Partial Differential Equations (PDEs) – Complete Formulas and Concepts with Clear Explanation
1. Origin of Partial Differential Equations
PDEs arise when a physical phenomenon depends on more than one independent variable (e.g., time t and space x, y, z).
Common examples:
- Wave equation (vibration of string, sound waves):
$$\dfrac{\partial^2 u}{\partial t^2} = c^2 \dfrac{\partial^2 u}{\partial x^2}$$ - Heat conduction equation:
$$\dfrac{\partial u}{\partial t} = \alpha \nabla^2 u = \alpha \left( \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} + \dfrac{\partial^2 u}{\partial z^2} \right)$$ - Laplace equation (steady-state temperature, electrostatics):
$$\nabla^2 u = 0 \Rightarrow \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} + \dfrac{\partial^2 u}{\partial z^2} = 0$$ - Poisson equation: $$\nabla^2 u = f(x,y,z)$$
2. Classification: Linear, Semi-linear, Quasi-linear, Fully Nonlinear (First-Order PDEs)
A general first-order PDE in two independent variables:
$$F(x, y, u, p, q) = 0$$, where $$p = u_x$$, $$q = u_y$$.
| 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 nonlinear | Depends nonlinearly on $p$ and $q$ | $p^2 + q^2 = 1$ (Eikonal equation) |
3. Lagrange’s Method (Linear First-Order PDE)
Standard form:
$$P(x,y,u) \dfrac{\partial u}{\partial x} + Q(x,y,u) \dfrac{\partial u}{\partial y} = R(x,y,u)$$
or $$P p + Q q = R$$
Method of solution: Solve the auxiliary (characteristic) equations
$$\dfrac{dx}{P} = \dfrac{dy}{Q} = \dfrac{du}{R}$$
The general solution is $f(u_1, u_2) = 0$, where $u_1(x,y,u) = c_1$, $u_2(x,y,u) = c_2$ are two independent integrals.
Example: $x p + y q = z$ (homogeneous)
$$\dfrac{dx}{x} = \dfrac{dy}{y} = \dfrac{dz}{z} \Rightarrow \ln x = \ln y + c \Rightarrow \dfrac{x}{y} = c_1$$, and $$\dfrac{y}{z} = c_2$$
Solution: $$\dfrac{x}{y} = \phi\left(\dfrac{y}{z}\right)$$ or $$u = y\,\phi\left(\dfrac{x}{y}\right)$$
4. Charpit’s Method (Non-linear First-Order PDE)
For $f(x,y,u,p,q) = 0$
Charpit’s auxiliary equations:
$$\dfrac{dx}{f_p} = \dfrac{dy}{f_q} = \dfrac{du}{p f_p + q f_q} = \dfrac{dp}{-(f_x + p f_u)} = \dfrac{dq}{-(f_y + q f_u)} = -\dfrac{dp}{ds} \quad \text{(parameter } s\text{)}$$
Key step: Solve
$$\dfrac{dp}{f_p} = \dfrac{-dp}{(f_x + p f_u)} \Rightarrow f_p \, dp + (f_x + p f_u) \, dp = 0$$
i.e., $$f_p \, dp = -(f_x + p f_u) \, dp$$ → find relation between $p$ and other variables (say $p = \phi(x,y,u,a)$).
Then integrate again to eliminate $p$ and get complete integral.
Standard complete integral form: $$u = g(x,y,a,b)$$, containing two arbitrary constants $a,b$.
