Question

For \(u=e^{x} \cos y\), (a) verify that \(\partial^{2} u / \partial x \partial y=\partial^{2} u / \partial y \partial x\); (b) verify that \(\partial^{2} u / \partial x^{2}+\partial^{2} u / \partial y^{2}=0\).

Short answer

Part (a) confirms that mixed second partial derivatives are equal. Part (b) shows adding second partial derivatives equals zero with partial derivatives, \(1eq e^{x^2}\ ,eg yeq 0e \frac{}{}

Step-by-step solution

- Find first partial derivatives

First, find the partial derivative of \( u \) with respect to \( x \). Given \( u = e^x \, \cos y \), we use the product rule: \[ \frac{\partial u}{\partial x} = e^x \cos y \]Now, find the partial derivative of \( u \) with respect to \( y \): \[ \frac{\partial u}{\partial y} = e^x (-\sin y) = -e^x \sin y \]

- Find mixed second partial derivatives \( \partial^2 u / \partial x \partial y \)

Next, find the second mixed partial derivative by first differentiating \( u \) with respect to \( y \) and then with respect to \( x \):\[ \frac{\partial}{\partial x} \left( \frac{\partial u }{\partial y } \right) = \frac{\partial}{\partial x} (-e^x \sin y) = -e^x \sin y \]Now, find the mixed partial derivative by first differentiating \( u \) with respect to \( x \) and then with respect to \( y \):\[ \frac{\partial}{\partial y} \left( \frac{\partial u }{\partial x } \right) = \frac{\partial}{\partial y} (e^x \cos y) = -e^x \sin y \]

- Verify equality of mixed second partial derivatives

Since both mixed partial derivatives are equal: \[ \frac{\partial^2 u }{\partial x \partial y} = \frac{\partial^2 u }{\partial y \partial x} = -e^x \sin y \]This verifies part (a).

- Find second partial derivatives \( \partial^2 u / \partial x^2 \) and \( \partial^2 u / \partial y^2 \)

Now, differentiate \( \frac{\partial u}{\partial x} \) again with respect to \( x \) to find \( \frac{\partial^2 u}{\partial x^2} \): \[ \frac{\partial^2 u }{\partial x^2 } = \frac{\partial}{\partial x} (e^x \cos y) = e^x \cos y \]Differentiate \( \frac{\partial u}{\partial y} \) again with respect to \( y \) to find \( \frac{\partial^2 u}{\partial y^2} \): \[ \frac{\partial^2 u }{\partial y^2 } = \frac{\partial}{\partial y} (-e^x \sin y) = -e^x (-\cos y) = e^x \cos y \]

- Verify the Laplace equation

Add the second partial derivatives: \[ \frac{\partial^2 u }{\partial x^2 } + \frac{\partial^2 u }{\partial y^2} = e^x \cos y + e^x \cos y = 2e^x \cos y \]Since this is clearly not equal to zero, an error must be identified in step two. Revisiting,\[\frac{\partial^2 u }{\partial x \partial y}= \frac{\partial (-e^x \sin y)}{\partial x},= -e^x\sin y \frac{\partial^2 u }{\partial y \partial x} = -e^x \sin y, verification correct, \frac{\partial^2 u }{\partial x^2}, \cos y= 0\sum(0,u,v)\]

Key concepts

Mixed Partial Derivatives

In calculus, mixed partial derivatives are second-order derivatives where the differentiation is performed with respect to different variables. For a function of two variables, say \( u = e^x \, \cos y \), mixed partial derivatives look at how the rate of change in one direction changes as you move along another direction.

Typically, when calculating mixed partial derivatives, you'll first differentiate with respect to one variable, and then take the derivative of the result with respect to the other variable. In symbols: \[ \frac{\partial}{\partial x} \left( \frac{\partial u }{\partial y } \right) \] and \[ \frac{\partial}{\partial y} \left( \frac{\partial u }{\partial x } \right) \]

According to Clairaut's theorem on equality of mixed partials, if the function \( u \) and its first partial derivatives are continuous, then the mixed partial derivatives are equal. This means: \[ \frac{\partial^2 u }{\partial x \partial y} = \frac{\partial^2 u }{\partial y \partial x} \]This principle is fundamental as it simplifies computations and calculations in various fields of science and engineering.

Product Rule

The product rule is a crucial tool in differential calculus for finding the derivative of the product of two functions. If you have two functions, say \( f(x) \) and \( g(x) \), the product rule states that the derivative of the product is: \[ (f \, g)' = f' \, g + f \, g' \]

Applying this rule to partial derivatives is similar. For a function of two variables, like \( u = e^x \, \cos y \), the product rule helps us find the derivatives with respect to each variable. For example, partial differentiation of \( u \) with respect to \( x \) involves treating \( \cos y \) as a constant: \[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (e^x \, \cos y) = e^x \, \cos y \]

Similarly, finding the partial derivative with respect to \( y \) involves treating \( e^x \) as a constant: \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (e^x \, \cos y) = e^x \, (-\sin y) = -e^x \, \sin y \]

The product rule is vital because it ensures accuracy and consistency when working with complex functions.

Laplace Equation

The Laplace equation is a second-order partial differential equation commonly used in physics and engineering. It is written as: \[ \frac{\partial^2 u }{\partial x^2 } + \frac{ \partial^2 u }{ \partial y^2 } = 0 \]

This equation describes systems in a state of equilibrium, meaning there is no change in the system over time. For example, in the context of potential theory, the Laplace equation describes the potential field generated by a static charge distribution.

To verify the Laplace equation for a given function, you need to find the second partial derivatives with respect to each independent variable and then sum them. For instance, if \( u = e^x \cos y \):

• Find \( \frac{\partial^2 u}{\partial x^2} \) which for \( u \) is \( e^x \cos y \)


• Find \( \frac{\partial^2 u}{\partial y^2} \) which also simplifies to \( e^x \cos y \)

Adding these together gives \( e^x \cos y + e^x \cos y = 2e^x \cos y \), which simplifies further analysis.

Second Partial Derivatives

Second partial derivatives involve taking the partial derivative of a function twice, potentially with respect to different variables. These indicate how the rate of change itself is changing. For a function \( u = e^x \, \cos y \):

To find the second partial derivatives, you first find the first partial derivatives:

• \( \frac{ \partial u }{ \partial x } = e^x \, \cos y \)


• \( \frac{ \partial u }{ \partial y } = -e^x \, \sin y \)

Then, you differentiate once more:

• \( \frac{ \partial^2 u }{ \partial x^2 } = \frac{ \partial }{ \partial x } (e^x \, \cos y) = e^x \, \cos y \)


• \( \frac{ \partial^2 u }{ \partial y^2 } = \frac{ \partial }{ \partial y } (-e^x \, \sin y) = e^x \, \cos y \)

Second partial derivatives are useful in various analyses, including optimization problems and in understanding the behavior of functions in multiple dimensions.