Inverse Laplace Transform. Existence Laplace Transform.

Beware that, when fighting monsters, you yourself do not become a monster… for when you gaze long into the abyss. The abyss gazes also into you, Friedrich W. Nietzsche

Recall

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

• Dependent variables: Variables that depend on one or more other variables (y).
• Independent variables: Variables upon which the dependent variables depend (x).
• Derivatives: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

• y = y(x) is the unknown function of the independent variable x.
• y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
• a(x), b(x), and c(x) are known functions of x.

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

• The function f(x, y) (the right-hand side of the ODE) in the equation y’ = f(x, y) is continuous in some neighborhood around a point (x₀, y₀) and
• Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it satisfies the initial condition y(x₀) = y₀.

This theorem ensures that under these conditions, the solution exists and is unique near x = x₀.

Laplace Transform

The Laplace Transform of a function f(t), where t ≥ 0, is defined as $\mathcal{L}(f(t)) = \int_{0}^{∞} f(t)e^{-st}dt = F(s)$.

One of the most important properties of the Laplace Transform is linearity, which states: $\mathcal{L}(af(t)+bg(t)) = a\mathcal{L}(f(t))+b\mathcal{L}(g(t))$

Table of Common Laplace Transforms

The last entry illustrates the Exponential Shift Theorem, indicating that multiplying a function by an exponential term e^at shifts its Laplace Transform.

Additionally, the Laplace Transforms of derivatives are given by: $\mathcal{L}(f’(t)) = sF(s)-f(0), \mathcal{L}(f’’(t)) = s^2F(s)-sf(0)-f’(0)$

• $\mathcal{L}(f(at)), \mathcal{L}(sin(bt))$

The Laplace transform of f(at) is given by: $\mathcal{L}(f(at)) = \int_{0}^{∞} e^{-st}f(at)dt =$ We perform a change of variables, letting u = at. Then: du = adt. When t = 0, u = 0. When t = ∞, u = ∞.

Substituting these into the integral, we get: $\int_{0}^{∞} e^{-s\frac{u}{a}}f(u)\frac{du}{a} = \frac{1}{a} \int_{0}^{∞} e^{-\frac{s}{a}u}f(u)du$ =[Recognizing that the integral on the right is F(s) evaluated at ^s⁄_a] $\frac{1}{a}F(\frac{s}{a})$

$\mathcal{L}(sin(bt)) =[\text{To find the Laplace transform of sin(bt), we use the result from the previous section } \mathcal{L}(f(at)) = \frac{1}{a}F(\frac{s}{a})] \frac{1}{b}F(\frac{s}{b}) = [sin(t) \leadsto \frac{1}{s^2+1}] \frac{1}{b}·\frac{1}{(\frac{s}{b})^2+1} = \frac{1}{b}·\frac{b^2}{s^2+b^2} = \frac{b}{s^2+b^2}$

• $\mathcal{L}(\frac{f(t)}{t})$

We aim to calculate $\mathcal{L}(\frac{f(t)}{t}) = \int_{0}^{∞} e^{-st}\frac{f(t)}{t}dt$

We can express $\frac{e^{-st}}{t}$ as an integral over u: $\int_{s}^{∞} e^{-ut}du = \frac{1}{-t}e^{-ut}\bigg|_{s}^{∞} = 0 -(\frac{1}{-t})e^{-st} = \frac{e^{-st}}{t}$

$\mathcal{L}(f(t)) = \int_{0}^{∞} e^{-st}\frac{f(t)}{t}dt = \int_{0}^{∞} (\int_{s}^{∞} e^{-ut}du)f(t)dt = [\text{Interchange the Order of Integration}] \int_{s}^{∞} (\int_{0}^{∞} e^{-ut}f(t)dt)du = \int_{s}^{∞} F(u)du$

• $\mathcal{L}(\frac{sin(t)}{t})$

We know that $\mathcal{L}(sin(t)) = \frac{1}{s^2+1} =[\text{Using a dummy variable}] \frac{1}{u^2+1}$, and from the previous exercise $\mathcal{L}(\frac{f(t)}{t}) = \int_{s}^{∞} F(u)du$

$\mathcal{L}(\frac{sin(t)}{t}) = \int_{s}^{∞} \frac{1}{u^2+1}du = tan^{-1}(u)\bigg|_{s}^{∞} = tan^{-1}(∞)-tan^{-1}(s) = \frac{π}{2}-tan^{-1}(s)$

Inverse Laplace Transform

• We aim to find the inverse Laplace Transform of the following expression: ¹⁄_s(s+3).

