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The Diversity of Solutions Satisfied Partial Boundary Conditions for a Partial Differential Equation
DOI: 10.12677/AAM.2018.75073, PDF, HTML, XML, 下载: 1,268  浏览: 1,599  科研立项经费支持

Abstract: For (initial) boundary value problems of partial differential equations, most of current methods are based on their partial (initial) boundary conditions. Then the solution obtained in this way could satisfy all the boundary conditions? For this reason, we based on Adomian decomposition method to solve the Dirichlet boundary value problem of groundwater recharge effect model on triangular area. We find that: 1) the solution satisfies all boundary conditions sometimes, sometimes not satisfies; 2) the solution satisfied partial boundary conditions is not unique; 3) the solution obtained by the Adomian decomposition method is a particular solution that satisfies partial boundary conditions.

1. 引言

$\frac{{\partial }^{2}h}{\partial {x}^{2}}+\frac{{\partial }^{2}h}{\partial {y}^{2}}=-\frac{Rg}{T},\text{}0\le x\le 600,x\le y\le 600-x$ (1.1)

$h\left(x,x\right)={f}_{1}\left(x\right),$ (1.2)

$h\left(x,600-x\right)={f}_{2}\left(x\right),$ (1.3)

$h\left(x,0\right)={f}_{3}\left(x\right).$ (1.4)

${f}_{1}\left(x\right)=100+\frac{2x}{125}-\frac{{x}^{2}}{50000},$ (1.5)

${f}_{2}\left(x\right)=\frac{448}{5}+\frac{103x}{1500}-\frac{{x}^{2}}{12500},$ (1.6)

${f}_{3}\left(x\right)=100+\frac{17x}{3750}-\frac{{x}^{2}}{500000}$ (1.7)

${L}_{x}h\left(x,y\right)+{L}_{y}h\left(x,y\right)=-\frac{Rg}{T}$ (1.8)

${L}_{1}^{-1}={\int }_{y}^{x}{\int }_{y}^{x}\cdot \text{d}x\text{d}x-\frac{x-y}{600-2y}{\int }_{y}^{600-y}{\int }_{y}^{x}\cdot \text{d}x\text{d}x$

$\underset{n=0}{\overset{\infty }{\sum }}{h}_{n}=\left(1-\frac{x-y}{600-2y}\right){f}_{1}\left(y\right)+\frac{x-y}{600-2y}{f}_{2}\left(600-y\right)-{L}_{1}^{-1}\frac{Rg}{T}-{L}_{1}^{-1}{L}_{y}\underset{n=0}{\overset{\infty }{\sum }}{h}_{n}$ (2.1)

${h}_{0}\left(x,y\right)=\left(1-\frac{x-y}{600-2y}\right){f}_{1}\left(y\right)+\frac{x-y}{600-2y}{f}_{2}\left(600-y\right)-{L}_{x}^{-1}\frac{rg}{t},$

${h}_{n}\left(x,y\right)=-{L}_{x}^{-1}{L}_{y}{h}_{n-1}$ (2.2)

${h}_{0}\left(x,y\right)=100-\frac{{x}^{2}}{20000}+x\left(\frac{1}{30}+\frac{3y}{100000}\right)-\frac{13y}{750},{h}_{n}=0,\left(n\ge 1\right)$

${H}_{1}\left(x,y\right)=100-\frac{{x}^{2}}{20000}+x\left(\frac{1}{30}+\frac{3y}{100000}\right)-\frac{13y}{750}$ (2.3)

${L}_{2}^{-1}={\int }_{0}^{y}{\int }_{0}^{y}\cdot \text{d}y\text{d}y-\frac{y}{600-x}{\int }_{0}^{600-x}{\int }_{0}^{y}\cdot \text{d}y\text{d}y$

$\underset{n=0}{\overset{\infty }{\sum }}{h}_{n}=\left(1-\frac{y}{600-x}\right){f}_{3}\left(x\right)+\frac{y}{600-x}{f}_{2}\left(x\right)-{L}_{2}^{-1}\frac{Rg}{T}-{L}_{2}^{-1}{L}_{x}\underset{n=0}{\overset{\infty }{\sum }}{h}_{n},$ (3.1)

${h}_{0}\left(x,y\right)=\left(1-\frac{y}{600-x}\right){f}_{3}\left(x\right)+\frac{y}{600-x}{f}_{2}\left(x\right)-{L}_{y}^{-1}\frac{rg}{t},$

${h}_{n}\left(x,y\right)=-{L}_{y}^{-1}{L}_{x}{h}_{n-1}$ (3.2)

${h}_{0}\left(x,y\right)=\frac{-3{x}^{2}+x\left(6800+42y\right)+25\left(6000000+760y-3{y}^{2}\right)}{1500000}$ (3.3)

${h}_{1}\left(x,y\right)=\frac{y\left(-600+x+y\right)}{500000}$ (3.4)

${h}_{n}\left(x,y\right)=0,\left(n\ge 2\right)$ (3.5)

${H}_{2}\left(x,y\right)=\frac{-3{x}^{2}+5x\left(1360+9y\right)+8\left(18750000+2150y-9{y}^{2}\right)}{1500000}$ (3.6)

4. 总结

NOTES

*通讯作者。

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