Question

Berechnen Sie die FRÉCHET-Ableitung des Funktionals $$ f(u)=\int_{0}^{\frac{\pi}{2}} \cos u(\phi) d \phi $$ wobei \(f\) auf dem BANACH-Raum der stetigen Funktionen \(u(\phi)\) über dem Intervall \(\left[0, \frac{\pi}{2}\right]\) mit der Maximum-Norm definiert ist.

Short answer

The Fréchet derivative is the operator \( L(h) = -\int_{0}^{\frac{\pi}{2}} \sin u(\phi) \cdot h(\phi) \, d\phi \).

Step-by-step solution

Understand the Functional

The functional provided is \( f(u) = \int_{0}^{\frac{\pi}{2}} \cos u(\phi) \, d\phi \). This integral maps a continuous function \( u(\phi) \) to a real number by integrating over the interval \( \left[0, \frac{\pi}{2}\right] \). We need to find its Fréchet derivative.

Define Fréchet Derivative

The Fréchet derivative of a functional \( f \) at a point \( u \) is a continuous linear operator \( L \) such that for a small perturbation \( h \), the following holds: \[ f(u+h) = f(u) + L(h) + o(\|h\|) \] where \( o(\|h\|) \) denotes the little-o notation indicating terms that tend to zero faster than \( \|h\| \) as \( \|h\| \to 0 \).

Compute the Functional Increment

Substitute \( u + h \) into the functional: \[ f(u+h) = \int_{0}^{\frac{\pi}{2}} \cos(u(\phi) + h(\phi)) \, d\phi \]Use the series expansion for cosine: \[ \cos(a+b) = \cos a \cos b - \sin a \sin b \]This gives:\[ \cos(u+h) \approx \cos u - \sin u \cdot h \quad \text{(considering only first order terms in \( h \))} \]

Apply Linearization

From the cosine expansion, the linearized form is:\[ \int_{0}^{\frac{\pi}{2}} \cos(u(\phi) + h(\phi)) \, d\phi \approx \int_{0}^{\frac{\pi}{2}} \cos u(\phi) \, d\phi - \int_{0}^{\frac{\pi}{2}} \sin u(\phi) \cdot h(\phi) \, d\phi \]Here, the term \( -\int_{0}^{\frac{\pi}{2}} \sin u(\phi) \cdot h(\phi) \, d\phi \) is the linear part \( L(h) \).

Identify Fréchet Derivative Operator

The Fréchet derivative of \( f \) at \( u \) is the linear operator which maps \( h \) to:\[ L(h) = -\int_{0}^{\frac{\pi}{2}} \sin u(\phi) \cdot h(\phi) \, d\phi \]Thus, the Fréchet derivative is given by the operator \( L(h) = -\langle \sin u, h \rangle \). This operator is continuous due to the boundedness and linearity imposed by the integral and sine function.

Key concepts

Banach Space

A Banach space is a complete normed vector space. This completeness means that if you have an infinite sequence of vectors in this space, whose lengths get closer and closer to zero, then these vectors will converge to a point also in the space.
This is very important for our problem, as we deal with continuous functions defined within a specific interval.

• Normed Space: Every vector has a length or norm assigned to it. In our problem, the maximum norm is used. It takes the supreme value (or the maximum) of the absolute values of the function within the interval.
• Vector Space: This space comprises functions that you can add or multiply by scalars while still belonging to this space.

Understanding these properties ensures that certain analytical techniques and operations we use will behave predictably. Banach spaces are foundational in various fields like quantum physics, as they help manage complex function spaces.

Continuous Functions

Continuous functions are functions where small changes in the input result in small changes in the output. This trait is crucial when dealing with integrals and derivatives. In our context, the function \( u(\phi) \) belongs to the space of continuous functions on \([0, \frac{\pi}{2}]\).
These functions possess some essential characteristics:

• Continuity: The function does not have sharp corners or jumps, ensuring a smooth curve.
• Integrability:: Since continuous functions on a closed interval like \([0, \frac{\pi}{2}]\) are integrable, we are guaranteed to be able to compute the integral, which often simplifies the process of finding derivatives.
• Boundedness:: Due to the closed and bounded interval, functions are bounded, meaning the maximum-norm can be applied, which simplifies analysis in Banach Spaces.

These properties simplify analysis, making it predictable and manageable within mathematical frameworks like Banach spaces.

Integral Calculus

Integral calculus allows us to find quantities when the rates of change are known, often represented by integrals. This is a crucial part of calculating functionals like the one introduced in the exercise.

• Definite Integral: The expression \( \int_{0}^{\frac{\pi}{2}} \cos u(\phi) d \phi \) is a definite integral that evaluates the area under the curve of the function\( \cos u(\phi) \) over the interval \([0, \frac{\pi}{2}]\).
• Linearization: When computing derivatives, we use integral calculus to handle small perturbations. By focusing on first-order terms, we separate the functional into a difference broken down by this operator.
• Operator Application: The Fréchet derivative relies on integral calculus to apply a linear operator to deviations in functions, calculated as an integral like \(- \int_{0}^{\frac{\pi}{2}} \sin u(\phi) \cdot h(\phi) d \phi \).

Integral calculus is the backbone of evaluating changes in functions and finding how these contribute to larger patterns or systems in mathematical analysis.