Partial Differential Equations and Formation Methods

1. Introduction to Partial Differential Equations (PDEs)

A Partial Differential Equation (PDE) is an equation that involves an unknown function of two or more independent variables and its partial derivatives with respect to those variables. Unlike Ordinary Differential Equations (ODEs) which involve functions of a single independent variable, PDEs are used to describe phenomena that vary in both space and time, or across multiple spatial dimensions.

For example, the temperature distribution in a room changes with position (x, y, z) and time (t). A PDE can model how this temperature T(x, y, z, t) evolves. Similarly, the displacement of a vibrating string depends on its position along the string (x) and time (t), which can be described by a PDE.

The order of a PDE is determined by the order of the highest partial derivative present in the equation. A PDE is classified as linear if the dependent variable and its partial derivatives appear only in the first degree and are not multiplied together. If these conditions are not met, the PDE is non-linear.

2. Classification of Second-Order Linear PDEs

Second-order linear PDEs are of fundamental importance and are classified into three main types based on their coefficients. This classification helps in choosing appropriate methods for solving them. The general form of a second-order linear PDE in two independent variables (say, x and y) is:

$A \frac{\partial^2 u}{\partial x^2} + B \frac{\partial^2 u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} + D \frac{\partial u}{\partial x} + E \frac{\partial u}{\partial y} + F u = G$

Where A, B, C, D, E, F, and G are functions of x and y. The classification depends on the discriminant $B^2 - 4AC$.

• Hyperbolic: If $B^2 - 4AC > 0$. These equations often describe wave-like phenomena, such as the one-dimensional wave equation.
• Parabolic: If $B^2 - 4AC = 0$. These equations typically model diffusion or heat transfer processes, like the one-dimensional heat equation.
• Elliptic: If $B^2 - 4AC < 0$. These equations are usually associated with steady-state problems, such as the two-dimensional Laplace equation.

Understanding this classification is crucial because the solution techniques and the nature of the solutions differ significantly for each type.

3. Formation of Partial Differential Equations

Partial Differential Equations can be formed in several ways, most commonly by eliminating arbitrary constants or arbitrary functions from a given equation.

3.1. Elimination of Arbitrary Constants

If we have an equation relating dependent variable $u$ and independent variables $x, y, \dots$, and it contains $n$ arbitrary constants, we can obtain a PDE of order $n$ by eliminating these constants. This involves differentiating the original equation partially with respect to each independent variable and then manipulating these equations to remove the constants.

Let's consider an example: Suppose we have a function $z = f(x, y)$ with two arbitrary constants, $a$ and $b$. $z = a x^2 + b y^2$ To form a PDE, we need to eliminate $a$ and $b$.

Step 1: Differentiate partially with respect to $x$: $\frac{\partial z}{\partial x} = 2ax$ Let's call this Equation (2).

Step 2: Differentiate partially with respect to $y$: $\frac{\partial z}{\partial y} = 2by$ Let's call this Equation (3).

Step 3: Differentiate Equation (2) partially with respect to $x$ again: $\frac{\partial^2 z}{\partial x^2} = 2a$ Let's call this Equation (4).

Step 4: Differentiate Equation (3) partially with respect to $y$ again: $\frac{\partial^2 z}{\partial y^2} = 2b$ Let's call this Equation (5).

Step 5: Differentiate Equation (2) partially with respect to $y$ (or Equation (3) with respect to $x$). This gives the mixed partial derivative: $\frac{\partial^2 z}{\partial y \partial x} = 0$ $\frac{\partial^2 z}{\partial x \partial y} = 0$ Let's call this Equation (6).