Example (Clairaut equation): $$u = x p + y q + f(p,q)$$
Complete integral: $$u = a x + b y + f(a,b)$$
5. Cauchy’s Method of Characteristics (General Non-linear First-Order)
Same as Charpit’s method in essence. The characteristic curves are given by the system:
$$\dfrac{dx}{dt} = f_p, \quad \dfrac{dy}{dt} = f_q, \quad \dfrac{du}{dt} = p f_p + q f_q, \quad \dfrac{dp}{dt} = -(f_x + p f_u), \quad \dfrac{dq}{dt} = -(f_y + q f_u)$$
6. Linear Homogeneous PDE with Constant Coefficients (Higher Order)
General form (two variables):
$$\left( D^m + a_1 D^{m-1} D' + a_2 D^{m-2} (D')^2 + \cdots + a_m (D')^m \right) u = 0$$
where $$D = \dfrac{\partial}{\partial x}$$, $$D' = \dfrac{\partial}{\partial y}$$
Solution method:
Replace $D \to m$, $D' \to 1$, get the auxiliary equation:
$$m^m + a_1 m^{m-1} + \cdots + a_m = 0$$
Let roots be $m_1, m_2, \dots$ (real or complex).
- If root $m_i$ real and distinct → term $f_i(y + m_i x)$
- Repeated root $m_i$ (multiplicity $k$) → $(y + m_i x)^{k-1} f(y + m_i x)$, etc.
- Complex roots $\alpha \pm i\beta$ → $e^{\alpha x} [f_1(y) \cos(\beta x) + f_2(y) \sin(\beta x)]$
Important particular cases:
- $$\dfrac{\partial^n u}{\partial x^n} = 0 \Rightarrow u = f_1(y) + x f_2(y) + \cdots + \dfrac{x^{n-1}}{(n-1)!} f_n(y)$$
- One-factor theorem (factors do not repeat):
If operator = $(D - m_1 D')(D - m_2 D')\cdots(D - m_n D')$
Solution: $$u = f_1(y + m_1 x) + f_2(y + m_2 x) + \cdots + f_n(y + m_n x)$$
7. Non-homogeneous Linear PDE with Constant Coefficients
Right-hand side = known function $f(x,y)$
Method:
- Find complementary function (CF) = solution of homogeneous equation (as above).
- Particular integral (PI) using standard rules:
| Form of $f(x,y)$ | Try PI of form | Multiply by $x^k$ if factor repeats |
|---|---|---|
| Polynomial in $x,y$ of degree $n$ | Same polynomial (unknown coefficients) | Yes |
| $e^{ax+by}$ | $A e^{ax+by}$ | If $(D - a - b D')^{k}$ factor |
| $\sin(ax+by)$ or $\cos(ax+by)$ | $A \cos(ax+by) + B \sin(ax+by)$ | Yes |
| $x^m (y + kx)^n$ | Use shift: replace operator by substituting $y + kx = \eta$ | — |
Operator rule: PI = $\dfrac{1}{f(D,D')} g(x,y)$
8. Equations Reducible to Linear with Constant Coefficients
Common cases:
- Change of independent variables to make coefficients constant
Example: Cauchy-Euler type in PDEs
$$\left( x^2 D^2 + 2x y D D' + y^2 (D')^2 \right) u = 0$$
Put $x = e^\xi$, $y = e^\eta$ → becomes constant coefficient. - Equations with variable coefficients reducible by substitution
- $x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} = u$ → put $u = v \ln x$ or use $x = e^\xi$ etc.
- General rule: If operator is function of $(x D + y D')$, assume $u = x^k v$ or use logarithmic coordinates.
Summary of Key Formulas
| Topic | Key Formula / Result |
|---|---|
| Lagrange linear PDE | $$\dfrac{dx}{P} = \dfrac{dy}{Q} = \dfrac{du}{R}$$ |
| Charpit’s relation for p | $$f_p \, dp + (f_x + p f_u) \, dp = 0$$ |
| Homogeneous linear constant coeff | Auxiliary: replace $D \to m$, solve $m$-equation |
| PI for $e^{ax+by} g(x,y)$ | $$\dfrac{1}{D + c D' - (a + b c)} e^{ax+by} g(x,y)$$ (if no repetition) |
| Repeated linear factor | Multiply PI by $x^k$ ($k$ = multiplicity) |
This covers the entire syllabus of Module I with all standard methods and formulas used in examinations. Practice solving previous years’ problems using these exact steps.