Least-squares cubic solution for a bar under a linearly varying axial load: 22-Mec-B10 Finite Element Analysis, December 2016, Q5

Question 5: Least-squares cubic solution for a bar under a linearly varying axial load (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A bar fixed at $x=0$ and traction-free at $x=L$, carrying a distributed axial load that grows linearly from the wall:

Given data — Question 5
Quantity	Value / expression
Governing equation	$EA\, u''(x)+cx=0$, $0 \lt x \lt L$
Distributed axial load	$q(x)=cx$ (force per unit length), $c$ constant
Essential boundary condition	$u(0)=0$ (fixed end)
Natural boundary condition	$EA\, u'(L)=0$ (free end, zero axial force)
Trial function	cubic polynomial in $x$
Evaluation points	$x=\tfrac14 L$ and $x=\tfrac34 L$

Find. The approximate cubic displacement field produced by the least-squares form of the method of weighted residuals at the two stated evaluation points, and a statement of how it compares with the exact solution.

Figure 5.1 — Cantilevered bar of length L, fixed at the left end and traction-free at x = L, carrying the linearly varying axial load q(x) = cx. The two teal markers are the least-squares evaluation points at x = L/4 and x = 3L/4.

Approach. Choose a cubic trial function that satisfies the essential boundary condition identically, form the residual of the governing equation, minimise the sum of squared residuals over the two evaluation points, and close the system with the natural boundary condition — three conditions for the three remaining coefficients.

Select an admissible cubic trial function. The essential condition $u(0)=0$ must be satisfied exactly by any admissible trial function, so the constant term is dropped: $$u_a(x)=a_1x+a_2x^{2}+a_3x^{3}$$ which gives $u_a(0)=0$ for any coefficients. Three unknowns $a_1,a_2,a_3$ remain, so three independent conditions are required.
Form the residual. Differentiating twice, $u_a''(x)=2a_2+6a_3x$, so substituting into the governing equation leaves $$\mathcal{R}(x)=EA\, u_a''(x)+cx = EA\big(2a_2+6a_3x\big)+cx$$ Note that $a_1$ does not appear in the residual at all: it is a rigid-stretch coefficient that only the boundary condition can determine.
State the least-squares criterion. The least-squares method minimises the squared residual, using the residual's own sensitivity as the weight function, $w_j = \partial\mathcal{R}/\partial a_j$. Evaluated discretely at the two points specified by the question, the object to minimise is $$S(a_2,a_3)=\sum_{k=1}^{2}\Big[\mathcal{R}(x_k)\Big]^{2},\qquad x_1=\tfrac14L,\;\; x_2=\tfrac34L$$ and the stationarity conditions are $$\frac{\partial S}{\partial a_2}=2\sum_k \mathcal{R}(x_k)\frac{\partial \mathcal{R}(x_k)}{\partial a_2}=0,\qquad \frac{\partial S}{\partial a_3}=2\sum_k \mathcal{R}(x_k)\frac{\partial \mathcal{R}(x_k)}{\partial a_3}=0$$ With $\partial\mathcal{R}/\partial a_2 = 2EA$ and $\partial\mathcal{R}/\partial a_3 = 6EAx$, these are two linear equations in $a_2$ and $a_3$.
Reduce the normal equations. Because there are exactly two free coefficients in the residual and exactly two evaluation points, the $2\times 2$ normal-equation system is non-singular and its solution drives $S$ to its absolute floor of zero, which happens only if the residual vanishes at both points: $$\mathcal{R}\big(\tfrac14L\big)=0 \quad\text{and}\quad \mathcal{R}\big(\tfrac34L\big)=0$$ Writing them out, $$2EA\, a_2 + \tfrac{3}{2}EA\, L\, a_3 = -\frac{cL}{4},\qquad 2EA\, a_2 + \tfrac{9}{2}EA\, L\, a_3 = -\frac{3cL}{4}$$
Solve for the residual coefficients. Subtracting the first equation from the second eliminates $a_2$ and leaves $3EA\, L\, a_3 = -cL/2$, hence $$a_3=-\frac{c}{6EA}$$ Back-substituting into the first equation, $2EA\, a_2 + \tfrac32 EAL\left(-\dfrac{c}{6EA}\right)=-\dfrac{cL}{4}$ gives $2EA\, a_2 - \dfrac{cL}{4} = -\dfrac{cL}{4}$ and therefore $$a_2=0$$
Close the system with the natural boundary condition. The traction-free end requires $EA\, u_a'(L)=0$, that is $a_1+2a_2L+3a_3L^{2}=0$. With $a_2=0$ and $a_3=-c/(6EA)$, $$a_1+3L^{2}\left(-\frac{c}{6EA}\right)=0 \quad\Longrightarrow\quad a_1=\frac{cL^{2}}{2EA}$$
Assemble the approximate solution. Collecting the three coefficients, $$\boxed{\, u_a(x)=\frac{cL^{2}}{2EA}\, x-\frac{c}{6EA}\, x^{3}=\frac{c}{6EA}\Big(3L^{2}x-x^{3}\Big)\,}$$ The tip displacement follows as $u_a(L)=\dfrac{c}{6EA}\big(3L^{3}-L^{3}\big)=\dfrac{cL^{3}}{3EA}$, and the axial force distribution is $N(x)=EA\, u_a'(x)=\tfrac{c}{2}\big(L^{2}-x^{2}\big)$, which correctly equals the total applied load $cL^{2}/2$ at the wall and zero at the free end.
Check the residual and compare with the exact solution. Substituting the coefficients back, $$\mathcal{R}(x)=EA\big(0+6x\big)\left(-\frac{c}{6EA}\right)+cx = -cx+cx = 0\quad\text{for all }x$$ so the residual vanishes identically, not merely at the two evaluation points, and $S_{\min}=0$. Integrating the governing equation directly, $u''=-cx/(EA)$ gives $u'=-cx^{2}/(2EA)+K_1$; the condition $u'(L)=0$ fixes $K_1=cL^{2}/(2EA)$, and $u(0)=0$ fixes the second constant at zero, so $$u_{\text{exact}}(x)=\frac{c}{6EA}\Big(3L^{2}x-x^{3}\Big)$$ The least-squares cubic is the analytical solution.

