Unit 1, lesson 1

Differential Equations and Their Language

What a differential equation and its unknown function are, how ordinary and partial equations differ, and how to read order, dependent and independent variables, linearity and the meanings of homogeneous.

  • Video 1
  • Equations, Solutions and Calculus Tools

Water flows into a tank through one pipe and out through another. At any moment you can read the water level and how fast it is changing, but neither number by itself tells you the level ten minutes from now. What the physics supplies is a rule for the rate of change, and the level over time is a function that has to obey that rule. This lesson sets up the words used to describe such equations, from ordinary and partial to order, linearity and the two meanings of homogeneous, and shows how to check that a proposed function really is a solution.

Key terms

Differential equation

Definition A differential equation is an equation that relates an unknown function to one or more of its derivatives, and possibly to the independent variable and to known functions. The unknown function is the function the equation is solved for, so the answer is a whole function rather than a number. A solution on an interval II is a function that has every derivative appearing in the equation at each point of II and makes the equation true at every point of II.

In plain words The equation is a rule that the unknown function and its rates of change must obey everywhere on an interval. Solving it means finding functions that follow the rule, and checking a candidate means substituting it into the equation.

Example Take the equation y′=−2yy' = -2y for an unknown function y(t)y(t), where y′=dy/dty' = dy/dt. The candidate y(t)=3e−2ty(t) = 3e^{-2t} has derivative −6e−2t-6e^{-2t} by the chain rule, and −2y-2y is also −6e−2t-6e^{-2t}, so

y′=−6e−2t=−2yy' = -6e^{-2t} = -2y

holds for every real tt. The same function has y(0)=3y(0) = 3. As a model of decay it would be used for t≥0t \ge 0, but as a mathematical solution it works on the whole real line. The candidate y=3+ty = 3 + t fails. Its derivative is 11, while −2y=−6−2t-2y = -6 - 2t, and the two agree only at t=−72t = -\tfrac{7}{2}. Agreement at a single point is not agreement on an interval, so 3+t3 + t is not a solution.

Ordinary and partial differential equations

Definition An ordinary differential equation, or ODE, is a differential equation whose unknown function or functions depend on a single independent variable, so only ordinary derivatives such as dy/dtdy/dt and d2y/dt2d^2y/dt^2 appear. A partial differential equation, or PDE, involves an unknown function of two or more independent variables together with some of its partial derivatives. For a function u(x,t)u(x, t), the partial derivative ∂u/∂t\partial u/\partial t is the rate of change in tt while xx is held fixed, and ∂u/∂x\partial u/\partial x is the rate of change in xx while tt is held fixed.

In plain words Count the inputs of the unknown function. One input, often time, gives an ODE. Several inputs, such as position and time, give a PDE, and each partial derivative says which input moves while the others stay put.

Example A falling object has height y(t)y(t) at time tt, measured upward, and y′=dy/dty' = dy/dt. Its velocity is y′y' and its acceleration is y′′y'', so ignoring air resistance it obeys the ODE y′′=−gy'' = -g, where gg is the gravitational acceleration. The temperature u(x,t)u(x, t) at position xx along a thin rod at time tt depends on two inputs, and a standard model for it is the heat equation

∂u∂t=κ∂2u∂x2\frac{\partial u}{\partial t} = \kappa \frac{\partial^2 u}{\partial x^2}

with a constant diffusivity κ>0\kappa > 0. It is a PDE, and both of its derivatives come from the same function uu.

Order of an equation

Definition The order of a differential equation is the order of the highest derivative of the unknown function that appears in it. For a PDE it is the highest total order of the partial derivatives that appear. The power to which a derivative is raised has no effect on the order, so the order of an equation is a different thing from the degree of a polynomial.

In plain words Look for the derivative taken the most times. How often the unknown has been differentiated sets the order, while squares or cubes of terms say nothing about it.

Example For y(t)y(t) with y′=dy/dty' = dy/dt, the equation (y′)2+y3=t(y')^2 + y^3 = t is first order, even though it contains a square and a cube. The equation y′′+3y′+2y=0y'' + 3y' + 2y = 0 is second order, even though every term appears only to the first power. The heat equation above is second order, because ∂2u/∂x2\partial^2 u/\partial x^2 appears. An equation of order nn usually needs nn pieces of starting data to single out one solution, such as a starting position and a starting velocity for a second-order equation of motion.

