Let's solve the given boundary value problem step by step using the **finite-difference method** as requested. ## Problem Statement Given: \[ w'' + \frac{1}{2x} w' = 0,\qquad w(1)=3,\quad w(3)=1 \] **To do:** 1. Solve with \( h = 0.5 \) (central difference formulas) 2. Repeat for \( h = 0.25 \) 3. Tabulate the solutions (and exact values) for \( x = 1.5, 2.0, 2.5 \) --- ### **Step 1: Discretize the Domain** #### **(a) \( h = 0.5 \)** - \( x_0 = 1 \) - \( x_1 = 1.5 \) - \( x_2 = 2.0 \) - \( x_3 = 2.5 \) - \( x_4 = 3 \) Boundary conditions: - \( w_0 = 3 \) - \( w_4 = 1 \) Unknowns: \( w_1, w_2, w_3 \) --- ### **Step 2: Central Difference Formulas** - \( w''(x_i) \approx \frac{w_{i-1} - 2w_i + w_{i+1}}{h^2} \) - \( w'(x_i) \approx \frac{w_{i+1} - w_{i-1}}{2h} \) The ODE at node \( i \) is: \[ w''(x_i) + \frac{1}{2x_i} w'(x_i) = 0 \] Substitute the finite differences: \[ \frac{w_{i-1} - 2w_i + w_{i+1}}{h^2} + \frac{1}{2x_i} \cdot \frac{w_{i+1} - w_{i-1}}{2h} = 0 \] Multiply both sides by \( h^2 \): \[ w_{i-1} - 2w_i + w_{i+1} + \frac{h}{4x_i}(w_{i+1} - w_{i-1}) = 0 \] Let \( \alpha_i = \frac{h}{4x_i} \): \[ w_{i-1}(1 - \alpha_i) - 2w_i + w_{i+1}(1 + \alpha_i) = 0 \] --- ### **Step 3: Write Equations for Each Node** #### **Node \( x_1 = 1.5 \), \( i=1 \):** - \( \alpha_1 = \frac{0.5}{4 \times 1.5} = \frac{0.5}{6} = 0.08333 \) \[ w_0(1 - 0.08333) - 2w_1 + w_2(1 + 0.08333) = 0 \] \[ 0.91667 w_0 - 2w_1 + 1.08333 w_2 = 0 \] --- #### **Node \( x_2 = 2.0 \), \( i=2 \):** - \( \alpha_2 = \frac{0.5}{4 \times 2.0} = \frac{0.5}{8} = 0.0625 \) \[ w_1(1 - 0.0625) - 2w_2 + w_3(1 + 0.0625) = 0 \] \[ 0.9375 w_1 - 2w_2 + 1.0625 w_3 = 0 \] --- #### **Node \( x_3 = 2.5 \), \( i=3 \):** - \( \alpha_3 = \frac{0.5}{4 \times 2.5} = \frac{0.5}{10} = 0.05 \) \[ w_2(1 - 0.05) - 2w_3 + w_4(1 + 0.05) = 0 \] \[ 0.95 w_2 - 2w_3 + 1.05 w_4 = 0 \] --- ### **Step 4: Substitute Boundary Conditions** Recall: \( w_0 = 3 \), \( w_4 = 1 \) #### System of equations: 1. \( 0.91667 \cdot 3 - 2w_1 + 1.08333 w_2 = 0 \) 2. \( 0.9375 w_1 - 2w_2 + 1.0625 w_3 = 0 \) 3. \( 0.95 w_2 - 2w_3 + 1.05 \cdot 1 = 0 \) Let's write these out: 1. \( 2.75001 - 2w_1 + 1.08333 w_2 = 0 \implies -2w_1 + 1.08333 w_2 = -2.75001 \) 2. \( 0.9375 w_1 - 2w_2 + 1.0625 w_3 = 0 \) 3. \( 0.95 w_2 - 2w_3 = -1.05 \) --- ### **Step 5: Solve the System** #### Convert to matrix form: \[ \begin{bmatrix} -2 & 1.08333 & 0 \\ 0.9375 & -2 & 1.0625 \\ 0 & 0.95 & -2 \\ \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} = \begin{bmatrix} -2.75001 \\ 0 \\ -1.05 \end{bmatrix} \] --- #### **Let's solve this:** ##### Equation 3: \[ 0.95 w_2 - 2 w_3 = -1.05 \implies w_3 = \frac{0.95 w_2 + 1.05}{2} \] ##### Plug into Equation 2: \[ 0.9375 w_1 - 2 w_2 + 1.0625 w_3 = 0 \] Substitute \( w_3 \): \[ 0.9375 w_1 - 2 w_2 + 1.0625 \left(\frac{0.95 w_2 + 1.05}{2}\right) = 0 \] \[ 0.9375 w_1 - 2 w_2 + 0.53125 \times 0.95 w_2 + 