problemset3

Lrho dxmxwi0 rwimxdxrrho wimxdx0 forj0jaj

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Unformatted text preview: ot; " << float(as[i])∗blocks/2/mx/(iter−100) << endl; } } C Simple C++ code for simulating model B #include <string> #include <iostream> #include <cstdio> #include <cmath> using namespace std; const const const const const const int iter=400; float dx=0.2; float rhomax=100; float lrho=25; float rrho=75; float mx=4; //Total number of iterations //Width for calculating local density //Value of rho max //Value of rho for x<−4 //Value of rho for x>4 //Half−width of simulation region 14 M. Z. Bazant – 18.366 Random Walks and Diffusion – Problem Set 3 Solutions 15 const float udt=0.01; //u max ∗ dt; const float ddt=0.0025; //D ∗ dt; const int blocks=40; //Number of blocks to save out const const const const const const const float pi=3.1415926535897932384626433832795; int lwalk=int(mx ∗ lrho); //Total initial walkers on LHS int twalk=int(mx ∗ (lrho+rrho)); //Total initial walkers float lspeed=udt∗(1−lrho/rhomax); float lrate=lspeed ∗ lrho; //L. walker intro rate int mem=twalk∗2; //Total memory float isp=float(blocks)/mx∗0.5; //Inverse block spacing inline float rnd() { return (float(rand())+0.5)/RAND MAX; } inline int poisson(float l) { if (l>20) { int r=int(sqrt(−2∗log(rnd())∗l) ∗ cos(2∗pi ∗ rnd())+0.5+l); return r>0?r:0; } else { double a=rnd()∗exp(l)−1,b=1;int r=0; while(a>0) { a−=b ∗ =l/++r; } return r; } } int main() { int a=0,i,j=0,t;float w[mem],v[mem],as[blocks],r,s,x; while(j++<lwalk) w[a++]=−mx ∗ rnd(); while(j++<twalk) w[a++]=mx ∗ rnd(); for(i=0;i<blocks;i++) as[i]=0; for(t=0;t<iter;t++) { for(i=0;i<a;i++) { j=int((w[i]+mx) ∗ isp); if (j>=0&&j<blocks&&t>100) as[j]++; s=r=w[i]<dx−mx?lrho ∗ (dx−mx−w[i]):0; r+=w[i]>mx−dx?rrho ∗ (w[i]−mx+dx):0; for(j=0;j<a;j++) { x=w[j]−w[i]; if (x>−dx) { if (x<dx) { r++;if (x<0) s++; } } } s=(r−2∗s)/(dx ∗ dx); r/=2∗dx; v[i]=udt∗(1−r/rhomax)−ddt ∗ s/r; M. Z. Bazant – 18.366 Random Walks and Diffusion – Problem Set 3 Solutions 16 }...
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