# [C++] Crank-Nicolson advection of gaussian pulse.

1. May 29, 2013

### randombill

I tried to used the Crank-Nicolson method to solve the advection equation for a Gaussian pulse using the code on this page and I'm lost at trying to setup the initial values for the Gaussian which is passed as an array into the advection matrix.

The Gaussian pulse is defined

initial condition: u(x,0) = exp[-100(x-0.5)^2] as on this page written for the Lax method.

The initial velocity is v = 1, timestep = 9.95x10^(-3), N = 200 and C = 2.

In the following code the matrix u is initialized to 202 elements and I try to initialize the matrix u
with the gaussian pulse for each index between 1 and 200 but something is very wrong. For one I'm not sure how to put the timestep into the gaussian equation. I suppose if x/t = v then tv = x and (9.95x10^(-3))*(velocity = 1) = distance then each timestep solved with advect1D would move the Gaussian to the right a small amount.

The thing I'm not sure about is this part in my code

Code (Text):

for (int index = 1; index < 201; index++)
{
u(index) = exp(-100*(distance - 0.5)*(distance-0.5));
}
I initialize every element of u(index) with the same gaussian? Well hopefully someone can help me heres the complete code.

Code (Text):

// Function to evolve advection equation in 1-d:

//  du / dt + v du / dx = 0  for  xl <= x <= xh

//  u = 0  at x=xl and x=xh

// Array u assumed to be of extent N+2.

// Now, ith element of array corresponds to

//  x_i = xl + i * dx    i=0,N+1

// Here, dx = (xh - xl) / (N+1) is grid spacing.

// Function evolves u by single time-step.

// C = v dt / dx, where dt is time-step.

// Uses Crank-Nicholson scheme.

#include <blitz/array.h>
#include <iostream>
#include <cmath>

using namespace blitz;

void Tridiagonal (Array<double,1> a, Array<double,1> b, Array<double,1> c, Array<double,1> w, Array<double,1>& u);
void Advect1D (Array<double,1>& u, double C);

int main()
{
Array<double,1> u(202);
u = 0;

double distance = 9.95e-3;

for (int index = 1; index < 201; index++)
{
u(index) = exp(-100*(distance - 0.5)*(distance-0.5));
}

double C = 2;

Advect1D ( u,  C);
std::cout<<u<<std::endl;

return 0;
}
void Advect1D (Array<double,1>& u, double C)
{
// Find N. Declare local arrays.
int N = u.extent(0) - 2;
std::cout<<"N = "<<N<<std::endl;

Array<double,1> a(N+2), b(N+2), c(N+2), w(N+2);

a = 0;
b = 0;
c = 0;
w = 0;

// Initialize tridiagonal matrix
for (int i = 2; i <= N;   i++)
a(i) = - 0.25 * C;

for (int i = 1; i <= N; i++)
b(i) = 1.;

for (int i = 1; i <= N-1; i++)
c(i) = + 0.25 * C;

// Initialize right-hand side vector
for (int i = 1; i <= N; i++)
w(i) = u(i) - 0.25 * C * (u(i+1) - u(i-1));

std::cout<<a<<std::endl;
std::cout<<b<<std::endl;
std::cout<<c<<std::endl;
std::cout<<w<<std::endl;
// Invert tridiagonal matrix equation
Tridiagonal (a, b, c, w, u);

// Calculate i=0 and i=N+1 values
u(0) = 0.;
u(N+1) = 0.;
}
void Tridiagonal (Array<double,1> a, Array<double,1> b, Array<double,1> c,  Array<double,1> w, Array<double,1>& u)
{
// Find N. Declare local arrays.
int N = a.extent(0) - 2;
Array<double,1> x(N), y(N);

// Scan up diagonal from i = N to 1
x(N-1) = - a(N) / b(N);
y(N-1) = w(N) / b(N);

for (int i = N-2; i > 0; i--)
{
x(i) = - a(i+1) / (b(i+1) + c(i+1) * x(i+1));
y(i) = (w(i+1) - c(i+1) * y(i+1)) / (b(i+1) + c(i+1) * x(i+1));
}

x(0) = 0.;
y(0) = (w(1) - c(1) * y(1)) / (b(1) + c(1) * x(1));

// Scan down diagonal from i = 0 to N-1
u(1) = y(0);
for (int i = 1; i < N; i++)
u(i+1) = x(i) * u(i) + y(i);
}

2. May 31, 2013

### randombill

I made some changes to the code and I think I solved it but I'm not sure the solution makes sense. Basically I create a Gaussian pulse function with

width (standard deviation) = 0.247
height = 1;
position = .78

the actual width of the Gaussian is 1.55 from tail to tail and I divide that number by 200 to get the space discretization which is used to find the value of the Gaussian at specific intervals between 0 and 1.55. I output the original Gaussian and the advection equation into a dat file and use gnuplot. Now I'm not sure if the solution I found is correct. I think it makes sense, any ideas?

Heres the source and the gnuplot .plt file below it plus a photo of the beast.

