Math 555: Differential Equations

2.4 Linear and Nonlinear Equations

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2.4.1 Existence and Uniqueness of Linear and Nonlinear ODEs

So far we have written a first order ordinary differential equation in general form

$$\dfrac{dy}{dt} = f(t, y)$$
and used software to draw the direction field of the differential equation. Then we were able to use that same software to draw several integral curves or graphs of solutions to our differential equation. We need to talk about the differences between linear and nonlinear equations. In Chapter 3, we will discuss the topic of linearity at length. For now we need to recognize what makes a differential equation linear or nonlinear. We also need to understand how to obtain enough information about a differential equation to know that there is only one integral curve that represents the solution to our initial value problem that passes through a given point in our direction field.

Current differential equations used by engineers and scientists are too complex for one person or even a team of people to make all the computations necessary to solve them numerically. We rely on computers to make these computations for us instead of working them by hand. Therefore, it is necessary for us to understand more about our differential equation so that we know which algorithms and programs we may use to perform our computations. Differential equations are too vast a subject and we cannot expect a single system or program, even Wolfram Alpha , to be able to solve all of them without our direction. Fortunately, most of this course provides information about how to classify differential equations and recognize those properties that allow us to choose for the computer, " How to computationally solve the problem. "

2.4.2 Linearity

Linearity is used frequently in the study of differential equations because the operations we use in calculus are in fact, linear . In the beginning of our calculus instruction we learned that the derivative operation and the integral operation were both linear. When we multiply a mathematical object by real (or complex) numbers we call the multiplication scaling the object. If we multiply it by 2 we make it twice as big; if we multiply by $\frac{1}{2}$ we make it half the original size, and if we multiply by -1, we change the direction of the original object. We call the real number we use to scale the object a coefficient or a scalar .

When we multiply like mathematical objects by scalars and add them together we call the result

$$ a\mathbf{x} + b\mathbf{y} + c\mathbf{z} $$
a linear combination . We can create linear combination of vectors as in the example above. We can create linear combinations of matrices

$$ \alpha A + \beta B + \gamma C.$$
We can also create linear combinations of functions

$$ a\,f(x) + b\,g(x) + c\,h(x).$$
We learned early in our study of calculus that we can compute derivatives and integrals of complicated functions using the linearity property of the derivative operation and the integral operation. We compute the following derivative "term by term:"

$$\begin{align*} \dfrac{d}{dt}\left( 3t^2 + \log(t) - 2\sin(t)\right) &= \dfrac{d}{dt}\,3t^2 + \dfrac{d}{dt}\log(t) + \dfrac{d}{dt}\left(-2\sin(t)\right) \\ \\ &= 3\dfrac{d}{dt}t^2 + \dfrac{d}{dt}\log(t) - 2\dfrac{d}{dt}\sin(t)\\ \\ &= 3\left(2t\right) + \left(\dfrac{1}{t}\right) - 2\left(\cos(t)\right) \\ \\ &= 6t + \dfrac{1}{t} -2\cos(t). \end{align*}$$
Likewise, we compute integrals in a similar manner:

$$\begin{align*} \displaystyle\int 3t^2 + \dfrac{1}{t} - 2\sin(t)\,dt &= \displaystyle\int 3t^2\,dt + \displaystyle\int\dfrac{1}{t}\,dt + \displaystyle\int -2\sin(t)\,dt \\ \\ &= \displaystyle\int 3t^2\,dt + \displaystyle\int\dfrac{1}{t}\,dt + 2\displaystyle\int -\sin(t)\,dt \\ \\ &= t^3 + \log|t| + 2\left(\cos(t)\right) + C \\ \\ &= t^3 + \log|t| + 2\cos(t) + C \end{align*}$$
We memorized these linearity properties

$$ \begin{align*} \dfrac{d}{dt}\left(\alpha f(t)\right) &= \alpha\dfrac{d}{dt}\,f(t) \\ \\ \dfrac{d}{dt}\,\left( f(t) + g(t)\right) &= \dfrac{d}{dt}\,f(t) + \dfrac{d}{dt}\,g(t) \\ \\ \displaystyle\int\alpha f(t)\,dt &= \alpha\displaystyle\int f(t)\,dt \\ \\ \displaystyle\int \left(f(t) + g(t)\right)\,dt &= \displaystyle\int f(t)\,dt + \displaystyle\int g(t)\,dt \end{align*}$$