Step 1: Partial Fraction Decomposition. Express the rational function as a sum of simpler fractions: $\frac{1}{s(s+3)} =[\text{Using fractions decomposition}] = \frac{A}{s} + \frac{B}{s+3}$

Now, we need to find A and B. To do this, we rewrite the expression: $\frac{1}{s(s+3)} = \frac{A(s+3)+Bs}{s(s+3)} = \frac{As + 3A + Bs}{s(s+3)} =[\text{Grouping terms}] \frac{(A + B)s + 3A}{s(s+3)}$

Step 2: Solve for A and B

Expand and group like terms: 1 = (A + B)s + 3A ⇒ A + B = 0 (coefficient of s), 3A = 1 (constant term). Solving for A and B: $A = \frac{1}{3}, B = -\frac{1}{3}$

Step 3. Write the Decomposed Form and Apply Inverse Laplace Transform

Using standard inverse Laplace Transforms: $\frac{1}{s(s+3)} = \frac{\frac{1}{3}}{s} + \frac{\frac{-1}{3}}{s+3} \leadsto_\mathcal{L}^{-1} \leadsto \frac{1}{3} -\frac{1}{3}e^{-3t}$ where we are using that $1 \leadsto \frac{1}{s}, e^{at} \leadsto \frac{1}{s-a}$ where s > a. The inverse Laplace Transform of $\frac{1}{s(s+3)}$ is: $\frac{1}{s(s+3)} \leadsto_\mathcal{L}^{-1} \frac{1}{3}(1 -e^{-3t}) $

• Compute the inverse Laplace Transform of the following expression: ²⁄_(s+1)³.

Recall that $\mathcal{L}(e^{at}f(t)) = F(s-a), \mathcal{L}(t^{n}) = \frac{n!}{s^{n+1}}$. Our example is a translation with a = -1 of ¹⁄_s³ and n = 2.

$\mathcal{L}^{-1}(\frac{2}{(s+1)^3}) = 2·\mathcal{L}^{-1}(\frac{1}{(s+1)^3}) = 2·e^{-t}\frac{t^{2}}{2!} = e^{-t}t^2$

• Compute the inverse Laplace Transform of the following expression: $\frac{s}{s^2-2s+5}$

$\frac{s}{s^2-2s+5} = \frac{s}{(s-1)^2+2^2} = \frac{(s-1)+1}{(s-1)^2+2^2} = \frac{s-1}{(s-1)^2+2^2} + \frac{+1}{(s-1)^2+2^2}$

Recall $\mathcal{L}(sin(wt)) = \frac{w}{s^2+w^2}, \mathcal{L}(cos(wt)) = \frac{s}{s^2+w^2}, \mathcal{L}(e^{at}f(t)) = F(s-a)$

Applying the translation property again (with a = 1) and adjusting for the coefficient of the sine term: $\mathcal{L}^{-1}(\frac{s-1}{(s-1)^2+2^2} + \frac{+1}{(s-1)^2+2^2})= e^tcos(2t)+ \frac{1}{2}e^tsin(2t)$

Laplace Transforms of Polynomials

This is a fundamental result in Laplace Transform theory and is widely used in solving differential equations and analyzing systems.

Let’s calculate the Laplace Transform of tⁿ, where n is a non-negative integer. Recall that the Laplace Transform of a function f(t) is defined as:

$\mathcal{L}(f(t)) = F(s) = \int_{0}^{∞} f(t)e^{-st}dt$ where s is a complex number, typically with Re(s) > 0 to ensure convergence.

For f(t) = tⁿ, the Laplace Transform becomes:

$\mathcal{L}(t^n) = \int_{0}^{∞} t^ne^{-st}dt$

To compute the integral, we will use the method of integration by parts, which is based on the formula: ∫udv =uv − ∫vdu.

A common strategy is to choose u as a function that becomes simpler when differentiated, and dv as a function that remains manageable when integrated, e.g., u = tⁿ. We use the method of integration by parts repeatedly to reduce the power of t

u = tⁿ ⇒ du = nt^n-1dt, and dv = e^-stdt, so $v = \frac{e^{-st}}{-s}$.

