Discussion
The Green's function for the following boundary value problem \(\frac{d^{2}y}{dx^{2}} = f(x)\) with boundary conditions : y(0) = 0, y(1) = 1, is :
Answer: A
The Green's function is \(G(x, x_{0}) = x_{0}(x - 1)\) for \(0 \le x_{0} \le x\) and \(x(x_{0} - 1)\) for \(x \le x_{0} \le 1\).
The Green's function solves \(\frac{d^{2}G}{dx^{2}} = \delta(x - x_{0})\) with homogeneous conditions \(G = 0\) at \(x = 0\) and \(x = 1\) (the non-zero condition \(y(1) = 1\) is met by adding a solution of the homogeneous equation, here \(y = x\)). Away from \(x_{0}\), \(G\) is linear: \(Ax\) on the left so that \(G(0) = 0\), and \(B(x - 1)\) on the right so that \(G(1) = 0\).
Continuity at \(x = x_{0}\) and a unit jump in slope give \(A = x_{0} - 1\) and \(B = x_{0}\), which is the first option. It is symmetric in \(x\) and \(x_{0}\), as it should be. The quotient forms are not linear in \(x\), and \(x_{0}(x + 1)\) does not vanish at \(x = 1\).
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development