Lagrange's Method: Solving First Order PDEs Explained

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Lagrange's Method: Solving First Order PDEs Explained

Hey guys! Today, we're diving into a super useful technique for tackling first-order partial differential equations (PDEs) called Lagrange's Method. If you've ever felt lost in the world of PDEs, don't worry; we'll break it down step by step with an example that’ll make everything crystal clear. So, buckle up, and let's get started!

What is Lagrange's Method?

Lagrange's method, also known as the method of characteristics, is a powerful technique used to solve first-order linear and quasi-linear partial differential equations (PDEs). These types of PDEs pop up in various fields, including fluid dynamics, gas dynamics, and traffic flow. The method's beauty lies in its ability to transform a PDE into a system of ordinary differential equations (ODEs), which are often much easier to solve. Instead of directly solving the original PDE, we find characteristic curves along which the solution remains constant. This simplifies the problem and allows us to find a general solution more straightforwardly.

At its heart, Lagrange's method exploits the geometry of the PDE. Think of the solution to a PDE as a surface in a multi-dimensional space. Lagrange's method identifies curves on this surface along which the solution doesn't change. These curves are called characteristic curves, and they provide a pathway to understanding the overall behavior of the solution. By integrating along these characteristic curves, we can determine the solution's value at any point, provided we know its value at some initial point. The power of this method comes from its ability to reduce a complex PDE problem into a set of simpler ODE problems, making it more manageable and intuitive to solve.

The method is applicable to first-order PDEs of the form:

a(x, y, u) * ∂u/∂x + b(x, y, u) * ∂u/∂y = c(x, y, u)

Where u is the unknown function of x and y, and a, b, and c are given functions. Lagrange's method provides a systematic way to find the general solution u(x, y) for such equations.

The Auxiliary Equations

The cornerstone of Lagrange's method is the set of auxiliary equations, also known as characteristic equations. These equations are derived directly from the coefficients of the PDE and form a system of ODEs that define the characteristic curves. The auxiliary equations are given by:

dx/a(x, y, u) = dy/b(x, y, u) = du/c(x, y, u)

Each term represents a ratio, and the equality between these ratios allows us to form two independent ODEs. Solving these ODEs will give us two independent integrals (solutions), which are crucial for constructing the general solution of the original PDE. These integrals represent the characteristic curves along which the solution u(x, y) remains constant.

The auxiliary equations essentially tell us how x, y, and u are related along the characteristic curves. By integrating these equations, we find relationships between these variables that describe the paths along which the solution remains constant. These paths are the characteristic curves, and they provide the framework for constructing the general solution of the PDE. The beauty of this approach is that it transforms the problem from solving a PDE directly to solving a set of ODEs, which are often much easier to handle analytically.

Steps to Solve a First Order PDE Using Lagrange's Method

Alright, let's break down the process into manageable steps:

  1. Identify the PDE: Make sure the given PDE is in the form:

    a(x, y, u) * ∂u/∂x + b(x, y, u) * ∂u/∂y = c(x, y, u)

    where a, b, and c are functions of x, y, and u.

  2. Write the Auxiliary Equations: Form the auxiliary equations:

    dx/a(x, y, u) = dy/b(x, y, u) = du/c(x, y, u)

  3. Solve the Auxiliary Equations: Obtain two independent solutions (integrals) from these equations. Let's call them:

    f(x, y, u) = C1
g(x, y, u) = C2

    where C1 and C2 are arbitrary constants.

  4. Write the General Solution: The general solution u(x, y) can be expressed as:

    F(f(x, y, u), g(x, y, u)) = 0

    or equivalently:

    f(x, y, u) = G(g(x, y, u))

    where F and G are arbitrary functions.

Example: Solving a First Order PDE Using Lagrange's Method

Let's illustrate Lagrange's method with a concrete example. Consider the following first-order PDE:

y * ∂u/∂x + x * ∂u/∂y = 0

Here, a(x, y, u) = y, b(x, y, u) = x, and c(x, y, u) = 0.

Step 1: Write the Auxiliary Equations

The auxiliary equations are:

dx/y = dy/x = du/0

Step 2: Solve the Auxiliary Equations

From dx/y = dy/x, we have:

x * dx = y * dy

Integrating both sides, we get:

∫ x dx = ∫ y dy

(x^2)/2 = (y^2)/2 + C1'

Simplifying, we obtain the first integral:

x^2 - y^2 = C1

From du/0, we have du = 0. Integrating, we get:

u = C2

Thus, the second integral is:

u = C2

Step 3: Write the General Solution

The general solution can be expressed as:

F(x^2 - y^2, u) = 0

Or, equivalently:

u = G(x^2 - y^2)

where G is an arbitrary function. This means the solution u is constant along the curves x^2 - y^2 = C1.

