Partial Differential Equations (PDEs) model a wide range of physical phenomena, including heat conduction, wave propagation, fluid flow, and quantum mechanics. While solving PDEs analytically can be challenging, Laplace Transforms provide a powerful technique that simplifies complex PDEs—especially those involving time-dependent processes—by converting them into easier Ordinary Differential Equations (ODEs) in the spatial variable. Once solved, the inverse Laplace Transform is used to retrieve the original solution. This chapter explores the application of Laplace Transform methods in solving PDEs, focusing on linear PDEs with initial and boundary conditions.

Let 𝑓(𝑡) be a function defined for 𝑡 ≥ 0. The Laplace Transform of 𝑓(𝑡) is defined as:

ℒ{𝑓(𝑡)}= 𝐹(𝑠) = ∫ 𝑒^{−𝑠𝑡} 𝑓(𝑡) 𝑑𝑡

Common Properties:
• Linearity:
ℒ{𝑎𝑓(𝑡)+ 𝑏𝑔(𝑡)}= 𝑎𝐹(𝑠) +𝑏𝐺(𝑠)
• Derivatives:
ℒ{𝑓′(𝑡)}= 𝑠𝐹(𝑠) −𝑓(0)
ℒ{𝑓″(𝑡)} = 𝑠^2𝐹(𝑠)− 𝑠𝑓(0)− 𝑓′(0)

These properties are instrumental in transforming time-dependent PDEs.

Laplace Transforms are particularly effective for solving linear PDEs with constant coefficients and initial value problems (IVPs), because:
• Time derivatives become algebraic terms in 𝑠
• Initial conditions are automatically embedded in the transformed equations
• The PDE is converted into an easier-to-solve ODE

1. Heat Equation (One-Dimensional):
   ∂𝑢/∂𝑡 = 𝛼^2 ∂²𝑢/∂𝑥² with 𝑢(𝑥,0) = 𝑓(𝑥)
2. Wave Equation:
   ∂²𝑢/∂𝑡² = 𝑐^2 ∂²𝑢/∂𝑥² with initial conditions 𝑢(𝑥,0) = 𝑓(𝑥), 𝑢(𝑡,0) = 𝑔(𝑥)
3. Laplace’s Equation (in non-time domain, but in spatial PDEs)

Example: Solve the heat equation ∂𝑢/∂𝑡 = 𝛼² ∂²𝑢/∂𝑥², with 𝑢(𝑥,0) = 𝑓(𝑥), 𝑢(0,𝑡) = 0.

Step 1: Take Laplace Transform (w.r.t. 𝑡)
Let ℒ{𝑢(𝑥,𝑡)} = 𝑢‾(𝑥,𝑠)
Apply Laplace to both sides:
ℒ{∂𝑢/∂𝑡} = 𝛼²ℒ{∂²𝑢/∂𝑥²}
Using properties:
𝑠𝑢‾(𝑥,𝑠) − 𝑢(𝑥,0) = 𝛼² 𝑑²𝑢‾/𝑑𝑥²
Substitute 𝑢(𝑥,0) = 𝑓(𝑥):
𝛼² −𝑠𝑢‾(𝑥,𝑠) = −𝑓(𝑥)
This is now a second-order Ordinary Differential Equation in 𝑥.

Step 2: Solve the resulting ODE
In many standard problems, 𝑓(𝑥) is simple (e.g., 𝑓(𝑥) = sin𝑥, 𝑒^{−𝑥}, etc.). Let’s assume 𝑓(𝑥) = 0 for simplicity:

𝛼² −𝑠𝑢‾ = 0
Solve this ODE:
𝑑²𝑢‾/𝑑𝑥² − 𝑢‾ = 0
This is a linear second-order equation with constant coefficients. Let 𝜆² = 𝑠/𝛼², the general solution is:
𝑢‾(𝑥,𝑠) = 𝐴𝑒^{𝜆𝑥} + 𝐵𝑒^{−𝜆𝑥}
From boundary condition 𝑢(0,𝑡) = 0, so 𝑢‾(0,𝑠) = 0 implies 𝐴 + 𝐵 = 0 → 𝐴 = −𝐵
Thus, 𝑢‾(𝑥,𝑠) = 𝐴(𝑒^{𝜆𝑥} − 𝑒^{−𝜆𝑥}) = 𝐴sinh(𝜆𝑥).

Step 3: Take Inverse Laplace Transform
Once you have 𝑢‾(𝑥,𝑠), use inverse Laplace transform techniques (by table or complex inversion) to obtain 𝑢(𝑥,𝑡).

• Handles initial conditions naturally.
• Converts PDEs to simpler ODEs.
• Avoids use of complex separation of variables or infinite series.
• Effective for semi-infinite or infinite domains.
• Especially useful in engineering problems with time dependence (e.g., transient heat conduction).

• Only works for linear PDEs with constant coefficients.
• Requires initial value problems (not always applicable to boundary-only problems).
• May be difficult to find inverse Laplace for complex expressions.

Laplace Transform is a powerful technique for solving linear partial differential equations, particularly those involving time-dependent processes. By transforming the PDE into an ODE, the problem becomes more manageable, and the solution can be transformed back using the inverse Laplace transform. This method finds extensive application in physics and engineering, such as heat conduction, wave propagation, and electrical circuits.