We recall that the reason for creating the idea of function $f : D \to \mathbb{R}$ where $D$ is the domain is to understand better a relationship between two sets of quantities: for each $x$ in $D$, there is a $y$, which is also labelled by $y = f(x)$. So people are interested to see how $y$ changes against the corresponding change in $x$.
Scientists already needed the idea of functions in the seventeenth century in the works of Newton and Leibniz. Let us recall that the derivative of $f$ at $x$ is given by
$$f'(x) = \frac{dy}{dx}\bigg|_x = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}.$$
The notation $y = f'(x)$ indicates that we treat the derivative also as a function of $x$. We also recall that the derivative $f'$ of a function $y = f(x)$ can be interpreted as the instantaneous rate of change of $y$ with respect to $x$.
We start with a quadratic function
$$y = 2x^2 + 3x - 4.$$For each given $y = k$, one can solve this equation to obtain
$$x = \frac{-3 \pm \sqrt{9 + 8(4 + k)}}{4}.$$So the unknowns or the solutions are two fixed numbers. Similarly one can have an equation for which the unknown is a function $y = f(x)$. Such an equation is called a differential equation. Just as a quadratic equation may arise from solving a practical problem, one can also formulate a differential equation when the rate of change of a quantity is involved.
A differential equation is an equation involving (higher) derivatives of an unknown function $y(t)$ defined in a domain about $t$.
We briefly introduce some of the simplest differential equations below.
We often hear people stating that certain quantities “growth exponentially” or “growth geometrically”. Here is a precise mathematical description. Indeed such growth/decay rate appears commonly in the natural world.
Let $y = f(x)$. Then the differential equation $$\frac{dy}{dx} = ky, \quad k \neq 0 \text{ constant}$$ describes that $$(\text{the rate of change of } y) \propto y.$$
In ordinary language, the above differential equation is interpreted as: (the rate of change of $y$) is in direct proportion “$\propto$” to $y$, at any $x$ with the proportionality constant $k$. The mathematical interpretation is that the rate of change of $f(x)$ is proportional to $f(x)$.
The differential equation $$\frac{dy}{dx} = ky, \quad k \neq 0 \text{ constant} \tag{1}$$ admits a solution of the unique form $$y = f(t) = Ce^{kt}, \quad C \neq 0,$$ apart from the non-vanishing constant $C$.
Here is another approach. We rearrange the “$dx$”, “$dy$” of the differential equation (1) to obtain
$$\frac{dy}{y} = k\,dx.$$This is called the method of separation of variables. We then integrate both sides with respect to their respective variables:
$$\int \frac{dy}{y} = \int k\,dx.$$So we obtain $\ln|y| = kx + c$, where $c$ denotes an unknown constant. Suppose we know that $y$ is always positive. Then we have
$$y = e^{kx + c} = C\,e^{kx}, \quad C = e^c.$$The above calculation shows that the new unknown $C \neq 0$.
The (1) is a first-order linear differential equation. A solution to (1) would be a function $y = f(x)$. The equation is called first order because it involves no higher derivatives $y'', y''', $ etc and it is linear because it involves no $y^2, y^3,$ etc and $y^2, y'^2, yy', y^2y',$ etc.
Let $y = f(x)$. Then the differential equation $$\frac{dy}{dx} = ay - b, \quad b,\; a \neq 0 \text{ constants} \tag{2}$$ describes that $(\text{the rate of change of } y) + b \propto y$.
The differential equation (2) possesses a solution of the form $$y(x) = b/a + ce^{at} \tag{3}$$ with a non-vanishing constant $c$.
One can verify that (3) is a solution to (2) by direct verification.
Special cases of the DE (2) include models describing:
Example (B&D). Write down a differential equation that describes an object of mass $m$ falls under the Earth's gravitational force near sea level (assume $9.8\,m/s^2$ near the Earth's surface) and air drag (resistance) with drag coefficient $\gamma (> 0)$.
