| #### D.11.4.1 findifs_example Procedure from library `findifs.lib` (see findifs_lib). Usage: findifs_example(); Return: nothing (demo) Purpose: demonstration of our approach and this library Example:
    LIB "findifs.lib";
    findifs_example();
    ==> * Equation: u_tt - A^2 u_xx -B^2 u_yy = 0; A,B are constants
    ==> * we employ three central differences
    ==> * the vector we act on is (u_xx, u_yy, u_tt, u)^T
    ==> * Set up the ring:
    ==> ring r = (0,A,B,dt,dx,dy),(Tx,Ty,Tt),(c,dp);
    ==> * Set up the matrix with equation and approximations:
    ==> matrix M[4][4] =
    ==>     // direct equation:
    ==>     -A^2, -B^2, 1, 0,
    ==>     // central difference u_tt
    ==>     0, 0,  -dt^2*Tt, (Tt-1)^2,
    ==>     // central difference u_xx
    ==>     -dx^2*Tx, 0, 0, (Tx-1)^2,
    ==>     // central difference u_yy
    ==>     0, -dy^2*Ty, 0, (Ty-1)^2;
    ==> * Print the differential form of equations:
    ==> (-A^2)*Uxx+(-B^2)*Uyy+Utt,
    ==> (-dt^2*Tt)*Utt+(Tt^2-2*Tt+1)*U,
    ==> (-dx^2*Tx)*Uxx+(Tx^2-2*Tx+1)*U,
    ==> (-dy^2*Ty)*Uyy+(Ty^2-2*Ty+1)*U
    ==> * Perform the elimination of module components:
    ==>  module R = transpose(M);
    ==>  module S = std(R);
    ==>  * See the result of Groebner bases: generators are columns
    ==>  print(S);
    ==> 0,     0,           0,     0,           0,         (A^2),
    ==> 0,     0,           0,     (dy^2)*Ty,   S[2,5],    (B^2),
    ==> 0,     (dt^2)*Tt,   S[3,3],0,           (-dx^2)*Tx,-1,
    ==> S[4,1],-Tt^2+2*Tt-1,S[4,3],-Ty^2+2*Ty-1,S[4,5],    0
    ==>  * So, only the first column has its nonzero element in the last componen\
       t
    ==>  * Hence, this polynomial is the scheme
    ==>  poly   p = S[4,1];
    ==>  print(p);
    ==> (A^2*dt^2*dy^2)*Tx^2*Ty*Tt+(B^2*dt^2*dx^2)*Tx*Ty^2*Tt+(-dx^2*dy^2)*Tx*Ty*\
       Tt^2+(-2*A^2*dt^2*dy^2-2*B^2*dt^2*dx^2+2*dx^2*dy^2)*Tx*Ty*Tt+(-dx^2*dy^2)\
       *Tx*Ty+(B^2*dt^2*dx^2)*Tx*Tt+(A^2*dt^2*dy^2)*Ty*Tt
    ==> * Create the nodal of the scheme in TeX format:
    ==>  ideal I = decoef(p,dt);
    ==>  difpoly2tex(I,L);
    ==> \frac{1}{\tri t^{2}}\cdot (u^{n+2}_{j+1,k+1}+(-2) u^{n+1}_{j+1,k+1}+u^{n}\
       _{j+1,k+1})+ \frac{-\lambda^{2}}{\tri x^{2}}\cdot (u^{n+1}_{j+2,k+1}+\fra\
       c{B^{2}\tri x^{2}}{\lambda^{2}\tri y^{2}} u^{n+1}_{j+1,k+2}+\frac{-(2\lam\
       bda^{2}\tri y^{2}+2B^{2}\tri x^{2}}{\lambda^{2}\tri y^{2}} u^{n+1}_{j+1,k\
       +1}+\frac{B^{2}\tri x^{2}}{\lambda^{2}\tri y^{2}} u^{n+1}_{j+1}+u^{n+1}_{\
       j,k+1})
    ==> * Preparations for the semi-factorized form:
    ==>  poly pi1 = subst(I[2],B,0);
    ==>  poly pi2 = I[2] - pi1;
    ==> * Show the semi-factorized form of the scheme: 1st summand
    ==>  factorize(I[1]);
    ==> [1]:
    ==>    _[1]=(-dx^2*dy^2)
    ==>    _[2]=Tx
    ==>    _[3]=Ty
    ==>    _[4]=Tt-1
    ==> [2]:
    ==>    1,1,1,2
    ==> * Show the semi-factorized form of the scheme: 2nd summand
    ==>  factorize(pi1);
    ==> [1]:
    ==>    _[1]=(A^2*dt^2*dy^2)
    ==>    _[2]=Ty
    ==>    _[3]=Tt
    ==>    _[4]=Tx-1
    ==> [2]:
    ==>    1,1,1,2
    ==> * Show the semi-factorized form of the scheme: 3rd summand
    ==>  factorize(pi1);
    ==> [1]:
    ==>    _[1]=(B^2*dt^2*dx^2)
    ==>    _[2]=Tx
    ==>    _[3]=Tt
    ==>    _[4]=Ty-1
    ==> [2]:
    ==>    1,1,1,2

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