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I'm trying to re-solve the governing equations in hydraulic fracturing modeling as instructed step by step in a paper.

After (A-9), the author stated that by substituting A-6, A-8 and A-9 into equation A-4 we obtain a linear tridiagonal system which is easily solved for $∆W_i^m$ at m =1, 2,… Could anyone please help me at this step?

The governing equation:

$\frac{∂q}{∂x}+\frac{2hC}{\sqrt{t-τ(x)}}+\frac{∂A}{∂t}=0, 0<x<L (A-1)$

$q(0,t)=q_i (A-2)$

$\frac{∆t_m}{2}\frac{∂}{∂x}({q^{m+1}+q^m})+[{\frac{4hC}{\sqrt{t-τ(x)}}+A}]_{t_m}^{t_{m+1}} =0 (A-3)$

$∆t_m=t_{m+1}-t_m, q^m=q(x,t_m)$

Integration A-3 with respect to x from $x_{i-1/2}$ to $x_{i+1/2}$ (with the use of $A=π/4 Wh$):

$\frac{∆t_m}{2∆x}(q_{i+1/2}^{m+1}-q_{i-1/2}^{m+1}+q_{i+1/2}^m-q_{i-1/2}^m )+4hC(\sqrt{t_{m+1}-τ_i}-\sqrt{t_m-τ_i})+\frac{πh}{4}(W_i^{m+1}-W_i^m)=0 (A-4)$

where:

$∆x=x_{i+1/2}-x_{i-1/2}$

$W_i^m=W(x_i,t_m)$

$x_i=\frac{1}{2}(x_{i+1/2}+x_{i-1/2})$

Take $x_{1/2}=0$ then by A-2:

$q_{1/2}^m=q_i (A-5)$

$W_i^0=0(A-6)$

$q=\frac{-πG}{256(1-ϑ)μ∆x}\frac{∂}{∂x}W^4$, by finite difference analog gives:

$q_{i+1/2}=\frac{-πG}{256(1-ϑ)μ∆x}[{(W_{i+1}^m)}^4-{(W_i^m )}^4 ] (A-7)$

$q_{i+1/2}^m=\frac{-πG}{256(1-ϑ)μ∆x}[{(W_{i+1}^m)}^4-{(W_i^m )}^4 ]$ (m was missing??)

Take:

$W_i^{m+1}=W_i^m+∆W_i^m (A-8)$

${(W_i^{m+1} )}^4={(W_i^m)}^4+4{(W_i^m )}^3{∆W}_i^m (A-9)$

This is what I get after substituting A-7 into A-4:

$\frac{πG∆t_m}{256(1-ϑ)μ}[{(W_{i+1}^m)}^4-2{(W_i^m)}^4+{(W_{i-1}^m)}^4+2{(W_{i+1}^m)}^3{∆W}_{i+1}^m-4{(W_i^m)}^3{∆W}_i^m+2{(W_{i-1}^m)}^3{∆W}_{i-1}^m]+4hC\sqrt{t_{m+1}-τ_i}-\sqrt{t_m-τ_i}+\frac{πh}{4}{∆W}_i^m=0 (A-10)$

How to get the linear tridiagonal system from the above equation?

Thank you very much

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Welcome to the site. You are actually virtually finished, but may not have realised it yet. A tridiagonal linear system is another name for a matrix problem which only has non zero entries on the leading diagonal and the one above and below it, so lets try writing your problem like that. We want a form $$ \mathbf{A} (\mathbf{\Delta W}^m) = \mathbf{b},$$

where $\mathbf{A}$ is a matrix with values we know, $\mathbf{\Delta W}^m$ is the (unknown) vector of updates we want to solve for and $\mathbf{b}$ is a known vector of values on the right hand side.

So now, reading off the terms in $\Delta W$ from your equation (A-10) we have

$$A_{ij}=\begin{cases} 2(W^m_{i-1})^3 &j=i-1\\ -4(W^m_i)^3 & j=i\\ 2(W^m_{i+1})^3 & j=i+1\\ 0& \mbox{otherwise}\end{cases}$$

meanwhile $\mathbf{b}$ collects all the other terms which don't have a $\Delta W$ in. So, this is indeed a linear tridiagonal system. Since we known $\mathbf{W}^0$, we can invert $\mathbf{A}$ for $\mathbf{\Delta W}^0$, which gives us $\mathbf{W}^1$ and so on.

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