JustToThePoint English Website Version
JustToThePoint en español
Colaborate with us

Partial differential equations

Irony is wasted on the stupid, Oscar Wilde

Recall

Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x0, y0) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y0. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x0

$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.

The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.

The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.

The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point. It is perpendicular (orthogonal) to the level surfaces of the function. This means that the gradient points in the direction of the steepest ascent or increase, and there is no change in the function’s value along the level surface, making the gradient perpendicular to it.

Image

Partial differential equations

A partial differential equation (PDE) is a type of mathematical equation that involves a function of severable variables and its partial derivatives with respect to those variables, e.g., $\frac{∂f}{∂t} = \frac{∂^2f}{∂x^2}, \frac{∂w}{∂t}-\frac{∂^2w}{∂x^2} = 0, etc.$ Partial differential equations are incredibly important in many fields of science and engineering.

Classification of differential equations

Differential equations can be classified in different ways based on various characteristics.

  1. Ordinary differential equations or (ODEs) are equations where the derivatives are taken with respect to a single independent variable, e.g., $\frac{d^2y}{dx^2}+\frac{dy}{dx} = 3xsin(y)$. This equation involves derivatives with respect to x only, so it’s an ODE.
  2. Partial differential equations or (PDEs) are equations that depend on partial derivatives with respect to multiple independent variables, e.g., $\frac{∂y}{∂t}+z\frac{∂y}{∂z} = \frac{x-t}{x+t}$. This equation involves partial derivatives with respect to both t and z, so it’s a PDE.
  1. A linear PDE is one in which the dependent variable (the function we are solving for) and its derivatives appear linearly. In simple terms, this means that the function and its derivatives are not raised to any powers, multiplied by each other, or involved in any nonlinear functions like sine or exponential, e.g., $\frac{∂^2f}{∂x^2} =\frac{∂f}{∂t}.$
  2. Otherwise, the equation is called nonlinear, e.g., $f\frac{∂^2f}{∂x^2} =\frac{∂f}{∂t}$ because f is multiplied by its second derivative .

Homogeneity. A PDE is homogeneous if every term in the equation only involve the dependent variable and/or its derivatives, e.g., $f\frac{∂^2f}{∂x^2} =\frac{∂f}{∂t}+ x^2 + tan(t)$ is not homogeneous, but $4\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0$ is homogeneous.

Example: The Heat Equation

One of the most well-known and classic PDEs is the heat equation. It describes how heat (or temperature) spreads out over time within a medium.

The heat equation in three dimensions is written as: $\frac{∂f}{∂t} = k(\frac{∂^2f}{∂x^2}+\frac{∂^2f}{∂y^2}+\frac{∂^2f}{∂z^2})\text{where}\frac{∂f}{∂t}$ where:

The heat equation says that the rate at which the temperature changes over time at any point is proportional to the sum of these second derivatives with respect to space. Futhermore, it tells us that heat flows from hotter regions to cooler regions, smoothing out temperature differences over time.

Ordinary Differential Equations

An Ordinary Differential Equations (ODE) is an equation that involves an unknown function and its derivatives. The “ordinary” in ODEs means that the function depends on just one independent variable, unlike partial differential equations (PDEs) that involve functions of multiple variables.

Understanding ODEs with a Simple Example

Let’s start with a simple example. the equation that describes the motion of a particle moving a straight path at a constant velocity. The position of this particle as a function of time t can be described by the ODE: $\frac{d}{dt}x(t) = v$.

This equation means that the rate of change of the particle’s position x(t) with respect to time t is equal to its velocity v, here:

Solving the ODE

To find the position x(t) at any time t, we need to solve this differential equation. This involves integrating the equation: $\frac{d}{dt}x(t) = v$. The solution is x(t) = v·t + x0 where v·t represents the distance the particle has traveled after certain time t at a constant velocity v and x0 is a constant that represents the initial position of the particle when t = 0.

This solution shows that the position of the particle increases linearly with time.

Moving to More Complex ODEs: The Harmonic Oscillator

Now, let’s explore a more complex example, a second-order ODE. A second-order ODE involves the second derivative of the function, which often represents acceleration in physical systems.

The harmonic oscillator equation is a classic example: $\frac{d^2}{dt^2}x(t) = -w^2x(t)$. This equation describes the motion of a system that oscillates back and forth, like a mass on a spring or a pendulum where

The general solution of the equation is x(t) = Acos(wt) + Bsin(wt) where:

This solution tells us that the position x(t) of the particle oscillate back and forth in time in a pattern described by a combination of sine and cosine waves.

