%%?一維熱傳導方程差分格式求解實例
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%?求解的邊值問題為
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%?????u（x0）=sin（pi*x）??????0%%
%?????u（0t）=u（1t）=0?????0%%
%?所求問題的精確解為
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%????????????????u（xt）=sin（pi*x）*exp（-t）????其中a=1/（pi^2）f（x）=0
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%????網比：tao<0.5*h^2*pi^2=?0.5*h^2*9.8696???
%取?h=0.1??tao=0.005????????????2.9419?????0.0479;???????h=0.1???tao=0.01??????0.9578?????0.0180;???h=0.1??tao=0.02?????????0.6440?????0.0144
%取?h=0.2??tao=0.005????????????0.0116?????0.0044;???????h=0.2???tao=0.01??????0.0116?????0.0044;???h=0.2??tao=0.02?????????0.0118?????0.0045
%取?h=0.5??tao=0.005????????????0.0767?????0.0177?????????h=0.5??tao=0.01?????0.0767??????0.0176????h=0.5??tao=0.02?????????0.0767?????0.0176???
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%?此處所用的是直接差分化格式
clc
clear?
syms?temp;????

d=15;
h=0.2;
tao=0.01;
m=5;
n=100;
a=1/（pi^2）;
r=（a*tao）/

?屬性????????????大小?????日期????時間???名稱
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?????文件???????1577??2014-05-10?22:44??一維熱傳導\Richardson\Richardson.m

?????文件????????737??2014-05-10?22:42??一維熱傳導\Richardson\wucha.m

?????文件???????1493??2014-05-11?10:38??一維熱傳導\六點差分\liudian2.m

?????文件????????997??2014-05-11?10:34??一維熱傳導\六點差分\wucha.m

?????文件???????1121??2014-05-11?10:05??一維熱傳導\向前差分\wucha.m

?????文件???????1437??2014-05-11?10:04??一維熱傳導\向前差分\xiangqian2.m

?????文件????????818??2014-05-11?10:21??一維熱傳導\向后差分\wucha.m

?????文件???????1395??2014-05-11?10:15??一維熱傳導\向后差分\xianghou2.m

?????目錄??????????0??2014-05-10?11:41??一維熱傳導\Richardson

?????目錄??????????0??2014-05-10?10:49??一維熱傳導\六點差分

?????目錄??????????0??2014-05-11?09:59??一維熱傳導\向前差分

?????目錄??????????0??2014-05-11?10:20??一維熱傳導\向后差分

?????目錄??????????0??2014-05-10?11:20??一維熱傳導

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?????????????????9575????????????????????13