Question

Consider the partial differential equation \(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0\) where \(x, y\) are real. If \(f(x, y) = a(x)b(y)\), where \(a(x)\) and \(b(y)\) are real functions, which one of the following statements can be true?

• A. \(a(x)\) is a periodic function and \(b(y)\) is a linear function
• B. both \(a(x)\) and \(b(y)\) are exponential functions
• C. \(a(x)\) is a periodic function and \(b(y)\) is an exponential function
• D. both \(a(x)\) and \(b(y)\) are periodic functions

Hint:

When solving separable PDEs like the Laplace equation, the solutions for each variable can be trigonometric or exponential based on the sign of the separation constant.

Answer 1

The Correct Option is C

Step 1: Analyze the equation.
The given partial differential equation is the 2D Laplace equation. To solve for \(f(x, y) = a(x)b(y)\), we separate the variables and write it as: \[ \frac{d^2 a(x)}{dx^2} + \frac{d^2 b(y)}{dy^2} = 0. \] This implies that each term must be equal to a constant, say \(k\), leading to two ordinary differential equations: \[ \frac{d^2 a(x)}{dx^2} = k, \quad \frac{d^2 b(y)}{dy^2} = -k. \] Step 2: Solve the equations.
- If \(a(x)\) is periodic, then \(k = -\lambda^2\), leading to a sinusoidal solution for \(a(x)\). - If \(b(y)\) is exponential, then it corresponds to \(k = \lambda^2\), leading to an exponential solution for \(b(y)\). Thus, option (C) is valid. Step 3: Analyze options.
Option (A): Incorrect, as \(b(y)\) cannot be linear with the given equation.
Option (B): Exponential functions are valid for both \(a(x)\) and \(b(y)\), so this is also correct.
Option (D): Incorrect because \(a(x)\) can be periodic while \(b(y)\) needs to be exponential.