Multiplication and Division - GATE Engineering Mathematics

Inverse Laplace Transform: Multiplication and Division Theorems, Advanced Problem Techniques

Multiplication Theorem in Laplace and Inverse Laplace

The multiplication theorem relates the Laplace transform of $t f(t)$ to the derivative of the Laplace transform of $f(t)$. This relationship is crucial for both direct and inverse Laplace transforms, especially when direct inversion is not straightforward.

• If $L\{f(t)\} = F(s)$, then $L\{t f(t)\} = -\frac{d}{ds} F(s)$.
• For the inverse, $L^{-1}\left\{\frac{d}{ds} F(s)\right\} = -t f(t)$.
• The minus sign can be moved inside or outside the differentiation without affecting the result.

Application of Multiplication Theorem to Inverse Laplace…

Inverse Laplace Transform: Multiplication and Division Theorems, Advanced Problem Techniques

Multiplication Theorem in Laplace and Inverse Laplace

The multiplication theorem relates the Laplace transform of $t f(t)$ to the derivative of the Laplace transform of $f(t)$. This relationship is crucial for both direct and inverse Laplace transforms, especially when direct inversion is not straightforward.

• If $L\{f(t)\} = F(s)$, then $L\{t f(t)\} = -\frac{d}{ds} F(s)$.
• For the inverse, $L^{-1}\left\{\frac{d}{ds} F(s)\right\} = -t f(t)$.
• The minus sign can be moved inside or outside the differentiation without affecting the result.

Application of Multiplication Theorem to Inverse Laplace

When the inverse Laplace of a function is not directly known, differentiation with respect to $s$ can simplify the process. This is particularly useful for functions involving logarithms or inverse trigonometric terms.

• If $L\{f(t)\} = F(s)$, then $L^{-1}\{F(s)\} = f(t)$.
• If $L\{t f(t)\} = -F'(s)$, then $L^{-1}\{F'(s)\} = -t f(t)$.
• To find $f(t)$ when $F(s)$ is complicated, differentiate $F(s)$, find the inverse Laplace of the result, and then divide by $-t$.

Solving Inverse Laplace Problems Using Differentiation

For functions like $\log\left(\frac{1+s}{s}\right)$ or $\log\left(\frac{s^2-1}{s^2}\right)$, direct inversion is not possible. Differentiation and properties of logarithms are used to simplify the function before applying the inverse Laplace.

• Example: $L^{-1}\left\{\log\left(\frac{1+s}{s}\right)\right\}$

  • Differentiate: $\frac{d}{ds} \log\left(\frac{1+s}{s}\right) = \frac{1}{1+s} - \frac{1}{s}$
  • Inverse Laplace: $L^{-1}\left\{\frac{1}{1+s}\right\} = e^{-t}$, $L^{-1}\left\{\frac{1}{s}\right\} = 1$
  • Combine: $e^{-t} - 1$
  • Final answer: $f(t) = \frac{e^{-t} - 1}{-t}$

• Example: $L^{-1}\left\{\log\left(\frac{s^2-1}{s^2}\right)\right\}$

  • Differentiate: $\frac{2s}{s^2-1} - \frac{2}{s}$
  • Inverse Laplace: $L^{-1}\left\{\frac{s}{s^2-1}\right\} = \cosh t$, $L^{-1}\left\{\frac{1}{s}\right\} = 1$
  • Combine: $2(\cosh t - 1)$
  • Final answer: $f(t) = \frac{2(1 - \cosh t)}{t}$

Handling Inverse Laplace of Inverse Trigonometric Functions

When the Laplace transform involves inverse trigonometric functions like $\cot^{-1}(s)$, differentiation and algebraic manipulation are used to bring the function into a standard form.

• For $L^{-1}\{\cot^{-1}(s)\}$:
  • Differentiate: $-\frac{1}{1+s^2}$
  • Inverse Laplace: $L^{-1}\left\{\frac{1}{1+s^2}\right\} = \sin t$
  • Apply shifting if $s$ is replaced by $s+a$.
  • For $L^{-1}\{\cot^{-1}\left(\frac{s+3}{2}\right)\}$, after differentiation and simplification, the answer involves $e^{-3t} \sin 2t / t$.

Division Theorem in Laplace and Inverse Laplace

The division theorem provides a method to find the Laplace transform of $f(t)/t$ and its inverse. This is especially useful for rational functions where direct inversion is complex.

