How to find the general term of the following sequence?

Question

Consider the following recurrence problem: \begin{align} d_{i-1} &= 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1} - F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , \\ \varphi_{i-1} &= -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) + F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , \end{align} where $d_i, i \in \{2, \cdots , N+1\}$ represent displacements, $\varphi_i$ inclinations, and $F$ is a known force acting at the nodes $N$ and $N+1$.

We require by the system symmetry that $d_{N+1}=d_N = d_\mathrm{C}$ and $\varphi_N = -\varphi_{N+1}= \varphi_\mathrm{C}$, where $d_\mathrm{C}$ and $\varphi_\mathrm{C}$ are still to be determined from the boundary conditions:

$d_1 = 0$ (zero displacement) and $2\varphi_1+\varphi_2 = 3d_2$ (zero torque)

In order to proceed, i have tried to first determine $d_{N-1}$ and $\varphi_{N-1}$ from the system above, and then $d_{N-2}$ and $\varphi_{N-2}$, etc... in a recursive way and then try to find out the general term of the resulting sequences.

For the term $N-1$, we obtain \begin{align} d_{N-1} &= d_\mathrm{C}+2\varphi_\mathrm{C}-F \, \\ \varphi_{N-1} &= -9\varphi_\mathrm{C}+3F \, . \end{align}

Analogously, we get for the term $N-2$ \begin{align} d_{N-2} &= d_\mathrm{C}-18\varphi_\mathrm{C}+4F \, \\ \varphi_{N-2} &= 89\varphi_\mathrm{C}-24F \, . \end{align}

I was wondering whether there is a particular way to figure out the general term of such sequence.

Any help or suggestions are most welcome.

Answer 1

Thanks again to Yuri for the helpful insights. I can say that I have now found the solution to my problem.

First of all, let us define $D_i = d_i-d_{i-1}$, in the same way as Yuri did, and let us get a recurrence equation that involves $\varphi_i$ only.

Actually, original problem (which is equivalent to the one stated above) was \begin{align} d_{i+1}-2d_i+d_{i-1} - \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1}\right) + F \left( \delta_{i, N} + \delta_{i, N+1} \right) &=0\, , \\ d_{i+1}-d_{i-1} - \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) &= 0 \, . \end{align}

Accordingly, it can easily be shown that for $0 <i <N$, \begin{align} D_{i+1}-D_i &= \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1} \right) \, , \\ D_{i+1}+D_i &= \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) \, . \end{align}

This leads to \begin{align} D_i = \frac{7}{24} \varphi_i + \frac{1}{24} \varphi_{i-2} + \frac{2}{3} \varphi_{i-1} = \frac{1}{24} \varphi_{i+1}+\frac{7}{24} \varphi_{i-1} + \frac{2}{3} \varphi_i \, , \tag{1} \label{1} \end{align} or $$ \frac{1}{24} \left( \varphi_{i+1}-\varphi_{i-2} \right) +\frac{3}{8} \left( \varphi_i-\varphi_{i-1} \right) =0\, . $$

Using Yuri's powerful approach, and posing $\varphi_i = A/p^i$ yields $$ p^3+9p^2-9p-1=0 \, , $$ whose solutions are $p=1$ and $p=-5\pm 2 \sqrt{6}$.

Finally, $$ \varphi_i = C_1 + C_2 \left(-5-2\sqrt{6}\right)^i + C_3 \left(-5+2\sqrt{6}\right)^i \, . $$ (again, by noting that $\left(-5-2\sqrt{6}\right)\left(-5+2\sqrt{6}\right)=1$.)

Therefore, $$ d_i =D_i + D_{i-1} + \cdots + D_2 = \sum_{n=2}^{n=i} D_n \, , $$ where the expression of $D_n$ follows readily from Eq. \eqref{1}, and using the fact that $d_1=0.$

The final expression read $$ D_i = C_1 + C_2 \left( -1+\frac{\sqrt{6}}{2} \right) \left( -5-2\sqrt{6} \right)^i + C_3 \left( -1-\frac{\sqrt{6}}{2} \right) \left( -5+2\sqrt{6} \right)^i \, , $$ and $$ d_i = (i-1)C_1 + C_2 \left( -1+\frac{5\sqrt{6}}{12} \right) \left[ 49+20\sqrt{6}-\left( -5-2\sqrt{6} \right)^{i+1} \right] + C_3 \left( 1+\frac{5\sqrt{6}}{12} \right) \left[ -49+20\sqrt{6}+\left( -5+2\sqrt{6} \right)^{i+1} \right] \, . $$

Clearly, $d_1=0$. Finally, at this point, the 3 unknown coefficients can readily be determined from the 3 boundary conditions.

I have checked that the solution is very well behaved and is in full agreement with the numerical solution.