Let us assume that the downward direction is the positive direction. According to Newton's second law of motion, the gravitational force $F$ acting on the mass is $F = mg$, $g = 9.8$. However, there is an air-resistance force acting on the mass in exactly the opposite direction. It is known that such air-resistance force is directly proportional to velocity (speed with direction) of the object falling down. So we may write
$$\text{air-resistance} = -\gamma\,v$$where $\gamma > 0$ is the air-resistance constant of the object (found by experiment) and $v$ its velocity. Thus the total force acting on the object is the sum of the two forces:
$$\text{Total force} = mg - \gamma v.$$But acceleration $= dv/dt$, so that
This is a first order DE in $v$.
Example (B&D). Suppose we have $m = 10$ kg, $\gamma = 2$ kg/s. Then
$$10\frac{dv}{dt} = 10 \cdot 9.8 - 2\,v$$or
$$\frac{dv}{dt} = 9.8 - \frac{v}{5}.$$It turns out that separation of variables method again works for this DE. Then
$$\frac{dv}{9.8 - \frac{v}{5}} = dt.$$Hence
$$\int \frac{dv}{9.8 - \frac{v}{5}} = \int dt.$$Hence
$$\ln\left|9.8 - \frac{v}{5}\right| = -\frac{1}{5}t + c$$for some constant $c$. We can rewrite this as
$$\left|9.8 - \frac{v}{5}\right| = e^{-\frac{1}{5}t} + c' = Ce^{-\frac{1}{5}t}.$$Thus we could have
$$9.8 - \frac{v}{5} = Ce^{-\frac{1}{5}t},$$or
$$\frac{v}{5} - 9.8 = Ce^{-\frac{1}{5}t}.$$The former gives
$$v(t) = 5(9.8 - Ce^{-\frac{1}{5}t})$$and the latter gives
$$v(t) = 5(9.8 + Ce^{-\frac{1}{5}t}).$$Which one of the two solutions to choose depends on the initial condition which would allow one to determine the unknown constant $C$ too. We will study this later. We notice that
$$v(t) \to 5(9.8) = 49, \qquad t \to \infty.$$Not all DEs can be solved. In fact, we could only solve very few DEs from applications although those tend to be very important equations. If we could express the solution of a DE in terms of known functions, then we call the equation exactly solvable and the corresponding solutions exact solutions. We can extract some information from the DE without even solving the equation (which may not be easy or possible).
Example (B&D). Let us revisit the DE
$$\frac{dv}{dt} = 9.8 - \frac{v}{5}. \tag{4}$$Suppose we put $v = 40$ into the equation. Then we have $\frac{dv}{dt} = 1.8$. In fact,
$$v = 40, \qquad \frac{dv}{dt} = 1.8$$ $$v = 50, \qquad \frac{dv}{dt} = -0.2$$ $$\cdots\cdots$$So if we use data to plot $(v, \frac{dv}{dt})$ in the $tv$-plane, then we could see that there is a difference: $v'(t)$ changes its sign when $v = 40$ and $v = 50$. So it is worthwhile to look at when $v'(t) = 0$. That is, when $v = 49$ that we encountered earlier. It turns out that $v(t) \equiv 49$ is a solution to the DE too (check!). It is called equilibrium solution.
We notice that the “field lines” converge to the solution which is represented by the horizontal line $v(t) = 49$ as $t \to \infty$ even for the non-equilibrium solutions. This observation conforms with our earlier derivation that $v(t) \to 49$ as $t \to \infty$ for any solution of the DE.
In general, we can consider direction field of DE of the form
$$\frac{dy}{dx} = f(x, y)$$where $f$ is a fixed function in $x$ and $y$.
Example. Plot the direction field of
$$\frac{dy}{dt} = 3 - 2y.$$We see that the equilibrium solution is when $y(t) = 1.5$.
Match each of the following six equations to exactly one of the six direction field figures below.
Hint: Analyse each equation's equilibrium points (where $y' = 0$) and the sign of $y'$ above and below each equilibrium. For example, $y' = 2 - y$ has equilibrium at $y = 2$ and is stable (solutions converge to $y = 2$). $y' = y - 2$ has equilibrium at $y = 2$ and is unstable (solutions diverge from $y = 2$). The quadratics $y(y-3)$ and $y(3-y)$ have equilibria at $y = 0$ and $y = 3$, with opposite stability characteristics.
