Let's break down the **step-by-step solution** for the PDE using **central differences**. --- ### **Given PDE:** \[ u_{xx} + y u_{yy} - 4u = 0 \] where \(u_{xx}\) and \(u_{yy}\) are the second partial derivatives of \(u\) with respect to \(x\) and \(y\), respectively. --- ### **Boundary & Initial Conditions:** 1. \(u_y = 0\) on \(y = \pm 0.8\) 2. \(u = 4\) on the circle \(x^2 + y^2 = 0.16\) 3. \(u = 1\) on the straight lines (not fully specified, but use as boundary wherever straight lines are indicated) --- ### **Grid Spacing:** - \(h = 0.5\) (in \(x\)-direction) - \(k = 0.6\) (in \(y\)-direction) Let \(u_{i,j}\) denote the value of \(u\) at grid point \((x_i, y_j)\). --- ## **1. Central Difference Approximations** - Second derivative w.r.t. \(x\): \[ u_{xx} \approx \frac{u_{i+1, j} - 2u_{i, j} + u_{i-1, j}}{h^2} \] - Second derivative w.r.t. \(y\): \[ u_{yy} \approx \frac{u_{i, j+1} - 2u_{i, j} + u_{i, j-1}}{k^2} \] --- ## **2. Discretized PDE** Plug these into the PDE: \[ \frac{u_{i+1, j} - 2u_{i, j} + u_{i-1, j}}{h^2} + y_j \frac{u_{i, j+1} - 2u_{i, j} + u_{i, j-1}}{k^2} - 4u_{i, j} = 0 \] --- ### **Substitute \(h = 0.5\), \(k = 0.6\):** - \(h^2 = 0.25\) - \(k^2 = 0.36\) So, \[ \frac{u_{i+1, j} - 2u_{i, j} + u_{i-1, j}}{0.25} + y_j \frac{u_{i, j+1} - 2u_{i, j} + u_{i, j-1}}{0.36} - 4u_{i, j} = 0 \] --- ### **3. Rearranged Discrete Equation:** \[ 4 \big(u_{i+1, j} - 2u_{i, j} + u_{i-1, j}\big) + \frac{y_j}{0.36} \big(u_{i, j+1} - 2u_{i, j} + u_{i, j-1}\big) - 4u_{i, j} = 0 \] Or fully expanded: \[ 4u_{i+1, j} + 4u_{i-1, j} + \frac{y_j}{0.36} u_{i, j+1} + \frac{y_j}{0.36} u_{i, j-1} - \left[8 + \frac{2y_j}{0.36} + 4 \right] u_{i, j} = 0 \] Or, simplified: \[ 4u_{i+1, j} + 4u_{i-1, j} + \frac{y_j}{0.36} (u_{i, j+1} + u_{i, j-1}) - (12 + \frac{2y_j}{0.36}) u_{i, j} = 0 \] --- ## **4. Boundary Conditions in Discrete Form:** ### (a) \(u_y = 0\) at \(y = \pm 0.8\) \[ u_y \approx \frac{u_{i, j+1} - u_{i, j-1}}{2k} = 0 \implies u_{i, j+1} = u_{i, j-1} \] So at \(y = 0.8\) or \(-0.8\), enforce this condition. ### (b) \(u = 4\) on \(x^2 + y^2 = 0.16\) Any grid point on this circle: \(u_{i, j} = 4\) ### (c) \(u = 1\) on straight lines Any grid point on these lines: \(u_{i, j} = 1\) --- ## **5. **Final Discretized System to Solve:** For all interior points: \[ 4u_{i+1, j} + 4u_{i-1, j} + \frac{y_j}{0.36} (u_{i, j+1} + u_{i, j-1}) - (12 + \frac{2y_j}{0.36}) u_{i, j} = 0 \] **Boundary points:** apply the corresponding boundary conditions as described above. --- ### **Summary Table** | Condition | Discretization | |----------------------------|-------------------------------------------------------------------| | Interior Points | See equation above | | \(u_y = 0\) at \(y = \pm0.8\) | \(u_{i, j+1} = u_{i, j-1}\) | | \(u = 4\) on \(x^2 + y^2 = 0.16\) | \(u_{i, j} = 4\) | | \(u = 1\) on straight lines | \(u_{i, j} = 1\) | --- ## **Final Answer** ### **The equations to solve are:** \[ 4u_{i+1, j} + 4u_{i-1, j} + \frac{y_j}{0.36} (u_{i, j+1} + u_{i, j-1}) - (12 + \frac{2y_j}{0.36}) u_{i, j} = 0 \] for all interior grid points, **with the given boundary conditions**: - \(\frac{u_{i, j+1} - u_{i, j-1}}{2k} = 0\) at \(y = \pm 0.8\) - \(u_{i, j} = 4\) on \(x^2 + y^2 = 0.16\) - \(u_{i, j} = 1\) on straight lines as specified. --- **Let me know if you want the equations at specific points or a sample grid!**