How to solve this infinite system of equations? I'm trying to solve an infinite set of coupled equations.  j is a real integer index where $-\infty<j<\infty$. {a,b,c,d,f} are variables, and K's are constants.  I need to find closed form solutions for $\{a_{j},b_{j},c_{j}\}$ in terms of only the K's. I believe the respective solutions for $\{a_{j},b_{j},c_{j}\}$ should have the same form for all j.  I don't need explicit forms for the d's or f's.
$$a_{j}+b_{j}+c_{j}=K_{j}$$
$$3a_{j}+2b_{j}+c_{j}=c_{j+1}$$
$$6a_{j}+3b_{j}+d_{j}+3f_{j}=0$$
$$3a_{j}+2b_{j}+d_{j}+2f_{j}=0$$
$$d_{j}+f_{j}-f_{j-1}=0$$
Can this be done?
 A: You can certainly obtain a formal solution by treating the variables as Fourier coefficients. Multiplying every equation by $e^{ijx}$ and summing over $j$ after defining the functions
$$\{a,b,c,d,f,K\}(x)=\sum_{j=-\infty}^\infty\{a,b,c,d,f,K\}_j e^{ijx}$$
one obtains the linear system
$$\begin{pmatrix}1&1&1&0&0\\ 
3&2 &1-e^{-ix}&0&0\\6&3&0&1&3\\3&2&0&1&2\\0&0&0&1&1-e^{ix}\end{pmatrix}\begin{pmatrix}a(x)\\b(x)\\c(x)\\d(x)\\f(x)\end{pmatrix}=\begin{pmatrix}K(x)\\0\\0\\0\\0\end{pmatrix}$$
with the solution
$$\begin{pmatrix}a(x)\\ b(x)\\c(x)\\d(x)\\f(x)\end{pmatrix}=\frac{K(x)}{2(\cos x+2)}\begin{pmatrix}2(\cos x-1)\\ 3(1-e^{ix})\\ 3(1+e^{ix})\\ 6(\cos x-1)\\3(1-e^{ix})\end{pmatrix}$$
Now it isn't very hard to undo the Fourier series and obtain expressions for the unknown variables in terms of sums over the integers. As an example, let us find a formal expression for $a_j$.
First, note that
$$K(x)(\cos x -1)=\sum_{j=-\infty}^{\infty}(K_{j+1}+K_{j-1}-2K_j)e^{ij x}$$
It remains to extract the Fourier series of $(\cos x +2)^{-1}$. A standard complex analysis argument shows that
$$(\cos x+2)^{-1}=-\frac{\pi}{\sqrt{3}}\sum_{k=-\infty}^{\infty}(2-\sqrt{3})^{|k|}e^{ikx}:=\sum_j {W}_j e^{ijx}$$
Multiplying the two series together we get the expression
$$a_j=\sum_{\ell=-\infty}^{\infty}K_\ell(W_{j-\ell+1}+W_{j-\ell-1}-2W_{j-\ell})=2\pi(1-1/\sqrt{3})K_j+\frac{6\pi}{\sqrt{3}}\sum_{\ell\neq j}K_{\ell}(2-\sqrt{3})^{|j-\ell|}$$
Similarly we obtain solutions for the other coefficients as well.
A: Using the $Z$ transform we have
$$
\cases{
A(z)+B(z)+C(z)=K(z)\\
3A(z)+2B(z)+C(z)-zC(z)=zc_{-\infty}\\
6A(z)+3B(z)+D(z)+3F(z)=0\\
3A(z)+2B(z)+D(z)+2F(z)=0\\
D(z)+F(z)-z^{-1}F(z)=f_{-\infty}
}
$$
and solving we have
$$
\left\{
\begin{array}{l}
 A(z)=\frac{c_{-\infty} z (z-1)+f_{-1} z (z+1)+K(z) (z-1)^2}{z^2+4 z+1} \\
 B(z)=\frac{3 c_{-\infty} z-f_{-1} z (z+2)+3 K(z) (z-1)}{z^2+4 z+1} \\
 C(z)=\frac{z (f_{-\infty}-c_{-\infty} (z+2))+3 K(z) (z+1)}{z^2+4 z+1} \\
 D(z)=\frac{3 \left(c_{-\infty} z (z-1)+f_{-\infty} z (z+1)+K(z) (z-1)^2\right)}{z^2+4 z+1} \\
F(z) = -\frac{z (3 c_{-\infty} z+2 f_{-\infty} z+f_{-\infty}+3 K(z) (z-1))}{z^2+4 z+1}
\end{array}
\right.
$$
now knowing $K(z)$ and $c_{-\infty},f_{-\infty}$ we can calculate the inverse.