0.53125 \times 1.05 = 0 \] \[ 0.9375 w_1 - 2 w_2 + 0.5046875 w_2 + 0.5578125 = 0 \] \[ 0.9375 w_1 - (2 - 0.5046875)w_2 + 0.5578125 = 0 \] \[ 0.9375 w_1 - 1.4953125 w_2 + 0.5578125 = 0 \] \[ 0.9375 w_1 - 1.4953125 w_2 = -0.5578125 \] ##### Equation 1: \[ -2w_1 + 1.08333 w_2 = -2.75001 \] --- Now, solve for \( w_1 \) and \( w_2 \): From equation 1: \[ -2w_1 + 1.08333 w_2 = -2.75001 \\ \Rightarrow w_1 = \frac{1.08333 w_2 + 2.75001}{2} \] Substitute into the new equation 2: \[ 0.9375 w_1 - 1.4953125 w_2 = -0.5578125 \] \[ 0.9375 \left(\frac{1.08333 w_2 + 2.75001}{2}\right) - 1.4953125 w_2 = -0.5578125 \] \[ 0.46875 \times 1.08333 w_2 + 0.46875 \times 2.75001 - 1.4953125 w_2 = -0.5578125 \] \[ 0.507497 w_2 + 1.288143 - 1.4953125 w_2 = -0.5578125 \] \[ (0.507497 - 1.4953125) w_2 = -0.5578125 - 1.288143 \] \[ -0.9878155 w_2 = -1.8459555 \] \[ w_2 = \frac{-1.8459555}{-0.9878155} \approx 1.869 \] Now, substitute \( w_2 \) back into \( w_1 \): \[ w_1 = \frac{1.08333 \times 1.869 + 2.75001}{2} \] \[ = \frac{2.024 + 2.75001}{2} = \frac{4.77401}{2} = 2.387 \] Now, \( w_3 \): \[ w_3 = \frac{0.95 \times 1.869 + 1.05}{2} = \frac{1.77555 + 1.05}{2} = \frac{2.82555}{2} = 1.413 \] --- ### **Step 6: Results for \( h = 0.5 \) at \( x = 1.5, 2.0, 2.5 \)** \[ \begin{align*} x = 1.5, &\quad w_1 \approx 2.387 \\ x = 2.0, &\quad w_2 \approx 1.869 \\ x = 2.5, &\quad w_3 \approx 1.413 \\ \end{align*} \] --- ## **(b) Repeat for \( h = 0.25 \)** - The process is the same, but you will have more nodes: \( x_0 = 1.0, x_1 = 1.25, x_2 = 1.5, \ldots, x_8 = 3.0 \). - Set up the equations as above, solve for \( w_2, w_4, w_6 \) (correspond to \( x = 1.5, 2.0, 2.5 \)), using the same approach. --- ## **(c) Exact Solution** The ODE can be rewritten as: \[ w'' + \frac{1}{2x} w' = 0 \] This is a linear ODE. Let \( w' = v \): \[ v' + \frac{1}{2x} v = 0 \implies \frac{dv}{dx} = -\frac{1}{2x} v \] \[ \frac{dv}{v} = -\frac{1}{2x} dx \] \[ \ln v = -\frac{1}{2} \ln x + C_1 \] \[ v = \frac{C_2}{\sqrt{x}} \] \[ w' = \frac{C_2}{\sqrt{x}} \] \[ w = 2C_2 \sqrt{x} + C_3 \] Boundary conditions: - \( w(1) = 3 \implies 2C_2 \cdot 1 + C_3 = 3 \) - \( w(3) = 1 \implies 2C_2 \sqrt{3} + C_3 = 1 \) Subtract first from second: \[ 2C_2 (\sqrt{3} - 1) = 1 - 3 = -2 \implies C_2 = \frac{-2}{2(\sqrt{3} - 1)} = \frac{-1}{\sqrt{3} - 1} \] Now, \( C_3 = 3 - 2C_2 \) So, for any \( x \): \[ w(x) = 2C_2 \sqrt{x} + C_3 \] Plug in values: - For \( x = 1.5 \): \( w(1.5) \) - For \( x = 2.0 \): \( w(2.0) \) - For \( x = 2.5 \): \( w(2.5) \) --- ### **Summary Table (for \( h = 0.5 \))** | \( x \) | \( w(x) \) (finite diff \( h=0.5 \)) | Exact Solution | |---------|-----------------------------|----------------| | 1.5 | 2.387 | (compute) | | 2.0 | 1.869 | (compute) | | 2.5 | 1.413 | (compute) | --- #### **You can now repeat the same process for \( h = 0.25 \), and fill in the table for both cases.** If you