Inverse Laplace Transform Lecture

Introduction

• 13th lecture in a series on complex analysis.
• Goal: Solve inverse Laplace transform using complex analysis tools.
• Involves understanding a complex integral in the complex plane.

Why Laplace Transform Matters

• Laplace Transform: Converts time domain functions to a generalized frequency domain.
• Inverse Laplace Transform: Converts the frequency domain function back to the time domain.
• Applications:
  • Solving ODEs and PDEs.
  • Especially useful for parabolic PDEs (e.g., heat equation).
  • Used in control theory for analyzing input-output systems.
  • Common in engineering for transforming differential equations into easier algebraic forms.

Inverse Laplace Transform Formula

• Defined as:

  L^(-1){F(s)} = 1 / (2πi) ∫{γ-i∞, γ+i∞} F(s) e^(st) ds
  

  • γ: Real part greater than all poles of F(s)
  • t: Non-negative time values.

Example Problem

• Inverse Laplace transform of F(s) = 1 / (s - a).
• Laplace Transform Steps:
  1. Take a differential equation, e.g., dx/dt = Ax. Transform to: sX(s) - x(0) = AX(s).
  2. Solving gives: X(s) = x(0) / (s - A).
  3. Find the inverse Laplace transform using the complex integral.

Contour Integral - Bromwich Contour

• Contour setup in the complex plane to facilitate Cauchy's Integral Formula application.
• Main Idea: Convert complex integrals into a solvable form.

Bromwich Contour Elements

• Create a vertical line (integral path γ) through the complex plane where real part is greater than poles.
• Close the contour with semicircles (R) and horizontal lines (C+ and C-).
• Overall contour path is a combination of γ, C+, C-, and R paths.

Cauchy Integral Formula Recap

• Result: Closed contour integral of an analytic function divided by (s-a) simplifies by evaluating the function at a.
• Formula: ∮{C} f(z)/(z-a) dz = 2πi f(a)

Solving the Integral

• Objective: Show the complex contour integral equals the known solution e^(at).
• Break the contour into four integrals: ∮{C} = ∫{γ} + ∫{C+} + ∫{C-} + ∫{R}
• Prove integrals over C+, C-, and R go to zero as radius R goes to infinity.

ML Bound Method (C+ and C- Paths)

• Length L of paths C+ and C- is always γ.
• Show numerator integral e^(xt + iRt) magnitude bounded by e^(γt).
• Denominator term shows length dominated by R, hence ratio tends to zero as R goes to infinity.

Polar Coordinates Method (R Path)

• Switch to polar coordinates to handle large radius integral.
• Norms of numerator (e^(-R cos θ t)) and denominator ((R cos θ + a)^2 + (R sin θ)^2)^0.5) bounds calculated.
• Combine results to show limit behavior proves integral goes to zero.

Conclusion

• Summarize the procedure: Using Cauchy's Integral Formula and proving boundary integrals tend to zero proves the desired inverse Laplace transform result.

• Key Result: ∫{γ} of (e^(st) / (s - a)) ds = e^(at).

• Armed with this method, complex integrals and inverse transforms can be tackled methodically.

Final Notes

• This concludes an intricate application of complex function theory demonstrating the power of analytical techniques to solve challenging real-world problems.