),S\left(\rho \right)$ 和非线性项 $S\left(\rho \right)\phi \left(\rho \right),{\left(\phi \left(\rho \right)\right)}^2$ 分解成无穷项级数，无量纲横向均布载荷P分解为有限级数：

$\phi \left(\rho \right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n},S\left(\rho \right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n},S\left(\rho \right)\phi \left(\rho \right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{F}_n},\frac{1}{2}{\left(\phi \left(\rho \right)\right)}^2={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{G}_n},P={\displaystyle \underset{i=0}{\overset{n}{\sum }}{P}_i}.$ (13)

其中

$\begin{array}{l}{F}_n=\frac{1}{n!}\frac{{\text{d}}^n}{\text{d}{\lambda }^n}{\left[\left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\lambda }^n{S}_n}\right)\cdot \left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\lambda }^n{\phi }_n}\right)\right]}_{\lambda =0},n=0,1,2,\cdots ,\\ G_n=\frac{1}{n!}\frac{{\text{d}}^n}{\text{d}{\lambda }^n}{\left[\frac{1}{2}{\left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\lambda }^n{\phi }_n}\right)}^2\right]}_{\lambda =0},n=0,1,2,\cdots .\end{array}$ (14)

由(14)可知

$\begin{array}{l}{F}_0=S_0{\phi }_0,\\ F_1=S_1{\phi }_0+S_0{\phi }_1,\\ F_2=S_2{\phi }_0+S_1{\phi }_1+S_0{\phi }_2,\end{array}$

$\begin{array}{l}{G}_0=\frac{1}{2}{\left({\phi }_0\right)}^2,\\ G_1={\phi }_0{\phi }_1,\\ G_2=\frac{1}{2}{\left({\phi }_1\right)}^2+{\phi }_0{\phi }_2,\\ ⋮\end{array}$

将(13)~(14)代入方程(11)和(12)中得到

$\{\begin{array}{l}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n}=L_1^{-1}\left({\rho }^2{\displaystyle \underset{i=0}{\overset{n}{\sum }}{P}_i}-K\rho {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n}-{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{F}_n}\right),(\text{15})\\ {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n}=\frac{\rho \alpha }{1-\mu }+L_2^{-1}\left(K\rho {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n}+{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{G}_n}\right).(\; 16\; )\end{array}$

根据上面提到的公式，对多层扁球壳的大挠度问题可以构造出如下递推公式：

$\{\begin{array}{l}{\phi }_0=L_1^{-1}\left(P_0{\rho }^2\right),(\text{17})\\ S_0=\frac{\rho \alpha }{1-\mu }+L_2^{-1}\left(K\rho {\phi }_0\right)+L_2^{-1}\left(G_0\right).(\; 18\; )\end{array}$

$\{\begin{array}{l}{\phi }_n=L_1^{-1}\left(P_n{\rho }^2\right)-L_1^{-1}\left(K\rho S_{n-1}\right)-L_1^{-1}\left(F_{n-1}\right),n\ge 1,(\text{19})\\ S_n=L_2^{-1}\left(K\rho {\phi }_n\right)+L_2^{-1}\left(G_n\right),n\ge 1.(\; 20\; )\end{array}$

根据(17)，我们可以计算出

${\phi }_0=\frac{P_0}{8}\rho \left({\rho }^2-1\right),$ (21)

考虑无量纲变换与固定夹紧边界条件可以得到挠度y与 $\phi$ 的关系式：

$y={\displaystyle {\int }_1^0\phi \text{d}\rho }=0,$ (22)

则有

${\displaystyle {\int }_1^0{\phi }_0\text{d}\rho }={\displaystyle {\int }_1^0{\phi }_1\text{d}\rho }=\cdots ={\displaystyle {\int }_1^0{\phi }_n\text{d}\rho }=0,$ (23)

由(21)~(23)可得：

$P_0=32y_0,$ (24)

