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1 changed files with 29 additions and 13 deletions
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@ -7,22 +7,35 @@ using namespace std;
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int main()
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{
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int l = 1925; // Edge length
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int dx = 40; // Spatial step
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int dt = 100000; // Euler's step
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int kmax = l / dx; // Number of spatial steps
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int l = 1925; // Edge length
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int dx = 5; // Spatial step
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int kmax = l / dx; // Number of spatial steps
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// Parameters
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double alpha = 0.093;
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/**
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* Coefficients alpha and gamma are different for different
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* edges. It depends on the edge's lenght, cross-sectional area etc.
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*/
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double alpha = 0.744;
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double beta = 0.004;
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double gamma = 728.6;
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double gamma = 5829;
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/**
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* dt - the number of approximation in Euler method
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* h - discretization step in Euler method
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*
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* Those two values work well with all existing steps and give us
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* result that satisfies the initial condition pretty close.
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*
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* Changing of any of them might result in NaN error.
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*/
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int dt = 100000;
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double h = 0.007;
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// Arrays to store data
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// Stores the approximated values for each step
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double Q[kmax][dt];
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double P[kmax][dt];
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// Loop over time steps
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// Prepare initial condition
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for (int i = 0; i < dt; i++)
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{
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// Loop over spatial steps
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@ -41,7 +54,11 @@ int main()
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}
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}
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// Main calculation loop
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/**
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* In fact, we have two 'steps'. The first step represents the approximation
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* criteria in Euler's method (dt) and the second one represents the actual
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* spatial step along the edge (dx).
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*/
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for (int i = 2; i < dt; i++)
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{
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// Loop over spatial steps
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@ -55,7 +72,7 @@ int main()
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}
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}
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// Print out the first 10 of Q for each spatial step
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// Print out the first 10 values of Q for each spatial step
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std::cout << "FIRST 10 VALUES\n";
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for(int i = 0; i < kmax; i++)
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{
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@ -68,7 +85,7 @@ int main()
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std::cout << std::endl;
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}
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// Print out the first 10 of Q for each spatial step
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// Print out the last 10 values of Q for each spatial step
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std::cout << "LAST 10 VALUES:\n";
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for(int i = 1; i < kmax; i++)
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{
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@ -81,6 +98,5 @@ int main()
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std::cout << std::endl;
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}
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return 0;
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}
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