Dependent and independent variables

Definition In an equation for an unknown function y(t)y(t), the input tt is the independent variable and the output yy is the dependent variable, since its value depends on the chosen tt. A parameter is a constant that specifies the model, such as a mass, a damping coefficient or a cross-sectional area. It stays fixed while tt varies, though it may differ from one version of the model to another. Primes always mean derivatives with respect to the independent variable, so that variable has to be named, as in y′=dy/dty' = dy/dt or y′=dy/dxy' = dy/dx.

In plain words The independent variable is the input you choose, often time. The dependent variable is the quantity you track. Parameters are the fixed settings of the model.

Example Return to the tank. Let y(t)y(t) be the water level in meters at time tt in minutes, let QinQ_{\text{in}} and QoutQ_{\text{out}} be the volume flow rates in cubic meters per minute, and let the tank have a constant cross-sectional area AA in square meters. A change in volume spread over the area AA changes the level by that volume divided by AA, so

dydt=Qin−QoutA.\frac{dy}{dt} = \frac{Q_{\text{in}} - Q_{\text{out}}}{A}.

Both sides are measured in meters per minute. Here tt is the independent variable, yy is the dependent variable and AA is a parameter. If the outflow through a drain at the bottom depends on the level, for example Qout=kyQ_{\text{out}} = k\sqrt{y} with a constant k>0k > 0, the equation still has the single dependent variable yy, but it becomes nonlinear.

Linear and nonlinear equations

Definition An ODE of order nn for y(t)y(t) is linear if it can be written as

∑k=0nak(t) y(k)=g(t),\sum_{k=0}^{n} a_k(t)\, y^{(k)} = g(t),

that is, an(t)y(n)+⋯+a1(t)y′+a0(t)y=g(t)a_n(t)y^{(n)} + \cdots + a_1(t)y' + a_0(t)y = g(t), where y(0)=yy^{(0)} = y, the coefficients aka_k and the right side gg are given functions of tt alone, and ana_n is not the zero function. An equation that cannot be written this way is nonlinear. Coefficients may vary with tt, but yy and its derivatives may not appear to powers other than the first, in products with each other or inside other functions. A PDE is linear when it has the same form in uu and its partial derivatives, with coefficients that depend only on the independent variables.

In plain words In a linear equation the unknown and its derivatives enter in the simplest possible way, each multiplied by a known function of the independent variable and then added up. Linearity is a property of the equation, not of the graphs of its solutions.

Example The equations y′′+3y′+2y=0y'' + 3y' + 2y = 0 and y′+(cos⁡t) y=0y' + (\cos t)\,y = 0 are linear, and so is the heat equation. The cosine in the second one is a known coefficient, because it depends only on tt. The equations y′=cos⁡yy' = \cos y, y′+y2=0y' + y^2 = 0 and yy′=tyy' = t are nonlinear, because the unknown sits inside a cosine, is squared, or multiplies its own derivative. Linear does not mean that the solution graphs are straight lines. The curve y=e−ty = e^{-t} solves the linear equation y′+y=0y' + y = 0, since y′=−e−t=−yy' = -e^{-t} = -y.

Homogeneous and nonhomogeneous linear equations

Definition A linear equation as above is homogeneous if its right side g(t)g(t) is zero for every tt in the interval, and nonhomogeneous if gg is not the zero function there. The function gg is called the forcing term or input. For a homogeneous linear equation the zero function is a solution, and if y1y_1 and y2y_2 are solutions, then so is c1y1+c2y2c_1y_1 + c_2y_2 for all constants c1c_1 and c2c_2. For a nonhomogeneous one the zero function is not a solution, and the difference of two solutions solves the homogeneous equation with the same left side.

A second, unrelated meaning applies to first-order equations y′=f(x,y)y' = f(x, y) with y′=dy/dxy' = dy/dx. Such an equation is called homogeneous when its right side depends only on the ratio y/xy/x, so that y′=F(y/x)y' = F(y/x) for x≠0x \ne 0. An equation of this kind can be nonlinear.

