Solving a Quasi-Linear PDE with the Method of Characteristics

In this article, we will explore the process of solving a given partial differential equation (PDE) using the method of characteristics. Specifically, we will tackle a quasi-linear PDE, focusing on a practical example and detailing the steps to find a solution. This method is widely used in solving various types of PDEs and is particularly effective for quasi-linear equations.

Problem Statement

The problem we are given is to solve the partial differential equation (PDE):

(u_x sin u u_t) at the point (u(7,0) frac{pi}{2}).

Solution Approach

Identify the Type of PDE

The given PDE is a quasi-linear first-order PDE. Quasi-linear PDEs are characterized by their ability to be solved using the method of characteristics, which involves transforming the PDE into a system of ordinary differential equations (ODEs).

Applying the Method of Characteristics

The method of characteristics requires us to solve the following system of ODEs:

[ frac{du}{dx} 0, quad frac{dt}{dx} -sin u cdot frac{dt}{dx} -sin u quad text{and} quad frac{dw}{dx} -sin u cdot frac{ dw}{dx} -sin u ]

General Solution

By solving the ODEs, we can find the general solution. We start with the ODE for (u) and (t):

[ frac{du}{dx} 0 quad Rightarrow quad u u_0 ]

Next, we solve the ODE for (t):

[ frac{dt}{dx} -sin u_0 quad Rightarrow quad sin u_0 x t C ]

Using algebra, we express (u_0) in terms of the constant (C):

[ sin u_0 x t C quad Rightarrow quad u_0 sin^{-1} left( frac{C - t}{x} right) ]

Thus, the general solution to the PDE is given by:

[ u(x, t) sin^{-1} left( frac{C - t}{x} right) ]

Determining the Constant (C)

Using the initial condition (u(7,0) frac{pi}{2}), we find the value of (C):

[ u(7, 0) sin^{-1} left( frac{C - 0}{7} right) frac{pi}{2} quad Rightarrow quad C 7 sin left( frac{pi}{2} right) 7 ]

Substituting (C 7) into the general solution, we get:

[ u(x, t) sin^{-1} left( frac{7 - t}{x} right) ]

Discussion

The solution given here is based on the method of characteristics, and it provides a way to solve the quasi-linear PDE. However, when analyzing the problem, we noticed that the equations can also be interpreted as (sin u) being constant along characteristics. This leads us to another potential solution:

[ sin u sin u_0 ]

Integrating this, we find:

[ u F sin u t ]

With the initial condition (u(7,0) frac{pi}{2}), we have:

[ F(7,0) frac{pi}{2} quad Rightarrow quad F frac{pi}{2} ]

Thus, another form of the solution is:

[ u frac{pi}{2} sin u t ]

While this solution is valid, it leaves the value of (u) at other points undefined without additional information about the values of (u) along other characteristics.

Therefore, the solution derived using the method of characteristics is the one that provides a complete determination of (u) along a single characteristic curve. This highlights the importance of using the method of characteristics for solving quasi-linear PDEs.

Conclusion

In conclusion, the method of characteristics is a powerful tool for solving quasi-linear PDEs. By transforming the given PDE into a system of ODEs, we can derive the general solution and determine the specific solution using initial conditions.

Understanding and applying this method is crucial for solving similar PDEs in various fields, including physics, engineering, and mathematics. If you need further assistance with PDEs or any other mathematical concepts, feel free to reach out!