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Engineering

Finite Element Methods

Practice Finite Element Methods with 115 published problems. Preview a problem, explore the course outline, and work through one problem at a time.

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Course outline

6 topics

  1. Weak formulations, Galerkin projection, and variational structure

  2. Reference elements, interpolation, and numerical quadrature

  3. Global assembly and essential or natural boundary conditions

  4. Stability, consistency, a priori error, and adaptivity

  5. Time-dependent and eigenvalue finite-element systems

  6. Nonlinear, constrained, and mixed formulations

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Proof: make a limit precise

For each positive integer \(n\), let \(a_n=n/(n+2)\). Prove from the definition of convergence that \(a_n\to1\).

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The definition asks: given any \(\varepsilon>0\), can you make \(|a_n-1|<\varepsilon\) for every sufficiently large \(n\)? First simplify \(|a_n-1|\).

Show worked explanation

Let \(\varepsilon>0\). For every positive integer \(n\), \[ \left|\frac{n}{n+2}-1\right|=\left|\frac{-2}{n+2}\right|=\frac{2}{n+2}. \] Choose a positive integer \(N>2/\varepsilon\). Such an integer exists because the positive integers are unbounded. If \(n\ge N\), all denominators are positive, so \[ \frac{2}{n+2}\le\frac{2}{N+2}<\frac{2}{N}<\varepsilon. \] Thus, for every positive tolerance \(\varepsilon\), this \(N\), chosen for the given \(\varepsilon\), works for every \(n\ge N\). This is exactly the definition of \(a_n\to1\).

Calculation: complete the square

Solve \(x^2-6x+5=0\) over the real numbers by completing the square. Check each solution in the original equation.

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Move \(5\) to the other side. Half of \(-6\) is \(-3\), and its square is \(9\). Add \(9\) to both sides to form \((x-3)^2\).

Show worked explanation

The identity \( (x-3)^2=x^2-6x+9 \) explains the completed square. Each step below preserves equality: \[ x^2-6x=-5,\qquad x^2-6x+9=4,\qquad (x-3)^2=4. \] A real number whose square is \(4\) is either \(2\) or \(-2\). Therefore \(x-3=2\) or \(x-3=-2\), giving \(x=5\) or \(x=1\). Check both values: \[ 5^2-6(5)+5=25-30+5=0,\qquad 1^2-6(1)+5=1-6+5=0. \] Both satisfy the original equation. The two square-root cases cover all real possibilities, so the solution set is \(\{1,5\}\).

Physical model: distance and displacement

A cart travels along a straight line: \(18\,\mathrm{m}\) east in \(5\,\mathrm{s}\), then \(12\,\mathrm{m}\) west in \(3\,\mathrm{s}\), with no pause between the two legs. Find its average velocity and average speed for the whole trip.

Show a hint

Take east as positive. Average velocity uses signed displacement divided by total time. Average speed uses total distance traveled divided by total time.

Show worked explanation

With east positive, the displacements of the two legs are \(+18\,\mathrm{m}\) and \(-12\,\mathrm{m}\). The net displacement is \(18-12=6\,\mathrm{m}\) east, while the total distance traveled is \(18+12=30\,\mathrm{m}\). Total time is \(5+3=8\,\mathrm{s}\). \[ \text{Average velocity}=\frac{6\,\mathrm{m}}{8\,\mathrm{s}}=0.75\,\mathrm{m/s}\text{ east}, \] \[ \text{Average speed}=\frac{30\,\mathrm{m}}{8\,\mathrm{s}}=3.75\,\mathrm{m/s}. \] Check: the total distance is at least the magnitude of the displacement, so average speed must be at least the magnitude of average velocity. Here \(3.75\ge0.75\), as expected.