Partial Differential Equations

Partial Differential Equations (PDEs) are mathematical equations that involve multiple independent variables, their partial derivatives, and an unknown function or we can say that an equation which involves partial derivatives of a function of two or more independent variables is called Partial Differential Equation.

If z = f(x, y)

Then $p=\frac{\partial z}{\partial x},q=\frac{\partial z}{\partial y},r=\frac{{\partial }^{2}z}{\partial x^2},t=\frac{{\partial }^{2}z}{\partial y^2},s=\frac{{\partial }^{2}z}{\partial x^{2}y}$

1.0Order and Degree of Partial Differential Equations

The order of a PDF is the order of the highest order partial derivative present in the equation.

Order 1: $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0{\left(\frac{\partial z}{\partial x}\right)}^2+\left(\frac{\partial z}{\partial y}\right)=0$

Order 2: $\frac{{\partial }^2\mathrm{z}}{\partial {\mathrm{t}}^2}={\mathrm{k}}^2\left(\frac{{\partial }^2\mathrm{z}}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2\mathrm{z}}{\partial {\mathrm{y}}^2}\right)$

The degree of a PDE is the power of the highest order partial derivative present in the equation.

2.0Types of Partial Differential Equations

Partial Differential Equations can be categorized into several types based on their order and linearity:

1. First-Order Partial Differential Equations: These equations involve the first partial derivatives of the unknown function. A classic example is the transport equation, which models the distribution of a quantity over time and space.
2. Second-Order Partial Differential Equations: These equations involve second-order partial derivatives. The wave equation, heat equation, and Laplace's equation are notable examples of second-order PDEs. They are crucial in fields like acoustics, thermodynamics, and electromagnetism.
3. Higher-Order Partial Differential Equations: Although less common, higher-order PDEs are used in advanced applications. Their complexity often requires specialized methods for solving problems.
4. Linear vs. Nonlinear Partial Differential Equations: Linear PDEs have solutions that can be superimposed, making them easier to analyze. Nonlinear PDEs, on the other hand, are more complex and can exhibit phenomena like shock waves and solitons.

3.0Solving Partial Differential Equations

Solving PDEs can be challenging, but several methods are commonly used:

1. Separation of Variables: This technique assumes that the solution can be written as a product of functions, each depending on a single independent variable. It is particularly effective for linear PDEs with homogeneous boundary conditions.
2. Fourier Transform: The Fourier transform converts PDEs into algebraic equations, which are easier to solve. This method is particularly useful in solving problems involving periodic boundary conditions.
3. Numerical Methods: For complex partial differential equations that cannot be solved analytically, numerical methods like finite difference, finite element, and finite volume techniques are employed. These techniques approximate the solution by discretizing the domain.

4.0First Order Partial Differential Equations 

One of the simplest examples of a first-order partial differential equation is the linear transport equation:

$\frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x}=0$

Here, u(x, t) is the unknown function, and c is a constant representing the wave speed. The solution to this equation describes how a wave propagates through a medium without changing shape.

Lagrange's form

Pp + Qq = R

Lagrange's Auxiliary equation

$\frac{dx}{p}=\frac{dy}{Q}=\frac{dz}{R}$

It is also called as subsidiary equation.

5.0Second Order Partial Differential Equations

Second-order PDEs are more complex and can describe a wider range of physical phenomena. The heat equation, for example, is a second-order PDE given by:

$\frac{\partial u}{\partial t}=\alpha \frac{{\partial }^{2}u}{\partial x^2}$

This equation models the distribution of temperature u(x, t) in a rod over time, with α being the thermal diffusivity.

Consider a general PDE of second order in two variables.

Rr + Ss + Tt + f (x, y, z, p, q) = 0

Where R, S, T are continuous function and.

$r=\frac{{\partial }^2\mathrm{u}}{\partial {\mathrm{x}}^2},\mathrm{S}=\frac{{\partial }^2\mathrm{u}}{\partial {\mathrm{x}}^2\mathrm{y}},\mathrm{t}=\frac{{\partial }^2\mathrm{u}}{\partial {\mathrm{y}}^2}$

then the PDE is

1. Hyperbolic of a point (x, y) in D if S² – 4RT > 0
2. Parabolic of a point (x, y) in D if S² – 4RT = 0
3. Elliptic at a point (x, y) in D if S² – 4RT < 0

6.0Difference Between Ordinary and Partial Differential Equations

The primary difference between Ordinary Differential Equations (ODEs) and Partial Differential Equations (PDEs) lies in the number of independent variables. Ordinary Differential Equations involve derivatives with respect to a single independent variable, while PDEs involve partial derivatives with respect to multiple independent variables. This distinction makes PDEs more versatile in modeling scenarios where the change in a quantity depends on several factors, such as time and space.