Answer 2

Except for i=N and i=N+1, you can write the recursion for a 4-vector $$\vec{v}_i=(d_i,d_{i-1},\phi_i,\phi_{i-1})^t$$ and a 4x4 matrix $A$. Write $d_i$ and $\phi_i$ in terms of the eigenvalues and eigenvectors of $A$, then deal with the boundary conditions at $0,N$ and $N+1$

Answer 3

$$\mathbf{\color{green}{The\ linear\ approach}}$$

As it follows from the comments, the issue task can be detalized in the form of \begin{cases} d_{i-1} = 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1}\\[4pt] \varphi_{i-1} = -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) \\[4pt] i=2,3\dots N-1\\[4pt] d_1 = 0\\[4pt] 2\varphi_1+\varphi_2 = 3d_2\\[4pt] d_{N-1} = 2\varphi_{N+1}+4\varphi_N + 8d_N-7d_{N+1} - F\\[4pt] \varphi_{N-1} = -7\varphi_{N+1}-16\varphi_{N} + 24 \left( d_{N+1}-d_{N}\right) + F\\[4pt] d_{N} = 2\varphi_{N-1}+4\varphi_{N+1} + 8d_{N+1}-7d_{N-1} - F\\[4pt] \varphi_{N} = -7\varphi_{N-1}-16\varphi_{N+1} + 24 \left( d_{N-1}-d_{N+1}\right) + F\\[4pt] d_{N+1}=d_N\\[4pt] \varphi_{N+1} = -\varphi_{N},\tag{I} \end{cases} or \begin{cases} d_{i-1} - 8d_i - 4\varphi_i + 7d_{i+1} - 2\varphi_{i+1} = 0\\[4pt] \varphi_{i-1} + 24d_i + 16\varphi_{i} - 24d_{i+1} + 7\varphi_{i+1} = 0 \\[4pt] i=2,3\dots N-1\\[4pt] d_1 = 0\\[4pt] 2\varphi_1 - 3d_2 + \varphi_2 = 0\\[4pt] d_{N-1} - d_N - 2\varphi_N = - F\\[4pt] \varphi_{N-1} + 9\varphi_{N} = F.\tag{II} \end{cases} This gives the linear system $$A\overrightarrow v=\overrightarrow f,\tag{III}$$ where $$A= \begin{pmatrix} 1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&2&-3&1&0&0&0&0&0&0&0&0&0&0&0&0\\ 1&0&-8&-4&7&-2&0&0&0&0&0&0&0&0&0&0\\ 0&1&24&16&-24&7&0&0&0&0&0&0&0&0&0&0\\ 0&0&1&0&-8&-4&7&-2&0&0&0&0&0&0&0&0\\ 0&0&0&1&24&16&-24&7&0&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&-8&-4&7&-2&0&0&0&0&0&0\\ 0&0&0&0&0&1&24&16&-24&7&0&0&0&0&0&0\\ &&&&&&&&&\dots&&&&&&\\ 0&0&0&0&0&0&0&0&0&0&1&0&-8&-4&7&-2\\ 0&0&0&0&0&0&0&0&0&0&0&1&24&16&-24&7\\ 0&0&0&0&0&0&0&0&0&0&0&0&1&0&-1&-2\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&9\\ \end{pmatrix},$$ $$ \overrightarrow v=\begin{pmatrix} d_1\\\varphi_1\\d_2\\\varphi_2\\\vdots\\d_{N-1}\\\varphi_{N-1}\\d_N\\\varphi_N \end{pmatrix}, \overrightarrow f=\begin{pmatrix} 0\\0\\0\\0\\\vdots\\0\\0\\-F\\F \end{pmatrix}. $$ Methods for solving $n$-diagonal SLAE $(n=6)$ are well known.

$$\mathbf{\color{green}{Common\ solution}}$$

Let \begin{cases} d_{i-1} - 8d_i + 7d_{i+1} = 4\varphi_i + 2\varphi_{i+1}\\[4pt] \varphi_{i-1} + 16\varphi_{i} + 7\varphi_{i+1} = -24d_i + 24d_{i+1}\\[4pt] i=2,3\dots N-1\\[4pt] d_1 = 0\\[4pt] 2\varphi_1 - 3d_2 + \varphi_2 = 0\\[4pt] d_{N-1} - d_N - 2\varphi_N = - F\\[4pt] \varphi_{N-1} + 9\varphi_{N} = F,\tag1 \end{cases} Looking for the solution in the form of $$d_i = Ap^{-i},\quad\varphi_i=Bq^{-i}.\tag2$$ we get the system \begin{cases} Ap^{-i-1}(p^2-8p+7) = Bq^{-i-1}(4q+2) \\ 24Ap^{-i-1}(1-p) = Bq^{-i-1}(q^2+16q+7),\tag3 \end{cases}

which leads to the relation $$\dfrac{d_i}{\varphi_i} = \frac pq\frac{4q+2}{p^2-8p+7} = \frac pq\frac{q^2+16q+7}{24(1-p)},\tag4$$ where $$(p^2-8p+7)(q^2+16q+7) + 48(p-1)(2q+1)=0,$$ or $$(p-1)\left((p-7)q^2+16(p-1)q + 7p-1\right) = 0,$$ with the solutions $$\left[\begin{align} &p=1\\ &p=7,\quad q=-\frac12\\[4pt] &p=q=-5\pm2\sqrt6, \end{align}\right.\tag5$$ wherein third and fourth solutions obtained, using additional condition $p=q$, which allows to hold the ratio $(4).$

