Let's solve each part step-by-step, as requested. --- ### (a) **Sketch the three quadratic shape functions on bi-unit domain.** Recall the shape functions: \[ N_1(x) = \frac{1}{2}x(x-1) \] \[ N_2(x) = 1 - x^2 \] \[ N_3(x) = \frac{1}{2}x(x+1) \] - For \( N_1 \): At \( x = -1 \): \( N_1(-1) = 1 \), at \( x = 0 \): \( N_1(0) = 0 \), at \( x = 1 \): \( N_1(1) = 0 \) - For \( N_2 \): At \( x = -1 \): \( N_2(-1) = 0 \), at \( x = 0 \): \( N_2(0) = 1 \), at \( x = 1 \): \( N_2(1) = 0 \) - For \( N_3 \): At \( x = -1 \): \( N_3(-1) = 0 \), at \( x = 0 \): \( N_3(0) = 0 \), at \( x = 1 \): \( N_3(1) = 1 \) **Sketch:** - \( N_1(x) \): Starts at 1 at \( x = -1 \), goes to 0 at \( x = 0 \), stays at 0 at \( x = 1 \) - \( N_2(x) \): 0 at \( x = -1 \), 1 at \( x = 0 \), 0 at \( x = 1 \) - \( N_3(x) \): 0 at \( x = -1 \), 0 at \( x = 0 \), 1 at \( x = 1 \) --- ### (b) **Evaluate the terms of the stiffness matrix** The stiffness matrix is: \[ K_{ij} = \int_{-1}^1 N'_{i}(x) N'_{j}(x) dx \] Compute the derivatives: - \( N_1'(x) = x - \frac{1}{2} \) - \( N_2'(x) = -2x \) - \( N_3'(x) = x + \frac{1}{2} \) Now, compute the integrals for each entry. Let's denote the nodes as \( i, j = 1, 2, 3 \): #### \( K_{11} \): \[ K_{11} = \int_{-1}^1 (x - \frac{1}{2})^2 dx \] \[ = \int_{-1}^1 (x^2 - x + \frac{1}{4}) dx \] \[ = \int_{-1}^1 x^2 dx - \int_{-1}^1 x dx + \int_{-1}^1 \frac{1}{4} dx \] \[ = \left[ \frac{x^3}{3} \right]_{-1}^1 - \left[ \frac{x^2}{2} \right]_{-1}^1 + \frac{1}{4} [x]_{-1}^1 \] \[ = \left(\frac{1}{3} - (-\frac{1}{3})\right) - ( \frac{1}{2} - \frac{1}{2}) + \frac{1}{4}(1 - (-1)) \] \[ = \frac{2}{3} + 0 + \frac{1}{2} = \frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6} \] #### \( K_{12} \): \[ K_{12} = \int_{-1}^1 (x - \frac{1}{2})(-2x) dx \] \[ = \int_{-1}^1 (-2x^2 + x) dx \] \[ = -2 \int_{-1}^1 x^2 dx + \int_{-1}^1 x dx \] \[ = -2 \cdot \frac{2}{3} + 0 = -\frac{4}{3} \] #### \( K_{13} \): \[ K_{13} = \int_{-1}^1 (x - \frac{1}{2})(x + \frac{1}{2}) dx \] \[ = \int_{-1}^1 (x^2 - \frac{1}{4}) dx \] \[ = \int_{-1}^1 x^2 dx - \int_{-1}^1 \frac{1}{4} dx \] \[ = \frac{2}{3} - \frac{1}{2} \] \[ = \frac{2}{3} - \frac{3}{6} = \frac{4-3}{6} = \frac{1}{6} \] #### \( K_{22} \): \[ K_{22} = \int_{-1}^1 (-2x)^2 dx = 4 \int_{-1}^1 x^2 dx = 4 \cdot \frac{2}{3} = \frac{8}{3} \] #### \( K_{23} \): \[ K_{23} = \int_{-1}^1 (-2x)(x+\frac{1}{2}) dx = \int_{-1}^1 (-2x^2 - x) dx = -2 \int_{-1}^1 x^2 dx - \int_{-1}^1 x dx = -\frac{4}{3} + 0 = -\frac{4}{3} \] #### \( K_{33} \): \[ K_{33} = \int_{-1}^1 (x+\frac{1}{2})^2 dx = \int_{-1}^1 (x^2 + x + \frac{1}{4}) dx = \int_{-1}^1 x^2 dx + \int_{-1}^1 x dx + \int_{-1}^1 \frac{1}{4} dx \] \[ = \frac{2}{3} + 0 + \frac{1}{2} = \frac{7}{6} \] **Stiffness Matrix:** \[ K = \begin{bmatrix} \frac{7}{6} & -\frac{4}{3} & \frac{1}{6} \\ -\frac{4}{3} & \frac{8}{3} & -\frac{4}{3} \\ \frac{1}{6} & -\frac{4}{3} & \frac{7}{6} \end{bmatrix} \] --- ### (c) **Evaluate the force vector in terms of \( f, h, g \).