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If I have the following problem $$-\mu u'' + u' = 1$$ with boundary conditions $u(0) = u'(1) = 1$ in the interval $\Omega = (0,1)$. The exact solution is $$u(x) = x + 1$$ Will the FEM approximation with piecewise linear functions ($\mathbb{P}^1$) have a very small (or no) error? If so, how to think about it to have an image in my mind? |
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By Céa's lemma, the finite element discretization error is bounded by the best approximation error: $$\|u-u_h\|_{H^1(0,1)} \leq C \min_{v_h\in \mathbb{P}^1} \|u-v_h\|_{H^1(0,1)},$$ where $u\in H^1(0,1)$ is the exact solution, $u_h\in\mathbb{P}^1$ is the finite element solution. Since in your case the exact solution $u$ is actually in $\mathbb{P}^1$, the minimum is attained for $v_h = u$, and so the error is zero. Basically, you are solving the equation by projection onto a finite-dimensional subspace. If your solution actually lies in this subspace, you will find it. EDIT: The key idea is the following. You are looking for a weak solution, i.e., $u\in V$ such that $$a(u,v) = f(v) \quad\text{for all} \quad v\in V.\tag{1}$$ (In your case, $a(u,v) = \int_0^1 \mu u'v' + \int_0^1 u' v$ and $f(v) = \int_0^1 v + v(1)$.) The finite element approximation consists in choosing $V_h\subset V$ (in your case, $V_h = \mathbb{P}^1$) and finding $u_h\in V_h$ such that $$a(u_h,v_h) = f(v_h) \quad\text{for all} \quad v_h\in V_h.\tag{2}$$ Setting $v=v_h\in V$ in $(1)$ and subtracting $(2)$ gives $$a(u-u_h,v_h) = 0 \quad \text{for all}\quad v_h\in V_h.$$ This tells you that your error $u-u_h$ is orthogonal to $V_h$ with respect to the "inner product" defined by $a$, or that $u_h$ is the orthogonal projection of $u$ on $V_h$ (again, with respect to $a$). And orthogonal projections on a subspace are characterized by best approximation properties with respect to appropriate norms (in this case, the $H^1$ norm since $a$ happens to be coercive in this norm). |
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