1. **Problem statement:** We have the boundary value problem (BVP) $$u''''(x) = f(x),$$ with boundary conditions $$u(0)=0,$$ $$u'(0)=0,$$ $$u''(1)=0,$$ and $$u'''(1)=0.$$ We want to discretize this using finite differences: central differences for the interior points, forward differences at the left boundary ($x=0$), and backward differences at the right boundary ($x=1$), without introducing pseudo mesh points. 2. **Discretization setup:** Let the mesh points be $$x_i = ih$$ for $$i=0,1,\ldots,N$$ with mesh size $$h=\frac{1}{N}$$. 3. **Central difference for interior points:** The fourth derivative at interior points $$x_i$$ ($i=2,\ldots,N-2$) is approximated by the central difference formula: $$ u''''(x_i) \approx \frac{u_{i-2} - 4u_{i-1} + 6u_i - 4u_{i+1} + u_{i+2}}{h^4} = f_i. $$ 4. **Left boundary conditions using forward differences:** - For $$u(0)=0$$: $$u_0=0.$$ - For $$u'(0)=0$$, approximate first derivative forward difference: $$ u'(0) \approx \frac{-3u_0 + 4u_1 - u_2}{2h} = 0 \implies -3u_0 + 4u_1 - u_2 = 0. $$ Since $$u_0=0$$, this reduces to: $$4u_1 - u_2 = 0.$$ 5. **Right boundary conditions using backward differences:** - For $$u''(1)=0$$, approximate second derivative backward difference: $$ u''(1) \approx \frac{u_N - 2u_{N-1} + u_{N-2}}{h^2} = 0. $$ - For $$u'''(1)=0$$, approximate third derivative backward difference: $$ u'''(1) \approx \frac{-u_N + 3u_{N-1} - 3u_{N-2} + u_{N-3}}{h^3} = 0. $$ 6. **Matrix form:** Collecting all equations for $$u_1, u_2, \ldots, u_N$$ (note $$u_0=0$$ known), we get a linear system: $$ A \mathbf{u} = \mathbf{f}, $$ where $$\mathbf{u} = [u_1, u_2, \ldots, u_N]^T$$ and $$\mathbf{f} = [f_1, f_2, \ldots, f_{N-1}, 0, 0]^T$$ (last two zeros correspond to boundary conditions at $$x=1$$). The matrix $$A$$ is banded with the following structure: - Row 1 (from $$4u_1 - u_2=0$$): $$[4, -1, 0, \ldots, 0]$$ - Rows 2 to $$N-2$$ (central difference): $$[1, -4, 6, -4, 1] / h^4$$ pattern shifted along the diagonal - Row $$N-1$$ (second derivative BC): $$[\ldots, 1, -2, 1]/h^2$$ - Row $$N$$ (third derivative BC): $$[\ldots, -1, 3, -3, 1]/h^3$$ 7. **Error dependence on mesh size $$h$$:** - Central difference for fourth derivative is $$O(h^2)$$ accurate. - Forward and backward differences used for boundary conditions are first order accurate $$O(h)$$. Since the boundary approximations dominate the global error, the overall error behaves like $$O(h)$$. Therefore, $$m=1$$. **Final answers:** (a) The system can be written as $$A\mathbf{u} = \mathbf{f}$$ with $$A$$ constructed as above using forward differences at left boundary, central differences inside, and backward differences at right boundary. (b) The maximum absolute error depends on mesh size $$h$$ as $$O(h^1)$$, so $$m=1$$.