Question

Untersuchen Sie die auf \(D=\left\\{(x, y)^{T} \mid x^{2}+y^{2} \leq 1\right\\}\) definierte Funktion $$ f(x, y)=\left\\{\begin{array}{lll} \frac{\sin \left(\sqrt{x^{2}+y^{2}}\right)}{\sqrt{x^{2}+y^{2}}} & \text { für } & x^{2}+y^{2} \neq 0 \\ 1 & \text { für } & x^{2}+y^{2}=0 \end{array}\right. $$ auf lokale und globale Extrema.

Short answer

The global maximum is 1 at \((0, 0)\).

Step-by-step solution

Understand the Domain and Function

The function is defined on the domain \(D = \{(x, y)^T \mid x^2 + y^2 \leq 1\}\). This means it is defined within and on the boundary of the unit circle centered at the origin in the xy-plane. Inside this domain, the function is defined piecewise. If \(x^2 + y^2 eq 0\), then \(f(x, y) = \frac{\sin(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}}\); if \(x^2 + y^2 = 0\), \(f(x, y) = 1\). This takes care of the indeterminate form at the origin.

Analyze the Boundary Conditions

Since \(f(x, y)\) depends on \(\sqrt{x^2 + y^2}\), we analyze what happens around the boundary, \(x^2 + y^2 = 1\). On this boundary, \(f(x, y) = \frac{\sin(1)}{1} = \sin(1)\). We must later compare this value to that of any critical points within the domain.

Check for Critical Points

To find critical points, we consider the derivatives. Because the function is radially symmetric, substitution into polar coordinates \((r, \theta)\) with \(r = \sqrt{x^2 + y^2}\) is useful. The function simplifies to \(f(r) = \frac{\sin(r)}{r}\) with \(f(0) = 1\). We then calculate its derivative \(f'(r)\), which is needed for critical points in the domain excluding the boundary.

Derivative Calculation

Compute the derivative of \(f(r)\), which is \(f'(r) = \frac{r\cos(r) - \sin(r)}{r^2}\). Setting \(f'(r) = 0\) leads to \(r\cos(r) - \sin(r) = 0\). Solving this gives possible critical values for \(r\).

Evaluate Critical Points

Upon solving the equation \(r \cos(r) = \sin(r)\), we need to find values of \(r\) that yield critical points within the interval \([0, 1]\). Consider values \(r \approx 0.876\), where \(f'(r)\) changes sign.

Compare Function Values

Evaluate the function at \(r = 0\) (i.e., origin), \(r = 1\) (on the boundary), and any other critical points found. We have \(f(0) = 1, f(1) = \sin(1)\). Compare these to find extrema. \(f(0) = 1\) is larger than \(f(1)\) and any found critical points within \(r = [0, 1]\).

Conclusion

Since \(f(x, y)\) achieves its maximum value of \(1\) at the origin and maintains lesser values within \(r = 0.876\) and boundary, the function's local and global maximum is at the origin.

Key concepts

Critical Points in Multivariable Calculus

Critical points are significant in understanding the behavior of a function, especially in multivariable calculus. These are points where the function's gradient—representing the slope in multiple dimensions—is zero or undefined. For our example function \(f(x, y)\), a critical point occurs when the derivative in terms of \(r\) (where \(r = \sqrt{x^2 + y^2}\)) is zero.

When examining \(f(r) = \frac{\sin(r)}{r}\) in polar coordinates, we calculate its derivative as \(f'(r) = \frac{r\cos(r) - \sin(r)}{r^2}\). This derivative helps us find where changes in the function's direction occur. To find critical points, solve \(r\cos(r) - \sin(r) = 0\), identifying where the function's rate of change, or slope, is zero.

Discovering critical points is pivotal as they indicate where a function might have local maxima, minima, or saddle points. In the case of our function, the solution to this equation, pinpointing potential critical values for \(r\), reveals local behavior across the domain excluding the origin and boundaries.

Understanding Extrema

Extrema refer to the function's minimum and maximum values, which can be local (in the nearby region) or global (overall highest or lowest value). With our piecewise function \(f(x, y)\), finding both local and global extrema entails evaluating the function at critical points as well as on the boundary.

Our function \(f(x, y)\) is defined over a radial domain \((r = \sqrt{x^2 + y^2})\). We assess the function's value at critical points within this domain and compare these with the function's boundary value, \( f(1) = \sin(1)\), and the origin \(f(0) = 1\).

The process involves:

• Checking the critical points, where derivative \(f'(r)\) equals zero.
• Evaluating function values at known critical points (0, and boundary points).
• Comparing these values to identify extremum points.

In this particular function, despite many critical points, the global maximum occurs at the origin, \((0,0)\), where \(f(0) = 1\). Knowing how to effectively determine extrema is critical in multivariable optimization and problem solving.

Analyzing Piecewise Functions

Piecewise functions are functions that have different expressions or rules for different intervals of their domain. They can seem daunting but are very useful in modeling situations where behavior changes over different conditions.

For our given function \(f(x, y)\), the definition is piecewise:

• \( \frac{\sin(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} \) for \( x^2 + y^2 eq 0 \)
• \( 1 \) for \( x^2 + y^2 = 0 \)

This approach elegantly handles the problematic \( \frac{0}{0} \) form by assigning a defined value at the origin.

When analyzing piecewise functions, it is vital to:

• Understand each piece's domain and expression.
• Compute limits to handle indeterminate forms and ensure continuity.
• Analyze behavior at boundary conditions (like \(x^2 + y^2 = 1 \) here).

Piecewise functions allow flexibility, capturing complex behaviors of real-world models, and their analysis is foundational in understanding diverse mathematical and physical concepts.