Exercise: Verifying a Solution to a PDE

Now, let’s apply what we’ve learned to a specific problem. We want to show that the function u(x,t) = sin(at)cos(x) is a solution to the following PDE: $\frac{∂^2u}{∂t^2}=a\frac{∂^2u}{∂x^2}$

Step 1: Compute the Second Partial Derivative with Respect to Time t. Since a is a constant, $\frac{∂u}{∂t} = acos(at)cos(x)$. Now, differentiate again to find the second derivative with respect to t: $\frac{∂^2u}{∂t^2} = -a^2sin(at)cos(x).$

Step 2: Compute the Second Partial Derivative with Respect to x. Next, let’s calculate the first and second partial derivatives of u(x,t) with respect to the spatial variable x:

$\frac{∂u}{∂x} = -sin(at)sin(x), \frac{∂^2u}{∂x^2} = -sin(at)cos(x).$

Step 3: Substitute into the PDE. Substitute the expressions for $\frac{∂^2u}{∂t^2}$ and $\frac{∂^2u}{∂x^2}$ into the original partial differential equation:

$-a^2sin(at)cos(x) = a^2(-sin(at)cos(x))$. Since both sides of the equation are equal, we have verified that u(x, t) = sin(at)cos(x) is indeed a solution to the partial differential equation.

First-order differential equations

First-order differential equations can be classified into several types, each with specific characteristics and solution methods:

  1. Proposed Solution. We propose that the solution is of the form y = t3/3 + t + C where C is an arbitrary constant (the constant of integration).
  2. Verification. To verify that this is indeed a solution, we differentiate y(t) with respect to t and see if we get the original differential equation: $\frac{d}{dt}(\frac{t^3}{3}+t+C) = t^2 +1$.
  3. General Solution. The general solution to this type of equation can be found by integrating both sides: y(t) = $\int (t^2+1)dt = \frac{t^3}{3}+ t + C$. Therefore, all solutions to this equation are of the form $\frac{t^3}{3}+ t + C$.

This can be rewritten as: $\frac{1}{g(y)}dy = f(t)dt$. Here, we’ve separated the y-terms on the left side and the t-terms on the right side, and we can attempt to solve it by converting to the form: $\int \frac{1}{g(y)}dy = \int f(t)dt$

Let’s consider a specific separable equation:

y’ + 2xy = x ⇒$\frac{dy}{dx} + 2xy = x ⇒[\text{Start by isolating}~ \frac{dy}{dx}] \frac{dy}{dx} = x -2xy ⇒[\text{Factor the right-hand side}] \frac{dy}{dx} = x(1 -2y) ⇒[\text{Separate the y-terms and x-terms}] \frac{dy}{1-2y} = xdx$.

Now integrate both sides with respect to their respective variables: $\int \frac{dy}{1-2y} = \int xdx ⇒[\text{For the left side, use the substitution}~u = 1 -2y, du = -2dy, dy = \frac{-1}{2}du] = \frac{-1}{2}\int \frac{du}{u} = \frac{1}{2}x^2 + C ⇒\frac{-1}{2}ln|u| = \frac{1}{2}x^2 + C ⇒ \frac{-1}{2}ln|1-2y| = \frac{1}{2}x^2 + C ⇒[\text{Multiplying by -2}] ln|1-2y| = -x^2 + C_1$ where C1 = -2C is another constant.

Remove the logarithm by exponentiating both sides:

$e^{ln|1-2y|} = e^{-x^2+C_1} ⇒[e^{a+b} = e^a·e^b] 1-2y = C_2e^{-x^2}$ where C2 = eC1 is another constant

Finally, solve for y: $-2y = C_2e^{-x^2} -1 ⇒ y = -\frac{C_2e^{-x^2} -1}{2} = \frac{1-C_2e^{-x^2}}{2}$

The term “homogeneous” means that the equation is set equal to zero, and “linear” means that the function y and its derivative y′ appear in the equation in a linear way (i.e., no powers or products of y and y′).

Since first order homogeneous linear differential equations are separable, we can solve then in the usual way:

First, we isolate the derivative y′ on one side: y’ = -p(t)y ⇒[Separate the variables] $\frac{y’}{y} = -p(t)⇒[\text{Integrate both sides}] \int \frac{y’}{y} = \int -p(t)dt ⇒ ln|y| = P(t) + C$ where P(t) is an antiderivative of -p(t) and C is the constant of integration.

Solve for y. To solve for y, exponentiate both sides to remove the natural logarithm: $y = ±e^{P(t)+C} ⇒ y = Ae^{P(t)}$ where P(t) is an antiderivative of -p(t) and A = eC is just a new constant.

Let’s solve the specific first-order homogeneous linear differential equation: y’ + ycos(t) = 0, with the initial condition y(0) = 1/2.

Rewrite the equation. Start by isolating y′ on one side: y’ = -cos(t)y⇒[Separate both sides] $\frac{y’}{y} = -cos(t)⇒[\text{Integrate both sides}]\int \frac{y’}{y} = \int -cos(t)dt ⇒ ln|y| = -sin(t) + C.$

Solve for y. Exponentiate both sides to remove the natural logarithm: $y = e^{-sin(t)+C} ⇒y = Ae^{-sin(t)}$ where A = eC is a constant.

