Question

From the definition of the Laplace transform, and using integration by parts, show that, $$ \mathcal{L}\left\\{f^{\prime \prime}(t)\right\\}=s^{2} F(s)-s f(0)-f^{\prime}(0) $$

Short answer

Answer: The Laplace Transform of the second derivative of a function f(t) is given by the equation: $$ \mathcal{L}\{f''(t)\} = s^2 F(s) - sf(0) - f'(0) $$ Here, F(s) represents the Laplace Transform of the function f(t).

Step-by-step solution

Recall the Definition of the Laplace Transform

The Laplace Transform of a function f(t) is defined as: $$ \mathcal{L} \{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt $$ We will apply this definition to the function f''(t) to find the Laplace Transform of the second derivative.

Apply Integration by Parts

Let's use integration by parts to evaluate the integral. Integration by parts states that for functions u(t) and v(t), the integral of their product is: $$ \int u \ dv = uv - \int v \ du $$ For our problem, let's choose: $$ u = e^{-st} \\ dv = f''(t) dt $$ Now we need to find du and v from u and dv: $$ du = -se^{-st} dt \\ v = \int f''(t) dt = f'(t) $$

Apply Integration by Parts Twice

Now we will apply the integration by parts formula twice. First time: $$ \mathcal{L}\{f''(t)\} = \int_{0}^{\infty} e^{-st} f''(t) dt = \left[-e^{-st} f'(t)\right]_{0}^{\infty} + s \int_{0}^{\infty} e^{-st} f'(t) dt $$ Second time, use the same substitution u and dv to the new integral: $$ u = e^{-st} \\ dv = f'(t) dt $$ $$ du = -se^{-st} dt \\ v = \int f'(t) dt = f(t) $$ Apply integration by parts again: $$ \mathcal{L}\{f''(t)\} = \left[-e^{-st} f'(t)\right]_{0}^{\infty} + s\left[\left(-e^{-st}f(t)\right)_{0}^{\infty} + s\int_{0}^{\infty} e^{-st} f(t) dt\right] $$

Evaluate the Integrals

Now, we have to evaluate the limits of integration. Note that \(\lim_{t\to\infty} e^{-st}=0\) for \(s>0\), so we have: $$ \mathcal{L}\{f''(t)\} = \lbrace s[(-e^{-s(0)}f(0)) - 0] - sf'(0)\rbrace + s^2 \mathcal{L}\{f(t)\} $$ Simplify and rewrite the equation: $$ \mathcal{L}\{f''(t)\} = s^2 F(s) - sf(0) - f'(0) $$ This equation shows the Laplace Transform of the second derivative of a function f(t).

Key concepts

Integration by Parts

When faced with the challenging task of integrating the product of two functions, a technique known as integration by parts comes to the rescue. This method is derived from the product rule for differentiation and provides a way to simplify complex integrals. Consider two functions, u(t) and its differential du, and another function, v(t) with its differential dv. Integration by parts states that the integral of u dv is given by:
\[ \int u dv = uv - \int v du \]
In simpler terms, we're breaking down the original integral into more manageable parts that can be easily evaluated. Our formula takes the integral of one part of the function and multiplies it with another, then subtracts the integral of a new function derived from the parts we initially chose. For successful application, the key is to correctly choose u and dv such that the resulting integral of v du is easier to evaluate than the original integral. This method is widely used in various disciplines of calculus, including solving differential equations and finding the Laplace transforms, as seen in our textbook problem.

Second Derivative

The concept of the second derivative is deeply connected to the curvature and acceleration of functions. It refers to the derivative of the derivative, and it gives us information about the rate at which the first derivative changes. If we consider a function f(t), its first derivative f'(t) represents the rate of change of f(t) at any given point, or graphically, the slope of the tangent line to the function at that point.

The second derivative, denoted as f''(t), then provides insight into how this slope itself changes. When f''(t) is positive, the slope of f(t) is increasing, implying the function is curving upwards, and conversely, a negative f''(t) indicates a downward curvature. This second-level derivative is essential in physics for understanding acceleration, as well as in mathematics for analyzing the concavity of graphs and identifying points of inflection. In our Laplace transform exercise, understanding the second derivative is crucial as it is central to finding the transform of this higher-order derivative.

Laplace Transform of Derivatives

The Laplace transform of derivatives is a powerful tool used to convert differential equations into algebraic equations, which are often much easier to solve. When dealing with a function's derivatives, the Laplace transform provides a method to handle these derivatives as algebraic terms. For instance, the Laplace transform of a first derivative will involve the initial value of the function as well as its transform. This property is extended to higher-order derivatives such as the second derivative, where the Laplace transform incorporates both the initial value of the function and its first derivative, alongside the second power of the complex frequency variable, s.

For a second derivative, the Laplace transform is given by:\[ \mathcal{L}\left\{f^{''}(t)\right\} = s^2 F(s) - s f(0) - f^{'}(0) \]
This equation elegantly encapsulates the second derivative's influence, integrating it with the initial conditions of the original function. It demonstrates the leverage that Laplace transform provides in simplifying the process of solving differential equations by transforming time domain operations into simpler algebraic forms in the s-domain.