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Can any on help me understand how from the simple delta function on the right hand and performing the integration we can get values as in the left? Note that $h = x_i + 1 - x_i$

The diffusion and mass matrices are given by $$D_{ij} = \int_0^L \frac{d\phi_i}{dx}\frac{d\phi_j}{dx} dx = \left\{\begin{matrix} \frac{1}{h_1} &\text{if } i=j=1,\\ \frac{1}{h_{i-1}} + \frac{1}{h_{i}} &\text{if } i=j\neq 1 \text{ and } N_E +1,\\ \frac{1}{h_{N_E}} &\text{if } i=j=N_E+1,\\ -\frac{1}{h_{i}} &\text{if } j=i+1,\\ -\frac{1}{h_{i-1}} &\text{if } j=i-1,\\ 0 &\text{otherwise.} \end{matrix}\right.$$

and

$$M_{ij} = \int_0^L \phi_i \phi_j dx = \left\{\begin{matrix} \frac{1}{3}h_1 &\text{if } i=j=1,\\ \frac{1}{3}(h_{i-1} + h_i) &\text{if } i=j\neq 1 \text{ and } N_E +1,\\ \frac{1}{3}h_{N_E} &\text{if } i=j=N_E+1,\\ \frac{1}{6}h_{i} &\text{if } j=i+1,\\ \frac{1}{6}h_{i-1} &\text{if } j=i-1,\\ 0 &\text{otherwise.} \end{matrix}\right.$$

And the shape functions satisfy the condition $\phi_i(x_j) = \delta_{i,j}$ is the Kronecker delta.

References

  1. Pozrikidis, Constantine. Introduction to finite and spectral element methods using MATLAB. CRC Press, 2005.
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1 Answer 1

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The basis functions seem to be piecewise linear with $\phi_i(x_j) = \delta_{ij}$, so $$ \phi_i(x) = \begin{cases} \frac{1}{h_{i-1}}(x-x_{i-1}) \qquad &x\in[x_{i-1},x_i]\,,\\ \frac{-1}{h_{i}}(x-x_{i+1}) \qquad &x\in[x_{i},x_{i+1}]\,,\\ 0 & \text{otherwise}\,, \end{cases} $$ and as such $$ \frac{\mathrm{d}\phi_i}{\mathrm{d}x}(x) = \begin{cases} \frac{1}{h_{i-1}} \qquad &x\in[x_{i-1},x_i]\,,\\ \frac{-1}{h_{i}} \qquad &x\in[x_{i},x_{i+1}]\,,\\ 0 & \text{otherwise}\,. \end{cases} $$

As an example to get you started, the first integral, we have to show $$ \int_0^L \frac{\mathrm{d}\phi_i}{\mathrm{d}x} \frac{\mathrm{d}\phi_j}{\mathrm{d}x} \mathrm{d}x = \frac{1}{h_{i-1}} + \frac{1}{h_{i}} \quad \text{when } i=j\neq 1 \text{ and } N_E+1\,. $$ With $i=j$ and with the above definition $$ \begin{split} \int_0^L \left(\frac{\mathrm{d}\phi_i}{\mathrm{d}x}\right)^2 \mathrm{d}x &= 0 + \int_{x_{i-1}}^{x_i} \frac1{h_{i-1}^2}\mathrm{d}x + \int_{x_{i}}^{x_{i+1}} \frac1{h_{i}^2}\mathrm{d}x + 0 \\ &= \left[\frac{x}{{h_{i-1}}^2}\right]_{x_{i-1}}^{x_i} + \left[\frac{x}{{h_{i}}^2}\right]_{x_{i}}^{x_{i+1}} \\ &= \frac{1}{h_{i-1}} + \frac{1}{h_{i-1}}\,. \end{split} $$

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