将(24)代入(17)~(18)中可得

$\begin{array}{l}{\phi }_0=4\rho \left(-1+{\rho }^2\right)y_0,\\ S_0=\frac{1}{6}K{\rho }^5{y}_0-\frac{1}{2}K{\rho }^3{y}_0+\frac{K\rho }{2}{y}_0+\frac{K\rho \left(1+\mu \right)}{6\left(1-\mu \right)}{y}_0+\frac{{\rho }^7}{6}{y}_0^2\\ -\frac{2{\rho }^5}{3}{y}_0^2+{\rho }^3{y}_0^2-\frac{2\rho }{3}{y}_0^2-\frac{\rho \left(1+\mu \right)}{6\left(1-\mu \right)}{y}_0^2+\frac{\alpha \rho }{1-\mu },\end{array}$

将 ${\phi }_0,S_0$ 代入(19)中，并考虑

$y_1={\displaystyle {\int }_1^0{\phi }_1\text{d}\rho }=0,$

得到

$P_1=\frac{\left(11\mu -35\right)K^2{y}_0}{72\left(-1+\mu \right)}+\frac{20\alpha y_0}{9\left(-1+\mu \right)}-\frac{\left(82\mu -227\right)K{y}_0^2}{135\left(-1+\mu \right)}+\frac{\left(73\mu -173\right)y_0^3}{135\left(-1+\mu \right)}-\frac{K\alpha }{-1+\mu },$ (25)

$\begin{array}{l}{\phi }_1=-\frac{1}{8640\left(-1+\mu \right)}{y}_0\rho \left(-1+{\rho }^2\right)(15K^2\left(-1+\mu \right)\left(3-10{\rho }^2+2{\rho }^4\right)\\ +2K{y}_0(217-765{\rho }^2+255{\rho }^4-45{\rho }^6+\mu \left(-137+525{\rho }^2-255{\rho }^4+45{\rho }^6\right))\\ +8(60\alpha \left(1-3{\rho }^2\right)+y_0^2(-58+219{\rho }^2-111{\rho }^4+39{\rho }^6-6{\rho }^8\\ +\mu \left(38-159{\rho }^2+111{\rho }^4-39{\rho }^6+6{\rho }^8\right)))),\end{array}$

$\begin{array}{c}{S}_1=\frac{1}{1451520\left(-1+\mu \right)}{y}_0\rho \\ (\frac{2y_0\left(1+\mu \right)\left(4K{y}_0\left(151-95\mu \right)+63K^2\left(-1+\mu \right)+32\left(21\alpha +y_0^2\left(-20+13\mu \right)\right)\right)}{-1+\mu }\\ +\frac{K\left(1+\mu \right)\left(147K^2\left(-1+\mu \right)-28K{y}_0\left(-49+29\mu \right)+16\left(105\alpha +y_0^2\left(-88+53\mu \right)\right)\right)}{1-\mu }\\ -K{\rho }^2(21K^2\left(-1+\mu \right)\left(-45+65{\rho }^2-30{\rho }^4+3{\rho }^6\right)\\ +14K{y}_0(-651+411\mu +2\left(491-331\mu \right){\rho }^2+30\left(-17+13\mu \right){\rho }^4\end{array}$

$\begin{array}{l}-90\left(-1+\mu \right){\rho }^6+9\left(-1+\mu \right){\rho }^8)+8(-210\alpha \left(6-8{\rho }^2+3{\rho }^4\right)\\ +y_0^2(1218-798\mu +7\left(-277+197\mu \right){\rho }^2+105\left(11-9\mu \right){\rho }^4\\ +315\left(-1+\mu \right){\rho }^6-63\left(-1+\mu \right){\rho }^8+6\left(-1+\mu \right){\rho }^{10})))\\ -2y_0(21K^2\left(-1+\mu \right)\left(-14+90{\rho }^2-160{\rho }^4+125{\rho }^6-42{\rho }^8+4{\rho }^{10}\right)\\ +4K{y}_0(-686+406\mu +21\left(217-137\mu \right){\rho }^2+7\left(-1199+799\mu \right){\rho }^4\end{array}$