In plain words For a linear equation, homogeneous means that nothing drives the system from outside. In the ratio sense, it means the slope stays the same when xx and yy are scaled by the same nonzero factor. When you meet the word, ask which of the two structures is meant.

Example The equation y′+2y=0y' + 2y = 0 is homogeneous. Its solutions Ce−2tCe^{-2t} include the zero function, and sums and multiples of them have the same form. The equation y′+2y=sin⁡ty' + 2y = \sin t is nonhomogeneous, and the zero function fails it, because sin⁡t\sin t does not vanish on any interval. The two meanings can disagree. The equation

y′=1+yx,x≠0,y' = 1 + \frac{y}{x}, \qquad x \ne 0,

is homogeneous in the ratio sense, with F(v)=1+vF(v) = 1 + v. Written as the linear equation y′−y/x=1y' - y/x = 1, it is nonhomogeneous, because its forcing term is 11. The equation y′=(y/x)2y' = (y/x)^2 is homogeneous in the ratio sense and nonlinear.

Check yourself

  1. Classify the equation y′′′+t2y=cos⁡ty''' + t^2y = \cos t for y(t)y(t) and the equation ut=uxx+u2u_t = u_{xx} + u^2 for u(x,t)u(x, t), where ut=∂u/∂tu_t = \partial u/\partial t and uxx=∂2u/∂x2u_{xx} = \partial^2 u/\partial x^2. For each, say whether it is ordinary or partial, give its order, and say whether it is linear.
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The first is an ODE, because yy depends only on tt. Its highest derivative is y′′′y''', so it is third order. It is linear, with coefficients 11 and t2t^2 and forcing term cos⁡t\cos t, and since the forcing term is not zero it is nonhomogeneous.

The second is a PDE, because uu depends on xx and tt and partial derivatives appear. Its highest derivative uxxu_{xx} is second order, and the term u2u^2 makes it nonlinear.

  1. A classmate says that (y′)2=4y(y')^2 = 4y is second order because of the square, and that y′+(sin⁡t) y=0y' + (\sin t)\,y = 0 is nonlinear because it contains a sine. Here y′=dy/dty' = dy/dt. Correct both claims.
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The highest derivative in (y′)2=4y(y')^2 = 4y is y′y', so the equation is first order. The square does not change the order. It makes the equation nonlinear, which is a separate property.

In y′+(sin⁡t) y=0y' + (\sin t)\,y = 0 the sine is a coefficient that depends only on tt, and yy and y′y' appear to the first power, so the equation is linear, first order and homogeneous. By contrast, the equation y′+sin⁡y=0y' + \sin y = 0 is nonlinear, because there the unknown is inside the sine.

  1. Check that y1=2y_1 = 2 and y2=2+e−2ty_2 = 2 + e^{-2t} both solve y′+2y=4y' + 2y = 4. Is y1+y2y_1 + y_2 a solution? Which equation does y2−y1y_2 - y_1 solve?
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For y1=2y_1 = 2 the derivative is 00, and 0+2⋅2=40 + 2 \cdot 2 = 4. For y2y_2 the derivative is −2e−2t-2e^{-2t}, and adding 2y2=4+2e−2t2y_2 = 4 + 2e^{-2t} gives 44. Both are solutions on the whole real line.

The sum y=4+e−2ty = 4 + e^{-2t} gives y′+2y=−2e−2t+8+2e−2t=8y' + 2y = -2e^{-2t} + 8 + 2e^{-2t} = 8, which is not 44, so the sum is not a solution. Each solution already produces the forcing term 44, so together they produce 88. That is the practical meaning of nonhomogeneous.

The difference y2−y1=e−2ty_2 - y_1 = e^{-2t} gives y′+2y=−2e−2t+2e−2t=0y' + 2y = -2e^{-2t} + 2e^{-2t} = 0, so it solves the homogeneous equation y′+2y=0y' + 2y = 0. For that equation sums of solutions do work. If z1z_1 and z2z_2 solve it, then (z1+z2)′+2(z1+z2)(z_1 + z_2)' + 2(z_1 + z_2) splits into z1′+2z1z_1' + 2z_1 plus z2′+2z2z_2' + 2z_2, which is 0+00 + 0.

How this lesson connects

It also uses derivatives of elementary functions.