The discrete differential operator $L_h$ is defined as: $(L_h v)(x_j)=-\frac{v(x_{j+1})-2v(x_j)+v(x_{j-1})}{\Delta x^2}$ (Equation 1). The contineous problem had solution $v(x_j)=sin(\beta x_j)$. It is therefore tested if this is also a solution to the discrete case. The expression for $v(x_j)$ is then inserted into the equation for $(L_h v)(x_j)$ as follows below: $L_h u=\frac{-sin(\beta x_j +\beta \Delta x)+2sin(\beta x_j)-sin(\beta x_j -\Delta x)}{\Delta x^2}v(x)$ (Equation 2). By using the two trigonometric identities below, this can expression can be changed to Equation 3:
$1-cos(y)=2sin^2(\frac{y}{2})$
$sin(x+y)+sin(x-y)=2cos(y)sin(x)$
$L_h u= \left.\frac{-2cos(\beta \Delta x)sin(\beta x_j)+2sin(\beta x_j)}{\Delta x^2}v(x)\right\vert \cdot \frac{1}{sin(\beta x_j)}$
$L_h u=\frac{2-2cos(\beta \Delta x)}{\Delta x^2}v(x)=\frac{2}{\Delta x^2}[1-cos(\beta \Delta x)]v(x)$ (Equation 3)
My book goes from equation 1 to 3, without showing the steps between, but I believe the steps I have made between is correct when equation 3 is correct. My problem is that I do not understand why equation 2 and 3 has to contain a $v(x)$ multiplied on the right side of the equal sign. In my mind, equation 2 and 3 are equal to $(Equation1*v(x)) $. Does anyone understand why equation 2 and three has to contain $v(x)$ multiplied with the fraction in front of it? David
I don't think the $v(x)$ in equation 2 should have been there. You only want to see that (actually, $v(x_j)$) at the end. In other words you should get
$$\frac{-\sin(\beta x_j + \beta \Delta x) + 2 \sin(\beta x_j) - \sin(\beta x_j - \beta \Delta x)}{\Delta x^2} = C \sin(\beta x_j)$$
for some $C$ independent of $j$, for $j=2,\dots,N-1$.