In English, these may be stated as

• The derivative of a scalar times a function is the scalar times the derivative of that function.
• The derivative of a sum (of functions) is the sum of the derivatives (of those functions).
• The integral of a scalar times a function is the scalar times the integral of that function.
• The integral of a sum (of functions) is the sum of the integrals (of those functions).

2.4.3 Operators

To a mathematician derivative and integral are not just operations; they are operators . At first our functions were functions of real variables. They had one real input and one real output. As our study of mathematics matured, we realized functions can have several real inputs and even several real outputs. In our general form of a differential equation

$$ \dfrac{dy}{dt} = f(t, y)$$
the function $f$ on the right has two real inputs and one real output. To create the direction field we use the same function above and define a new one that takes our variables $t$ and $y$ as inputs and outputs a unit vector that has slope $\frac{dy}{dt}$.

$$ \mathbf{v} = \mathbf{F}(t, y)$$
where now function $\mathbf{F}$ has two real inputs and a vector output that is graphed on the $ty$-plane.

Derivative can be thought of as a function that has a differentiable function as an input and the derivative of that function as the output

$$ D\left[f(x)\right] = f'(x).$$
At this point we need some vocabulary to distinguish the normal functions with scalar inputs and scalar outputs from the ones with functions as inputs. We call a function that has a function as its input an operator . We can define integral to be an operator that has an integrable function as its input and outputs a family of anti-derivatives.

$${\mathscr{I}}\left[f(t)\right] = F(t) + C.$$
The First Fundamental Theorem of Calculus says that if $F$ is any anti-derivative of $f$; that is if $F'(t) = f(t)$, then all anti-derivatives of $f$ look like $F(t) + C$. The definite integral can be defined as an operator with an integrable function as an input and a real number as an output

$$ \text{Area} = \mathscr{I}_a^b\left[f(t)\right] = \displaystyle\int_a^b f(t)\,dt.$$
Notice we also use square brackets instead of round brackets for the inputs of an operator to further distinguish it from normal functions.

2.4.4 First Order Linear Operators

If we write an first order, linear ordinary differential equation in standard form; that is with all terms that contain the dependent variable on the left-hand side of the equations, and any remaining terms on the right-hand side of the equation then it will look like this

$$ \dfrac{dy}{dt} + p(t)y = g(t).$$
Notice that in addition to displaying any term with the dependent variable on the left-hand side of the equation, we divide if necessary to make sure that the coefficient of $\frac{dy}{dt}$ is 1. In this standard form of a first order linear ordinary differential equation $p(t)$ is called the coefficient function . We can define a new linear differential operator ; one that takes a differentiable function as its input and outputs a function. The rule for computing the output from the input is multiply the input by $p(t)$ and add it to the input's derivative . We usually call this linear differential operator $L$ and

$$ L[y] = \dfrac{dy}{dt} + p(t)y.$$

The first two questions we have about all this new vocabulary and the new operator is,

• What is linear about it?
• Why is that important?