** The force vector components: \[ F_i = \int_{-1}^1 N_i(x) f(x) dx \] Also, due to the Neumann BC at \( x = -1 \), the weak form introduces a boundary term for the test function: \[ F_i = \int_{-1}^1 N_i(x) f(x) dx + N_i(-1) h \] Since only \( N_1(-1) = 1 \), \( N_2(-1) = 0 \), \( N_3(-1) = 0 \): So, \[ F_1 = \int_{-1}^1 N_1(x) f(x) dx + h \] \[ F_2 = \int_{-1}^1 N_2(x) f(x) dx \] \[ F_3 = \int_{-1}^1 N_3(x) f(x) dx \] --- ### (d) **Compare the finite element solution with the provided exact solution for each \( f \):** | Forcing Function | Exact Solution | |------------------|---------------| | \( f = 0 \) | \( u = g + (1-x)h \) | | \( f = p \) | \( u = g + (1-x)h + \frac{p}{2}(3-x^2-2x) \) | | \( f = qx \) | \( u = g + (1-x)h + \frac{q}{6}(3x-x^3-2) \) | - For \( f = 0 \): The FE solution **will be exact** if you use quadratic shape functions, since the solution is linear and quadratic shape functions can represent linear functions exactly. - For \( f = p \): The exact solution is quadratic in \( x \), so the quadratic FE solution **will be exact**. - For \( f = qx \): The exact solution is cubic in \( x \), quadratic shape functions **cannot capture cubic terms exactly**, so the FE solution will be approximate, not exact. --- ### (e) **What order polynomial shape functions are needed for \( f = 2x^2 - 4x \)?** Plug \( f = 2x^2 - 4x \) into the exact solution formula: \[ \int_{-1}^y f(z)dz = \int_{-1}^y (2z^2-4z)dz = [\frac{2}{3}z^3 - 2z^2]_{-1}^y \] The exact solution will involve terms up to quartic (\( x^4 \)). **Therefore, quartic (degree 4) shape functions are needed.** --- ### (f) **If \( f \) includes a trigonometric term (e.g., \( \sin(\pi x) \)), can you get the exact solution by increasing polynomial order?** **No**, you cannot, because polynomials cannot represent trigonometric functions exactly (trigonometric functions are not polynomials). Even increasing the polynomial order will only provide an approximation, not the exact solution. --- ## **Summary Table Answers:** | Forcing Function | Exact Solution | FE Solution with Quadratics | |------------------|-----------------------------------------------------|-----------------------------| | \( f = 0 \) | \( u = g + (1-x)h \) | Exact | | \( f = p \) | \( u = g + (1-x)h + \frac{p}{2}(3-x^2-2x) \) | Exact | | \( f = qx \) | \( u = g + (1-x)h + \frac{q}{6}(3x-x^3-2) \) | Approximate (not exact) | --- **If you need explicit calculations or further clarification on any step, let me know!**

Certainly! Let's go through each part step-by-step with detailed explanations, calculations, and final answers. --- ## Part (a): Sketch the three quadratic shape functions on the bi-unit domain ### Given shape functions: \[ N_1(x) = \frac{1}{2}x(x - 1) \] \[ N_2(x) = 1 - x^2 \] \[ N_3(x) = \frac{1}{2}x(x + 1) \] ### Step 1: Find key values at nodes - At \( x = -1 \): \[ N_1(-1) = \frac{1}{2}(-1)(-1 - 1) = \frac{1}{2}(-1)(-2) = 1 \] \[ N_2(-1) = 1 - (-1)^2 = 1 - 1 = 0 \] \[ N_3(-1) = \frac{1}{2}(-1)(-1 + 1) = \frac{1}{2}(-1)(0) = 0 \] - At \( x = 0 \): \[ N_1(0) = \frac{1}{2} \times 0 \times (0 - 1) = 0 \] \[ N_2(0) = 1 - 0^2 = 1 \] \[ N_3(0) = \frac{1}{2} \times 0 \times (0 + 1) = 0 \] - At \( x = 1 \): \[ N_1(1) = \frac{1}{2} \times 1 \times (1 - 1) = 0 \] \[ N_2(1) = 1 - 1^2 = 0 \] \[ N_3(1) = \frac{1}{2} \times 1 \times (1 + 1) = \frac{1}{2} \times 1 \times 2 = 1 \] ### **Sketch summary:** - \( N_1 \): peaks at 1 at \( x = -1 \), zero at \( x=0, 1 \) - \( N_2 \): peaks at 1 at \( x=0 \), zero at \( x = -1, 1 \) - \( N_3 \): peaks at 1 at \( x=1 \), zero at \( x=-1, 0 \) --- ## Part (b): Evaluate the terms of the stiffness matrix ### Step 1: Derivatives of shape functions \[ N_1(x) = \frac{1}{2}x(x - 1) \Rightarrow