Using the integration by parts formula ∫udv =[Part A] uv − [Part B] ∫vdu, we get:

[Part A] $t^n\frac{e^{-st}}{-s}\bigg|_{0}^{∞} $

[Part B] $- \int_{0}^{∞} nt^{n-1}\frac{e^{-st}}{-s}dt = [1]$

[Part A] $\lim_{t \to ∞} t^n\frac{e^{-st}}{-s} = \frac{1}{-s}\lim_{t \to ∞} \frac{t^n}{e^{st}} = 0$ by n applications of L’Hospital’s rule (s > 0). As t → ∞, e^-st decay exponentially (faster) while tⁿ grows potentially! At t = 0, the term is also zero: $0^n·\frac{e^0}{-s} = 0$

[1] = $ 0 -0 [\text{ Part A }] + \frac{n}{s}\int_{0}^{∞} t^{n-1}e^{-st}dt = \frac{n}{s}\mathcal{L}(t^{n-1})$. Thus, we have a recursive relation.

We can apply the recursive formula repeatedly: $\mathcal{L}(t^n) = \frac{n}{s}\mathcal{L}(t^{n-1}) = \frac{n}{s}\frac{n-1}{s}\mathcal{L}(t^{n-2}) = ··· = \frac{n(n-1)···1}{s^n}\mathcal{L}(t^0) = \frac{n!}{s^n}\mathcal{L}(1) = \frac{n!}{s^n}\frac{1}{s} = \frac{n!}{s^{n+1}}$ for s > 0.

This result is fundamental and provides a direct way to compute the Laplace Transform of any non-negative integer power of t.

Alternative Method Using the Gamma Function Laplace Transform of tⁿ

Recall that the Gamma function is defined as: $Γ(z) = \int_{0}^{∞} e^{-t}t^{z-1}dt$ for Re(z) > 0

$Γ(1) = \int_{0}^{∞} e^{-t}1^{z-1}dt = 1, Γ(n+1) = \int_{0}^{∞} e^{-t}t^{n}dt$

Integrating by parts. Let u = tⁿ ⇒ du = nt^n-1dt, dv = e^-tdt, v = -e^-t

$Γ(n+1) = \int_{0}^{∞} e^{-t}t^{n}dt = -e^{-t}t^n\bigg|_{0}^{∞}$

This first part ($-t^ne^{-t}\bigg|_{0}^{∞}$) is zero because the exponential decays exponentially, meaning faster than the polynomial grows.

$-\int_{0}^{∞} -e^{-t}·nt^{n-1}dt = n\int_{0}^{∞} e^{-t}·t^{n-1}dt = nΓ(n)$.

It can be recursively applied to get, Γ(n+1) = nΓ(n) = n(n-1)Γ(n-1) = ··· = n!

$\mathcal{L}(f(t)) = F(s) = \int_{0}^{∞} f(t)e^{-st}dt$ where s is a complex number, typically with Re(s) > 0 to ensure convergence.

By substituting st=u ⇒ sdt = du ⇒$dt = \frac{du}{s} $, we can write the integral as:

$\int_{0}^{∞} f(t)e^{-st}dt = \int_{0}^{∞} t^ne^{-st}dt = \int_{0}^{∞} e^{-u}\frac{u^n}{s^n}\frac{du}{s} = \frac{1}{s^{n+1}}\int_{0}^{∞} e^{-u}·u^ndu = \frac{1}{s^{n+1}}Γ(n+1) = \frac{n!}{s^{n+1}}$

Growth condition for the Laplace Transform

Recall that the Laplace Transform of a function f(t) is given by: $\mathcal{L}(f(t)) = F(s) = \int_{0}^{∞} f(t)e^{-st}dt$

To ensure the Laplace Transform of a function f(t) exists (i.e., the integral converges to a finite value), we need to impose certain conditions on the behavior of f(t) as t grows. Specifically, the function should not grow too rapidly for large t, otherwise, the exponential decay of e^-st may not be enough to ensure convergence, i.e., the integral involved in the Laplace Transform may not converge.

A function f(t) is said to be of exponential rate α if there exist positive constants M, α, and t₀ such that |f(t)| ≤ Me^αt, for all t ≥ t₀. This condition means that f(t) does not grow faster than an exponential function.