2.4.5 Question: What is Linear About a Linear Operator

Notice that we defined our operator $L$ using derivative and scalar multiplication. The function $p(t)$ is one of our familiar normal functions with one scale input and one scalar output so we can call this function a scalar function . What is linear about it is that we have our familiar linearity properties from algebra and differential calculus

$$\begin{align*} L\left[\alpha y\right] &= \dfrac{d}{dt}\left(\alpha y\right) + p(t)\left(\alpha y\right) \\ \\ &= \alpha \dfrac{d}{dt}\,y + \alpha\,p(t) y \\ \\ &= \alpha\left( \dfrac{dy}{dt} + p(t)y \right) \\ \\ &= \alpha\,L[y]. \end{align*}$$
If $y$ and $z$ are both differential functions then

$$\begin{align*} L\left[ y + z \right] &= \dfrac{d}{dt}\left( y + z \right) + p(t)\left( y + z \right) \\ \\ &= \dfrac{dy}{dt} + \dfrac{dz}{dt} + p(t)y + p(t)z \\ \\ &= \dfrac{dy}{dt} + p(t) y + \dfrac{dz}{dt} + p(t) z \\ \\ &= L[y] + L[z] \end{align*}$$

Answer:

If we input a linear combination of differential functions $\alpha f(t) + \beta g(t)$ then the output we get is the same linear combination of the individual inputs $f(t)$ and $g(t)$

$$\begin{align*} L\left[\alpha f(t) + \beta g(t)\right] &= \dfrac{d}{dt}\left(\alpha f(t) + \beta g(t)\right) + p(t)\left(\alpha f(t) + \beta g(t)\right) \\ \\ &= \dfrac{d}{dt}\left(\alpha f(t)\right) + \dfrac{d}{dt}\left(\beta g(t)\right) + p(t)\left(\alpha f(t)\right) + p(t)\left(\beta g(t)\right) \\ \\ &= \alpha\dfrac{df}{dt} + \beta\dfrac{dg}{dt} + \alpha p(t)f + \beta p(t)g \\ \\ &= \alpha\left(\dfrac{df}{dt} + p(t)f\right) + \beta\left(\dfrac{dg}{dt} + p(t)g\right) \\ \\ &= \alpha L[f] + \beta L[g] \end{align*}$$

Show that a Differential Operator is a Linear Operator

2.4.6 Why is Linearity Important?

In chapter one we wrote the differential equation in general form

$$ y'(t) = f(t, y) $$
and differentiated both side of the equation to determine that if two solutions shared even one point in common, then they were the same solution and their graph was the same integral curve. In other words we determined that integral curves could neither touch nor cross each other. The conclusion relied on an assumption about the differential equation that I did not mention at the time. What are these assumptions. It turns out that the assumption I need to make about the differential equation is different for linear and nonlinear differential equations. If the differential equation is linear

$$y' + p(t)y = g(t)$$
then the conclusion that integral curves never cross or touch holds on any interval where both $p(t)$ and $g(t)$ are continuous.

Theorem 2.4.1

Existence and Uniqueness for First Order Linear Differential Equations

Given the first order linear initial value problem

$$y' + p(t)y = g(t),\qquad y(t_0) = y_0$$
If the functions $p$ and $g$ are both continuous on an open interval $I: a < t < b$ containing the point t = t_0, that is $a < t_0 < b$, then there is a unique function $y = \phi(t)$ that solves the initial value problem. That means $\phi(t)$ solves the differential equation

$$\phi'(t) + p(t)\,\phi(t) = g(t)$$
for every point $t$ in $I$ and $\phi$ satisfies the initial condition

$$\phi(t_0) = y_0.$$

Example 2.4.1

Determine without solving the differential equation an interval in which a unique solution of the given initial value problem is certain to exist

$$ ty' + 2ty = 4,\qquad y(1) = 2$$

Solution:

First this differential equation is linear but not in standard form. If we divide both sides by $t$ we obtain

$$ y' + 2y = \dfrac{4}{t} $$
The coefficient function $p(t) = 2$ is a constant function and continuous everywhere on the real line but the right-hand side $g(t) = \frac{4}{t}$ has a vertical asymptote at $t=0$. This breaks up the real line into two intervals, $(-\infty,0)$ and $(0,\infty)$. Only one of those intervals has the initial point $t_0 = 1$, $(0,\infty)$. Therefore our interval of solution is $(0,\infty)$.
∎

