Question

For a twice continuously differentiable function 𝑔: ℝ → ℝ, define
\(u_g(x,y)=\frac{1}{y}\int^y_{-y}g(x+t)dt\ \ \ \text{for}(x,y)\in \R^2, \ \ \ y \gt0.\)
Which one of the following holds for all such 𝑔 ?

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

Answer 1

The Correct Option is A

To determine which given equation holds for the defined function \( u_g(x, y) \), we begin by calculating the partial derivatives of \( u_g \). The function \( u_g \) is given as:

\(u_g(x,y) = \frac{1}{y} \int_{-y}^{y} g(x+t)\,dt \)

Steps:

1. First, find the partial derivative of \( u_g \) with respect to \( x \):
2. Now, differentiate again with respect to \( x \) to find \( \frac{\partial^2 u_g}{\partial x^2} \):
3. Next, calculate the partial derivative of \( u_g \) with respect to \( y \):
4. Differentiate the expression for \( \frac{\partial u_g}{\partial y} \) with respect to \( y \) to find \( \frac{\partial^2 u_g}{\partial y^2} \):
5. To show that the relationship \( \frac{\partial^2 u_g}{\partial x^2} \) equals the right expression, substitute the derived formulas in the options and simplify:

Hence, the correct option is: \(\frac{∂^2u_g}{∂x^2}=\frac{2}{y}\frac{∂u_g}{∂y}+\frac{∂^2u_g}{∂y^2}\).