Problem 6 - Differential Equations and Linear Algebra

Question

Define \(T_{1}: C^{1}[a, b] \rightarrow C^{0}[a, b]\) and \(T_{2}\) \(C^{0}[a, b] \rightarrow C^{1}[a, b]\) by $$\begin{aligned} T_{1}(f) &=f^{\prime} \\ \left[T_{2}(f)\right](x) &=\int_{a}^{x} f(t) d t, \quad a \leq x \leq b \end{aligned}$$ (a) If \(f(x)=\sin (x-a),\) find \(\left[T_{1}(f)\right](x) \quad\) and \(\quad\left[T_{2}(f)\right](x)\) and show that, for the given function, $$\left[T_{1} T_{2}\right](f)=\left[T_{2} T_{1}\right](f)=f$$ (b) Show that for general functions \(f\) and \(g\) $$\left[T_{1} T_{2}\right](f)=f,\left\\{\left[T_{2} T_{1}\right](g)\right\\}(x)=g(x)-g(a)$$

Step-by-Step Solution

Answer

For the given function \(f(x) = \sin(x-a)\), we have: 1. \(\displaystyle T_1(f)(x) = \cos(x-a)\) 2. \(\displaystyle T_2(f)(x) = \int_{a}^{x} \sin(t-a) dt\) We showed that: 3. \(T_1 T_2(f) = \sin(x-a) = f(x)\) 4. \(T_2 T_1(f) = \sin(x-a) = f(x)\) For general functions, we proved: 5. \(T_1 T_2(f) = f\) 6. \(T_2 T_1(g)(x) = g(x) - g(a)\)

Step 1: Recall the definitions of \(T_1\) and \(T_2\) operators.

First, note that: - \(T_1(f) = f'\) - \(T_2(f)(x) = \int_{a}^{x} f(t) dt\)

Step 2: Calculate \(T_1(f)\) and \(T_2(f)\) for \(f(x) = \sin(x-a)\)

For \(f(x) = \sin(x - a)\): (a) Calculate \(T_1(f)(x)\): \[T_1(f)(x) = f'(x) = \left(\sin(x - a)\right)' = \cos(x - a)\] (b) Calculate \(T_2(f)(x)\): \[T_2(f)(x) = \int_{a}^{x} \sin(t-a) dt\]

Step 3: Verify that \(T_1 T_2(f) = T_2 T_1(f) = f\)

(a) Calculate \(T_1 T_2(f)(x)\): \[T_1 T_2(f)(x) = T_1 \left( \int_{a}^{x} \sin(t-a) dt \right)\] Now, take the derivative with respect to x: \[\frac{d}{dx} \left(\int_{a}^{x} \sin(t-a) dt \right) = \sin(x-a)\] So, \(T_1 T_2(f)(x) = \sin(x-a) = f(x)\) (b) Calculate \(T_2 T_1(f)(x)\): \[T_2 T_1(f)(x) = T_2(\cos(x - a))\] We already found \(T_2(f)(x)\), so now compute the integral with \(\cos(x - a))\): \[T_2 T_1(f)(x) = \int_{a}^{x} \cos(t - a) dt\] Applying the boundary limits, we have: \[\int_{a}^{x} \cos(t - a) dt = \left[ \sin (t-a) \right]_{a}^{x} = \sin(x-a) - \sin(a-a)\] So, \(T_2 T_1(f)(x) = \sin(x-a) = f(x)\)

Step 4: Prove the general results for \(T_1 T_2(f) = f\) and \(T_2 T_1(g)(x) = g(x) - g(a)\)

(a) Proof for \(T_1 T_2(f) = f\): \[T_1 T_2(f)(x) = \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)\] (b) Proof for \(T_2 T_1(g)(x) = g(x) - g(a)\): \[T_2 T_1(g)(x) = \int_{a}^{x} g'(t) dt = \left[g(t)\right]_{a}^{x} = g(x) - g(a)\] That is the general properties for linear operators \(T_1\) and \(T_2\).

Key Concepts

Differential Operators

Differential operators are mathematical tools used to compute the derivative of a function. In functional analysis, these operators play a crucial role in understanding the behavior of functions. Think of them as tools that help us analyze how functions change.In the given problem, the operator \(T_1\) is defined as a differential operator that takes a function \(f\) and returns its derivative \(f'\). For example, if \(f(x) = \sin(x-a)\), then applying \(T_1\) results in the derivative, which is \(\cos(x-a)\). Differential operators have many applications, including solving differential equations, finding rates of change, and understanding physical phenomena.

Integral Operators

Integral operators transform a function by integrating it, essentially finding the "area under the curve." In functional analysis, integral operators are frequently used to reconstruct functions from their rate of change or to smooth data.In our exercise, \(T_2\) is an integral operator that applies the definite integral to a function from \(a\) to \(x\). Specifically, for \(f(x) = \sin(x-a)\), \(T_2\) represents the computation \(\int_{a}^{x} \sin(t-a) \, dt\). This operation accumulates "summing" the values of \(f\) from \(a\) to \(x\), which is essential in understanding and reconstructing functions from their derivatives.

Function Spaces

Function spaces are collections of functions that share certain properties and follow specific rules. These spaces help us understand the structure and behavior of functions as a group rather than individually. Examples include spaces of continuous functions and differentiable functions.In our problem, \(C^1[a,b]\) and \(C^0[a,b]\) are function spaces:

• \(C^1[a,b]\) includes all functions that are continuous and have continuous first derivatives on the interval \([a, b]\).
• \(C^0[a,b]\) contains all continuous functions on \([a, b]\).

These spaces allow operators like \(T_1\) and \(T_2\) to act on functions within them, ensuring continuity and differentiability where necessary. Function spaces are central to numerous areas of math and physics, offering a framework for analysis.

Inverse Operators

Inverse operators essentially "reverse" the effect of another operator. They are critical in solving equations and understanding transformations in functional analysis.In this exercise, the concept of inverse operators emerges when we explore \(T_1 \circ T_2\) and \(T_2 \circ T_1\):

• \(T_1 \circ T_2(f) = f\), meaning applying \(T_2\) followed by \(T_1\) recovers the original function \(f\). This makes \(T_2\) an inverse in a certain sense to \(T_1\), but only when applied in this order.
• \(T_2 \circ T_1(g) = g(x) - g(a)\) means \(T_2\) followed by \(T_1\) does not fully return \(g\), as there is an extra term \(g(a)\). Hence, \(T_1\) and \(T_2\) are not true inverses when applied in this order.

Understanding these relationships helps in numerous applications like solving integral and differential equations, where finding the inverse operation is key.