Example 2.4.2

Determine without solving the differential equation an interval in which a unique solution of the given initial value problem is certain to exist

$$ y' + (\tan(t))y = \sin(t),\qquad y(\pi) = 0 $$

Video Solution

2.4.7 First Order Nonlinear Differential Equations

If a differential operator is not linear then we simply say it is nonlinear. If a differential equation is not linear then it is a nonlinear differential equation. We looked at some rules for recognizing nonlinear differential equations. Nonlinear differential equations are usually much harder to solve. A theorem that guarantees us a unique solution will be similarly more difficult. We do not have a standard form for all possible nonlinear differential equations so the general form of a first order initial value problem will be used

$$y' = f(t,y),\qquad y(t_0) = y_0.$$
Notice that $f$ has two inputs and one output. The function $f$ doesn't need to take into consideration that $y$ is a function of $t$ or that its output is actually the derivative $\frac{dy}{dt}$. The function is only concerned with its two real number inputs and one real number output. We are going to need some multi-variate calculus to determine an interval over which we can find a unique solution.

2.4.8 Partial Derivatives

How do we compute the derivative of a function with two inputs? We resort to the definition of derivative

$$ f'(x) = \lim_{h\rightarrow 0} \dfrac{f(x+h) - f(x)}{h}$$
If $f(t,y)$ is a function of two inputs we can picture the domain of $f$ as a rectangle on the $ty$-plane

We use the cross-product symbol $\times$ to represent the rectangle formed by an interval $(a,b)$ on the horizontal axis and the interval $(c,d)$ on the vertical axis denoting it $R = (a,b)\times(c,d)$. Keep in mind that any or all of the endpoints may by infinity. Using this notation the entire $ty$-plane is denoted $(-\infty,\infty)\times(-\infty,\infty)$.

How do we compute a derivative? We compute two different derivatives with respect to each variable one at a time. We compute the limit

$$\lim_{h\rightarrow 0}\dfrac{f(t+h,y)-f(t,y)}{h}$$
and the limit

$$\lim_{h\rightarrow 0}\dfrac{f(t, y+h)-f(t,y)}{h}.$$
If you start at a point inside the domain with two inputs you can move away from it in an infinite number of different directions.

Since these two directions are only a part of all possible directions we travel and compute a derivative they are called partial derivatives and they are denoted with a special symbol call the partial , $\partial$. So, the partial derivative with respect to the first input $t$ is given by

$$f_{t}(t,y) = \dfrac{\partial f}{\partial t}(t,y) = \lim_{h\rightarrow 0}\dfrac{f(t+h,y)-f(t,y)}{h}$$
and the partial derivative with respect to the second input $y$ is given by

$$f_{y}(t,y) = \dfrac{\partial f}{\partial y}(t,y) = \lim_{h\rightarrow 0}\dfrac{f(t,y+h)-f(t,y)}{h}.$$
The first notation for partial derivatives is Newtonian notation $f_y$ and the second is Leibniz notation $\frac{\partial f}{\partial y}$.

How does one compute partial derivatives?

When you want to compute the partial derivative of a function with respect to one variable, pretend the other variable is a constant.

Let's compute some partial derivatives:

$$\begin{align*} \dfrac{\partial}{\partial t} \left(ty^2\right) &= y^2\dfrac{\partial}{\partial t} \left(t\right) = y^2 \\ \\ \dfrac{\partial}{\partial y} \left(ty^2\right) &= t\dfrac{\partial}{\partial y} \left(y^2\right) = t(2y) = 2ty \\ \\ \dfrac{\partial}{\partial t} \left(t^2e^{ty}\right) &= \dfrac{\partial}{\partial t}\left(t^2\right)e^{ty} + t^2\dfrac{\partial}{\partial t}\left(e^{ty}\right) = 2te^{ty} + t^2\left(ye^{ty}\right) = 2te^{ty} + t^2ye^{ty} \\ \\ \dfrac{\partial}{\partial y} \left(t^2e^{ty}\right) &= t^2\dfrac{\partial}{\partial y} \left(e^{ty}\right) = t^2\left(te^{ty}\right) = t^3e^{ty} \\ \end{align*}$$

2.4.9 Determining the Interval of Solution of a Nonlinear IVP

Determining the interval of solution is more complicated too. To understand this let's look at the example of a curling wave at the shoreline of a beach.

