微分方程上機作業(yè)new

上機作業(yè)一姓名：學(xué)號：班級：時間：2012年3月26日1：作業(yè)題目0≤x≤1,t>0u（x,0）=sinx0≤x≤1u(0,t)=u(1,t)=0t>0已知：h=0.1;r=0.5;由r=,所以解得2：離散方程顯示格式：k=1,2,3,4,5,6,7,8,9j=0=(1-2r)+r(+)=(1-2r)+r(+)=(1-2r)+r(+)=(1-2r)+r(+)=(1-2r)+r(+)=(1-2r)+r(+)=(1-2r)+r(+)=(1-2r)+r(+)=(1-2r)+r(+)依次將j取到20，即可得離散方程！3：所得結(jié)果0 0.309017 0.587785 0.809017 0.951057 1 0.951057 0.809017 0.587785 0.309017 0 0 0.293893 0.559017 0.769421 0.904508 0.951057 0.904508 0.769421 0.559017 0.293893 0 0 0.279508 0.531657 0.731763 0.860239 0.904508 0.860239 0.731763 0.531657 0.279508 0 0 0.265828 0.505636 0.695948 0.818136 0.860239 0.818136 0.695948 0.505636 0.265828 0 0 0.252818 0.480888 0.661886 0.778093 0.818136 0.778093 0.661886 0.480888 0.252818 0 0 0.240444 0.457352 0.629491 0.740011 0.778093 0.740011 0.629491 0.457352 0.240444 0 0 0.228676 0.434967 0.598681 0.703792 0.740011 0.703792 0.598681 0.434967 0.228676 0 0 0.217484 0.413679 0.56938 0.669346 0.703792 0.669346 0.56938 0.413679 0.217484 0 0 0.206839 0.393432 0.541512 0.636586 0.669346 0.636586 0.541512 0.393432 0.206839 0 0 0.196716 0.374176 0.515009 0.605429 0.636586 0.605429 0.515009 0.374176 0.196716 0 0 0.187088 0.355862 0.489802 0.575797 0.605429 0.575797 0.489802 0.355862 0.187088 0 0 0.177931 0.338445 0.46583 0.547616 0.575797 0.547616 0.46583 0.338445 0.177931 0 0 0.169223 0.32188 0.44303 0.520813 0.547616 0.520813 0.44303 0.32188 0.169223 0 0 0.16094 0.306126 0.421347 0.495323 0.520813 0.495323 0.421347 0.306126 0.16094 0 0 0.153063 0.291144 0.400725 0.47108 0.495323 0.47108 0.400725 0.291144 0.153063 0 0 0.145572 0.276894 0.381112 0.448024 0.47108 0.448024 0.381112 0.276894 0.145572 0 0 0.138447 0.263342 0.362459 0.426096 0.448024 0.426096 0.362459 0.263342 0.138447 0 0 0.131671 0.250453 0.344719 0.405241 0.426096 0.405241 0.344719 0.250453 0.131671 0 0 0.125226 0.238195 0.327847 0.385408 0.405241 0.385408 0.327847 0.238195 0.125226 0 0 0.119097 0.226537 0.311801 0.366544 0.385408 0.366544 0.311801 0.226537 0.119097 0 0 0.113268 0.215449 0.296541 0.348604 0.366544 0.348604 0.296541 0.215449 0.113268 0 圖形分析：從正面看圖形近似為一個立體三角函數(shù)圖像；從側(cè)面看10層以前的結(jié)果變化比較明顯，10層以后趨于平緩，證明結(jié)果已經(jīng)接近準(zhǔn)確值！4：程序c.cpp程序#include<iostream>#include<fstream>#include<string>#include<cmath>#include"guoyiling.h"usingnamespacestd;doubleinitializing(doublex){returnsin(3.14159265358979323846*x);}//初始化條件doubleleftboundary(doublet){return0;}//左邊界條件doublerightboundary(doublet){return0;}//右邊界條件intmain(intargc,char**argv){ double**m=0; unsignedintrow=21;//=t方向的間隔點個數(shù) unsignedintcol=11;//=x方向的間隔點個數(shù) doubletao=0.005;//=時間步長 doublex1=0; doublex2=1; doubler=0.5;//=? alloc(m,row,col); inti(m,initializing,col,x1,x2); boun1(m,leftboundary,row,tao); boun2(m,rightboundary,row,col,tao); compute(m,row,col,r); show(m,row,col); return0;}//====================================================guoyilin.h程序#ifndefGUOYILIN#defineGUOYILIN#include<iostream>usingnamespacestd;//函數(shù)的申明voidalloc(double**&m,unsignedintrow,unsignedintcol);//分配row*col的矩陣給m，采用引用格式voidshow(double**&m,unsignedintrow,unsignedintcol);//輸出矩陣m的內(nèi)容voidinti(double**&m,double(*f)(double),unsignedintcol,doublex1,doublex2);//初始化mvoidboun1(double**&m,double(*f)(double),unsignedintrow,doubletao);//邊界條件x1(左邊界)voidboun2(double**&m,double(*f)(double),unsignedintrow,unsignedintcol,doubletao);//邊界條件x2(右邊界)voidcompute(double**&m,unsignedintrow,unsignedintcol,doubler);//計算u的值//==========================================//函數(shù)的定義voidalloc(double**&m,unsignedintrow,unsignedintcol){ m=newdouble*[row]; for(unsignedinti=0;i<row;++i) m[i]=newdouble[col];}voidshow(double**&m,unsignedintrow,unsignedintcol){ unsignedinti=0; unsignedintj=0; for(i=0;i<row;++i){ for(j=0;j<col;++j)cout<<m[i][j]<<"\t"; cout<<endl; }}voidinti(double**&m,double(*f)(double),unsignedintcol,doublex1,doublex2){ if(x1>x2){doubletemp=x1;x1=x2;x2=temp;} for(unsignedinti=0;i<col;++i) m[0][i]=f(x1+i*(x2-x1)/(col-1));}voidboun1(double**&m,double(*f)(double),unsignedintrow,doubletao){ for(unsignedinti=0;i<row;++i) m[i][0]=f(i*tao);}voidboun2(double**&m,double(*f)(double),unsignedintrow,unsignedintcol,doubletao){ for(unsignedinti=0;i<row;++i) m[i][col-1]=f(i*tao);}voidcompute(double**&m,unsignedintrow,unsignedin