Greens Functions Questions and Answers

Answer: D

The Green's function is \(\frac{1}{2L}(x_{0} - L)(x + L)\) for \(x \le x_{0}\) and \(\frac{1}{2L}(x_{0} + L)(x - L)\) for \(x \ge x_{0}\).

On each side of \(x_{0}\), G'' = 0, so G is linear. To vanish at x = -L and x = L, take \(G = A(x + L)\) on the left and \(G = B(x - L)\) on the right. G must be continuous at \(x_{0}\), and integrating the delta function shows that its slope jumps by 1:

\[ \begin{aligned} A(x_{0} + L) &= B(x_{0} - L) \\ B - A &= 1 \end{aligned} \]

Solving, \(A = (x_{0} - L)/2L\) and \(B = (x_{0} + L)/2L\), which is the fourth option. The first option has the opposite overall sign (its slope falls by 1 at \(x_{0}\)), the second is not continuous at \(x_{0}\), and the third has no kink at \(x_{0}\) at all.

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 \(x = 1\).