Conditions for the Existence of the Laplace Transform

For $\mathcal{L}(f(t))$ to exist:

1. Piecewise Continuity: f(t) must be piecewise continuous on the interval [0, ∞). This means that f(t) is continuous except at a finite number of points where it may have finite jump discontinuities (no vertical asymptotes).
2. Exponential order. A common condition that ensures the Laplace Transform is that f(t) must be of exponential order α.

Assuming f(t) is of exponential order α, we have: $|f(t)| ≤ Me^{αt}⇒ |f(t)e^{-st}| ≤ Me^{αt}e^{-st} = Me^{-(s-α)t}$ for t ≥ t₀

• If s > α, then (s -α)> 0, and $e^{-(s-α)t}$ decays exponentially as t → ∞. Consequently, the integrand $f(t)e^{-st}$ approaches zero, and the improper integral converges.
• However, if s ≤ α, the exponent -(s -α)t is non-negative or zero, and the integrand may not decay, or might even grow, and this can cause the integral to diverge.

$\int_{t_0}^{∞} |e^{-st}f(t)|dt ≤ \int_{t_0}^{∞} e^{-st}Me^{αt}dt = M\int_{t_0}^{∞} e^{ -(s-α)t}dt = M\frac{e^{-(s-α)t}}{-(s-α)}\bigg|_{t_0}^{∞} =[\text{s> α}] M(0-(\frac{e^{-(s-α)t_0}}{-(s-α)})) = M\frac{e^{-(s-α)t_0}}{(s-α)}$. This upper bound is finite when s > α, ensuring the integral converges on [t₀,∞).

$|F(s)| = |\int_{0}^{∞} e^{-st}f(t)dt| ≤ |\int_{0}^{t_0} e^{-st}f(t)dt| + |\int_{t_0}^{∞} e^{-st}f(t)dt|$

1. The first integral $|\int_{0}^{t_0} e^{-st}f(t)dt|$ is finite because f(t) is piecewise continuous on [0, t₀] and e^-st is bounded on this interval.
2. The second integral $|\int_{t_0}^{∞} e^{-st}f(t)dt|$ is finite for s > α as shown above.

Therefore, the Laplace Transform F(s) exists and is finite when s > α.

Examples and Counterexamples of Functions of Exponential Type

• Sine Function: The sine function oscillates between -1 and 1, so it is bounded. Therefore, |sin(t)| ≤ 1. So we can write: |sin(t)| ≤ 1·e^0·t. In this case, the sine function is of exponential type with α = 0, indicating that it does not grow exponentially at all. Laplace Transform of sin(t) exists for all s > 0.
• Polynomial functions: tⁿ is of exponential type α: tⁿ ≤ Me^αt for some M and all t > t₀.

Observation: For any positive α, the exponential function e^αt eventually outgrow any polynomial tⁿ as t → ∞.

$\frac{t^n}{e^{αt}}≤ M$ for some M, and this is true because $\lim_{t \to ∞} \frac{t^n}{e^{αt}} = 0$ by n applications of L’Hospital’s rule (Each differentiation reduces the degree of the numerator, it becomes a constant n! and the denominator remains an exponential function multiplied by a polynomial term, which still grows exponentially) -since e^αt grow faster than any polynomial tⁿ!

• Reciprocal Function. 1/t is not of exponential type because as t → 0, 1/t grows unboundedly. Additionally, near t = 0, the integrand behaves like 1/t, which is not integrable on [0, ε], the integral for its Laplace Transform diverges: $\int_{0}^{∞} e^{-st}\frac{1}{t}dt$.

  This integral diverges because ¹⁄_t behaves too wildly near t = 0, so the Laplace Transform of ​¹⁄_t does not exist. ¹⁄_t is not piecewise continuous on [0,∞) due to the infinite discontinuity at t = 0.

• Functions Growing Faster Than Exponential

If f(t) grows faster than any exponential function (e.g., f(t)= $e^{t^2}$), the Laplace Transform may not exist.

$f(t)e^{-st} = e^{t^2-st} = e^{t(t-s)}$. As t → ∞, t(t-s) → ∞, so $f(t)e^{-st}$ grows with bond, hence the Laplace Transform does not exist.

In practice, the exponential growth condition ensures that the Laplace Transform can handle a wide variety of useful functions, including polynomials, exponentials, and trigonometric functions, all of which appear frequently in solving differential equations and modeling physical systems.

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