7.0Solved Examples of Partial Differential Equation

Example 1: p + 3q = 5z + tan(y – 3x)

Solution: 

p = 1, Q = 3, R = 5z + tan(y –3x)

$\frac{dx}{p}=\frac{dy}{Q}=\frac{dz}{R}$

$\frac{dx}{1}=\frac{dy}{3}=\frac{dz}{5z+\tan (y-3x)}$

Solving: $\frac{dx}{1}=\frac{dy}{3}$

⇒ $\int 3\mathrm{d}x=\int \mathrm{dy}$

⇒ 3x = y + c.

⇒ 3x – y = c.

⇒ y – 3x = c.

Solving: $\frac{dx}{1}=\frac{dz}{5z+\tan (y-3x)}$

⇒ $\frac{dx}{1}=\frac{dz}{5z+\tan c}$

Integrating

⇒ $\mathrm{x}=\frac{1}{5}log(5z+tanc1)$ + log c₂

⇒ 5x = log (5z + tan c₁)c₂

⇒$\frac{{\mathrm{e}}^{5\mathrm{x}}}{5\mathrm{z}+\tan {\mathrm{c}}_1}={\mathrm{c}}_2$

F (c₁, c₂) = 0

$\left[y-3x,\frac{e^{5x}}{5z+\tan (y-3x)}\right]=0$

Example 2: (y² + z² – x²) p – 2xyq + 2xz = 0

Solution: 

p = y² + z² – x², Q = –2xy, R = – 2xz

$\frac{dx}{y^2+z^2-x^2}=\frac{dy}{-2xy}=\frac{dz}{-2xz}$

Solving $\frac{dy}{-2xy}=\frac{dz}{-2xz}$

⇒ $\int \frac{dy}{y}=\int \frac{dz}{z}$

⇒ log y = log z + log c₁

⇒ log y = log zc₁

⇒ $\frac{\mathrm{y}}{\mathrm{z}}={\mathrm{c}}_1$

Now, multiplying x to dx, y to dy and z to dz and add

$\frac{dx}{y^2+z^2-x^2}=\frac{dy}{-2xy}=\frac{dz}{-2xz}=\frac{xdx+ydy+zdz}{x\left(y^2+z^2-x^2\right)-2x^2-2x{z}^2}$

Now, solving

$\frac{dz}{-2xz}=\frac{xdx+ydy+zdz}{-x\left(x^2+y^2+z^2\right)}$

$\int \frac{\mathrm{dz}}{\mathrm{z}}=\int \frac{2\mathrm{xdx}+2\mathrm{ydy}+2\mathrm{zdz}}{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}$

log z = log(x² + y² + z²) + log c₂.

log z = log (x² + y² + z²). c₂

$c_2=\frac{z}{x^2+y^2+z^2}$

f(c₁, c₂) = 0

$f\left(\frac{y}{z},\frac{z}{x^2+y^2+z^2}\right)=0$

Example 3: $\frac{2{\partial }^{2}u}{\partial x^2}+\frac{4{\partial }^{2}u}{\partial x\partial y}+\frac{3{\partial }^{2}y}{\partial y^2}=2$

Solution: 

Here R = 2, S = 4, T = 3

Now S² – RT

= 16 – 4(2) (3) = 16 – 24 = – 8 < 0

⇒ This PDE is elliptic.

Example 4: xyr – (x² – y²) s – xyt + py – qx = 2 (x² – y²)

Solution: 

Here R = xy, S = –(x² – y²), T = –xy

Now, S² – 4RT

= (x² – y²)² – 4(xy) (–xy)

= (x² – y²)² + 4x²y²

= x⁴ + y⁴ – 2x²y² + 4x²y²

⇒ x⁴ + y⁴ + 2x²y²

⇒ (x² + y²)² > 0

⇒ This PDE is Hyperbolic.

Example 5: y²r – 2xys + x²t = $\frac{y^{2}p}{x}+\frac{x^{2}q}{y}$

Solution: 

Here R = y², S = –2xy, T = x²

Now, S² – 4RT

= 4x²y² – 4(y²)(x²) = 0

⇒ This PDE is parabolic.

8.0Practice Question on Partial Differential Equations

1. Solve: x q=y p+x e^{x^2+y^2}

2. Solve: z (p – q) = z² + (x + y)²

3. Classify the following PDE

x²(y – 1) r – x(y² – 1)s + y (y – 1)t + xyp – q = 0.

4. Classify: $\frac{{\partial }^2\mathbf{u}}{\partial x^2}+\frac{{\partial }^2\mathbf{u}}{\partial y^2}+\frac{2{\partial }^2\mathbf{u}}{\partial z^2}=0.$