Chapter 6: Problem 1
Let \(\mathrm{f}\) be the vector-valued function defined on $$ \begin{array}{c} R:|x| \leqq 1, \quad|\mathbf{y}| \leqq 1, \quad\left(\mathbf{y} \text { in } \mathrm{C}_{2}\right) \\ \mathbf{f}(x, \mathbf{y})=\left(y_{2}^{2}+1, x+y_{1}^{2}\right) \end{array} $$ (a) Find an upper bound \(M\) for \(|f(x, y)|\) for \((x, y)\) in \(R\). (b) Compute a Lipschitz constant \(K\) for \(\mathrm{f}\) on \(R\).
Short Answer
Step by step solution
Understand Vector-Valued Function
Calculate |f(x, y)|
Find Maximum of (y_2^2 + 1)
Find Maximum of (x + y_1^2)
Compute Upper Bound M
Derive Lipschitz Constant K
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Upper Bound
The goal is to ensure we're considering all possible values \(\mathbf{f}(x, \mathbf{y})\) can take within these bounds, and quantifying the highest possible magnitude.
- First, calculate each component of \(|\mathbf{f}(x, \mathbf{y})|\) separately, knowing that each is squared before added together. - The first component \((y_2^2 + 1)\) reaches a maximum of \(2\), since for \(|y_2| \leq 1\), \(y_2^2\) gets no larger than \(1\).- The second component \((x + y_1^2)\) also maxes out at \(2\), using similar logic where \(x\) can be \(1\) and \(y_1^2\) is also \(1\).Putting these together within the Euclidean framework, the combined maximum or upper bound \( M \) becomes the square root of the sum of their squares, yielding \( M = \sqrt{4 + 4} = 2\sqrt{2} \). This value provides the assurance that regardless of permissible \(x\) or \(y\), this function won't exceed that magnitude within the given range.
Lipschitz Constant
- To find \(K\), we examine the partial derivatives of the function. These derivatives show the rate of change of each component of \( \mathbf{f}(x, \mathbf{y}) \) relative to each variable.- For the function's first component \( f_1 \), partial derivatives with respect to \(x\) and \(y_1\) are zero because these variables do not appear in that component.- The partial derivative \( \frac{\partial f_1}{\partial y_2} = 2y_2 \) indicates how \( f_1 \) changes with \( y_2 \).For the second component \( f_2 = x + y_1^2 \), its partial derivatives as are:
- \( \frac{\partial f_2}{\partial x} = 1 \)
- \( \frac{\partial f_2}{\partial y_1} = 2y_1 \)
Partial Derivatives
- Consider the function \( \mathbf{f}(x, \mathbf{y}) = (y_2^2 + 1, x + y_1^2) \).For the first component:
- \( \frac{\partial f_1}{\partial x} = 0 \), since \(x\) does not influence \(f_1\).
- \( \frac{\partial f_1}{\partial y_1} = 0 \), because \(y_1\) also does not exist in \(f_1\).
- \( \frac{\partial f_1}{\partial y_2} = 2y_2 \), showing dependence solely on \(y_2\).
- \( \frac{\partial f_2}{\partial x} = 1 \), attributable entirely to \(x\).
- \( \frac{\partial f_2}{\partial y_1} = 2y_1 \), because \(y_1\) directly impacts \(f_2\).
- \( \frac{\partial f_2}{\partial y_2} = 0 \), as \(y_2\) doesn’t appear in this component.