Example. We recall that the falling body above sea level equation that we considered earlier
$$\frac{dv}{dt} = 9.8 - \frac{v}{5}.$$The separation of variables method allows us to derive solutions:
$$v(t) = 5(9.8 - Ce^{-\frac{1}{5}t})$$and
$$v(t) = 5(9.8 + Ce^{-\frac{1}{5}t}).$$But we were unable to decide on which of the above two is the solution we look for. In fact, there is insufficient information to pin down the solution. Suppose we know that the object is dropped from rest. Then this can be translated into $v(0) = 0$.
Substitute this information into the above potential solutions yields
$$0 = v(0) = 5(9.8 - Ce^{-\frac{1}{5}\cdot 0}) = 5(9.8 - C)$$and
$$0 = v(0) = 5(9.8 + Ce^{-\frac{1}{5}\cdot 0}) = 5(9.8 + C).$$Thus the first expression would give $C = 9.8$ and the second expression would give $C = -9.8$. So both expressions become one:
Clearly, $v(0) = 0$, and $v(t) \to 49$ as $t \to \infty$. Thus the object falls to a constant speed $49$ in the long run which we call the terminal velocity of the object. We note that $v(t) = 49$ was also the equilibrium solution that we discussed earlier. The velocity does not increase any further over time because our calculation shows that the air-resistance counters balance the gravitational force.
The condition $v(0) = 0$ is called a initial value condition of the differential equation. In general,
$$\frac{dy}{dx} = f(x, y), \quad y(0) = y_0$$ for some $y_0$, is called an initial value problem.
Example (Cont.). Find the positions of the object released from rest discussed in this section and measured at the initial position.
Let $x = x(t)$ be the position of the object measured from the height (in meters) where it is released from rest and measured at the initial position. We recall that we chose the downward direction is the positive direction. Now let us interpret
velocity = rate of change of distance.
This means on the one hand, that $x'(t) = v(t) = 49(1 - e^{-t/5})$. On the other hand, we have $x(0) = 0$. Thus we are led to another initial value problem:
$$\frac{dx}{dt} = 49(1 - e^{-t/5}), \quad x(0) = 0.$$Of course this DE is separable. But it may be easier to perform a simple integration of the above DE yields
$$x(t) = 49t + 245\,e^{-t/5} + c$$for some constant $c$. Substitute the initial condition $x(0) = 0$ into the solution further yields
$$0 = x(0) = 0 + 245 + c.$$Thus $c = -245$ and we finally obtain
Under this terminology, we immediately have
Let $k \neq 0$. The initial value problem $$\frac{dy}{dx} = ky, \quad y(0) = y_0 \tag{5}$$ admits the unique solution of the form $$y = f(t) = y_0\,e^{kt}.$$
Similarly,
Let $a \neq 0$. The initial value problem $$\frac{dy}{dx} = ay - b, \quad y(0) = y_0 \tag{6}$$ admits the unique solution of the form $$y = f(t) = b/a + [y_0 - b/a]\,e^{at}.$$
We have $$F\!\left(t,\, u(t),\, u'(t),\, u''(t),\, \ldots,\, u^{(n)}(t)\right) = 0$$ be a differential equation of order $n$, where $$F(t, x_1, x_2, \ldots, x_n)$$ is a multi-variable function in the variables $t, x_1, x_2, \ldots, x_n$.
Example. Consider the previously encountered examples:
| DE | DE rewritten as $F = 0$ | Explicit $F$ |
|---|---|---|
| $\dfrac{dy}{dx} = ky$ | $F(x,y,y') = \dfrac{dy}{dx} - ky = 0$ | $F(x,x_1,x_2) = x_2 - kx_1$ |
| $\dfrac{dv}{dt} = 9.8 - \dfrac{v}{5}$ | $F(t,v,v') = \dfrac{dv}{dt} + \dfrac{v}{5} - 9.8 = 0$ | $F(t,x_1,x_2) = x_2 + \dfrac{1}{5}x_1 - 9.8$ |
| $y' = \dfrac{3x^2}{y}$ | $F(x,y,y') = yy' - 3x^2 = 0$ | $F(x,x_1,x_2) = x_1 x_2 - 3x^2$ |
All the above examples are first order DEs since the highest order in those equations are first order derivative. The first two examples are linear first order DEs while the third example is non-linear since it involves the term $yy'$.
