# Symbolic solution of an iterative system

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

• For explicit n, U1, and some closed for expression for the Fj you might be able to use RSolve. – Daniel Lichtblau Mar 22 at 15:43
• Thank you for your help, I'll try it now – Almendrof66 Mar 22 at 16:11
• would you please explain bit more? I'm trying with RSolve butstill can't get results – Almendrof66 Mar 22 at 18:32
• It would help if you post an explicit (and small) example. This requires explicit forms of UU1 and the F vectors. – Daniel Lichtblau Mar 22 at 19:37
• the vector $U^{1}$ and $F^{j}$ are supposed known, and I need to determine explicitly $u_{1}^{2}, u_{2}^{2}, \cdots , u_{N}^{2}, u_{1}^{3}, \cdots \cdots u_{N-1}^{K},.u_{N}^{K},$ – Almendrof66 Mar 22 at 19:46

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}];

• 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. – Almendrof66 Mar 27 at 14:02