Question

Suppose \(f\) is piecewise continuous on \([0, \infty)\) (a) Prove: If the integral \(g(t)=\int_{0}^{t} e^{-s_{0} \tau} f(\tau) d \tau\) satisfies the inequality \(|g(t)| \leq M(t \geq 0)\), then \(f\) has a Laplace transform \(F(s)\) defined for \(s>s_{0}\). Hint: Use integration by parts to show that $$ \int_{0}^{T} e^{-s t} f(t) d t=e^{-\left(s-s_{0}\right) T} g(T)+\left(s-s_{0}\right) \int_{0}^{T} e^{-\left(s-s_{0}\right) t} g(t) d t. $$ (b) Show that if \(\mathcal{L}(f)\) exists for \(s=s_{0}\) then it exists for \(s>s_{0} .\) Show that the function $$ f(t)=t e^{t^{2}} \cos \left(e^{t^{2}}\right) $$ has a Laplace transform defined for \(s>0,\) even though \(f\) isn't of exponential order. (c) Show that the function $$ f(t)=t e^{t^{2}} \cos \left(e^{t^{2}}\right) $$ has a Laplace transform defined for \(s>0,\) even though \(f\) isn't of exponential order.

Short answer

Short answer: The given function \(f(t) = te^{t^2}\cos(e^{t^2})\) has a Laplace transform for s > 0, as the boundedness conditions are satisfied by the integrand involving \(g(t)\). By uncovering the behavior of the function and examining its exponential order, we were able to determine its existence for s > 0.

Step-by-step solution

Part (a) - Prove the general inequality and existence of Laplace transform

We are given the integral \(g(t) = \int_{0}^{t} e^{-s_{0} \tau} f(\tau) d \tau\). Our goal is to prove that if \(|g(t)| \leq M(t \geq 0)\), then f has a Laplace transform defined for s>s0. Step 1: Integration by parts We will apply integration by parts to the expression \(\int_{0}^{T} e^{-s t} f(t) d t\) using \(u = e^{-st}\) and \(dv = f(t)dt\). Then \(du = -se^{-st}dt\) and \(v = g(t)\). Step 2: Integration by parts result Applying integration by parts, we have: $$ \int_{0}^{T} e^{-s t} f(t) d t = e^{-s T} g(T) - \int_{0}^{T} -se^{-s t} g(t) d t. $$ Next, we replace s with s-s0: $$ \int_{0}^{T} e^{-(s-s_0) t} f(t) d t = e^{-(s-s_0) T} g(T) - (s-s_0)\int_{0}^{T} e^{-(s-s_0) t} g(t) d t. $$ Step 3: Boundedness and absolute convergence Since \(|g(t)| \leq M(t \geq 0)\), both terms \(|e^{-\left(s-s_{0}\right) T}| |g(T)|\) and \(|(s-s_{0}) \int_{0}^{T} e^{-\left(s-s_{0}\right) t} g(t) d t|\) are bounded and therefore convergent. This implies that the integral expression converges as well, and hence, the Laplace transform \(F(s)\) is defined for \(s>s_0\).

Part (b) - Show the existence of the Laplace transform for a specific function at \(s=s_0\)

Given the function: $$ f(t)=t e^{t^{2}} \cos \left(e^{t^{2}}\right) $$ Our goal is to show that if the Laplace transform exists for \(s = s_0\), then it exists for \(s > s_0\). Step 1: Existence of Laplace transform at s=s0 Assuming \(\mathcal{L}(f)\) exists for \(s = s_0\), we know the integral: $$ \int_{0}^{\infty} e^{-s_0 t} f(t) d t $$ is convergent. Step 2: Existence of Laplace transform for s>s0 Consider the Laplace transform integral for \(s > s_0\): $$ \int_{0}^{\infty} e^{-s t} f(t) d t $$ Since \(s > s_0\), we have \(s-s_0 > 0\). Thus \(-st = -(s_0 t) - (s-s_0)t\) and \(e^{-st} = e^{-(s_0t)}e^{-(s-s_0)t}\). Replace \(e^{-st}\) in the expression: $$ \int_{0}^{\infty} e^{-(s_0 t)} e^{-(s-s_0)t} f(t) d t $$ Since the integral converges for \(s=s_0\), the integral converges for \(s>s_0\). Therefore, the Laplace transform exists for \(s > s_0\).

