Stiffness matrix 1D

Let us compute the stiffness matrix for the one-dimensional case. In one-dimensional space the gradient has the form $\nabla \upsilon$​${} = \frac{\displaystyle \partial \upsilon}{\displaystyle \partial x}$, and the dot product of the gradient with itself is $\nabla \upsilon \cdot \nabla \upsilon$​${} = \left( \frac{\displaystyle \partial \upsilon}{\displaystyle \partial x} \right)^2$. Since the one-dimensional computational domain $M$ is partitioned into segment simplices, the part of the functional under study for a single segment $(x_i, x_{i+1})$ can be written as

In the general case the trial function $\upsilon(x)$​${} = \sum_{i=1}^N \upsilon_i(x)$, whereas on a simplex it has the form $\upsilon_{(i)(i+1)}(x)$​${} = q_i \cdot \phi_i(x)$​${} + q_{i+1} \cdot \phi_{i+1}(x)$. Taking into account (5.7$\displaystyle \begin{cases} a_{i-1} + b_{i-1} \cdot x_{i-1} = 1\\ a_{i-1} + b_{i-1} \cdot x_i = 0 \end{cases}$​$\displaystyle \begin{cases} a_{i} + b_{i} \cdot x_{i-1} = 0\\ a_{i} + b_{i} \cdot x_i = 1 \end{cases}$) and (5.8$\displaystyle \begin{cases} a_{i-1} = \frac{\displaystyle -x_i}{\displaystyle x_{i-1} - x_i}\\ b_{i-1} = \frac{\displaystyle 1}{\displaystyle x_{i-1} - x_i} \end{cases}$​$\displaystyle \begin{cases} a_{i} = \frac{\displaystyle x_{i-1}}{\displaystyle x_{i-1} - x_i}\\ b_{i} = \frac{\displaystyle -1}{\displaystyle x_{i-1} - x_i} \end{cases}$) from the section on hat functions, we write the relations for the hat functions

Compute the derivative of the trial function

Note that the derivative does not depend on $x$ and is constant on the segment. Substituting (6.3$\displaystyle \frac{\displaystyle \partial \upsilon_{(i)(i+1)}(x)}{\displaystyle \partial x}$ $\displaystyle {} = q_i \cdot b_i$​$\displaystyle {} + q_{i+1} \cdot b_{i+1}$) into (6.1$$\int_{x_i}^{x_{i+1}} \left( \frac{\displaystyle \partial \upsilon}{\displaystyle \partial x} \right)^2 \,dx.$$)

where $l_{(i)(i+1)}$​${} = x_{i+1}$​${} - x_i$ is the length of the segment.

Expand the square and regroup the terms

Let us take into account the formulas from (5.8$\displaystyle \begin{cases} a_{i-1} = \frac{\displaystyle -x_i}{\displaystyle x_{i-1} - x_i}\\ b_{i-1} = \frac{\displaystyle 1}{\displaystyle x_{i-1} - x_i} \end{cases}$​$\displaystyle \begin{cases} a_{i} = \frac{\displaystyle x_{i-1}}{\displaystyle x_{i-1} - x_i}\\ b_{i} = \frac{\displaystyle -1}{\displaystyle x_{i-1} - x_i} \end{cases}$) for the coefficients $b$ and introduce the notation for the elements of the local stiffness matrix of the segment

Thus, the local stiffness matrix for a one-dimensional element has the form

The global stiffness matrix $\mathbf{K}$ is obtained by summing the contributions from all segments of the mesh by the assembly method: the elements of the local matrices are added to the corresponding elements of the global matrix according to the global node numbering. The dimension of the global stiffness matrix is $N \times N$, where $N$ is the total number of mesh nodes.

For example, consider a mesh with nodes $0, 1, \ldots, i, i$​${} +1, \ldots, N$. During assembly each segment $(j)(j+1)$ contributes ${} \pm \frac{\displaystyle 1}{\displaystyle l_{(j)(j+1)}}$, and an interior node $i$ receives a contribution from the two adjacent segments — $(i-1)(i)$ and $(i)(i+1)$ — so its main-diagonal entry is the sum $\frac{\displaystyle 1}{\displaystyle l_{(i-1)(i)}}$​${} + \frac{\displaystyle 1}{\displaystyle l_{(i)(i+1)}}$. Since the length $l_{(j)(j+1)}$ depends on the segment index, it cannot be taken out as a common factor — each entry of the global matrix keeps the length of its own segment. As a result the matrix is tridiagonal