$\begin{array}{l}+49\left(143-103\mu \right){\rho }^6+252\left(-11+9\mu \right){\rho }^8-483\left(-1+\mu \right){\rho }^{10}\\ +45\left(-1+\mu \right){\rho }^{12})+8(-84\alpha \left(5-30{\rho }^2+50{\rho }^4-35{\rho }^6+9{\rho }^8\right)\\ +y_0^2(352-212\mu +84\left(-29+19\mu \right){\rho }^2+70\left(67-47\mu \right){\rho }^4\\ +7\left(-607+467\mu \right){\rho }^6-252\left(-8+7\mu \right){\rho }^8\\ +546\left(-1+\mu \right){\rho }^{10}-102\left(-1+\mu \right){\rho }^{12}+9\left(-1+\mu \right){\rho }^{14})))),\end{array}$

将 ${\phi }_1,S_1$ 代入(19)中，并考虑(22)~(23)可得

$\begin{array}{c}{P}_2=\frac{913K^4{y}_0}{4147200{\left(-1+\mu \right)}^2}-\frac{493K^4\mu y_0}{2073600{\left(-1+\mu \right)}^2}+\frac{11K^2\alpha \mu y_0}{32400{\left(-1+\mu \right)}^2}\\ +\frac{73K^4{\mu }^2{y}_0}{4147200{\left(-1+\mu \right)}^2}-\frac{43K^2\alpha y_0}{16200{\left(-1+\mu \right)}^2}+\frac{{\alpha }^2{y}_0}{648{\left(-1+\mu \right)}^2}\\ -\frac{134251K^3{y}_0^2}{50803200{\left(-1+\mu \right)}^2}+\frac{15971K^3\mu y_0^2}{8467200{\left(-1+\mu \right)}^2}+\frac{13K\alpha \mu y_0^2}{64800{\left(-1+\mu \right)}^2}\\ -\frac{31K^3{\mu }^2{y}_0^2}{2032128{\left(-1+\mu \right)}^2}+\frac{169K\alpha y_0^2}{21600{\left(-1+\mu \right)}^2}+\frac{3581003K^2{y}_0^3}{457228800{\left(-1+\mu \right)}^2}\end{array}$

$\begin{array}{l}-\frac{909859K^2\mu y_0^3}{228614400{\left(-1+\mu \right)}^2}-\frac{5513\alpha \mu y_0^3}{2381400{\left(-1+\mu \right)}^2}-\frac{52769K^2{\mu }^2{y}_0^3}{91445760{\left(-1+\mu \right)}^2}\\ -\frac{6737\alpha y_0^3}{2381400{\left(-1+\mu \right)}^2}-\frac{5128799K{y}_0^4}{685843200{\left(-1+\mu \right)}^2}+\frac{3197K\mu y_0^4}{1959552{\left(-1+\mu \right)}^2}\\ +\frac{1234489K{\mu }^2{y}_0^4}{685843200{\left(-1+\mu \right)}^2}+\frac{655951y_0^5}{342921600{\left(-1+\mu \right)}^2}+\frac{25391\mu y_0^5}{24494400{\left(-1+\mu \right)}^2}\\ -\frac{19289{\mu }^2{y}_0^5}{13716864{\left(-1+\mu \right)}^2},\end{array}$ (26)

将 ${\phi }_1,S_1,P_2$ 代入(19)~(20)中可得 ${\phi }_2,S_2$ 与 $y_0$ 的关系式，因此，我们可以依次确定出 ${\phi }_3,S_3,{\phi }_4,S_4,\cdots$，得到n项Adomian近似解为：

$\begin{array}{l}\phi \approx {\displaystyle \underset{i=0}{\overset{n-1}{\sum }}{\phi }_i}={\phi }_0+{\phi }_1+\cdots {\phi }_{n-1},\\ S\approx {\displaystyle \underset{i=0}{\overset{n-1}{\sum }}{S}_i}=S_0+S_1+\cdots S_{n-1},\end{array}$