Now, we need to eliminate $a$ and $b$ using Equations (2), (3), (4), and (5). From Equation (4), $a = \frac{1}{2} \frac{\partial^2 z}{\partial x^2}$. From Equation (5), $b = \frac{1}{2} \frac{\partial^2 z}{\partial y^2}$. Substitute these into Equation (2): $\frac{\partial z}{\partial x} = 2 \left( \frac{1}{2} \frac{\partial^2 z}{\partial x^2} \right) x = x \frac{\partial^2 z}{\partial x^2}$ Rearranging this gives: $x \frac{\partial^2 z}{\partial x^2} - \frac{\partial z}{\partial x} = 0$

Alternatively, substitute $a$ and $b$ into the original equation $z = ax^2 + by^2$: $z = \left( \frac{1}{2} \frac{\partial^2 z}{\partial x^2} \right) x^2 + \left( \frac{1}{2} \frac{\partial^2 z}{\partial y^2} \right) y^2$ $2z = x^2 \frac{\partial^2 z}{\partial x^2} + y^2 \frac{\partial^2 z}{\partial y^2}$ This is the PDE of order 2 formed by eliminating the constants. The order of the PDE is equal to the number of arbitrary constants eliminated.

3.2. Elimination of Arbitrary Functions

When an equation involves an arbitrary function, we can obtain a PDE of the first order by eliminating the function. If the function depends on a single variable, we can use ODE techniques. If the function depends on a combination of variables, we use partial differentiation.

Example 1: Eliminate the arbitrary function $f$ from $z = f(x+y)$. Let $u = x+y$. Then $z = f(u)$. Differentiate partially with respect to $x$: $\frac{\partial z}{\partial x} = \frac{df}{du} \frac{\partial u}{\partial x} = f'(u) \cdot 1 = f'(u)$ Differentiate partially with respect to $y$: $\frac{\partial z}{\partial y} = \frac{df}{du} \frac{\partial u}{\partial y} = f'(u) \cdot 1 = f'(u)$ Since both partial derivatives are equal to $f'(u)$, we can equate them: $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y}$ Rearranging gives: $\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = 0$ This is the first-order PDE obtained by eliminating the arbitrary function $f$.

Example 2: Eliminate the arbitrary function $\phi$ from $z = x \phi(y/x)$. Let $v = y/x$. Then $z = x \phi(v)$. Differentiate partially with respect to $x$: $\frac{\partial z}{\partial x} = 1 \cdot \phi(v) + x \cdot \frac{d\phi}{dv} \cdot \frac{\partial v}{\partial x}$ Here, $\frac{\partial v}{\partial x} = \frac{\partial}{\partial x} (y x^{-1}) = y (-1 x^{-2}) = -y/x^2$. So, $\frac{\partial z}{\partial x} = \phi(v) + x \phi'(v) (-y/x^2) = \phi(v) - (y/x) \phi'(v)$. $\frac{\partial z}{\partial x} = \phi(y/x) - (y/x) \phi'(y/x)$ (Equation A)

Differentiate partially with respect to $y$: $\frac{\partial z}{\partial y} = x \cdot \frac{d\phi}{dv} \cdot \frac{\partial v}{\partial y}$ Here, $\frac{\partial v}{\partial y} = \frac{\partial}{\partial y} (y x^{-1}) = 1/x$. So, $\frac{\partial z}{\partial y} = x \phi'(v) (1/x) = \phi'(v)$. $\frac{\partial z}{\partial y} = \phi'(y/x)$ (Equation B)

Now we need to eliminate $\phi$ and $\phi'$. From the original equation, $z/x = \phi(y/x)$. Substitute this into Equation A: $\frac{\partial z}{\partial x} = z/x - (y/x) \phi'(y/x)$ From Equation B, $\phi'(y/x) = \frac{\partial z}{\partial y}$. Substitute this into the modified Equation A: $\frac{\partial z}{\partial x} = z/x - (y/x) \frac{\partial z}{\partial y}$ Multiply by $x$ to clear the denominators: $x \frac{\partial z}{\partial x} = z - y \frac{\partial z}{\partial y}$ Rearranging gives: $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = z$ This is the first-order linear PDE.