Apply the initial condition: $\frac{1}{2} = Ae^{-sin(0)} = A$, so the solution is $y = \frac{1}{2}e^{-sin(t)}$.

Suppose we have a first-order linear differential equation of the form $\frac{dy}{dx} + Py = Q$ where P(x) and Q(x) are known functions of x. The idea behind the integrating factor method is to multiply the entire equation by a specially chosen function, called the integrating factor, which simplifies the equation into something that can be easily integrated.

We multiply both sides of the differential equation by an integrating factor I.

$I\frac{dy}{dx} + IPy = IQ ↭[\text{By the product rule} \frac{d}{dx}(Iy) = I\frac{dy}{dx}+\frac{dI}{dx}y = !!\text{That’s what we need 🚀}!! I\frac{dy}{dx}+IPy] \frac{d}{dx}(Iy) = IQ ⇒[\text{Integrate both sides}]\int \frac{d}{dx}(Iy) = \int IQdx ⇒Iy = \int IQdx ⇒[\text{Solve for y}] y = \frac{1}{I(x)}\int I(x)Q(x)dx$

As both I and Q are functions involving only x, $\int IQ dx$ can usually be found. So the general solution to the differential equation is found by integrating IQ and then re-arranging the formula to solve for y.

We just need to figure out (🚀): $\frac{dI}{dx}y = IPy$. But this is just a first order homogeneous linear equation, and we know a solution is I = $e^{\int Pdx}$: $\frac{dI}{dx}y = \frac{de^{\int Pdx}}{dx}y = e^{\int Pdx}Py = IPy$

Let’s solve the differential equation: $\frac{dy}{dx}+y=x$

Identify the Integrating Factor I = $e^{\int Pdx} = e^{\int dx}= e^x$.

Multiply the entire equation by ex: $ e^x\frac{dy}{dx} + e^xy = xe^x ⇒[\text{Recognize the left side as a derivative}]\frac{d}{dx}(e^xy) = xe^x ⇒[\text{Integrate both sides}]\int \frac{d}{dx}(e^xy) = \int xe^xdx ⇒ e^xy = \int xe^xdx $

To solve $\int xe^xdx$, we use integration by parts, $uv -\int vdu, u = x, du = dx, dv = e^xdx, v = e^x$:

$\int xe^xdx = xe^x-\int e^xdx = xe^x -e^x + C$

$e^xy = \int xe^xdx = xe^x -e^x + C⇒[\text{Solve for y}]y = x -1 + \frac{C}{e^x} = x -1 + Ce^{-x}.$

This means that the left-hand side of the equation can be interpreted as the differential df(x,y), where f(x,y) is a potential function.

Solve $(3x^2+1)+(3y^2+2y)y’=0$.

  1. Rewrite the Equation in Standard Form: $(3x^2+1)+(3y^2+2y)\frac{dy}{dx}=0 ⇒ (3x^2+1)dx +(3y^2+2y)dy = 0$
  2. Check if this is an exact differential equation by verifying the partial derivative of M(x, y) with respect to y must equal the partial derivative of N(x,y) with respect to x. In other words, My = Nx. $M_y = \frac{∂}{∂y}(3x^2+1)=0, N_y = \frac{∂}{∂x}(3y^2+2y) = 0.$ Since My = Nx = 0, the equation is exact.
  3. Find the Potential Function f(x, y). Since the equation is exact, there exist a function f(x, y) such that $\frac{∂f}{∂x} = M(x, y) = 3x^2 + 1$. To find f, we just need to integrate M(x, y) with respect to x: f(x, y) = $\int (3x^2 + 1)dx = x^3 + x + g(y)$ where g(y) is an arbitrary function of y that could have appeared during the integration since the derivative with respect to x does not affect y.
  4. Calculate g(y). Differentiate f(x, y) with respect to y to calculate g(y): $\frac{∂f}{∂y} = \frac{∂}{∂y}(x^3 + x + g(y)) = g’(y). \frac{∂f}{∂y} = N(x, y) = 3y^2+2y ⇒ g’(y) = 3y^2+2y ⇒ g(y) = \int (3y^2+2y)dy = y^3 + y^2 + C.$ Hence, the potential function is: f(x, y) = $x^3 + x + g(y) = x^3 + x + y^3 + y^2 + C$
  5. Write the general solution. The general solution to the exact differential equation is given by M(x, y)dx + N(x, y)dy = 0. We found that there exist a potential function f(x, y) such that $df = \frac{∂f}{∂x}dx + \frac{∂f}{∂y}dy = M(x, y)dx + N(x, y)dy = 0$. This is equivalent to df = 0 ↭ f(x, y) = C ↭ $x^3 + x + y^3 + y^2 = C$ where C is a constant.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
Bitcoin donation

JustToThePoint Copyright © 2011 - 2025 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us.

This website uses cookies to improve your navigation experience.
By continuing, you are consenting to our use of cookies, in accordance with our Cookies Policy and Website Terms and Conditions of use.