User:Eml4500.f08.FEABBQ.koby/HW6

November 10, 2008
Continuous Principle of Virtual Work to Discrete Principle of Virtual Work (Continued)

Lagrangian Interpolation

What is the motivation for the form of $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$? 1) $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$ are both linear, thus any linear combination of $$N_i\,\!$$ and $$N_{i+1}\,\!$$ is also linear.  In particular, the expression for $$u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}\,\!$$ must be linear.  Below is a demonstration:

Let: $$N_i(x)=\alpha_i + \beta_ix\,\!$$ -$$\alpha_i, \beta_i\,\!$$ are real numbers

$$N_{i+1}(x)=\alpha_{i+1} + \beta_{i+1}x\,\!$$ -$$\alpha_{i+1}, \beta_{i+1}\,\!$$ are real numbers

Now we can combine $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$ :

$$u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}\,\!$$

$$=(\alpha_i + \beta_ix)d_i+(\alpha_{i+1}, \beta_{i+1})d_{i+1}\,\!$$

$$=(\alpha_i+\alpha_{i+1}+(\beta_i+\beta_{i+1})x\,\!$$ < is a linear function

2) Recall the equation for $$u(x)\,\!$$, an interpolation of x:

$$u(x_i)=\underbrace{N_i(x_i)d_i}_{1}+N_{i+1}(x_i)d_{i+1} = d_i\,\!$$