Greens Functions Questions and Answers

Q: Consider the differential operator\[ \begin{aligned} L = a\frac{d^{3}}{dx^{3}} &+ b\frac{d^{2}}{dx^{2}} \\ &+ c\frac{d}{dx} \end{aligned} \]acting on the space of functions \(\{u_{i}(x) \,|\, 1 \le x \le 2\}\) satisfying the boundary conditions \(u_{i}(1) = 0 = u_{i}(2)\). If \(L = L^{\dagger}\), then :

The operator is self-adjoint for \(a = i, b = 1, c = i\).Integrating by parts with the boundary terms dropped, each derivative changes sign when moved onto the other function, and the inner product conjugates the coefficient:\[ \begin{aligned} \left(a\frac{d^{3}}{dx^{3}}\right)^{\dagger} &= -a^{*}\frac{d^{3}}{dx^{3}} \\ \left(b\frac{d^{2}}{dx^{2}}\right)^{\dagger} &= b^{*}\frac{d^{2}}{dx^{2}} \\ \left(c\frac{d}{dx}\right)^{\dagger} &= -c^{*}\frac{d}{dx} \end{aligned} \]\(L = L^{\dagger}\) therefore needs \(a = -a^{*}\) and \(c = -c^{*}\) (purely imaginary) and \(b = b^{*}\) (real). This is the familiar fact that \(-i\,d/dx\) is Hermitian while \(d/dx\) is not.Only \(a = i, b = 1, c = i\) meets all three conditions. Real values \(a = \pm 1\) fail the first condition, and \(c = 1\) fails the third.

The Green's function G(x, \(x_{0}\)) satisfying the equation \(\frac{d^{2}G(x, x_{0})}{dx^{2}} = \delta(x - x_{0})\) with boundary conditions G(-L, \(x_{0}\)) = G(L, \(x_{0}\)) = 0 is :

A.\(\frac{1}{2L}(L - x_{0})(x + L)\) for \(-L \le x \le x_{0}\); \(\frac{1}{2L}(x_{0} + L)(L - x)\) for \(x_{0} \le x \le L\)
B.\(\frac{1}{2L}(x_{0} + L)(x + L)\) for \(-L \le x \le x_{0}\); \(\frac{1}{2L}(x_{0} - L)(x - L)\) for \(x_{0} \le x \le L\)
C.\(\frac{1}{2L}(x - L)(x + L)\) for \(-L \le x \le L\)
D.\(\frac{1}{2L}(x_{0} - L)(x + L)\) for \(-L \le x \le x_{0}\); \(\frac{1}{2L}(x_{0} + L)(x - L)\) for \(x_{0} \le x \le L\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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.

Discuss in Forum

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 :

A.\(G(x, x_{0}) = x_{0}(x - 1)\) for \(0 \le x_{0} \le x\); \(x(x_{0} - 1)\) for \(x \le x_{0} \le 1\)
B.\(G(x, x_{0}) = x_{0}/(x - 1)\) for \(0 \le x_{0} \le x\); \(x/(x_{0} - 1)\) for \(x \le x_{0} \le 1\)
C.\(G(x, x_{0}) = (x_{0} - 1)/x\) for \(0 \le x_{0} \le x\); \((x - 1)/x_{0}\) for \(x \le x_{0} \le 1\)
D.\(G(x, x_{0}) = x_{0}(x + 1)\) for \(0 \le x_{0} \le x\); \(x(x_{0} + 1)\) for \(x \le x_{0} \le 1\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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 \(x = 1\).

Discuss in Forum

Consider the differential operator

\[ \begin{aligned} L = a\frac{d^{3}}{dx^{3}} &+ b\frac{d^{2}}{dx^{2}} \\ &+ c\frac{d}{dx} \end{aligned} \]

acting on the space of functions \(\{u_{i}(x) \,|\, 1 \le x \le 2\}\) satisfying the boundary conditions \(u_{i}(1) = 0 = u_{i}(2)\). If \(L = L^{\dagger}\), then :

A.\(a = -1, b = 1, c = i\)
B.\(a = i, b = 1, c = i\)
C.\(a = 1, b = 1, c = i\)
D.\(a = -1, b = 1, c = 1\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The operator is self-adjoint for \(a = i, b = 1, c = i\).

Integrating by parts with the boundary terms dropped, each derivative changes sign when moved onto the other function, and the inner product conjugates the coefficient:

\[ \begin{aligned} \left(a\frac{d^{3}}{dx^{3}}\right)^{\dagger} &= -a^{*}\frac{d^{3}}{dx^{3}} \\ \left(b\frac{d^{2}}{dx^{2}}\right)^{\dagger} &= b^{*}\frac{d^{2}}{dx^{2}} \\ \left(c\frac{d}{dx}\right)^{\dagger} &= -c^{*}\frac{d}{dx} \end{aligned} \]

\(L = L^{\dagger}\) therefore needs \(a = -a^{*}\) and \(c = -c^{*}\) (purely imaginary) and \(b = b^{*}\) (real). This is the familiar fact that \(-i\,d/dx\) is Hermitian while \(d/dx\) is not.

Only \(a = i, b = 1, c = i\) meets all three conditions. Real values \(a = \pm 1\) fail the first condition, and \(c = 1\) fails the third.

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