Consider the following boundary value problem:

$($p$($x$)$u$($x$))+$ q$($x$)$u$($x$)=$ f$($x$),$ a $<$ x $<$ b$,$ u$($a$)=\backslash$alpha $,$ u$($b$)=\backslash$beta $,$

where p in C$4([$a$,$ b$])$ and p$($x$)>=$ c $>0,$ q$($x$)$ in C$([$a$,$ b$]),$ q$($x$)>=0,$ and f$($x$)$ in C$([$a$,$ b$]).$ Divide the interval $[$a$,$ b$]$ into N $+1$ equal cells with cell size h $=($b $$ a$)/($N $+1).$ Define the grid points xi $=$ a$+$ih$,$ i $=0,...,$N $+1,$ with x$0=$ a$,$ xN$+1=$ b$.$ Define also the half grid points as xi$+1/2=$ xi $+$ h$/2.$ A natural finite difference scheme to approximate this problem is given as follows:

$1$ p$($xi$+1/2)$Ui$+1($p$($xi$+1/2)+$ p$($xi$1/2))$Ui $+$ p$($xi$1/2)$Ui$1+$ q$($xi$)$Ui $=$ f$($xi$),$ h$2$

fori$=1,...,$N$,$withU$0=\backslash$alpha andUN$+1=\backslash$beta $.$

$($a$)$ If we expand the left hand side as

$$p$($x$)$u$($x$)$ p$($x$)$u$($x$)+$ q$($x$)$u$($x$)=$ f$($x$),$ and then approximate by the finite difference scheme

$$p$($xi$)$Ui$+12$Ui $+$ Ui$1$ p$($xi$)$Ui$+1$ Ui$1+$ q$($xi$)$Ui $=$ f$($xi$).$ h$22$h

Show that this yields a tridiagonal but not symmetric matrix.

$($b$)$ For the equation

$$u$($x$)+$ a$($x$)$u$($x$)+$ b$($x$)$u$($x$)=$ f$($x$),$ a $<$ x $<$ b$,$ u$($a$)=\backslash$alpha $,$ u$($b$)=\backslash$beta $,$

where a$,$ b$,$ and f all belong to C$[$a$,$ b$],$ and b$($x$)>0.$ Formulate a similar finite difference scheme as in part $($a$)$ of this problem, and find a sufficient condition for the resulting linear system to have a unique solution.