Problem 67 - Calculus and its applications

Question

Consider \(f(x, y)=\ln \left(x^{2}+y^{2}\right) .\) Show that \(f\) is a solution of the partial differential equation \(\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0\).

Step-by-Step Solution

Answer

The function \(f(x, y) = \ln(x^2 + y^2)\) is a solution to the given PDE.

Step 1: Find the first partial derivatives

First, we find the partial derivatives of \(f(x, y) = \ln(x^2 + y^2)\) with respect to both \(x\) and \(y\). \[ \frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2} \] and \[ \frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2} \].

Step 2: Find the second partial derivatives with respect to x

Differentiate \(\frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2}\) with respect to \(x\) again to get \(\frac{\partial^2 f}{\partial x^2}\). Using the quotient rule, \[ \frac{\partial^2 f}{\partial x^2} = \frac{2(x^2 + y^2) - 4x^2}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2} \].

Step 3: Find the second partial derivatives with respect to y

Differentiate \(\frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2}\) with respect to \(y\) using the quotient rule to get \(\frac{\partial^2 f}{\partial y^2}\). \[ \frac{\partial^2 f}{\partial y^2} = \frac{2(x^2 + y^2) - 4y^2}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} \].

Step 4: Verify the partial differential equation

Add the second partial derivatives: \( \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \). \[ = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2} + \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} = \frac{0}{(x^2 + y^2)^2} = 0 \]. Thus, \(f(x, y)\) satisfies the partial differential equation.

Key Concepts

First Partial Derivatives

Understanding the concept of first partial derivatives is crucial for analyzing functions of several variables. When you have a function that depends on more than one variable, like our function \(f(x, y) = \ln (x^2 + y^2)\), you'll often want to understand how the function changes as each variable changes independently. This is exactly what first partial derivatives allow you to calculate.

For \(f(x, y)\), the first partial derivatives are found by differentiating \(f\) with respect to each variable independently, while treating the other variable as a constant.

• For \(x\), the first partial derivative is \(\frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2}\).
• For \(y\), the first partial derivative is \(\frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2}\).

These derivatives tell us how the function \(f(x, y)\) responds to small changes in \(x\) and \(y\). With these, we're one step closer to understanding how our multi-variable function behaves.

Second Partial Derivatives

Second partial derivatives give deeper insight into the curvature and concavity of a function's graph with respect to each variable. They are simply the derivatives of the first partial derivatives, taken again with respect to the same or a different variable. In our function \(f(x, y) = \ln (x^2 + y^2)\), we compute the second partial derivatives with respect to \(x\) and \(y\).

Here's how it's done:

• For \(\frac{\partial^2 f}{\partial x^2}\), we differentiate \(\frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2}\) with respect to \(x\) again using the quotient rule, resulting in \(\frac{\partial^2 f}{\partial x^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2}\).
• Similarly, for \(\frac{\partial^2 f}{\partial y^2}\), we differentiate \(\frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2}\) with respect to \(y\), which leads to \(\frac{\partial^2 f}{\partial y^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2}\).

By understanding the directions in which the function's curvature changes, these derivatives play a pivotal role in the analysis of many mathematical and physical phenomena.

Quotient Rule

The quotient rule is a fundamental tool for finding derivatives of ratios where both the numerator and the denominator are functions of a variable. This rule is essential when dealing with expressions like \[ \frac{u}{v} \] where both \(u\) and \(v\) are differentiable functions. For our function \(f(x, y) = \ln (x^2 + y^2)\), we use the quotient rule multiple times, particularly when calculating the second partial derivatives. It's applied during these differentiations due to the form of the first partial derivatives, which are fractions.

The quotient rule formula is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\) respectively. In applying this formula, attention to detail is necessary to avoid mistakes, especially in the sign and terms involved.

• For example, applying this to \(\frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2}\), where \(u = 2x\) and \(v = x^2 + y^2\), correctly yields the second partial derivative \(\frac{\partial^2 f}{\partial x^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2}\).

Mastery of the quotient rule is key in handling many calculus problems efficiently.