Table 5 coefficients used to capture the solution with the finite element approximation for the linear patch test

Linear \(\phi (x)\) and linear F(x)

The analytical solution cannot be represented. Incompatible with blending elements.

\(\left\{ \begin{array}{ll} a_1 &{}= 0\\ a_2 &{}= \bar{u}\\ b_1 &{}= b_2 = \frac{(E_2 - E_1)\bar{u}}{2(E_1 L + (E_2 - E_1)x_0)}\end{array}\right. \)

The analytical solution cannot be represented. Incompatible with blending elements.

Quadratic \(\phi (x)\) and quadratic F(x)

The analytical solution cannot be represented. Incompatible with blending elements.

\(\bullet \) if \(L/2 \ge x_0 \left\{ \begin{array}{ll}a_1 &{}= 0\\ a_2 &{}= \frac{\bar{u}}{2} + \frac{(E_2 - E_1)x_0\bar{u}}{2(E_1 L + (E_2 - E_1)x_0)} \\ a_3 &{}= \bar{u} \\ b_1 &{}= b_2 = b_3 = \frac{(E_2 - E_1)\bar{u}}{2(E_1 L + (E_2 - E_1)x_0)} \end{array}\right. \)

The analytical solution cannot be represented. Incompatible with blending elements.

\(\bullet \) if \(L/2 \le x_0 \left\{ \begin{aligned} a_1&= 0\\ a_2&= \frac{\bar{u}}{2} + \frac{(E_2 - E_1)(L - x_0)\bar{u}}{2(E_1 L + (E_2 - E_1)x_0)} \\ a_3&= \bar{u} \\ b_1&= b_2 = b_3 = \frac{(E_2 - E_1)\bar{u}}{2(E_1 L + (E_2 - E_1)x_0)} \end{aligned} \right. \)

Quadratic \(\phi (x)\) and linear F(x)

The analytical solution cannot be represented. Incompatible with blending elements.

\(\left\{ \begin{aligned} a_1&= 0\\ a_2&= \frac{(E_2 - E_1)x_0\bar{u} + E_1L\bar{u}}{2(E_1 L + (E_2 - E_1)x_0)} = \frac{\bar{u}}{2}\\ a_3&= \frac{(E_2 - E_1)x_0\bar{u} + E_1L\bar{u}}{(E_1 L + (E_2 - E_1)x_0)}= \bar{u}\\ b_1&= b_2 = b_3 = \frac{(E_2 - E_1)\bar{u}}{2(E_1 L + (E_2 - E_1)x_0)} \end{aligned}\right. \)

The analytical solution cannot be represented. Incompatible with blending elements.