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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 :)

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1 Answer 1

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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}

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