Example 3: Eliminate the arbitrary functions $f$ and $g$ from $z = f(x) + g(y)$. Differentiate partially with respect to $x$: $\frac{\partial z}{\partial x} = f'(x) + 0 = f'(x)$ Differentiate partially with respect to $y$: $\frac{\partial z}{\partial y} = 0 + g'(y) = g'(y)$ Differentiate $\frac{\partial z}{\partial x}$ with respect to $x$ again: $\frac{\partial^2 z}{\partial x^2} = f''(x)$ Differentiate $\frac{\partial z}{\partial y}$ with respect to $y$ again: $\frac{\partial^2 z}{\partial y^2} = g''(y)$ Differentiate $\frac{\partial z}{\partial x}$ with respect to $y$: $\frac{\partial^2 z}{\partial y \partial x} = 0$ This is the PDE of order 2. The presence of two arbitrary functions leads to a second-order PDE.

Example 4: Eliminate the arbitrary functions $f$ and $g$ from $z = f(x^2+y) + g(x^2-y)$. Let $u = x^2+y$ and $v = x^2-y$. So $z = f(u) + g(v)$. Differentiate partially with respect to $x$: $\frac{\partial z}{\partial x} = \frac{df}{du} \frac{\partial u}{\partial x} + \frac{dg}{dv} \frac{\partial v}{\partial x}$ $\frac{\partial u}{\partial x} = 2x$, $\frac{\partial v}{\partial x} = 2x$. $\frac{\partial z}{\partial x} = f'(u)(2x) + g'(v)(2x) = 2x [f'(x^2+y) + g'(x^2-y)]$

Differentiate partially with respect to $y$: $\frac{\partial z}{\partial y} = \frac{df}{du} \frac{\partial u}{\partial y} + \frac{dg}{dv} \frac{\partial v}{\partial y}$ $\frac{\partial u}{\partial y} = 1$, $\frac{\partial v}{\partial y} = -1$. $\frac{\partial z}{\partial y} = f'(u)(1) + g'(v)(-1) = f'(x^2+y) - g'(x^2-y)$

Now we need to find a second-order PDE. Let's differentiate $\frac{\partial z}{\partial y}$ with respect to $y$ again. $\frac{\partial^2 z}{\partial y^2} = \frac{d}{dy} [f'(x^2+y) - g'(x^2-y)]$ $\frac{\partial^2 z}{\partial y^2} = f''(u) \frac{\partial u}{\partial y} - g''(v) \frac{\partial v}{\partial y}$ $\frac{\partial^2 z}{\partial y^2} = f''(x^2+y)(1) - g''(x^2-y)(-1)$ $\frac{\partial^2 z}{\partial y^2} = f''(x^2+y) + g''(x^2-y)$

This doesn't seem to directly eliminate $f$ and $g$. Let's try differentiating $\frac{\partial z}{\partial x}$ with respect to $y$. $\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y} \{ 2x [f'(x^2+y) + g'(x^2-y)] \}$ $\frac{\partial^2 z}{\partial y \partial x} = 2x \left[ \frac{\partial}{\partial y} f'(x^2+y) + \frac{\partial}{\partial y} g'(x^2-y) \right]$ $\frac{\partial^2 z}{\partial y \partial x} = 2x \left[ f''(x^2+y)(1) + g''(x^2-y)(-1) \right]$ $\frac{\partial^2 z}{\partial y \partial x} = 2x [f''(x^2+y) - g''(x^2-y)]$

Let's try another approach. Consider the original equation $z = f(x^2+y) + g(x^2-y)$. Let $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$. $p = 2x [f'(x^2+y) + g'(x^2-y)]$ $q = f'(x^2+y) - g'(x^2-y)$ From $p$, we have $\frac{p}{2x} = f'(x^2+y) + g'(x^2-y)$. We have a system of two equations with $f'$ and $g'$: 1) $q = f' - g'$ 2) $\frac{p}{2x} = f' + g'$ Adding (1) and (2): $q + \frac{p}{2x} = 2f'$ $f' = \frac{1}{2} (q + \frac{p}{2x})$ Subtracting (1) from (2): $\frac{p}{2x} - q = 2g'$ $g' = \frac{1}{2} (\frac{p}{2x} - q)$

