Given $-u''(x) + \alpha u(x) = f(x)$ ($0\leq x\leq 1$, $\alpha > 0$) with $u'(0) = A$, $u'(1) = B$.

(a) Find the finite element approximation using piecewise linear elements in $3$ equidistant points. Determine the convergence rate.

(b) Is $a>0$ essential for convergence in part (a)?

My attempt:

(a) The 3 equidistant points have $x_0 = 0$, $x_1 = \frac{1}{2}$, and $x_2 = 1$. Let $u(x) = \sum_{j=1}^{3} t_j\phi_{j}(x)$. Multiplying both sides of the ODE by the piecewise linear test function $\phi_{i}(x)$ and using integration by parts, we have:

$-u'(1)\phi_{i}(1) + u'(0)\phi_{i}(0) + \sum_{j=1}^{3} t_j\int_{0}^{1} \phi'_{j}(x)\phi'_{i}(x) = \int_{0}^{1} \phi_{i}(x)f(x)dx$ $\ \forall\ i=1,2,3$

After this step, I failed to construct a suitable piecewise linear polynomial that satisfies the inhomogeneous Dirichlet boundary condition to have the first $2$ terms on the LHS cancel each other.

Question: Can anyone please help with a suitable construction? Really appreciate it.

(b) We can rewrite the $2$nd-order as the system of $1$st-order by letting $u'(x) = y_1(x)$ and $u(x) = y_0(x)$. Then $[y'_0 y'_1] = [0 1, -\alpha 0]y - f(x)$. By using Variation of Parameters, and noting that the two eigenvalues of the matrix $[0 1, -\alpha 0]$ are $\pm i\sqrt{a}$, the solution is bounded regardless of the initial condition. If $a<0$, one of the eigenvalue is positive, so the solution is not bounded. Thus $a>0$ is essential for convergence (Q.E.D)

Question: Can someone help verify if my proof for part (b) is correct?

  • $\begingroup$ @Evgeny: can you please try out part (a) of this problem? $\endgroup$ – user177196 Jan 7 '16 at 20:15
  • $\begingroup$ nobody wants to help me with part (a)? I seriously got stucked on constructing the approriate piecewise linear function that satisfies the boundary condition:( $\endgroup$ – user177196 Jan 9 '16 at 2:57

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