Math Problem Statement
Solve -y''= cos 6 pi*x with fixed-fixed conditions u(0) = u(1) = 0 Use K3 to compute u[1],..., u[n]: u[i+1] +2u[t]+ u[i-1] = -h² cos 6(pi) (i) (h) with u[0] =u[n + 1] =0.
Solution
To solve the given boundary value problem:
with fixed-fixed boundary conditions , we will employ a discrete approximation using the finite difference method (FDM) as described by the equation:
[ u[i+1] + 2u[i] + u[i-1] = -h^2 \cos(6\pi i h) ] with .
Step-by-Step Solution
1. Discretization:
The domain is discretized into points, with a uniform grid size . For each grid point , we approximate the second derivative using finite differences, which leads to the central difference equation given above.
- The grid points are for .
- The boundary conditions imply .
2. System of Equations:
At each grid point , the finite difference equation can be written as:
[ u[i+1] + 2u[i] + u[i-1] = -h^2 \cos(6\pi i h) ]
This gives a system of linear equations for , which can be written in matrix form as:
where:
- is a tridiagonal matrix of size with on the diagonal and on the off-diagonals.
- is the vector of unknowns.
- is the right-hand side vector.
3. Matrix Setup:
The tridiagonal matrix for the system will look like this:
2 & -1 & 0 & 0 & \dots & 0 \\ -1 & 2 & -1 & 0 & \dots & 0 \\ 0 & -1 & 2 & -1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & -1 & 2 & -1 \\ 0 & 0 & \dots & 0 & -1 & 2 \end{bmatrix}$$ The right-hand side vector $$\mathbf{b}$$ is given by the discretized values of the function $$-h^2 \cos(6\pi x_i)$$. #### 4. **Numerical Solution**: You can now solve the linear system $$A \mathbf{u} = \mathbf{b}$$ using a numerical method such as Gaussian elimination, LU decomposition, or an iterative method like Jacobi or Gauss-Seidel. For example, if $$n = 4$$ (just for illustration purposes), the grid size is $$h = \frac{1}{5} = 0.2$$, and the system to solve would have the form: $$\begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} = \begin{bmatrix} -h^2 \cos(6\pi \cdot 0.2) \\ -h^2 \cos(6\pi \cdot 0.4) \\ -h^2 \cos(6\pi \cdot 0.6) \\ -h^2 \cos(6\pi \cdot 0.8) \end{bmatrix}$$ where $$h^2 = 0.04$$ and the right-hand side vector will contain the values of $$\cos(6\pi x_i)$$ evaluated at the grid points. #### 5. **Conclusion**: Once you have the system set up, you can solve for the unknowns $$u[1], u[2], \dots, u[n]$$, which approximate the solution to the boundary value problem at the discrete grid points. Would you like a specific example solution, or further details on the numerical method? --- ### Related Questions: 1. How do you solve tridiagonal systems efficiently using numerical methods? 2. What are the advantages and limitations of using finite difference methods? 3. Can the system be solved using iterative methods such as Jacobi or Gauss-Seidel? 4. How does the choice of $$n$$ affect the accuracy of the finite difference solution? 5. How would the solution change if we used a non-uniform grid? **Tip**: When solving boundary value problems numerically, always ensure that the grid size $$h$$ is small enough to capture the behavior of the solution, but not so small that the system becomes too computationally expensive.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Boundary Value Problems
Finite Difference Method
Tridiagonal Matrix
Cosine Function
Formulas
y'' = cos(6πx)
u[i+1] + 2u[i] + u[i-1] = -h² cos(6πih)
Matrix form: A * u = b
Theorems
Finite Difference Method for solving boundary value problems
Tridiagonal Matrix Systems
Numerical Linear Algebra
Suitable Grade Level
Undergraduate (Math/Engineering students)
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