Одномерное уравнение теплопроводности. Буй Ван Шань. 6 курс

Материал из Department of Theoretical and Applied Mechanics
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Задание

Решение краевой задачи для одномерного уравнения теплопроводности dU/dt-d2U/dx2=f(x,t), a<x<b, 0<t<Tk Начальное условие U(x,0)=U0(x) Конечные условия U(a,t)=M1(t); U(b,t)=M2(t); Где f(x,t), U0(x), M1(t), M2(t) - Заданные функции

  • Заданы: U(x,0)=cos(x+0,48);
         U(0,t)=6t+0,887;
         U(1,t)=0,0907
         0<x<1;0<t<1

Релизация MPI

  1. include <cstdlib>
  2. include <iostream>
  3. include <fstream>
  4. include <math.h>
  5. include <time.h>
  6. include "mpi.h"

using namespace std; double initial_condition ( double x, double time ); double boundary_condition ( double time,int flag); clock_t start,finish; int main(int argv, char** argc) { int my_id,p; //p: number of processes; my_id: id- process id MPI_Init(&argv,&argc); MPI_Comm_rank(MPI_COMM_WORLD, &my_id); /* get current process id */

   MPI_Comm_size(MPI_COMM_WORLD, &p);     /* get number of processes */

if(my_id==0) // Start recording time { start=clock(); } double x_min=0; //левый предел double x_max=1; // right limit double t_min=0; // starting time double t_max=1; // ending time double shagt=0.0000001; // time step int ShagovPoT=(t_max-t_min)/shagt; double shagx=0.0001; // space step int i_shagov=(x_max-x_min)/p/shagx;

double *T,*Tn; double *X; ofstream myFile,infoFile; char file_name[100]; sprintf(file_name, "result%i.txt", my_id); //file_name

X=new double[i_shagov+3]; for(int i=0;i<i_shagov+3;i++) { X[i]=x_min+(i-1)*shagx + my_id*i_shagov*shagx + my_id*shagx; } T=new double[i_shagov+3]; Tn=new double[i_shagov+3]; for(int i=0;i<i_shagov+3;i++) // Initial condition t=0 { T[i]=initial_condition(X[i],0); } /* myFile.open(file_name,ios::app);

for(int k=1;k<i_shagov+2;k++) { myFile<<T[k]<<"\t"; } myFile<< "\n"; myFile.close(); */ for(int j=1;j<ShagovPoT;j++) {

if(my_id==0) // The first process will only send and recive data with its right process { double right_send=T[i_shagov+1]; double right_recive; MPI_Send(&right_send,1,MPI_DOUBLE,1,0,MPI_COMM_WORLD); MPI_Recv(&right_recive,1,MPI_DOUBLE,1,0,MPI_COMM_WORLD,MPI_STATUSES_IGNORE); T[i_shagov+2]=right_recive; } else if(my_id==p-1) // The last process will only send and recive data with its left process { double left_send=T[1]; MPI_Send(&left_send,1,MPI_DOUBLE,my_id-1,0,MPI_COMM_WORLD); double left_recive; MPI_Recv(&left_recive,1,MPI_DOUBLE,my_id-1,0,MPI_COMM_WORLD,MPI_STATUSES_IGNORE); T[0]=left_recive; } else { //midle process send and recive message with left and right processes double left_send=T[1]; MPI_Send(&left_send,1,MPI_DOUBLE,my_id-1,0,MPI_COMM_WORLD); double left_recive; MPI_Recv(&left_recive,1,MPI_DOUBLE,my_id-1,0,MPI_COMM_WORLD,MPI_STATUSES_IGNORE); T[0]=left_recive;

double right_send= T[i_shagov+1]; double right_recive; MPI_Send(&right_send,1,MPI_DOUBLE,my_id+1,0,MPI_COMM_WORLD); MPI_Recv(&right_recive,1,MPI_DOUBLE,my_id+1,0,MPI_COMM_WORLD,MPI_STATUSES_IGNORE); T[i_shagov+2]=right_recive; } for(int i=1;i<i_shagov+2;i++) { //Tn[i]=T[i]+ shagt/shagx/shagx*(T[i+1]+T[i-1]-2*T[i]); Tn[i]=5; } if(my_id==0) //поправить самый левый и самый правый интервала по гранич. услов. { Tn[1]=boundary_condition(j*shagt,1); } else if(my_id==p-1) { Tn[i_shagov+1]=boundary_condition(j*shagt,2); } /* // print result to files myFile.open(file_name,ios::app);

for(int k=1;k<i_shagov+2;k++) { myFile<<Tn[k]<<"\t"; } myFile<< "\n"; myFile.close(); */ for(int i=0;i<i_shagov+3;i++) { T[i]=Tn[i]; } T[0]=T[1]; T[i_shagov+2]=T[i_shagov+1]; }

MPI_Finalize(); if(my_id==0) { finish=clock();// Stop recording time double second =finish-start; //myFile.open(file_name,ios::app); infoFile.open("infoSolution.txt"); infoFile<<"Calculating with "<<p<<" processes"<<endl; infoFile<<"It took "<< second/CLOCKS_PER_SEC<< " seconds"; //infoFile<<"" //printf ("It took me %d clicks (%f seconds).\n",second,((float)second)/CLOCKS_PER_SEC); } } double initial_condition ( double x, double time ) { double value; value=cos(x+0.48); return value; } double boundary_condition ( double t,int flag) { double value; if(flag==1) { value=6*t+0.886994923; } else if(flag==2) { value=0.090671624; } return value;

}

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