Example 2-51 Solve for $v_{j}, j=0,1,2, \cdots, n$ in the circuit network shown in Fig. 2-15. Assume that $v(t)=v$.

保留源文本的换行和格式，翻译结果如下：

Example 2-51 Solve for $v_{j}, j=0,1,2, \cdots, n$ in the circuit network shown in Fig. 2-15. Assume that $v(t)=v$.
According to Ohm's Law, we have
Figure 2-15
$$
\begin{array}{l}
4 i_{j+1}=v_{j+2}-v_{j+1} \\
4 i_{j}=v_{j+1}-v_{j}
\end{array}
$$

Therefore,
$$
\begin{array}{l}
i_{j+1}=\frac{1}{4}\left(v_{j+2}-v_{j+1}\right) \\
i_{j}=\frac{1}{4}\left(v_{j+1}-v_{j}\right)
\end{array}
$$

Since the algebraic sum of currents at each node is 0, we have
$$
\begin{array}{c}
i_{j+1}=i_{j}+\frac{1}{2} v_{j+1} \\
\frac{1}{4} v_{j+2}-v_{j+1}+\frac{1}{4} v_{j}=0
\end{array}
$$

Thus,
or
$$
v_{j+2}-4 v_{j+1}+v_{j}=0, j=0,1,2, \cdots, n-2
$$

From the algebraic sum of currents at the $v_{0}$ point being 0, we get
$$
\begin{array}{l}
\frac{1}{4}\left(v_{1}-v_{0}\right)=\frac{1}{2} v_{0} \\
v_{1}=3 v_{0}
\end{array}
$$

Therefore,
the characteristic equation is
$$
\begin{array}{l}
x^{2}-4 x+1=0 \\
x=2 \pm \sqrt{3}
\end{array}
$$

Assume
$$
\begin{array}{l}
v_{n}=k_{1}(2+\sqrt{3})^{n}+k_{2}(2-\sqrt{3})^{n} \\
k_{1}+k_{2}=v_{0} \\
k_{1}(2+\sqrt{3})+k_{2}(2-\sqrt{3})=3 v_{0} \\
k_{1}=\frac{1}{2 \sqrt{3}}(1+\sqrt{3}) v_{0} \\
k_{2}=\frac{-1}{2 \sqrt{3}}(1-\sqrt{3}) v_{0} \\
v_{n}=\frac{v_{0}}{2 \sqrt{3}}\left[(1+\sqrt{3})(2+\sqrt{3})^{n}-(1-\sqrt{3})(2-\sqrt{3})^{n}\right]
\end{array}
$$