Now, differentiate $q$ with respect to $y$: $\frac{\partial q}{\partial y} = \frac{\partial^2 z}{\partial y^2} = f''(x^2+y) + g''(x^2-y)$. Differentiate $p$ with respect to $y$: $\frac{\partial p}{\partial y} = \frac{\partial^2 z}{\partial y \partial x} = 2x [f''(x^2+y) - g''(x^2-y)]$. Differentiate $q$ with respect to $x$: $\frac{\partial q}{\partial x} = \frac{\partial^2 z}{\partial x \partial y} = 2x [f''(x^2+y) + g''(x^2-y)]$.

Notice that $\frac{\partial q}{\partial x} = 2x [f''(x^2+y) + g''(x^2-y)]$. And $\frac{\partial^2 z}{\partial y^2} = f''(x^2+y) + g''(x^2-y)$. So, $\frac{\partial q}{\partial x} = 2x \frac{\partial^2 z}{\partial y^2}$. This gives the PDE: $\frac{\partial^2 z}{\partial x \partial y} = 2x \frac{\partial^2 z}{\partial y^2}$ Rearranging: $\frac{\partial^2 z}{\partial x \partial y} - 2x \frac{\partial^2 z}{\partial y^2} = 0$

This PDE is of the second order. The general strategy for eliminating arbitrary functions is to differentiate the given equation with respect to the independent variables enough times to obtain a set of equations, and then eliminate the arbitrary functions from this set.

3.3. PDEs from Physical Principles

Many PDEs arise naturally from fundamental physical laws. For instance, conservation laws in physics and engineering often lead to differential equations.

Example: The One-Dimensional Heat Equation Consider the flow of heat in a thin rod. The temperature $u(x,t)$ at position $x$ and time $t$ is governed by the heat equation. This equation is derived from the principles of conservation of energy and Fourier's law of heat conduction. Fourier's Law states that the rate of heat flow is proportional to the negative temperature gradient: $q = -k \frac{\partial u}{\partial x}$, where $k$ is the thermal conductivity. Conservation of energy implies that the rate of change of heat energy in a small segment of the rod is equal to the net rate of heat flow into that segment. This leads to the PDE: $\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}$ where $\alpha = k/(\rho c)$ is the thermal diffusivity, $\rho$ is the density, and $c$ is the specific heat capacity.

Example: The One-Dimensional Wave Equation Consider the transverse vibrations of a stretched elastic string. The displacement $u(x,t)$ of the string at position $x$ and time $t$ is governed by the wave equation. This is derived from Newton's second law applied to a small element of the string, considering the tension and the forces acting on it. The resulting PDE is: $\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$ where $c$ is the wave speed, related to the tension and linear density of the string.

Example: Laplace's Equation In electrostatics, the electric potential $\phi$ in a charge-free region satisfies Laplace's equation. In two dimensions, it is: $\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0$ This is an example of an elliptic PDE.

4. Importance and Applications

Partial Differential Equations are fundamental tools in science and engineering for modeling a vast array of phenomena. They are used in:

• Physics: Describing wave propagation (sound, light, water waves), heat diffusion, quantum mechanics (Schrödinger equation), electromagnetism (Maxwell's equations).
• Engineering: Fluid dynamics (Navier-Stokes equations), structural analysis, heat transfer, signal processing.
• Biology: Modeling population dynamics, spread of diseases, reaction-diffusion processes.
• Finance: Pricing of financial derivatives (Black-Scholes equation).
• Computer Graphics: Image processing and rendering.

The ability to form and solve PDEs is essential for understanding and predicting the behavior of complex systems in the natural and engineered world.