The mathematical formulations of many problems in science and engineering reduce to study of first-order PDEs. The use of first-order PDEs was seen in gas flow problems, traffic flow problems, phenomenon of shock waves, the motion of wave fronts and quantum mechanics etc.. It is therefore essential to study the theory of first-order PDEs and the nature their solutions to analyze the related real-world problems.

We shall study first-order linear, quasi-linear and nonlinear PDEs and methods of solving these equations. An important method of characteristics is explained for these equations in which solving PDE reduces to solving an ODE system along a characteristics curve. Further, the Charpit’s method and the Jacobi’s method for nonlinear first-order PDEs are discussed.

## First Order Partial Differential Equation

A first order PDE in two independent variables $$x, y$$ and the dependent variable $$z$$ can be written in the form

$$f(x,y,z, \frac {\partial z} {\partial x}, \frac {\partial z} {\partial y} )= 0$$

For convenience, we set

Equation can be modified as $$f(x,y,z,p,q)=0$$

The equations of this type arise in many applications in geometry and physics. For
example, let us consider the following geometrical problem.

Problem 1: Find all functions $$z(x, y)$$ such that the tangent plane to the graph $$z = z(x, y)$$ at any arbitrary point $$(x_{0}, y_{0}, z(x_{0}, y_{0}))$$ passes through the origin characterized by the PDE $$xz_x + yz_y – z = 0$$.

Solution:    The equation of the tangent plane to the graph at $$(x_{0}, y_{0}, z(x_{0}, y_{0}))$$ is

$$z_x(x_0,y_0)(x-x_0)+z_y(x_0,y_0)-(z-z(x_0,y_0))=0$$

This plane passes through the origin $$(0, 0, 0)$$ and hence, we must have

$$-z_x(x_0,y_0)x_0-z_y(x_0,y_0+z(x_0,y_0)=0$$

For this equation to hold for all $$(x_{0}, y_{0})$$ in the domain of $$z, z$$ must satisfy

$$xz_x + yz_y – z = 0$$

which is a first-order PDE.

#### Classification of first-order PDEs

We classify the equation depending on the special forms of the function $$f$$.

1) If equation is of the form

$$a(x,y) \frac {\partial z} {\partial x} + b(x,y) \frac {\partial z} {\partial y} + c(x,y)z = d(x,y)$$

then it is called linear first-order PDE. Note that the function $$f$$ is linear in $$\frac {\partial z}{\partial x}, \frac {\partial z}{\partial y}$$ and $$z$$ with all coefficients depending on the independent variables $$x$$ and $$y$$ only.

2) If equation is of the form

$$a(x,y) \frac {\partial z} {\partial x} + b(x,y) \frac {\partial z} {\partial y} = c(x,y,z)$$

then it is called semi-linear because it is linear in the leading (highest-order) terms $$\frac {\partial z}{\partial x}, \frac {\partial z}{\partial y}$$. However, it need not be linear in $$z$$. Note that the coefficients of $$\frac {\partial z}{\partial x}, \frac {\partial z}{\partial y}$$ are functions of the independent variables only.

3) If equation is of the form

$$a(x,y,z) \frac {\partial z} {\partial x} + b(x,y,z) \frac {\partial z} {\partial y} = c(x,y,z)$$

then it is called quasi-linear PDE. Here the function $$f$$ is linear in the derivatives $$\frac {\partial z}{\partial x}, \frac {\partial z}{\partial y}$$ with the coefficients $$a, b and c$$ depending on the independent variables $$x$$ and $$y$$ as well as on the unknown $$z$$.

Note: Linear and Semi-linear equations are special cases of Quasi-linear equations.

Any equation that does not fit into one of these forms is called Non-linear.