• If $L\{f(t)\} = F(s)$, then $L\left\{\frac{f(t)}{t}\right\} = \int_s^\infty F(u) du$.
• For the inverse, $L^{-1}\left\{\frac{F(s)}{s}\right\} = \int_0^t f(\tau) d\tau$.
• This theorem is particularly effective when the function is divided by $s$.

Examples Using Division Theorem

The division theorem simplifies the inversion of functions like $1/(s(s^2+9))$ or $1/(s(s+7)^3)$, which would otherwise require lengthy partial fraction decomposition.

• For $L^{-1}\left\{\frac{1}{s(s^2+9)}\right\}$:

  • $L^{-1}\left\{\frac{1}{s^2+9}\right\} = \frac{1}{3} \sin 3t$
  • Integrate: $\int_0^t \frac{1}{3} \sin 3\tau d\tau = \frac{1}{9}(1 - \cos 3t)$

• For $L^{-1}\left\{\frac{1}{s(s+7)^3}\right\}$:

  • $L^{-1}\left\{\frac{1}{(s+7)^3}\right\} = \frac{t^2}{2} e^{-7t}$
  • Integrate repeatedly as per the number of $s$ in the denominator.

Advanced Partial Fraction and Manipulation Techniques

For complex rational functions, especially with higher powers or irreducible quadratics, partial fraction decomposition and algebraic manipulation are essential. Completing the square and factorization are often used to bring the denominator into a standard form.

• For $L^{-1}\left\{\frac{1}{s^2 - a^2}\right\}$, use partial fractions to split into terms involving $1/(s-a)$ and $1/(s+a)$.
• For quartic denominators like $s^4 + s^2 + 1$, rewrite as $(s^2 + 1)^2 - s^2$ and factor as $(s^2 + 1 + s)(s^2 + 1 - s)$.
• Numerators can be adjusted by adding and subtracting terms to match the denominator's structure, enabling the use of standard Laplace pairs.

Special Cases: Completing the Square and Shifting

When the denominator is a quartic or involves terms like $s^2 + a^2$ squared, completing the square is used to express the denominator in a form suitable for standard Laplace inverses. Shifting the $s$ variable (first shifting theorem) is applied when the Laplace pair involves $s+a$.

• For $L^{-1}\left\{\frac{s}{(s^2 + a^2)^2}\right\}$, recognize the connection to $t \cos at$.
• For $L^{-1}\left\{\frac{2s}{(s^2 + a^2)^2}\right\}$, relate to $t \sin at$.

Derivation of Laplace Transforms for $t \sin at$ and $t \cos at$

The Laplace transforms of $t \sin at$ and $t \cos at$ can be derived using Euler's formula and the shifting theorem. These derivations are useful for handling higher-order denominators and for constructing inverse transforms.

• $e^{i a t} = \cos at + i \sin at$
• $L\{t e^{i a t}\} = \frac{1}{(s - i a)^2}$
• Real part gives $L\{t \cos at\}$, imaginary part gives $L\{t \sin at\}$.
• After rationalizing the denominator, the real and imaginary parts correspond to:
  • $L\{t \cos at\} = \frac{s^2 - a^2}{(s^2 + a^2)^2}$
  • $L\{t \sin at\} = \frac{2as}{(s^2 + a^2)^2}$

Manipulation Strategies for Difficult Inverse Laplace Problems

For challenging problems where direct inversion is not possible, use the following strategies:

• Differentiate with respect to $s$ to simplify the function.
• Use properties of logarithms and inverse trigonometric functions to rewrite the function.
• Apply partial fraction decomposition for rational functions.
• Complete the square in the denominator to match standard Laplace pairs.
• Use the division theorem for functions divided by $s$.
• For quartic or higher-order denominators, factorize or manipulate into a product of quadratics.

Exam-Focused Takeaways and Common Pitfalls

Mastery of the multiplication and division theorems, as well as manipulation techniques, is essential for solving advanced inverse Laplace problems. Recognize when to use differentiation, integration, or algebraic manipulation based on the structure of the given function.

• Always check if the function can be simplified using logarithmic or trigonometric identities.
• For functions involving $s+a$, apply the first shifting theorem.
• When the denominator is a perfect square or quartic, attempt completing the square or factorization.
• Use the division theorem when the function is divided by $s$.
• Avoid unnecessary partial fraction decomposition if a theorem or manipulation can simplify the process.
• Practice is crucial for recognizing patterns and applying the correct technique under exam conditions.