Solving Linear Partial Differential Equations with Variable Coefficients Using the Method of Characteristics

Solving Linear Partial Differential Equations with Variable Coefficients Using the Method of Characteristics

Partial differential equations (PDEs) are widely used in various fields such as physics, engineering, and finance. This article focuses on solving a specific type of PDE with variable coefficients using the method of characteristics. We will guide you through the process with a detailed example and rigorous mathematical analysis.

Introduction

Partial differential equations (PDEs) are equations involving partial derivatives of an unknown function. In this article, we will solve the following PDE with variable coefficients:

[9t u_{xt} - u_{tt} 9u_x 0]

Step-by-Step Solution

Rewriting the PDE

We start by rewriting the given PDE to make it more manageable. By applying the product rule, we can rewrite the PDE as:

[u_{tt} 9tu_{xt} 9u_x]

Integrating Both Sides with Respect to t

Next, we integrate both sides of the equation with respect to t:

[u_t 9tu_x phi(x)]

where (phi(x)) is an arbitrary function of x.

Eliminating the Arbitrary Function

To eliminate the arbitrary function, we set (t 0) and use the initial condition (u_{x,t}(0) 0). This gives us:

[phi(x) 0]

Thus, we have reduced the PDE to:

[u_t 9tu_x]

Using the Method of Characteristics

To further simplify the PDE, we use the method of characteristics. The characteristic equations are given by:

[frac{dx}{t} 9] [frac{dt}{t} -1] [frac{du}{t} 9u_x]

From the first two characteristics, we get:

[9t , dt -dx implies 9t^2 -x C_1]

From the third characteristic, we have:

[frac{du}{t} 9u_x implies u - fu_x C_2]

Combining these, the general characteristic equations are:

[x frac{9}{2}t^2 C_1]

[u - f u_x C_2]

Final Solution

Using the initial condition (u(x, 0) 2x), we can determine the arbitrary function (psi). Setting (t 0) in the equation for (u_t), we get:

[2x psi(0) cdot 2x implies psi(0) 1]

This implies that (psi) is the identity function. Therefore, the solution to the original PDE is:

[u_t(x, t) 9t^2 - 2x]

Thus, the solution to the given PDE with the initial conditions is:

[boxed{u_t(x, t) 9t^2 - 2x}]

Conclusion

In this article, we demonstrated a systematic approach to solving a specific type of PDE with variable coefficients using the method of characteristics. This technique is particularly useful in various fields of applied mathematics and engineering. By understanding and applying these methods, you can solve more complex differential equations and gain valuable insights into the behavior of systems described by such equations.

Further Reading

For a deeper understanding of the method of characteristics and its applications, consider exploring the following resources:

Method of Characteristics (Wikipedia) Characteristics Method (SOSMath) Transversality and Characteristics (UCLA)

Keywords

Partial Differential Equation Variable Coefficients Method of Characteristics