Krdytkyu

Imagine we have the problem

begin{cases}
-frac{d^2u}{dx^2} = f(x), x in [0,L] \
u(0) = 0 \
u(L) = 0
end{cases}

We know that we can approximate the second derivative using this formula:$$ frac{d^2u(x)}{dx^2} approx frac{u(x+h)-2u(x)+u(x-h)}{h^2} $$

if we define $u_{k} := u(x_{k}) $; $ x_{k} = kh $ and $ k = 0,1,2,...,N$. $h$ is known as the mesh size or step size. We get:

$$ frac{d^2u_{k}}{dx^2} approx frac{u_{k+1}-2u_{k}+u_{k-1}}{h^2} = frac{u_{k-1}-2u_{k}+u_{k+1}}{h^2} $$ for $k=1,...,N-1$

Since $u(0) = u_{0} = 0$ and $ u(L) = u(x_{N}) = u_{N} = 0 $ we get the following matrix representation of the second derivative operator

begin{equation}
frac{d^2}{dx^2} approx L_{2} = frac{1}{h^2}left(begin{matrix}
-2 & 1 & & 0\
1 & ddots & ddots & \
& ddots & ddots & 1 \
0 & & 1 & -2 end{matrix} right)
end{equation}

Then to approximate the solution of the differental equation we solve the system:

$$-L_{2}hat{u} = hat{f}$$
where $ hat{f} = [ f(x_{1}) f(x_{2}) ... f(x_{N-1}) ]^T $ and $hat{u} = [ u_{1} ... u_{N-1} ]^T$

The matrix representation of the laplacian operator using the Kronecker product is: $$ Delta =nabla^2 = frac{partial^2}{partial x^2} + frac{partial ^2}{partial y^2} = L_{2}otimes I_{n_{x}} + I_{n_{y}} otimes L_{2} $$

$textbf{Note that I have used $L_{2}$ and kronecker product to get a matrix representation} $ of the laplacian operator.

With this in mind. I want to approximate the first derivative using central difference:

$$ frac{du_{k}}{dx} approx frac{ u_{k+1}-u_{k-1} }{ 2h } = frac{ -u_{k-1} +u_{k+1} }{ 2h } $$

Since $u(0) = u_{0} = 0$ and $ u(L) = u(x_{N}) = u_{N} = 0 $ we get the following matrix representation of the first derivative

begin{equation}
frac{d}{dx} approx L_{1} = frac{1}{2h}left(begin{matrix}
0 & 1 & & 0\
-1 & ddots & ddots & \
& ddots & ddots & 1 \
0 & & -1 & 0 end{matrix} right)
end{equation}

I want to use $L_{1}$ and kronecker product to get the matrix representation of

$$ frac{partial^2 }{partial y partial x } $$
and
$$ frac{partial^2 }{partial x partial y } $$

$textbf{My questions are:} $

1. Is this possible?




2. if question 1. is affirmative, what is the matrix representation of the mixed derivatives using $L_{1}$ and kronecker product?




3. if question 1. is negative, how can we get the matrix representation of the mixed derivatives using a simple matrix and kronecker product?




4. Do you know a book or document( article or other ) that explain in detail this or something similar?

$textbf{I want to use kronecker product because it is fast and easy to implement in} $ matlab or octave.

By the way I tried to use this formula

$$ frac{partial^2u_{k,j} }{partial x partial y } = frac{ u_{k+1,j+1}+u_{k-1,j-1}-u_{k+1,j-1}-u_{k-1,j+1} }{4h^2} $$

But it was hard to see a pattern. Thank you!

Once you have the matrix $bf L_1$ to represent the divided difference of $f(x)$ given as a vertical vector, then if you transpose the whole you'll get a horizontal vector for the difference.

So if $f(x,y)$ is represented as a matrix $bf F_{, x,, y}$ , the matrix representing ${partial over {partial xpartial y}}f(x,y)$ will be $bf L_1 bf F_{, x, ,y} bf L_1^T$

Now using $mathbf{L_{1}}$ and $textbf{kronecker product}$:

$$frac{partial^2 }{partial x partial y} = Big((L_{1})_{n_{x}} otimes I_{n_{y}} Big)Big( I_{n_{x}} otimes (L_{1})_{n_{y}} Big) = { (L_{1})_{n_{x}} otimes (L_{1})_{n_{y}} } $$

I have used the mixed-product property for kronecker product.
$$ (mathbf{A} otimes mathbf{B})(mathbf{C} otimes mathbf{D}) = (mathbf{AC}) otimes (mathbf{BD})$$.

In a similar way we get:

$$frac{partial^2 }{partial y partial x} = (L_{1})_{ n_{y} } otimes (L_{1})_{n_{x}} $$