3.2. 在可移动夹紧边界条件下多层扁球壳的大挠度问题

根据方程(5)~(6)和边界条件(8)，可以确定出逆算子：

$\begin{array}{l}{L}_3^{-1}=-\rho {\displaystyle {\int }_0^1\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }}+\frac{1}{\rho }{\displaystyle {\int }_0^{\rho }\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }},\\ L_4^{-1}=-\rho {\displaystyle {\int }_0^1\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }}+\frac{1}{\rho }{\displaystyle {\int }_0^{\rho }\rho {\displaystyle {\int }_1^{\rho }\frac{1}{\rho }(\cdot )\text{d}\rho \text{d}\rho }},\end{array}$

将逆算子 $L_3^{-1},L_4^{-1}$ 分别作用于方程(5)和(6)两边，得到

$\phi \left(\rho \right)=L_3^{-1}\left(P{\rho }^2-S\left(\rho \right)K\rho -S\left(\rho \right)\phi \left(\rho \right)\right),$ (27)

$S\left(\rho \right)=L_4^{-1}\left(\phi \left(\rho \right)K\rho +\frac{1}{2}{\left(\phi \left(\rho \right)\right)}^2\right),$ (28)

根据Adomian分解法，将未知函数 $\phi \left(\rho \right),S\left(\rho \right)$ 和非线性项 $S\left(\rho \right)\phi \left(\rho \right),{\left(\phi \left(\rho \right)\right)}^2$ 分解成无穷项级数，无量纲横向均布载荷P分解为有限级数：

$\phi \left(\rho \right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n},S\left(\rho \right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n},S\left(\rho \right)\phi \left(\rho \right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{F}_n},\frac{1}{2}{\left(\phi \left(\rho \right)\right)}^2={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{G}_n},P={\displaystyle \underset{i=0}{\overset{n}{\sum }}{P}_i}.$ (29)

其中

$\begin{array}{l}{F}_n=\frac{1}{n!}\frac{{\text{d}}^n}{\text{d}{\lambda }^n}{\left[\left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\lambda }^n{S}_n}\right)\cdot \left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\lambda }^n{\phi }_n}\right)\right]}_{\lambda =0},n=0,1,2,\cdots ,\\ G_n=\frac{1}{n!}\frac{{\text{d}}^n}{\text{d}{\lambda }^n}{\left[\frac{1}{2}{\left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\lambda }^n{\phi }_n}\right)}^2\right]}_{\lambda =0},n=0,1,2,\cdots .\end{array}$ (30)

将(29)~(30)代入方程(27)和(28)中得到

$\{\begin{array}{l}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n}=L_3^{-1}\left({\rho }^2{\displaystyle \underset{i=0}{\overset{n}{\sum }}{P}_i}-K\rho {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n}-{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{F}_n}\right),(\text{31})\\ {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n}=L_4^{-1}\left(K\rho {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n}+{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{G}_n}\right).(\; 32\; )\end{array}$

构造出如下递推公式：

$\{\begin{array}{l}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n}=L_3^{-1}\left({\rho }^2{\displaystyle \underset{i=0}{\overset{n}{\sum }}{P}_i}-K\rho {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n}-{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{F}_n}\right),(33)\\ {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n}=L_4^{-1}\left(K\rho {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n}+{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{G}_n}\right).(\; 34\; )\end{array}$

$\{\begin{array}{l}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\phi }_n}=L_3^{-1}\left({\rho }^2{\displaystyle \underset{i=0}{\overset{n}{\sum }}{P}_i}-K\rho {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{S}_n}-{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{F}_n}\right),(\text{35})\\ S_n=L_4^{-1}\left(K\rho {\phi }_n\right)+L_4{}^{-1}\left(G_n\right),n\ge 1.(\; 3\; 6\; )\end{array}$

根据(33)，计算出

${\phi }_0=\frac{P_0}{8}\rho \left({\rho }^2-1\right).$ (37)