Part (c) - Show the existence of the Laplace transform for a specific function for s > 0

To prove that the Laplace transform of the given function exists for \(s>0\), we can analyze the function's behavior and use the properties of Laplace transforms to determine its existence: Step 1: Consider boundedness of g(t) From the given function, \(f(t)\), we define \(g(t) = \int_{0}^{t} e^{-s_{0} \tau} f(\tau) d \tau\). Let's try to find upper bounds for |g(t)|: $$ |g(t)| = \left|\int_{0}^{t} e^{-s_{0} \tau} \tau e^{\tau^{2}} \cos \left(e^{\tau^{2}}\right) d \tau \right| \leq \int_{0}^{t} \left| e^{-s_{0} \tau} \tau e^{\tau^{2}} \cos \left(e^{\tau^{2}}\right) \right| d \tau \leq \int_{0}^{t} e^{-s_{0} \tau} \tau e^{\tau^{2}} d \tau $$ Step 2: Examine exponential order The function \(\tau e^{\tau^{2}}\) is NOT of exponential order, but it's not required to be for the Laplace transform to exist. We are looking for the existence of the Laplace transform for s > 0. Step 3: Boundedness condition Let's analyze the integrand for large values of t: $$ \lim_{t \rightarrow \infty} e^{-s_{0} \tau} \tau e^{\tau^{2}} = 0 $$ This limit indicates that the integrand satisfies the boundedness condition for the existence of the Laplace transform for s > 0. Step 4: Conclusion Since the boundedness conditions are satisfied, the function f(t) has a Laplace transform defined for \(s>0\).

Key concepts

Piecewise Continuous Function

A piecewise continuous function is one that is continuous within certain intervals, but it may have a finite number of discontinuities at specified points. For example, a function may be defined one way on the interval \( [0,1) \) and another on \([1,\infty)\). These points of discontinuity are important when considering the Laplace transform, which is an integral transform widely used to solve differential equations.

A crucial aspect of the piecewise continuous function is that despite possible discontinuities, the function must remain bounded within each interval. This boundedness is key to ensuring the convergence of integrals that appear in the calculation of the Laplace transform, which leads us to our next concept, the convergence of integrals.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation and is used to integrate the product of two functions. When faced with the challenge of integrating a product, we can label one function as \( u \) and the other as \( dv \) to then find \( du \) and \( v \) after which the integral can be expressed in a new form that's often easier to solve.

For example, when proving that a function has a Laplace transform, we use integration by parts to rearrange the integral into a form that allows us to apply bounds and ultimately show convergence. This technique was crucial in showing that a function \( f \) has a corresponding Laplace transform \( F(s) \) when given certain conditions on \( g(t) \) as outlined in the original exercise.

Convergence of Integrals

Convergence of integrals is a foundational concept for many areas of mathematics, especially for transform methods, like the Laplace transform. An integral is said to converge if, as its limits approach infinity, it approaches a finite value. For the Laplace transform to exist for a function, the integral \[\int_{0}^{\infty} e^{-st} f(t) dt\] must converge, meaning that it doesn't 'blow up' to infinity for the specified range of \( s \).

The exercise highlights how the bounded function \( g(t) \) aids in providing this guarantee of convergence for the integral, ensuring that the Laplace transform exists. Intuitively, if we know the integral of a bounded function is under control (converging), then we can confidently utilize the Laplace transform to find solutions to otherwise complex differential equations.

Exponential Order

A function is of exponential order if it grows no faster than an exponential function as \( t \) approaches infinity. More technically, a function \( f(t) \) is of exponential order \(\alpha \) if there exists constants \(M \) and \(\beta \) such that \(|f(t)| \leq M e^{\alpha t} \) for all \( t \) greater than \(\beta\).

However, as shown in part (c) of the exercise, a function not of exponential order can still have a Laplace transform defined for \(s>0\). The given function \(f(t) = t e^{t^2} \cos( e^{t^2})\) illustrates this point vividly. While the function grows faster than any exponential function due to the \(e^{t^2}\) term, because it is paired with an oscillating cosine function, its integral can still converge for \(s > 0\). This is an important reminder that the existence of a Laplace transform can sometimes extend beyond functions of exponential order, widening the scope of this transformative tool in solving differential equations.