Question

(a) Show that the equation $$ y^{\prime \prime}+a \delta(x) y+\lambda y=0 $$ with \(y(\pm \pi)=0\) and \(a\) real, has a set of eigenvalues \(\lambda\) satisfying $$ \tan (\pi \sqrt{\lambda})=\frac{2 \sqrt{\lambda}}{a} $$ (b) Investigate the conditions under which negative eigenvalues, \(\lambda=-\mu^{2}\) with \(\mu\) real, are possible.

Short answer

Eigenvalues \(\lambda\) satisfy \(\tan(\pi \sqrt{\lambda}) = \frac{2 \sqrt{\lambda}}{a}\). Negative eigenvalues ( \(\lambda = -\mu^2 \) real) are possible if is negative.

Step-by-step solution

Title - Rewrite the Differential Equation

Rewrite the differential equation considering the Dirac delta function, \[ y'' + a \delta(x) y + \lambda y = 0. \]

Title - Solve the Differential Equation Separately for x ≠ 0

Solve the differential equation for regions where \(x eq 0\): \[ y'' + \lambda y = 0. \]The general solution is \[ y = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} x). \]

Title - Apply Boundary Condition at \(y(\pm \pi) = 0\)

Applying the boundary condition \( y(\pm \pi) = 0 \): \[ A \sin(\sqrt{\lambda} \pi) + B \cos(\sqrt{\lambda} \pi) = 0. \]This implies either \(A = 0\) or \(\sin(\sqrt{\lambda} \pi) = 0\).

Title - Consider Continuity and Jump Conditions for Dirac Delta

Introduce the effect of the \(\delta(x)\) at \(x = 0\). The condition is \[ y'(0^+) - y'(0^-) = -a y(0). \]Using the general solution, the derivative jump condition gives \[ \sqrt{\lambda} \cos(\sqrt{\lambda} * 0) + \sqrt{\lambda} \cos(\sqrt{\lambda} * 0) = a \sin(0). \]

Title - Derive the Eigenvalue Condition

Combining the boundary conditions and solving, we get the condition for \(\lambda\): \[ \tan(\pi \sqrt{\lambda}) = \frac{2 \sqrt{\lambda}}{a}. \]

Title - Investigate Condition for Negative Eigenvalue

Set \( \lambda = -\mu^2\) with \(\mu\) real, giving the equation \[ \tan(\pi i\mu) = \frac{2 i\mu}{a}, \]which simplifies to \[ \tanh(\pi \mu) = -\frac{2 \mu}{a}. \]

Title - Condition Explanation for Negative Eigenvalue

For negative eigenvalues (\(\mu\) real) to occur, \(\tanh(\pi \mu) = -\frac{2 \mu}{a}.\)Given \(\tanh(\pi \mu)\) is always positive for ______ values of \(\mu\), negative eigenvalues are possible when \(a\) is negative.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They describe the rate of change of a quantity and are used in various fields like physics, engineering, and biology.
In this exercise, we are dealing with the differential equation:
\[ y^{\prime \prime}+a \delta(x) y+\lambda y=0\]
This equation includes a second derivative \(y''\), a Dirac delta function \(\delta(x)\), and an eigenvalue parameter \(\lambda\). The task is to solve this equation under the given boundary conditions \(y(\pm \pi)=0\) and find the eigenvalues \(\lambda\).
The solution approach involves solving the differential equation separately for regions excluding \(x = 0\), where the Dirac delta function is non-zero.

Dirac Delta Function

The Dirac delta function, \(\delta(x)\), is a special function used to model an idealized point charge or mass at a specific location - in this case, at \(x = 0\). It is not a function in the traditional sense but rather a distribution that is zero everywhere except at the origin.
Its key property is:
\[ \int_{-\infty}^{\infty} f(x) \delta(x) \, dx = f(0)\]
This property allows it to pick out the value of the function at a specific point. In our differential equation:
\[y^{\prime \prime}+a \delta(x)y+\lambda y=0,\]
the delta function introduces a discontinuity in the derivative of \(y\) at \(x = 0\). This contributes a 'jump' condition where the derivative changes abruptly, which needs to be considered while solving the equation. Specifically, we impose:
\[ y'(0^+) - y'(0^-) = -a y(0)\]
to account for this.

Boundary Conditions

Boundary conditions are constraints that are required to solve differential equations. They define the solution at specific points, which helps in determining the constants in the general solution.
In our problem, we are given:\[y(\pm \pi) = 0\]
This means that the function \(y\) must be zero at \(x = \pi\) and \(x = -\pi\). Such conditions are crucial in eigenvalue problems as they influence the values the eigenvalues \(\lambda\) can take. Applying these conditions, we find:
\[A \sin(\sqrt{\lambda} \pi) + B \cos(\sqrt{\lambda} \pi) = 0\]
This equation helps in determining the specific forms of solutions \(A\) and \(B\) given the eigenvalue \(\lambda\).

Negative Eigenvalues

Eigenvalues are specific values for which a differential equation has non-trivial solutions. In some cases, these values can be negative. For our problem, we need to investigate the conditions under which negative eigenvalues, \(\lambda = -\mu^2\) with \(\mu\) real, can exist.
If \(\lambda = -\mu^2\), the equation becomes:
\[\tanh(\pi \mu) = -\frac{2 \mu}{a} \]
The hyperbolic tangent function \(\tanh(\pi \mu)\) is always positive for real \(\mu\). Thus, for the equation to hold, \( a \) must be negative. This means that for negative eigenvalues to exist, the parameter \(a\) in our original equation must be negative, ensuring that \( \mu \) remains real, creating valid negative eigenvalues.