N_1' = \frac{1}{2}(2x - 1) = x - \frac{1}{2} \] \[ N_2(x) = 1 - x^2 \Rightarrow N_2' = -2x \] \[ N_3(x) = \frac{1}{2}x(x + 1) \Rightarrow N_3' = \frac{1}{2}(2x + 1) = x + \frac{1}{2} \] ### Step 2: Compute matrix entries \[ K_{ij} = \int_{-1}^1 N_i'(x) N_j'(x) dx \] --- ### Entry \( K_{11} \): \[ N_1' = x - \frac{1}{2} \] \[ K_{11} = \int_{-1}^1 (x - \frac{1}{2})^2 dx \] Expand: \[ (x - \frac{1}{2})^2 = x^2 - x + \frac{1}{4} \] Integrate term-by-term: \[ K_{11} = \int_{-1}^1 x^2 dx - \int_{-1}^1 x dx + \int_{-1}^1 \frac{1}{4} dx \] Calculate each: - \( \int_{-1}^1 x^2 dx = \left[\frac{x^3}{3}\right]_{-1}^1 = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3} \) - \( \int_{-1}^1 x dx = 0 \) (symmetric interval) - \( \int_{-1}^1 \frac{1}{4} dx = \frac{1}{4} \times (1 - (-1)) = \frac{1}{4} \times 2 = \frac{1}{2} \) **Sum:** \[ K_{11} = \frac{2}{3} + 0 + \frac{1}{2} = \frac{2}{3} + \frac{1}{2} \] Express with common denominator 6: \[ K_{11} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6} \] --- ### Entry \( K_{12} \): \[ N_1' = x - \frac{1}{2} \] \[ N_2' = -2x \] \[ K_{12} = \int_{-1}^1 (x - \frac{1}{2})(-2x) dx = -2 \int_{-1}^1 (x^2 - \frac{1}{2} x) dx \] Calculate: \[ \int_{-1}^1 x^2 dx = \frac{2}{3} \] \[ \int_{-1}^1 x dx = 0 \] \[ K_{12} = -2 \left( \frac{2}{3} - 0 \right) = -2 \times \frac{2}{3} = -\frac{4}{3} \] --- ### Entry \( K_{13} \): \[ N_1' = x - \frac{1}{2} \] \[ N_3' = x + \frac{1}{2} \] \[ K_{13} = \int_{-1}^1 (x - \frac{1}{2})(x + \frac{1}{2}) dx \] Note: \[ (x - \frac{1}{2})(x + \frac{1}{2}) = x^2 - (\frac{1}{2})^2 = x^2 - \frac{1}{4} \] Integrate: \[ K_{13} = \int_{-1}^1 x^2 dx - \int_{-1}^1 \frac{1}{4} dx = \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \] --- ### Entry \( K_{22} \): \[ N_2' = -2x \] \[ K_{22} = \int_{-1}^1 (-2x)^2 dx = 4 \int_{-1}^1 x^2 dx = 4 \times \frac{2}{3} = \frac{8}{3} \] --- ### Entry \( K_{23} \): \[ N_2' = -2x \] \[ N_3' = x + \frac{1}{2} \] \[ K_{23} = \int_{-1}^1 (-2x)(x + \frac{1}{2}) dx \] Expanding: \[ = -2 \int_{-1}^1 (x^2 + \frac{1}{2} x) dx = -2 \left( \int_{-1}^1 x^2 dx + \frac{1}{2} \int_{-1}^1 x dx \right) \] \[ = -2 \left( \frac{2}{3} + 0 \right) = -2 \times \frac{2}{3} = -\frac{4}{3} \] --- ### Entry \( K_{33} \): \[ N_3' = x + \frac{1}{2} \] \[ K_{33} = \int_{-1}^1 (x + \frac{1}{2})^2 dx = \int_{-1}^1 (x^2 + x + \frac{1}{4}) dx \] Calculate: \[ \int_{-1}^1 x^2 dx = \frac{2}{3} \] \[ \int_{-1}^1 x dx = 0 \] \[ \int_{-1}^1 \frac{1}{4} dx = \frac{1}{2} \] Sum: \[ K_{33} = \frac{2}{3} + 0 + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6} \] --- ## **Final stiffness matrix:** \[ \boxed{ K = \begin{bmatrix} \frac{7}{6} & -\frac{4}{3} & \frac{1}{6} \\ -\frac{4}{3} & \frac{8}{3} & -\frac{4}{3} \\ \frac{1}{6} & -\frac{4}{3} & \frac{7}{6} \end{bmatrix} } \] --- ## Part (c): Evaluate the force vector ### Step 1: Write the force vector components: \[ F_i = \int_{-1}^1 N_i(x) f(x) dx + N_i(-1) h \] - \( N_1(-1) = 1 \) - \( N_2(-1) = 0 \) - \( N_3(-1) = 0 \) ### Step 2: Express each component: \[ F_1 = \int_{-1}^1 N_1(x) f(x) dx + h \] \[ F_2 = \int_{-1}^1 N_2(x) f(x) dx \] \[ F_3 = \int_{-1}^1 N_3(x) f(x) dx \] where: \[ N_1(x) = \frac{1}{2} x (x - 1) \] \[ N_2(x) = 1 - x^2 \] \[ N_3(x) = \frac{1}{2} x (x + 1) \] --- ## Part (d): Comparing finite element solutions with exact solutions ### Exact solutions for different \( f \): | Forcing function | Exact solution \( u(x) \) | Comments | |-------------------|------------------------------|----------| | \( 0 \) | \( u = g + (1 - x) h \) | Linear, quadratic elements can represent exactly | | \( p \) (constant)| \( u = g + (1 - x) h + \frac{p}{2} (3 - x^2 - 2x) \) | Quadratic in \( x \), quadratic elements are exact | | \( qx \) (linear) | \( u = g + (1 - x) h + \frac{q}{6}(3x - x^3 - 2) \) | Cubic term, quadratic elements approximate | --- ## Part (e): Polynomial order needed for \( f = 2x^2 - 4x \) - When \( f = 2x^2 - 4x \), the integral for the solution involves terms like \( \int f(z) dz \), which produces cubic and quartic terms. - The solution will contain polynomial terms up to degree 4 (since integrating quadratic \( f \) twice produces quartic terms). **Therefore:** \[ \boxed{ \text{Polynomial degree 4 (quartic) shape functions are needed} } \] --- ## Part (f): Trigonometric terms in \( f \) - If \( f(z) = \sin(wz) \), the exact solution involves integrating sine functions, resulting in solutions with sine and cosine terms. - Polynomial shape functions **cannot** exactly represent sine functions. - Increasing polynomial order **won't** produce an exact solution; it will only improve approximation. - To exactly represent solutions with sine or cosine functions, you need **trigonometric basis functions** (not just polynomials). **Answer:** \[ \boxed{ \text{No, increasing polynomial order alone cannot yield the exact solution for } \sin(wz). } \] --- ## **Summary** - **Shape functions** are quadratic polynomials with known properties. - **Stiffness matrix** entries are obtained via integrals of derivatives. - **Force vector** involves boundary terms and integrals of \( N_i(x) f(x) \). - **Exact solutions** depend on the form of \( f \). For polynomial \( f \) up to degree 2, quadratic elements are sufficient; for higher degrees, higher polynomial order is needed. - **Trigonometric \( f \)** cannot be exactly captured by polynomial shape functions, regardless of order. --- Let me know if you'd like further clarifications or specific integral evaluations!

**(a)** Shape functions on \([-1,1]\): - \( N_1(x) = \frac{1}{2}x(x - 1) \) (peaks at \(x=-1\)) - \( N_2(x) = 1 - x^2 \) (peaks at \(x=0\)) - \( N_3(x) = \frac{1}{2}x(x + 1) \) (peaks at \(x=1\)) --- **(b)** Stiffness matrix \(K\): \[ \boxed{ K = \begin{bmatrix} \frac{7}{6} & -\frac{4}{3} & \frac{1}{6} \\ -\frac{4}{3} & \frac{8}{3} & -\frac{4}{3} \\ \frac{1}{6} & -\frac{4}{3} & \frac{7}{6} \end{bmatrix} } \] --- **(c)** Force vector components: \[ F_1 = \int_{-1}^1 N_1(x)f(x) dx + h,\quad F_2 = \int_{-1}^1 N_2(x)f(x) dx,\quad F_3 = \int_{-1}^1 N_3(x)f(x) dx \] --- **(d)** | Forcing \(f\) | Exact solution \(u(x)\) | FE with quadratic elements | |----------------|--------------------------|------------------------------| | \(0\) | \(g + (1 - x)h\) | Exact | | \(p\) (constant)| \(g + (1 - x)h + \frac{p}{2}(3 - x^2 - 2x)\)| Exact | | \(qx\) (linear)| \(g + (1 - x)h + \frac{q}{6}(3x - x^3 - 2)\)| Approximate | --- **(e)** To exactly solve for \(f = 2x^2 - 4x\), polynomial shape functions of **degree 4 (quartic)** are needed. --- **(f)** No, increasing polynomial order cannot exactly capture solutions involving \(\sin(wz)\). Trigonometric functions require basis functions that include sine/cosine, not just polynomials.