Formulas $(5)$ and $(4)$ lead to the common solution in the form of $$\begin{align} &\binom{d_i}{\varphi_i} = C_1\binom{1}{0}+C_27^{-i}\binom{7}{-96}\\[4pt] &+C_3(-5-2\sqrt6)^{i}\binom1{2\sqrt6}+C_4(-5+2\sqrt6)^{i}\binom1{-2\sqrt6}, \end{align}\tag6$$ (using the identity $(-5+2\sqrt6)(-5-2\sqrt6)=1$).

Coefficients $C_i$ can be defined from the boundary conditions.

$$\mathbf{\color{green}{Modified\ solution.}}$$

The previous model has a resonant solution $(p=1).$ To avoid this situation, should be used another basis.

Let $$D_i=d_{i}-d_{i-1},\quad d_1=0,\tag7$$ then \begin{cases} -D_{i} + 7D_{i+1} = 4\varphi_i + 2\varphi_{i+1}\\[4pt] \varphi_{i-1} + 16\varphi_{i} + 7\varphi_{i+1} = 24D_{i}\\[4pt] i=2,3\dots N-1\\[4pt] 2\varphi_1 - 3D_2 + \varphi_2 = 0\\[4pt] D_{N} + 2\varphi_N = + F\\[4pt] \varphi_{N-1} + 9\varphi_{N} = F,\tag8 \end{cases} Looking for the solution in the form of $$D_i = Ap^{-i},\quad\varphi_i=Bp^{-i}.\tag9$$ we get the system \begin{cases} A(7-p) = B(4p+2) \\ 24A = B(p^2+16p+7),\tag{10} \end{cases}

which leads to the relation $$\dfrac{D_i}{\varphi_i} = \frac{4p+2}{7-p},\tag{11}$$ where $$(p-7)(p^2+16p+7)+96p+48=0,$$ $$(p - 1) (p^2 + 10 p + 1) = 0,$$ with the solutions $$\left[\begin{align} &p_1=1,\quad D_i=\varphi_i\\ &p_2=-5+2\sqrt6,\quad D_i=\dfrac{\sqrt6-2}2\varphi_i\\ &p_3=-5-2\sqrt6,\quad D_i=-\dfrac{\sqrt6+2}2\varphi_i. \end{align}\right.\tag{12}$$

Formulas $(9)$ and $(12)$ lead to the common solution in the form of $$\begin{align} &\binom{D_i}{\varphi_i} = C_1\binom{1}{1}+C_2(-5-2\sqrt6)^{i}\binom{\sqrt6-2}2+C_3(-5+2\sqrt6)^{i}\binom{\sqrt6+2}{-2} \end{align}\tag{13}$$ (using the identity $(-5+2\sqrt6)(-5-2\sqrt6)=1$).

Coefficients $C_i$ can be defined from the boundary conditions and then the values of $d_i$ can be calculafed using $(7).$

$$\mathbf{\color{green}{Analogies}}$$

Are known the formulas $$1-\binom{n}{1}2^t+\binom{n}{2}n3^t+\dots+(-1)^n\binom{n}{n}n^t = 0,\quad t<n\tag{A1}$$ and $$\Delta^nf(x) = \sum\limits_{m=0}^n(-1)^{n-m}f(x+m)=\sum\limits_{k=0}^{r}(-1)^k \binom{r}{k}\Delta r^{n+k}f(x).\tag{A2}$$ This leads to the some analogies between the linear recurrence relations and the linear DEs.

$\mathbf{Analogy 1.}$

The ODE $$y'''+ay''+by'=R(x)\tag{A3}$$ does not contain the term with $cy$ and can be simplified using the substitution $Y=y'.$

This analogy is the hint to use the substitution $D=d_i-d_{i-1}.$

$\mathbf{Analogy 2.}$

Solution of the OOE $(A3)$ is the sum of the common solution of the homogenius ODE and the partial solution of the origin ODE.

The common solution can be defined using the substitution $y=e^{kx}.$

This analogy is the hint to use the approach with $D=q^n.$

Btw, the negative degree were used situatively and is not obligate.

$\mathbf{Analogy 3.}$

The partial solution of ODE $(3)$ with the polynomial RHS usually can be found as the same order polynomial with the unknown coefficients.

On the other hand, the ODE $$y'''+ay''+by'= F\tag{A4}$$ already has the common solution $y=C$ (the resonant case), so the form $y_p=Dx$ is usually applied.

Hint is clear, but the analogy isn't. Maybe, the formulas $(1)$ are the second hint?

This is the main.