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Lecture 24a, Computing partial derivatives


Extending derivative computation to partial derivatives uses
the previous work on computing derivatives. Here we consider
the Cartesian coordinate system in 2, 3, and 4 dimensions,
computing second, third and fourth partial derivatives

The algorithms and some code shown below will be used when the
function is unknown, when we are solving partial differential
equations.

Consider a function  f(x,y) that you can compute but do not know
a symbolic representation. To find the derivatives at a point
(x,y) we will use our previously discussed "nuderiv" nonuniform 
derivative function. 

To compute the numerical derivatives of f(x,y) at a set of points
we choose the number of points in the x and y directions, nx, ny.
Thus we have a grid of cells typically programmed as a
two dimensional array.



The partial derivative with respect to x at any (x1,y1) point
just uses points (x0,y1) (x1,y1) (x2,y1) ... (xn,y1)) 

The partial derivative with respect to y at any (x1,y1) point
just uses points (x1,y0) (x1,y1) (x1,y2) ... (x1,yn)) 

Similarly for second, third, fourth derivatives.

The partial derivative with respect to x and y is more complicated.
We first take the derivative with respect to x, same as calculus.
But, numerically, we need the value for every y.
Then, as in calculus, we take the derivative with respect to y
of the previously computed derivative with respect to x.
(Typically we save the x derivatives as a matrix because we
 typically need all the partial derivatives with respect to y.)

A code fragment could look like:

static int nx = 11;          /* includes boundary */
static int ny = 10;          /* includes boundary */
static double xg[11] = {-1.0, -0.9, -0.8, -0.6, -0.3, 0.0,
			 0.35, 0.65, 0.85, 1.0, 1.2};
static double yg[10] = {-1.1, -1.0, -0.9, -0.6, -0.2,
                         0.3,  0.8,  1.0,  1.2,  1.3};
                     /* nx-2 by ny-2  internal, non boundary cells */
static double cx[11];
static double cy[10];
static double cxy[11][10]; // d^2 u(x,y)/dx dy  or  Uxy

for(i=1; i<nx-1; i++)
{
  for(j=1; j<ny-1; j++)  // computing inside boundary
  {
    /* d^2 u(x,y)/dxdy */
    nuderiv(1,nx,i,xg,cx);
    nuderiv(1,ny,j,yg,cy);
    for(kx=0; k<nx; kx++)
    {
      for(ky=0; k<ny; ky++)
      {
        cxy[kx][ky] = cx[kx] * cy[ky]; // numerical derivative at each point
      }
    }
  }
}   // have cxy inside the boundary

∂2f(x,y)/∂x ∂y  at x=xg[i] y=yg[i] is
computed using coefficients
cxy[kx][ky] = ∂{∂f(x,y)/∂x}/∂y * cx[kx] * cx[ky]


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