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I am not an expert in Mathematica. I want to keep off from tedious calculation

I want to solve (in symbolic sens) this system: $\quad AU^{j+1}+BU^{j}=F^{j}$

where:

$*$ ${U}^{j}$ a $(N;1)$ vector $\quad\mathbf{U}^{j}=\left[\begin{array}{c}u_{1}^{j} \\ \vdots \\ u_{N}^{j}\end{array}\right]$ and $\quad\mathbf{U}^{1}=\left[\begin{array}{c}\phi(x_{2}) \\ \phi(x_{2}) \\\phi(x_{3}) \\ \vdots \\ \phi(x_{N-1}) \\ \phi(x_{N-1})\end{array}\right] $

$*$ ${F}^{j}$ a $(N-2;1)$ vector$\quad\mathbf{F}^{j}=\left[\begin{array}{c}kf_{2}^{j} \\ \vdots \\ kf_{N-1}^{j}\end{array}\right] $

$*$ $A$ a $(N-2;N)$ matrix $A=\left(\begin{array}{rrrrrr} -\lambda &-1& -\lambda&0&0&\cdots&0\\ 0&-\lambda &-1& -\lambda&0&\cdots&0\\ 0&0&-\lambda &-1& -\lambda&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\cdots&-1&-\lambda\\ \end{array}\right)$

$*$ $B$ a $(N-2;N)$ matrix $B=\left(\begin{array}{rrrrrr} 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots&0&0\\ 0&0&0&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\\ 0&0&0&\cdots&0&1&0\\ \end{array}\right)$

Thanks a lot for your time, you can ask me to clarify anything

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  • $\begingroup$ For explicit n, U1, and some closed for expression for the Fj you might be able to use RSolve. $\endgroup$ Mar 22, 2020 at 15:43
  • $\begingroup$ Thank you for your help, I'll try it now $\endgroup$ Mar 22, 2020 at 16:11
  • $\begingroup$ would you please explain bit more? I'm trying with RSolve butstill can't get results $\endgroup$ Mar 22, 2020 at 18:32
  • $\begingroup$ It would help if you post an explicit (and small) example. This requires explicit forms of UU1 and the F vectors. $\endgroup$ Mar 22, 2020 at 19:37
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    $\begingroup$ the vector $U^{1}$ and $F^{j}$ are supposed known, and I need to determine explicitly $u_{1}^{2}, u_{2}^{2}, \cdots , u_{N}^{2}, u_{1}^{3}, \cdots \cdots u_{N-1}^{K},.u_{N}^{K},$ $\endgroup$ Mar 22, 2020 at 19:46

1 Answer 1

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My intention is to share some ideas that can help in solving the problem.

Calling

$$ U_k = \left[ \mathbb{U}_k \ | \ \mathbb{U}_k^0\right]\\ A = \left[ \mathbb{A}\ |\ \mathbb{A}_0\right]\\ B = \left[ \mathbb{B} \ | \ \mathbb{B}_0\right]\\ $$

then

$$ A U_{k+1}+B U_k = F_k\Leftrightarrow \mathbb{A}\mathbb{U}_{k+1}+\mathbb{B}\mathbb{U}_k+\mathbb{A_0}\mathbb{U^0}_{k+1}+\mathbb{B_0}\mathbb{U^0}_k = F_k $$

so we can ask for solutions where

$$ \cases{\mathbb{A}\mathbb{U}_{k+1}+\mathbb{B}\mathbb{U}_k = F_k\\ \mathbb{A_0}\mathbb{U^0}_{k+1}+\mathbb{B_0}\mathbb{U^0}_k = 0 } $$

and for $m = n+2 = 4+2=6$ we have

$$ \cases{ \mathbb{A}=\left[ \begin{array}{cccc} -\lambda & -1 & -\lambda & 0 \\ 0 & -\lambda & -1 & -\lambda \\ 0 & 0 & -\lambda & -1 \\ 0 & 0 & 0 & -\lambda \\ \end{array} \right],\ \ \mathbb{A}_0 = \left[ \begin{array}{cc} -\lambda & 0 \\ -1 & -\lambda \\ \end{array} \right]\\ \mathbb{B} = \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right], \ \mathbb{B}_0 = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right]\\ \mathbb{U} = \{u_1,\cdots,u_4\},\ \mathbb{U_0} = \{u_5,u_6\} } $$

now $\mathbb{A},\ \mathbb{B}$ commute so we can develop some equivalent recurrences: from

$$ \mathbb{A}^2\mathbb{U}_{k+1}+\mathbb{A}\mathbb{B}\mathbb{U}_k = \mathbb{A}F_k\\ \mathbb{B}\mathbb{A}\mathbb{U}_{k}+\mathbb{B}^2\mathbb{U}_{k-1} = \mathbb{B}F_{k-1} $$

we have

$$ \mathbb{A}^2\mathbb{U}_{k+1}-\mathbb{B}^2\mathbb{U}_{k-1} = \mathbb{A}F_k-\mathbb{B}F_{k-1} $$

and proceeding

$$ \mathbb{A}^4\mathbb{U}_{k+1}-\mathbb{A}^2\mathbb{B}^2\mathbb{U}_{k-1} = \mathbb{A}^3F_k-\mathbb{A}^2\mathbb{B}F_{k-1}\\ \mathbb{B}^2\mathbb{A}^2\mathbb{U}_{k-1}-\mathbb{B}^4\mathbb{U}_{k-2} = \mathbb{B}^2\mathbb{A}F_{k-1}-\mathbb{B}^3F_{k-2} $$

or

$$ \mathbb{A}^4\mathbb{U}_{k+1}-\mathbb{B}^4\mathbb{U}_{k-2} = \cdots $$

note that $\mathbb{B}^n = 0$

etc.

NOTE

In MATHEMATICA you can represent $A$ and $B$ as follows

n = 6;
A = -Table[If[i == j, lambda, If[i + 1 == j, 1, If[i + 2 == j, lambda, 0]]], {i, 1, n}, {j, 1, n + 2}];
B = Table[If[i + 1 == j, 1, 0], {i, 1, n}, {j, 1, n + 2}];
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  • $\begingroup$ Thank you so much for your insights. I don't have now comment on your approach since I'm a newbie in Mathematica, but you have shown me where I need to start to tackle this kind of problem. $\endgroup$ Mar 27, 2020 at 14:02

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