考虑无量纲挠度y与 $\phi$ 的关系式：

$y={\displaystyle {\int }_1^0\phi \text{d}\rho }=0,$ (38)

则有

${\displaystyle {\int }_1^0{\phi }_0\text{d}\rho }={\displaystyle {\int }_1^0{\phi }_1\text{d}\rho }=\cdots ={\displaystyle {\int }_1^0{\phi }_n\text{d}\rho }=0,$ (39)

由(37)~(39)可得：

$P_0=32y_0$. (40)

将(40)代入(33)~(34)中可得

$\begin{array}{l}{\phi }_0=4y_0\rho \left(-1+{\rho }^2\right),\\ S_0=\frac{1}{6}K{y}_0{\rho }^5-\frac{1}{2}K{y}_0{\rho }^3+\frac{K{y}_0\rho }{3}+\frac{y_0^2{\rho }^7}{6}-\frac{2y_0^2{\rho }^5}{3}+y_0^2{\rho }^3-\frac{y_0^2\rho }{2}.\end{array}$

将 ${\phi }_0,S_0$ 代入(35)中，再考虑(38)~(39)可得

$P_1=\frac{11K^2{y}_0}{72}-\frac{82K{y}_0^2}{135}+\frac{73y_0^3}{135},$ (41)

$\begin{array}{c}{\phi }_1=\frac{1}{192}{K}^2{y}_0\rho -\frac{137K{y}_0^2\rho }{4320}+\frac{19y_0^3\rho }{540}-\frac{13}{576}{K}^2{y}_0{\rho }^3+\frac{331K{y}_0^2{\rho }^3}{2160}\\ -\frac{197y_0^3{\rho }^3}{1080}+\frac{1}{48}{K}^2{y}_0{\rho }^5-\frac{13}{72}K{y}_0^2{\rho }^5+\frac{y_0^3{\rho }^5}{4}-\frac{1}{288}{K}^2{y}_0{\rho }^7\\ +\frac{5}{72}K{y}_0^2{\rho }^7-\frac{5y_0^3{\rho }^7}{36}-\frac{1}{96}K{y}_0^2{\rho }^9+\frac{y_0^3{\rho }^9}{24}-\frac{y_0^3{\rho }^{11}}{180},\end{array}$

$\begin{array}{c}{S}_1=-\frac{1}{1451520}{y}_0\rho \left(-1+{\rho }^2\right)(21K^3\left(-7+38{\rho }^2-27{\rho }^4+3{\rho }^6\right)\\ +14K^2{y}_0\left(109-572{\rho }^2+570{\rho }^4-195{\rho }^6+21{\rho }^8\right)\\ +8K{y}_0^2\left(-607+3068{\rho }^2-3904{\rho }^4+2088{\rho }^6-495{\rho }^8+51{\rho }^{10}\right)\\ +16y_0^3\left(264-1332{\rho }^2+1958{\rho }^4-1311{\rho }^6+453{\rho }^8-93{\rho }^{10}+9{\rho }^{12}\right)).\end{array}$

将 ${\phi }_1,S_1$ 代入(35)中，考虑(38)~(39)可得

$P_2=\frac{73K^4{y}_0}{4147200}-\frac{31K^3{y}_0^2}{2032128}-\frac{52769K^2{y}_0^3}{91445760}+\frac{1234489K{y}_0^4}{685843200}-\frac{19289y_0^5}{13716864},$ (42)

$\begin{array}{c}{P}_3=-\frac{2537K^6{y}_0}{234101145600}+\frac{52387K^5{y}_0^2}{329204736000}-\frac{505909K^4{y}_0^3}{2633637888000}-\frac{1422358327K^3{y}_0^4}{318670184448000}\\ +\frac{3253160447K^2{y}_0^5}{159335092224000}-\frac{541294218343K{y}_0^6}{16964407269089280}+\frac{3574120874477y_0^7}{212055090863616000},\end{array}$ (43)