Example. The following are linear second order DEs
| DE | DE rewritten as $F = 0$ | Explicit $F$ |
|---|---|---|
| $y'' + 2y' - 3y = 0$ | $F(x,y,y',y'') = y'' + 2y' - 3y = 0$ | $F(x,x_1,x_2,x_3) = x_3 + 2x_2 - 3x_1$ |
| $x^2y'' - 2xy' + 3y = 0$ | $F(x,y,y',y'') = x^2y'' - 2xy' + 3y = 0$ | $F(x,x_1,x_2,x_3) = x^2x_3 - 2xx_2 + 2x_1$ |
We call the first example constant coefficient second order linear differential equation, and the second one second order linear differential equation with variable coefficients. It is called an Euler's equation.
The following differential equation
| DE | DE rewritten as $F = 0$ | Explicit $F$ |
|---|---|---|
| $\dfrac{d^2y}{dt^2} = 6y^2 + t$ | $F(t,y,y',y'') = \dfrac{d^2y}{dt^2} - 6y^2 - t = 0$ | $F(t,x_1,x_2,x_3) = x_3 - 6x_1^2 - t$ |
This DE is second order and nonlinear. It is called the Painlevé I equation.
Write down the $F$ involved in the DE $$\frac{d^2y}{dt^2} = 2y^3 + ty + \alpha,$$ where $\alpha$ is a fixed constant. This DE is called the Painlevé II equation.
Ans: $F(t, y, y', y'') = y'' - 2y^3 - ty - \alpha = 0$, i.e., $F(t, x_1, x_2, x_3) = x_3 - 2x_1^3 - tx_1 - \alpha$.
Example. Pendulum equation of a rod of length $L$ with an object of mass $m$ attached at the end:
$$\frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0.$$We see the appearance of the term $\sin\theta$ making the pendulum highly nonlinear.
When $\theta$ is small, then $\sin\theta \sim \theta$. So the pendulum equation is approximated by
$$\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0,$$which is a second order linear DE.
Determine the order and linear/nonlinearity of the following equations.
Construct a DE for each of the following $y$ for which the $y$ are their respective solutions.
Verify that the given $y$, or $y_1$ and $y_2$ are solutions of the corresponding DEs:
We assume an insulated metal rod of length $L$ that is heated at one of its end. Suppose that one keeps the temperatures at both ends of the rod, namely $x = 0$ and $x = L$ at constant temperatures $T_1$, and at $T_2$.
Let $u = u(x, t)$ denote the temperature of distribution of the metal rod at the location $0 \leq x \leq L$ and at time $t$. Suppose the thermal diffusivity
$$\alpha^2 = \frac{\kappa}{\rho s} = \frac{\text{thermal conductivity}}{(\text{metal density}) \times (\text{specific heat})}.$$Then it is known that the “heat” function $u$ satisfies, when considered as a kind of diffusion. Since there are more than one independent variables involved, so partial differentiations with respect to $x$ and $t$ are involved, hence partial differential equations are needed to describe heat flow.
$$\alpha^2 u_{xx} = u_t, \quad 0 < x < L, \quad t > 0;$$
$u(x,0) = f(x)$, $\;\;u(0,t) = T_1$, $\;\;u(L,t) = T_2,$
where $f(x)$ represents the initial temperature distribution of the metal rod.
Suppose we have an elastic string tight at both ends with a length $L$. Let $u = u(x, t)$ denote the vertical displacement of the string at the position $x$, $0 \leq x \leq L$ and at time $t$.
Suppose it has an initial position and initial velocity given by $u(x, 0) = f(x)$ and $u_t(x, 0) = g(x)$ respectively.
A population grows at a rate proportional to its current size. If $P(0) = 100$ and $P(5) = 200$, determine the governing DE, its solution, and the growth constant $k$.
What is the DE governing the population?
What is the general solution to $dP/dt = kP$?
Given $P(0) = 100$ and $P(5) = 200$, find $k$.
Classify differential equations by order and linearity.
What is the order of the DE: $y''' + xy'' + 3y(y')^2 + x^2 y = 0$?
Is the above DE linear or nonlinear?
Which of these is a linear second-order DE?
— End of What are Differential Equations? Notes —