Question

For a twice continuously differentiable function \(g: \mathbb{R} \to \mathbb{R}\), define \[ u_g(x, y) = \frac{1}{y} \int_{-y}^{y} g(x + t) \, dt \quad \text{for} \quad (x, y) \in \mathbb{R}^2, \, y>0. \] Which one of the following holds for all such \(g\)?

• A. \( \frac{\partial^2 u_g}{\partial x^2} = \frac{2}{y} \frac{\partial u_g}{\partial y} + \frac{\partial^2 u_g}{\partial y^2} \)
• B. \( \frac{\partial^2 u_g}{\partial x^2} = \frac{1}{y} \frac{\partial u_g}{\partial y} + \frac{\partial^2 u_g}{\partial y^2} \)
• C. \( \frac{\partial^2 u_g}{\partial x^2} = \frac{2}{y} \frac{\partial u_g}{\partial y} - \frac{\partial^2 u_g}{\partial y^2} \)
• D. \( \frac{\partial^2 u_g}{\partial x^2} = \frac{1}{y} \frac{\partial u_g}{\partial y} - \frac{\partial^2 u_g}{\partial y^2} \)

Hint:

Use Leibniz's rule for differentiating under the integral sign when computing derivatives of integrals with respect to parameters like \(x\) and \(y\).

Answer 1

The Correct Option is A

To compute the second partial derivative of \(u_g(x, y)\) with respect to \(x\), we apply Leibniz's rule for differentiating under the integral sign. First, calculate the first derivative with respect to \(x\): \[ \frac{\partial u_g}{\partial x} = \frac{1}{y} \int_{-y}^{y} \frac{\partial}{\partial x} g(x + t) \, dt = \frac{1}{y} \int_{-y}^{y} g'(x + t) \, dt. \] Now, differentiate again with respect to \(x\): \[ \frac{\partial^2 u_g}{\partial x^2} = \frac{1}{y} \int_{-y}^{y} g''(x + t) \, dt. \] Next, compute the derivative of \(u_g(x, y)\) with respect to \(y\). First, differentiate the original function with respect to \(y\): \[ \frac{\partial u_g}{\partial y} = -\frac{1}{y^2} \int_{-y}^{y} g(x + t) \, dt + \frac{1}{y} \left( g(x + y) - g(x - y) \right). \] Differentiate this result once more to obtain: \[ \frac{\partial^2 u_g}{\partial y^2} = \frac{2}{y} \frac{\partial u_g}{\partial y} + \frac{\partial^2 u_g}{\partial y^2}. \] Thus, the required identity holds as expressed in option (A).