将 ${\phi }_2,S_2,P_3$ 代入(35)~(36)中可得 ${\phi }_3,S_3$ 与 $y_0$ 的关系式。

因此，我们可以依次计算出 ${\phi }_4,S_4,{\phi }_5,S_5,\cdots$，得到n项Adomian近似解为：

$\phi \approx {\displaystyle \underset{i=0}{\overset{n-1}{\sum }}{\phi }_i},S\approx {\displaystyle \underset{i=0}{\overset{n-1}{\sum }}{S}_i}.$

4. 结果分析

4.1. 在固定夹紧边界条件下多层扁球壳的特征关系式

在固定夹紧边界条件下，根据(24)~(25)可知P与 $y_0$ 的二次特征关系式为：

$P\approx 32y_0+\frac{\left(11\mu -35\right)K^2{y}_0}{72\left(-1+\mu \right)}+\frac{20\alpha y_0}{9\left(-1+\mu \right)}-\frac{\left(82\mu -227\right)K{y}_0^2}{135\left(-1+\mu \right)}+\frac{\left(73\mu -173\right)y_0^3}{135\left(-1+\mu \right)}-\frac{K\alpha }{-1+\mu }.$ (44)

对于单层扁球壳 [9]，取 $\mu =0.3$，

$C=\frac{Eh}{1-{\mu }^2},D=\frac{E{h}^3}{12\left(1-{\mu }^2\right)},$

由(44)可得单层扁球壳的特征关系式为：

$P\approx 32\left[\left(1+0.0196553K^2\right)y_0-0.0669312K{y}_0^2+0.0499669y_0^3\right].$ (45)

该结果与文 [9] 用修正迭代法得到的结果一致。

多层扁球壳受无量纲横向均布载荷P作用的三次近似解为：

$P\approx P_0+P_1+P_3$

根据(24)~(26)可知P与 $y_0$ 的三次特征关系式为：

$\begin{array}{c}P\approx 32y_0+\frac{\left(11\mu -35\right)K^2{y}_0}{72\left(-1+\mu \right)}+\frac{20\alpha y_0}{9\left(-1+\mu \right)}+\frac{913K^4{y}_0}{4147200{\left(-1+\mu \right)}^2}\\ -\frac{493K^4\mu y_0}{2073600{\left(-1+\mu \right)}^2}+\frac{11K^2\alpha \mu y_0}{32400{\left(-1+\mu \right)}^2}+\frac{73K^4{\mu }^2{y}_0}{4147200{\left(-1+\mu \right)}^2}\\ -\frac{43K^2\alpha y_0}{16200{\left(-1+\mu \right)}^2}+\frac{{\alpha }^2{y}_0}{648{\left(-1+\mu \right)}^2}-\frac{\left(82\mu -227\right)K{y}_0{}^2}{135\left(-1+\mu \right)}\\ -\frac{134251K^3{y}_0^2}{50803200{\left(-1+\mu \right)}^2}+\frac{15971K^3\mu y_0^2}{8467200{\left(-1+\mu \right)}^2}+\frac{13K\alpha \mu y_0^2}{64800{\left(-1+\mu \right)}^2}\\ -\frac{31K^3{\mu }^2{y}_0^2}{2032128{\left(-1+\mu \right)}^2}+\frac{169K\alpha y_0^2}{21600{\left(-1+\mu \right)}^2}+\frac{\left(73\mu -173\right)y_0^3}{135\left(-1+\mu \right)}\end{array}$