Code (Text):

// Function to evolve advection equation in 1-d:

//  du / dt + v du / dx = 0  for  xl <= x <= xh

//  u = 0  at x=xl and x=xh

// Array u assumed to be of extent N+2.

// Now, ith element of array corresponds to

//  x_i = xl + i * dx    i=0,N+1

// Here, dx = (xh - xl) / (N+1) is grid spacing.

// Function evolves u by single time-step.

// C = v dt / dx, where dt is time-step.

// Uses Crank-Nicholson scheme.

#include <blitz/array.h>
#include <iostream>
#include <cmath>
#include <fstream>

using namespace blitz;

void Tridiagonal (Array<double,1> a, Array<double,1> b, Array<double,1> c, Array<double,1> w, Array<double,1>& u);
void Advect1D (Array<double,1>& u, double C);

int main()
{
Array<double,1> u(202);
std::ofstream gaussian;

gaussian.open("gaussian.dat");

u = 0;
//double distance = 9.95e-3;

double element = 1.55/200;//1.55 is the width of the gaussian
std::cout<<"element: "<<element<<std::endl;

double dx = 0;

double term = 0;
double numerator = 0;
double denominator = 0;

for (int index = 1; index < 201; index++)
{
numerator = -1*pow(dx - .78,2);
denominator = 2*pow(0.247,2);
term = numerator/denominator;
u(index) = exp(term);

gaussian<<u(index)<<"   "<<dx<<"\n";

dx += element;
}

double C = 2;

Advect1D ( u,  C);
std::cout<<u<<std::endl;
dx = 0;

std::ofstream gaussian_pde;
gaussian_pde.open("gaussian_pde.dat");

for (int index = 1; index < 201; index++)
{
gaussian_pde<<u(index)<<"   "<<dx<<"\n";
dx += element;
}

gaussian.close();
gaussian_pde.close();

return 0;
}
void Advect1D (Array<double,1>& u, double C)
{
// Find N. Declare local arrays.
int N = u.extent(0) - 2;
std::cout<<"N = "<<N<<std::endl;

Array<double,1> a(N+2), b(N+2), c(N+2), w(N+2);

a = 0;
b = 0;
c = 0;
w = 0;

// Initialize tridiagonal matrix
for (int i = 2; i <= N;   i++)
a(i) = - 0.25 * C;

for (int i = 1; i <= N; i++)
b(i) = 1.;

for (int i = 1; i <= N-1; i++)
c(i) = + 0.25 * C;

// Initialize right-hand side vector
for (int i = 1; i <= N; i++)
w(i) = u(i) - 0.25 * C * (u(i+1) - u(i-1));

std::cout<<a<<std::endl;
std::cout<<b<<std::endl;
std::cout<<c<<std::endl;
std::cout<<w<<std::endl;
// Invert tridiagonal matrix equation
Tridiagonal (a, b, c, w, u);

// Calculate i=0 and i=N+1 values
u(0) = 0.;
u(N+1) = 0.;
}
void Tridiagonal (Array<double,1> a, Array<double,1> b, Array<double,1> c,  Array<double,1> w, Array<double,1>& u)
{
// Find N. Declare local arrays.
int N = a.extent(0) - 2;
Array<double,1> x(N), y(N);

// Scan up diagonal from i = N to 1
x(N-1) = - a(N) / b(N);
y(N-1) = w(N) / b(N);

for (int i = N-2; i > 0; i--)
{
x(i) = - a(i+1) / (b(i+1) + c(i+1) * x(i+1));
y(i) = (w(i+1) - c(i+1) * y(i+1)) / (b(i+1) + c(i+1) * x(i+1));
}

x(0) = 0.;
y(0) = (w(1) - c(1) * y(1)) / (b(1) + c(1) * x(1));

// Scan down diagonal from i = 0 to N-1
u(1) = y(0);
for (int i = 1; i < N; i++)
u(i+1) = x(i) * u(i) + y(i);
}

Code (Text):
set output 'gaussian.png'
set key bmargin left horizontal Right noreverse enhanced autotitles box linetype -1 linewidth 1.000
plot 'gaussian.dat' using 2:1 with lines ,'gaussian_pde.dat' using 2:1 with lines

File size:
15.4 KB
Views:
497
3. Jun 8, 2013

### Solkar

What I find irritating at a first glance is that the arrays for the a,b,c diags are of equal size, thus two of them are padded; I'd doubt this improves efficiency or simplifies usage significantly.

---

Furthermore:

Let T be your Matrix with a,b,c diags.

From http://farside.ph.utexas.edu/teaching/329/lectures/node92.html
Code (Text):
// Invert tridiagonal matrix equation
Tridiagonal (a, b, c, w, u);
i take it one should solve

T * u = w

for u by inverting T in the Tridiagonal function.

boost::ublas has an adaptor for banded matrices
http://www.boost.org/doc/libs/1_53_0/libs/numeric/ublas/doc/banded.htm#banded_matrix

and LAPACK of course has this as well
http://www.netlib.org/lapack/lug/node125.html

Have you checked the correctness of your own Tridiagonal function with either of them?

Regards, Solkar