Origin of Partial Differential Equations (PDEs) + First-Order Linear & Non-Linear PDEs

1. Origin of Partial Differential Equations

Partial Differential Equations arise in physical problems where the unknown function u depends on two or more independent variables (typically space and time).

Common Classical PDEs and Their Physical Origins

PDE	Equation	Physical Meaning
Wave Equation	$$\frac{\partial^2 u}{\partial t^2} = c^2 \nabla^2 u$$	Vibration of string/membrane, sound waves, electromagnetic waves
Heat Equation	$$\frac{\partial u}{\partial t} = \alpha \nabla^2 u$$	Heat conduction, diffusion processes
Laplace Equation	$$\nabla^2 u = 0$$	Steady-state temperature, electrostatic potential
Poisson Equation	$$\nabla^2 u = f(x,y,z)$$	Electrostatics with charge density

2. First-Order PDEs – General Form

A first-order PDE in two independent variables x, y and dependent variable u(x,y) is

\[ F(x, y, u, u_x, u_y) = 0 \] where $ p = u_x = \frac{\partial u}{\partial x} $, $ q = u_y = \frac{\partial u}{\partial y} $.

3. Classification of First-Order PDEs

Type	Form	Example
Linear	$a(x,y,u)\, p + b(x,y,u)\, q = c(x,y,u)$	$x p + y q = u$
Semi-linear	$a(x,y)\, p + b(x,y)\, q = c(x,y,u)$	$x p + y q = u^2$
Quasi-linear	$a(x,y,u)\, p + b(x,y,u)\, q = c(x,y,u)$	$u p + q = 1$
Fully Non-linear	Non-linear in p and q	$p^2 + q^2 = 1$ (Eikonal equation)

4. Solution of Linear First-Order PDE (Lagrange’s Method)

Standard form:

\[ P(x,y,u) p + Q(x,y,u) q = R(x,y,u) \]

Method: Solve the characteristic equations (auxiliary equations)

\[ \frac{dx}{P} = \frac{dy}{Q} = \frac{du}{R} \]

Any two independent integrals give the general solution $ f(u_1, u_2) = 0 $.

Example 1 (Linear): Solve $x p + y q = z$ (where z = u(x,y))

Here P = x, Q = y, R = z

Characteristic equations:

\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z} \]

Integrating:

\[ \frac{dx}{x} = \frac{dy}{y} \implies \ln x - \ln y = c_1 \implies \frac{x}{y} = c_1 \] \[ \frac{dy}{y} = \frac{dz}{z} \implies \ln y - \ln z = c_2 \implies \frac{y}{z} = c_2 \]

General solution: $ \frac{x}{y} = \phi\left(\frac{y}{z}\right) $

or $ z = y \cdot \phi\left(\frac{x}{y}\right) $

5. Solution of Non-Linear First-Order PDE (Charpit’s Method)

For general non-linear PDE: $ f(x,y,u,p,q) = 0 $

Charpit’s auxiliary equations:

\[ \frac{dx}{f_p} = \frac{dy}{f_q} = \frac{du}{p f_p + q f_q} = \frac{dp}{-(f_x + p f_u)} = \frac{dq}{-(f_y + q f_u)} \]

Key step: From the last two,

\[ f_p \, dp + (f_x + p f_u) dp = 0 \quad \text{(solve for p in terms of x,y,u,a)}\]

Then substitute and integrate to get complete integral with two constants.

Example 2 (Non-linear – Clairautoc equation): Solve $u = x p + y q + f(p,q)$

Let the PDE be $f = u - x p - y q - f(p,q) = 0$

Charpit’s equations give a relation → the complete integral is

Complete integral: $ u = a x + b y + f(a,b) $

where a, b are arbitrary constants.

(This is the standard form for all Clairaut-type equations.)

Example 3 (Fully Non-linear): Solve $p^2 + q^2 = 1$ (Eikonal equation)

Here f = p² + q² – 1 = 0

Charpit:

\[ \frac{dp}{-(f_x + p f_u)} = \frac{dp}{0} \implies dp = 0 \implies p = a \text{ (constant)} \]

Then from f = 0: q = ±√(1 – a²) = b (say)

Now

\[ \frac{dx}{f_p} = \frac{dx}{2p} = \frac{dx}{2a}, \quad \frac{dy}{f_q} = \frac{dy}{2q} = \frac{dy}{2b} \]

Integrate:

\[ dx = 2a ds, \quad dy = 2b ds \implies x = 2a s + c_1, \quad y = 2b s + c_2 \]

Eliminate parameter s using du = p dx + q dy:

Complete integral: $ u = a x + b y + c $ where $ a^2 + b^2 = 1 $

Summary Table

Type	Method	Key Idea
Linear	Lagrange’s (Characteristics)	Solve $\frac{dx}{P} = \frac{dy}{Q} = \frac{du}{R}$
Non-linear	Charpit’s Method	Introduce parameter from $f_p dp + (f_x + p f_u)dp = 0$
Clairaut form	Direct	Solution $u = ax + by + f(a,b)$

These concepts and solved examples cover the entire topic of origin of PDEs and first-order linear/non-linear PDEs as asked in most university examinations.