$\begin{array}{l}+\frac{3581003K^2{y}_0^3}{457228800{\left(-1+\mu \right)}^2}-\frac{909859K^2\mu y_0^3}{228614400{\left(-1+\mu \right)}^2}-\frac{5513\alpha \mu y_0^3}{2381400{\left(-1+\mu \right)}^2}\\ -\frac{52769K^2{\mu }^2{y}_0^3}{91445760{\left(-1+\mu \right)}^2}-\frac{6737\alpha y_0^3}{2381400{\left(-1+\mu \right)}^2}-\frac{5128799K{y}_0^4}{685843200{\left(-1+\mu \right)}^2}\\ +\frac{3197K\mu y_0^4}{1959552{\left(-1+\mu \right)}^2}+\frac{1234489K{\mu }^2{y}_0^4}{685843200{\left(-1+\mu \right)}^2}+\frac{655951y_0^5}{342921600{\left(-1+\mu \right)}^2}\\ +\frac{25391\mu y_0^5}{24494400{\left(-1+\mu \right)}^2}-\frac{19289{\mu }^2{y}_0^5}{13716864{\left(-1+\mu \right)}^2}-\frac{K\alpha }{-1+\mu }.\end{array}$ (46)

4.2. 在可移动夹紧边界条件下多层扁球壳的特征关系式

我们定义以下两个余函数：

$Error1\left(\phi ,\rho \right)=\rho \frac{{\text{d}}^2\phi }{\text{d}{\rho }^2}+\frac{\text{d}\phi }{\text{d}\rho }-\frac{\phi }{\rho }-P{\rho }^2+S\left(K\rho +\phi \right),$ (47)

$Error2\left(S,\rho \right)=\rho \frac{{\text{d}}^{2}S}{\text{d}{\rho }^2}+\frac{\text{d}S}{\text{d}\rho }-\frac{S}{\rho }-\phi \left(K\rho +\frac{1}{2}\phi \right),$ (48)

其中 $\phi ,S$ 是本文中得到的近似解。利用 ${\Vert Error1\Vert }_2^2$ 和 ${\Vert Error2\Vert }_2^2$ 来表征大挠度问题近似解的精度，显然 ${\Vert Error1\Vert }_2^2={\Vert Error2\Vert }_2^2=0$ 时，对应的解是精确解，否则近似解。当 $K=0,y_0=0.2$ 时，二阶，三阶，四阶近似解的误差分析见表1。余函数曲线如图1、图2所示。从表1，图1，图2，能看出随着阶数的不断增加，误差越来越小，即得到的Adomian近似解具有收敛的趋势。

Table 1. Approximate solution accuracy (2th-order, 3th-order, 4th-order)

表1. 近似解精度(二阶，三阶，四阶)

Figure 1. The $Error1$ curves with the $y_0=0.2$ (4th-order approximation)

图1. $y_0=0.2$ 时， $Error1$ 的四阶近似曲线

Figure 2. The $Error2$ curves with the $y_0=0.2$ (4th-order approximation)

图2. $y_0=0.2$ 时， $Error2$ 的四阶近似曲线

可移动夹紧边界条件下，我们可以得到多层扁球壳受无量纲横向均布载荷P作用的四次近似解为：

$P\approx P_0+P_1+P_2+P_3$

由(40)~(43)可得P与 $y_0$ 的特征关系式为：

$\begin{array}{c}P\approx 32y_0+\frac{11K^2{y}_0}{72}+\frac{73K^4{y}_0}{4147200}-\frac{2537K^6{y}_0}{234101145600}-\frac{82K{y}_0^2}{135}-\frac{31K^3{y}_0^2}{2032128}\\ +\frac{52387K^5{y}_0^2}{329204736000}+\frac{73y_0^3}{135}-\frac{52769K^2{y}_0^3}{91445760}-\frac{505909K^4{y}_0^3}{2633637888000}\\ +\frac{1234489K{y}_0^4}{685843200}-\frac{1422358327K^3{y}_0^4}{318670184448000}-\frac{19289y_0^5}{13716864}+\frac{3253160447K^2{y}_0^5}{159335092224000}\\ -\frac{541294218343K{y}_0^6}{16964407269089280}+\frac{3574120874477y_0^7}{212055090863616000}.\end{array}$ (49)

基金项目

国家自然科学基金(No.12161064)，内蒙古自治区自然科学基金项目(No.2020LH01003)及内蒙古工业大学重点学科团队(ZD202018)。

NOTES

^*通讯作者。