NOTE
If $K(z)$ doesn't exists in closed form then we can use the convolution theorem which states that $K(z)G(z) \leftrightarrow \{k_j\}\circledast \{g_j\}$
A: Given
$$
\left\{ \begin{array}{l}
 a_j  + b_j  + c_j  = K_j  \\ 
 3a_j  + 2b_j  + c_j  = c_{j + 1}  \\ 
 6a_j  + 3b_j  + d_j  + 3f_j  = 0 \\ 
 3a_j  + 2b_j  + d_j  + 2f_j  = 0 \\ 
 d_j  + f_j  - f_{j - 1}  = 0 \\ 
 \end{array} \right.
$$
we can use 3rd and 4th rows to enucleate $d$ anf $f$ as a combination of  $a$ and $b$
$$
\left\{ \begin{array}{l}
 f_j  =  - 3a_j  - b_j  \\ 
 d_j  = 3a_j  \\ 
 \end{array} \right.
$$
Thereafter we can enucleate $a$ as
$$
\begin{array}{l}
 \left\{ \begin{array}{l}
 a_j  + b_j  + c_j  = K_j  \\ 
 3a_j  + 2b_j  + c_j  = c_{j + 1}  \\ 
 3a_{j - 1}  - b_j  + b_{j - 1}  = 0 \\ 
 \end{array} \right.\;\; \Rightarrow \;\left\{ \begin{array}{l}
 a_j  + b_j  + c_j  = K_j  \\ 
 3a_j  + 2b_j  + c_j  = c_{j + 1}  \\ 
 3a_j  + b_j  = b_{j + 1}  \\ 
 \end{array} \right.\;\; \Rightarrow  \\ 
  \Rightarrow \;\left\{ \begin{array}{l}
 a_j  = K_j  - b_j  - c_j  \\ 
 \left\{ \begin{array}{l}
 3K_j  - b_j  - 2c_j  = c_{j + 1}  \\ 
 3K_j  - 2b_j  - 3c_j  = b_{j + 1}  \\ 
 \end{array} \right. \\ 
 \end{array} \right. \\ 
 \end{array}
$$
where in the last row we have increased the index: we shall take that in due account, after solving the system, by accomodating the initial conditions.
So  we are left with
$$
\left\{ \begin{array}{l}
 3K_j  - b_j  - 2c_j  = c_{j + 1}  \\ 
 3K_j  - 2b_j  - 3c_j  = b_{j + 1}  \\ 
 \end{array} \right.\;\; \Rightarrow \left( {\begin{array}{*{20}c}
   {b_{j + 1} }  \\   {c_{j + 1} }  \\
\end{array}} \right) =  - \left( {\begin{array}{*{20}c}
   2 & 3  \\   1 & 2  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   {b_j }  \\   {c_j }  \\
\end{array}} \right) + 3K_j \left( {\begin{array}{*{20}c}
   1  \\   1  \\
\end{array}} \right)
$$
The matrix diagonalizes as
$$
\left( {\begin{array}{*{20}c}
   2 & 3  \\
   1 & 2  \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
   1 & {\sqrt 3 }  \\
   1 & { - \sqrt 3 }  \\
\end{array}} \right)^{ - 1} \left( {\begin{array}{*{20}c}
   {2 + \sqrt 3 } & 0  \\
   0 & {2 - \sqrt 3 }  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   1 & {\sqrt 3 }  \\
   1 & { - \sqrt 3 }  \\
\end{array}} \right)
$$
and thus finally we get the decoupled difference equations
$$
\left( {\begin{array}{*{20}c}
   {u_{j + 1} }  \\   {v_{j + 1} }  \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
   1 & {\sqrt 3 }  \\   1 & { - \sqrt 3 }  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   {b_{j + 1} }  \\   {c_{j + 1} }  \\
\end{array}} \right) =  - \left( {\begin{array}{*{20}c}
   {2 + \sqrt 3 } & 0  \\   0 & {2 - \sqrt 3 }  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   {u_j }  \\   {v_j }  \\
\end{array}} \right) + 3K_j \left( {\begin{array}{*{20}c}
   1  \\   1  \\
\end{array}} \right)
$$
which can be easily solved .
