# FEM 1D poisson substitution integral issue

I'm trying to solve $$\begin{cases} -u''=f \\ u(0)=0 \\ u(1)= \alpha \end{cases}$$ with FEM using reference elements and local coordinates.

So we have the global matrix $$K_{ij}=\int_\Omega N_i'(x) N_j'(x)$$.

Computing each local matrix for a 2noded element, I have $$K^e(\xi)=\begin{pmatrix} 1/2 & -1/2 \\ -1/2 & 1/2 \end{pmatrix}$$

To compute its global equivalent, I use the substitution rule $$\int_{\phi(a)=-1}^{\phi(b)=1} f(\xi)d\xi = \int_a^b f(\phi(x)) \phi'(x) dx$$ with $$\xi=\phi(x)=\frac{2}{h}(x-x_c)$$ and $$\phi'(x)=\frac{2}{h}$$.

So basically I have $$K^e=K^g\frac{2}{h}$$ so $$K^g=K^e\frac{h}{2}$$. Here this result is wrong, and I don't know where I missed up. I'm supposed to have $$K^g=K^e\frac{2}{h}$$

Thanks :)

## 1 Answer

On $$[x_i, x_{i+1}]$$, you can write $$\begin{equation} N_i(x) = \frac{x_{i+1}-x}{h} = 1 - \xi = \phi_i(\xi) \quad \mbox{where } \; \xi = \frac{x-x_i}{h} \end{equation}$$ So the derivatives satisfy $$\begin{equation} \frac{dN_i}{dx}(x) = - \frac{1}{h} = \frac{d\phi_i}{d\xi}\left(\xi (x) \right) \frac{d\xi}{dx}(x) = (-1) \left( \frac{1}{h} \right) \end{equation}$$ So the integral becomes $$\begin{equation} \int_{x_i}^{x_i+h} N_i'(x) N_i'(x) dx = \int_{x_i}^{x_i+h} \left( - \frac{1}{h} \right) \left( - \frac{1}{h} \right) dx = \frac{1}{h} \end{equation}$$ If we use the change of variables, we have $$\begin{multline} \int_{x_i}^{x_i+h} N_i'(x) N_i'(x) dx = \int_{x_i}^{x_i+h} \left( \frac{d\phi_i}{d\xi}\left(\xi (x) \right) \frac{d\xi}{dx}(x) \right)^2 dx = \int_{0}^{h} \left( \frac{d\phi_i}{d\xi}\left( \xi \right) \frac{d\xi}{dx}( \xi ) \right)^2 \left( h d\xi \right) \\ = \int_{0}^{1} \left[ \left( -1 \right) \left( \frac{1}{h} \right) \right]^2 \left( h d\xi \right) = \frac{1}{h} \end{multline}$$