Quasi-linear Partial Differential Equations

In calculus, we come across different differential equations, including partial differential equations and various forms of partial differential equations, one of which is the Quasi-linear partial differential equation. Before learning about Quasi-linear PDEs, let’s recall the definition of partial differential equations.

Partial Differential Equations

Differential equations involving partial derivatives of one or more dependent variables with more than one independent variable are known as partial differential equations (PDEs).

Linear Partial Differential Equations

If the dependent variable and its partial derivatives appear linearly in any partial differential equation, then the equation is said to be a linear partial differential equation; otherwise, it is a non-linear partial differential equation.

Click here to learn more about partial differential equations.

What are Quasi-linear Partial Differential Equations?

A partial differential equation is called a quasi-linear if all the terms with highest order derivatives of dependent variables appear linearly; that is, the coefficients of such terms are functions of merely lower-order derivatives of the dependent variables. In other words, if a partial differential equation is linear with respect to all the highest order derivatives of the unknown function, then it is called a quasi-linear partial differential equation.

The general form of a quasi-linear partial differential equation is given as:

Quasi-linear Partial Differential Equations Examples

Below are some examples of quasi-linear partial differential equations.

• xu_x + yu_y = x² + y² (or) x(∂u/∂x) + y(∂u/∂y) = x² + y²
• x(y² – u²)u_x + y(u² – x²)u_y = u(x² – y²)
• ux(∂²u/∂x²) + u²xy (∂²u/∂x∂y) + uy(∂²u/∂y²) + (∂u/∂x)² + (∂u/∂y)² = u³ = 0

First Order Quasi-linear Partial Differential Equations

The general form of the first order quasi-linear partial differential equation is given by:

L_{v(x, u)} [u] = v₁(x, u) (∂u/∂x₁) + v₂(x, u) (∂u/∂x₂) + …. + v_n(x, u) (∂u/∂x_n) = f(x, u)

We can also write the first order quasi-linear PDE with two independent variables x and y as:

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

Similarly, we can write the second-order quasi-linear PDE. This can be written as:

(∂u/∂x) (∂²u/∂x²) + (∂u/∂y) (∂²u/∂y²) + u² = 0

Semilinear Partial Differential Equations

Consider the first-order quasi-linear partial differential equation,

L_{v(x, u)} [u] = v₁(x, u) (∂u/∂x₁) + v₂(x, u) (∂u/∂x₂) + …. + v_n(x, u) (∂u/∂x_n) = f(x, u)

(or)

Here, if v_i = v_i(x), i.e., all the coefficients v_i’s are independent of u, then the obtained equation is called Semilinear partial differential equation.

How to Solve Quasi-linear Partial Differential Equations?

Let’s learn how to find the solution of a quasi-linear partial differential equation with the help of a solved example given below.

Question:

Find the general solution to the following quasi-linear partial differential equation.

(y + ux)u_x – (x + yu)u_y = x² – y²

Solution:

Given quasi-linear partial differential equation is:

(y + ux)u_x – (x + yu)u_y = x² – y²

i.e., (y + ux)(∂u/∂x) – (x + yu)(∂u/∂y) = x² – y²

On comparing with the standard form of quasi-linear PDE, we get;

a(x, y, u) = a = y + ux

b(x, y, u) = b = -(x + uy)

c(x, y, u) = c = x² – y²

The characteristic equations for the given PDE are given by:

dx/a = dy/b = du/c

dx/(y + ux) = dy/-(x + uy) = du/(x² – y²)………..(1)

This can be written as:

(xdx + ydy – udu)/ (xy + x²u – xy – y²u – ux² + uy²) = du/(x² – y²)

⇒ xdx + ydy − udu = 0 or x² + y² − u² = k₁

From equation (1), we can also get;

(ydx + xdy + udu)/ (y² + xyu – x² – xyu + x² – y²) = du/(x² – y²)

⇒ ydx + xdy + udu = 0

Thus, 2xy + u² = k₂

Therefore, the general solution is 2xy + u² = f(x² + y² − u² ) where f is an arbitrary differentiable function.

Practice Problems

1. Find the general solution of the PDE (3y – 2u)u_x + (u – 3x)u_y = 2x – y.
2. Determine the general solution of the quasi-linear partial differential equation (x² + 3y² + 3u²)u_x − 2xyuu_y + 2xu = 0.
3. Find the general solution to the quasi-linear partial differential equation (y − u)u_x + (u − x)u_y = x − y.