How to solve this infinite system of equations?

Question

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?

Answer 1

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.

Answer 2

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 $K(z)G(z) \leftrightarrow \{k_j\}\circledast \{g_j\}$

Answer 3

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 .