5.5.1. Example 1

Solve:

∂𝑧 ∂𝑧 𝑝 + 𝑞 = 𝑧 ⇒ + = 𝑧 ∂𝑥 ∂𝑦

In this chunk, we are presented with a partial differential equation (PDE) that we need to solve. Essentially, it's expressing a relationship between the rates of change of a function 'z' with respect to two variables, 'x' and 'y', along with the function 'z' itself on the right side of the equation. This type of equation falls under Lagrange's linear equations, which we will now solve using the method of characteristics.

Think of this PDE like a recipe where the ingredients ('x' and 'y') influence the final dish ('z'). The equation tells us how blending these variables together impacts the final outcome.

Solution: Here,

• 𝑃 = 1, 𝑄 = 1, 𝑅 = 𝑧

In this section, we identify the functions P, Q, and R from the PDE provided. Specifically, P and Q both equal 1, while R is equal to z. These values will be used in the next step where we set up our auxiliary (characteristic) equations. Identifying these correctly is vital, as it allows us to derive the relationships needed for finding the solution.

Imagine you're solving a puzzle. Identifying each piece and where it fits is crucial before you can see the complete picture. In our case, identifying P, Q, and R helps us clarify how they interact in the larger equation.

Auxiliary equations:

𝑑𝑥 𝑑𝑦 𝑑𝑧 = = 1 1 𝑧

Here, we create the auxiliary equations based on the identified P, Q, and R. This results in a system that allows us to analyze the relationships between the variables x, y, and z. The approach is to take the derivatives and represent them as ratios, revealing a pathway to express the relationships analytically.

Think of auxiliary equations like roads on a map that guide us through the landscape of our problem. They show us how to navigate from point A to point B in our PDE solution journey.

From 𝑑𝑥 = 𝑑𝑦 ⇒ 𝑥 − 𝑦 = 𝑐1 From 𝑑𝑥 = 𝑑𝑧 ⇒ ∫𝑑𝑥 = ∫ 𝑑𝑧 ⇒ 𝑥 = ln𝑧 +𝑐 ⇒ 𝑧 = 𝑒𝑥−𝑐2

This chunk involves integrating the auxiliary equations to derive expressions for the variables. By integrating, we find that x - y is a constant (c1), which acts as one solution. From the integration of dz, we discover another relationship, allowing us to express z in terms of x. Thus, we progress towards obtaining the general solution of this PDE.

Integration here is similar to putting together a jigsaw puzzle; each piece represents an equation that, when combined, reveals a part of the picture. The constants we encounter act like 'landmarks' on our puzzle map, guiding us where to place the next piece.

Now,

• 𝑢 = 𝑥 − 𝑦 = 𝑐1 • 𝑣 = 𝑧𝑒−𝑥 = 𝑐2

In this chunk, we define our two independent solutions, u and v. Each of these represents a characteristic curve obtained through earlier calculations. The equations for u and v are significant in the next stage, where we will formulate the general solution to the PDE. Recognizing these independent solutions is crucial as they guide the next steps in finding the overall function.

You can think of u and v as two separate paths leading to a common destination. Despite their differences, they both ultimately help us understand the overall landscape shaped by our PDE.

General solution:

𝜙(𝑥−𝑦,𝑧𝑒−𝑥) = 0 or 𝑧 = 𝑒𝑥𝑓(𝑥− 𝑦)

Finally, we arrive at the general solution of our PDE, expressed as a function 𝜙 in terms of the independent solutions u and v. This step synthesizes all previous calculations into a concise form that articulates how the variables interact. It's a pivotal moment because it encapsulates the findings into a single mathematical expression.

Arriving at the general solution is like completing the final drawing of an intricate landscape after assembling all the pieces. Once everything is in place, we can see how the variables interrelate and influence each other in the wider context of our problem.

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