19.2.3. Solving a PDE using Laplace Transform – Step-by-Step

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

In this section, we’re going to solve a specific type of partial differential equation (PDE), known as the heat equation, which models how heat diffuses through a medium over time. The equation is given as ∂𝑢/∂𝑡 = 𝛼² ∂²𝑢/∂𝑥². Here, 𝑢 represents the temperature at position 𝑥 and time 𝑡. We also have initial conditions: the temperature at time 0 is described by the function 𝑓(𝑥), and at position 0 for any time 𝑡, the temperature is held at 0.

Think of this situation like heating a metal rod. The equation describes how heat spreads along the rod over time. At the start, you may set a different temperature along the rod's length. But at one end, you cool it down to 0 degrees, modeling how the heat will change as time goes on.

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

In this step, we take the Laplace Transform of the heat equation. This transformation converts the problem from involving time (t) into one that involves a new variable (s). The property of the Laplace Transform simplifies the time derivative equations into algebraic expressions. After applying the transform, we substitute the initial condition into our transformed equation, which results in an ordinary differential equation (ODE) in terms of x.

Imagine turning an intricate recipe (the original PDE) into a simpler, more manageable list of steps (the ODE). By changing perspectives from time to a different variable, you simplify the process of understanding how all the ingredients fit together at once. It’s like cooking with a timer; once the timer is used, the cooking process becomes more straightforward to analyze.

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: 𝑑2𝑢‾ 𝛼2 −𝑠𝑢‾ = 0 𝑑𝑥2 Solve this ODE: 𝑑2𝑢‾ 𝑠 − 𝑢‾ = 0 𝑑𝑥2 𝛼2 This is a linear second-order equation with constant coefficients. Let 𝜆² = 𝑠/𝛼², the general solution is: 𝑢‾(𝑥,𝑠) = 𝐴𝑒𝜆𝑥 +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(𝑒𝜆𝑥 − 𝑒−𝜆𝑥) = 𝐴sinh(𝜆𝑥) If bounded as 𝑥 →∞, we choose only the decaying part → Final form depends on physical boundary conditions.

Now, we solve the ordinary differential equation we derived. This equation is linear and has constant coefficients, which allows us to apply standard techniques. We assume a specific form for our solution based on general principles of differential equations. Finally, by applying the boundary condition that the temperature at position 0 must be 0, we can find a specific solution that simplifies further analysis, focusing on terms that decay at infinity.

Think of it as trying to predict the temperature of a metal rod after heating one end. When analyzing the temperature, you realize that outwardly the effect of temperature diminishes as you move away from the heat source. Thus, you adjust your prediction (the solution) to incorporate only those parts that make sense. It’s like predicting how a flower will grow, knowing that it will wilt the further it gets from water.

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).

The final step involves applying the inverse Laplace Transform. This is where we convert our solution from the s-domain back to the time domain, giving us the temperature as a function of position and time. We can use tables of transforms or more complex methods to find this inverse. The goal is to return to the original variables of the problem, thereby providing a complete solution.

Imagine that you've baked a cake (the ODE solution) and now want to serve it as a beautiful dessert. The inverse transform is like frosting and decorating the cake; you’re bringing together all elements into a final product that can be enjoyed and understood. Just like that decorated cake, the final answer reflects all the work that went into it.

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