Few examples are listed below

1. $$xz_x + yz_y = z$$            (linear)
2. $$xz_x + yz_y = z^2$$             (semilinear)
3. $$z_x + (x + y)z_y = xy$$             (linear)
4. $$zz_x + z_y = 0$$             (quasilinear)
5. $$xz^{2}_{x} + yz^{2}_{y} = 2$$             (nonlinear)

#### Cauchy’s problem or IVP for first-order PDEs

Recall the initial value problem for a first-order ODE which ask for a solution of the
equation that takes a given value at a given point of $$R$$. The IVP for first-order PDE ask for a solution of which has given values on a curve in $$R^2$$. The conditions to be satisfied in the case of IVP for first-order PDE are formulated in the classic problem of Cauchy which may be stated as follows:

Let $$C$$ be a given curve in $$R^2$$ described parametrically by the equations

$$x = x_0(s), y = y_0(s)$$ ;        $$s$$    $$\epsilon$$   $$I$$

where $$x_0(s), y_0(s)$$ are in $$C^{1}(I)$$. Let $$z_0(s)$$ be a given function in $$C^{1}(I)$$. The IVP or Cauchy’s problem for first-order PDE

$$f(x, y, z, p, q) = 0$$

is to find a function $$u = u(x, y)$$ with the following properties:

• $$u(x, y)$$ and its partial derivatives with respect to x and y are continuous in a region
Ω of $$R^2$$ containing the curve $$C$$
• $$u = u(x, y)$$ is a solution of $$f(x, y, z, p, q) = 0$$ in Ω, i.e., $$f(x, y, u(x, y), u_x(x, y), u_y(x, y)) = 0$$ in Ω
• On the curve $$C$$     $$u(x_0(s), y_0(s)) = z_0(s)$$,  $$s$$    $$\epsilon$$   $$I$$
The curve $$C$$ is called the initial curve of the problem and the function $$z_0(s)$$ is called the initial data. This equation is called the initial condition of the problem.

## Linear First-Order PDE

Problem 2: Determine the solution the following IVP:

$$\frac{\partial z}{\partial y}+c \frac{\partial z}{\partial x}=0$$ ,         $$z(x,0)=f(x)$$,

where $$f(x)$$ is a given function and $$c$$ is a constant.

Solution:

To apply the method of characteristics, parameterize the initial curve $$C$$ as follows:

$$x = s, y = 0, z = f(s)$$

The family of characteristics curves $$x((s, t), y(s, t))$$ are determined by solving the ODEs

$$\frac{\text{d}}{\text{d}t}x(s,t)=c$$             $$\frac{\text{d}}{\text{d}t}y(s,t)=1$$

The solution of the system is  $$x(s,t) = ct+c_{1}(s)$$  and $$y(s,t) = t+c_{2}(s)$$.

Using initial conditions,

$$x(s,0)=s$$,    $$y(s,0)=0$$.

we find that $$c_{1}(s)=s$$, $$c_{2}(s)=0$$, and hence

$$x(s,t)=ct+s$$  and $$y(s,t)=t$$.

To modify equation in the parametric form of the solution, we have $$c(x,y)=0$$ and $$d(x,y)=0$$. Therefore, we find that $$d(s,t)=0,$$     $$\mu (s,t)=1$$.

Since $$z(x(s, 0), y(s, 0)) = z(s, 0) = g(s) = f(s)$$, we obtain $$z(s, t) = f(s)$$. Thus, the
parametric form of the solution of the problem is given by

$$x(s, t) = ct + s$$,    $$y(s, t) = t$$,    $$z(s, t) = f(s)$$.

To express $$s$$ and $$t$$ as $$s = s(x, y)$$ and $$t = t(x, y)$$, we have $$s =x-cy, t=y$$. We now write the solution in the explicit form as $$z(x,y)=z(s(x,y), y(x,y)) = f(x-cy)$$.

Clearly, if $$f(x)$$ is differentiable, the solution $$z(x,y) = f(x-cy)$$ satisfies given PDE as well as the initial condition.

NOTE:

This problem characterizes unidirectional wave motion with velocity $$c$$. If we consider the initial function $$z(x, 0) = f(x)$$ to represent a waveform, the solution $$z(x,y)=f(x-cy)$$ shows that a point $$x$$ for which $$x-cy = c$$ constant, will always occupy the same position on the wave form. If $$c > 0$$, the entire initial wave form $$f(x)$$ moves to the right without changing its shape with speed $$c$$ (if $$c < 0$$, the direction of motion is reversed).