The exactness deserves a sentence of its own, because it is the pedagogical point of the question rather than a coincidence. The exact solution of $EA\, u''+cx=0$ is itself a cubic polynomial, so the chosen trial space contains the exact solution. Any consistent weighted-residual method — least squares, collocation, subdomain or Galerkin — must then recover it, because the only member of the trial space with an identically zero residual and the correct boundary conditions is the exact solution itself. The choice of evaluation points at $L/4$ and $3L/4$ is therefore immaterial here: any two distinct points give the same coefficients.

Two features of the least-squares method are nevertheless worth naming, since they are what a marker is looking for beyond the algebra. First, the continuous form of the criterion — minimising $S=\int_0^L\mathcal{R}^{2}\, dx$ rather than a discrete sum — produces exactly the same coefficients on this problem, and it is the form to use when there is no reason to privilege particular stations. Second, least squares always yields a symmetric, positive-definite coefficient matrix, since the normal equations are of the form $\mathbf{B}^{\mathsf T}\mathbf{B}\,\mathbf{a}=\mathbf{B}^{\mathsf T}\mathbf{f}$; this is its advantage over collocation, which gives an unsymmetric system and no control over the residual between the chosen points. The price is that least squares needs second derivatives of the trial functions in the governing operator, so it demands higher continuity of the approximation than the Galerkin weak form does — which is precisely why the finite element method is built on Galerkin rather than on least squares.

Question 5 — final results
Quantity	Result
Trial function	$u_a=a_1x+a_2x^{2}+a_3x^{3}$ (satisfies $u(0)=0$)
Residual	$\mathcal{R}(x)=EA(2a_2+6a_3x)+cx$
$a_1$	$\dfrac{cL^{2}}{2EA}$
$a_2$	$0$
$a_3$	$-\dfrac{c}{6EA}$
Approximate solution	$u_a(x)=\dfrac{c}{6EA}\big(3L^{2}x-x^{3}\big)$
Minimum of $S$	$0$ — residual vanishes for all $x$
Comparison with exact	Identical (the exact solution is a cubic)
Tip displacement	$u_a(L)=\dfrac{cL^{3}}{3EA}$
Axial force	$N(x)=\tfrac{c}{2}\big(L^{2}-x^{2}\big)$, $N(0)=cL^{2}/2$

Topic. The least-squares method as a member of the weighted-residual family, applied to a second-order axial bar equation with one essential and one natural boundary condition.

Key relations. Residual $\mathcal{R}=\mathcal{L}(u_a)-f$; weighted-residual statement $\int_\Omega w_i\mathcal{R}\, d\Omega=0$; least-squares weights $w_i=\partial\mathcal{R}/\partial a_i$ giving $\partial S/\partial a_i = 0$ with $S=\sum_k\mathcal{R}(x_k)^{2}$ or $\int_\Omega\mathcal{R}^{2}d\Omega$; number of conditions = number of undetermined coefficients.

Why this works. Every weighted-residual method makes the same bargain: an approximation cannot satisfy the differential equation everywhere, so it is required instead to satisfy a finite number of weighted averages of the error. Least squares chooses the weights so that the criterion is the squared norm of the residual, which makes the resulting matrix symmetric and positive definite and gives a genuine minimisation property. When the trial space happens to contain the exact solution the minimum is zero and every weighted-residual method collapses to the same answer; when it does not, least squares distributes the error smoothly rather than pinning it to zero at isolated stations, which is why it is the method of choice for first-order systems and for problems with no available variational principle.

Common pitfall. Using only the two evaluation-point conditions and stopping there, leaving a one-parameter family of solutions: a cubic trial function that already satisfies the essential condition still has three free coefficients, so the natural boundary condition $EA\, u'(L)=0$ is mandatory. The mirror-image error is to force the natural condition into the trial function as well, leaving too few coefficients for the two conditions. A third slip is differentiating $S$ with respect to $a_1$, which appears nowhere in the residual and yields the vacuous identity $0=0$.

Source. Reddy, An Introduction to the Finite Element Method, 4th ed., §2.5.3 (the least-squares method) and Example 2.5.2; Logan, 6th ed., §3.12 (residual methods — collocation, subdomain, least squares and Galerkin compared).

Question 5 of 7: Least-squares cubic solution for a bar under a linearly varying axial load

Question 5: Least-squares cubic solution for a bar under a linearly varying axial load (20 marks)