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 $A=π/4 Wh$):

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





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

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


$q=\frac{-πG}{256(1-ϑ)μ∆x}\frac{∂}{∂x}W^4$, 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??)


$W_i^{m+1}=W_i^m+∆W_i^m (A-8)$

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

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


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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