Verification: To verify the solution, we compute the partial derivatives of u:

∂u/∂x = G'(x^2 - y^2) * 2x
∂u/∂y = G'(x^2 - y^2) * (-2y)

Substituting these into the original PDE:

y * (G'(x^2 - y^2) * 2x) + x * (G'(x^2 - y^2) * (-2y)) = 0

2xy * G'(x^2 - y^2) - 2xy * G'(x^2 - y^2) = 0

0 = 0

Since the equation holds true, our solution is verified.

Another Detailed Example

Let's consider another example to solidify our understanding. Solve the following PDE:

(x + u) * ∂u/∂x + y * ∂u/∂y = y + u

Step 1: Write the Auxiliary Equations

The auxiliary equations are:

dx/(x + u) = dy/y = du/(y + u)

Step 2: Solve the Auxiliary Equations

First, let's use dy/y = du/(y + u). Rearranging, we get:

du/dy = (y + u) / y = 1 + u/y

du/dy - u/y = 1

This is a first-order linear ODE. The integrating factor is:

IF = e^(∫(-1/y) dy) = e^(-ln|y|) = 1/y

Multiplying the ODE by the integrating factor:

(1/y) * (du/dy) - (1/y^2) * u = 1/y

d/dy (u/y) = 1/y

Integrating both sides with respect to y:

∫ d/dy (u/y) dy = ∫ (1/y) dy

u/y = ln|y| + C1'

Thus, the first integral is:

u - y * ln|y| = C1 * y

(u/y) - ln|y| = C1

Now, let's use dx/(x + u) = dy/y. We need to express u in terms of x and y using the first integral. From the first integral, we have:

u = y * ln|y| + C1 * y

Substituting this into dx/(x + u) = dy/y:

dx / (x + y * ln|y| + C1 * y) = dy / y

This is a bit tricky to solve directly. Instead, let’s manipulate the auxiliary equations differently. Consider:

dx/(x+u) = du/(y+u)

dx(y+u) = du(x+u)

y dx + u dx = x du + u du

y dx - x du + u dx - u du = 0

This doesn't seem to simplify easily. Let's try:

dx/(x + u) = dy/y

y dx = (x + u) dy

y dx - x dy = u dy

(y dx - x dy) / y^2 = (u/y^2) dy

d(x/y) = (u/y^2) dy

Now, substitute u = yln*|y| + C1*y:

d(x/y) = (y ln|y| + C1y)/y^2 dy

d(x/y) = (ln|y| + C1)/y dy

Integrating both sides:

x/y = ∫ (ln|y| + C1)/y dy

x/y = ∫ (ln|y|)/y dy + C1 ∫ (1/y) dy

Let v = ln|y|, then dv = (1/y) dy:

x/y = ∫ v dv + C1 ln|y|

x/y = (v^2)/2 + C1 ln|y| + C2'

x/y = (ln|y|)^2 / 2 + C1 ln|y| + C2'

Let's rewrite C1 as C3 for clarity:

x/y = (ln|y|)^2 / 2 + C3 ln|y| + C2'

Rearranging, we get the second integral:

x/y - (ln|y|)^2 / 2 - C3 ln|y| = C2'

Step 3: Write the General Solution

We have two integrals:

(u/y) - ln|y| = C1

x/y - (ln|y|)^2 / 2 - C3 ln|y| = C2'

The general solution can be expressed as:

F((u/y) - ln|y|, x/y - (ln|y|)^2 / 2 - C3 ln|y|) = 0

Or equivalently:

(u/y) - ln|y| = G(x/y - (ln|y|)^2 / 2 - C3 ln|y|)

where F and G are arbitrary functions.

Conclusion

So there you have it! Lagrange's method is a fantastic way to solve first-order PDEs. It might seem a bit complex at first, but with practice and a clear understanding of the steps, you'll be solving PDEs like a pro in no time. Remember, the key is to break down the problem into smaller, manageable parts and to carefully solve the auxiliary equations. Keep practicing, and you'll get the hang of it! Happy solving, guys!