A Breaking Wave Problem

Theorem 2.4.2

Existence and Uniqueness for First Order Nonlinear Differential Equations

Consider the possibly nonlinear first order initial value problem

$$y' = f(t,y),\qquad y(t_0) = y_0.$$
If the function $f$ and $\frac{\partial f}{\partial y}$ are both continuous in some rectangle $a < t < b$, $\ c < y < d$ containing the point $(t_0,y_0)$, then in some interval $t_0-h < t < t_0+h$ contained in $(a,b)$ for which there is a unique solution $y = \phi(t)$ that solves the initial value problem. That means $\phi$ solves the differential equation

$$\dfrac{d\phi}{dt} = f(t, \phi) $$
on the interval $(t_0-h, t_0+h)$, and $\phi$ satisfies the initial condition

$$\phi(t_0) = y_0.$$

Recall that since the point $(t_0,y_0)$ is contained in the rectangle $R = (a,b)\times(c,d)$ we have that $a < t_0 < b$ and $c < y_0 < d$. Further since the interval $(t_0-h, t_0+h)$ is contained in the interval $(a,b)$ we have that $a < t_0-h < t_0+h < b$.

Example 2.4.3

$$ y' = y^{1/3},\qquad y(0) = 0$$

This initial value problem in nonlinear because of the term $y^{1/3}$. We cannot use Theorem 2.4.1 so we will use the more difficult Theorem 2.4.2. Notice that the independent variable $t$ does not appear on the left-hand side of the equation. This means that the continuity of function $f(t,y)$ does not depend on $t$, only on $y$. Computing our partial derivative from Theorem 2.4.2 we obtain

$$ \dfrac{\partial f}{\partial y} = \dfrac{1}{3}y^{-2/3} = \dfrac{1}{3y^{2/3}}. $$

This means that $\frac{\partial f}{\partial y}$ has a vertical asymptote at the point $t_0=0$, our initial point! There is no interval of the form $(0-h,0+h)$ for which $\frac{\partial f}{\partial y}$ is continuous. Since we cannot apply Theorem 2.4.2 we can find no interval containing $t_0 = 0$ for which we have a unique solution.

This differential equation was constructed to illustrate how difficult nonlinear differential equation can get.

$$ \phi(t) = 0 $$
is a solution because $\phi'(t) = 0 = 0^{1/3} = \phi(t)$. Also $\phi(0) = 0$ so $\phi$ satisfies the initial condition as well. Our differential equation is separable so we can solve it as follows:

$$\begin{align*} y^{-1/3}\dfrac{dy}{dt} &= 1 \\ \\ \displaystyle\int y^{-1/3}\dfrac{dy}{dt}\,dt &= \displaystyle\int\,dt \\ \\ \dfrac{3}{2}y^{2/3} &= t + c \\ \\ y^{2/3} &= \left(y^{1/3}\right)^2 = \dfrac{2}{3}\left(t + c\right) \\ \\ \left|y^{1/3}\right| &= \left(\dfrac{2}{3}\left(t + c\right)\right)^{1/2} \\ \end{align*}$$
This gives us two more families of solutions

$$ y(t) = \pm\left(\dfrac{2}{3}\left(t + c\right)\right)^{3/2}. $$
For $c=0$ we get two solutions

$$ y(t) = \pm\left(\dfrac{2}{3}t\right)^{3/2}. $$
Since square root has no real value for $t < 0$ we define our solutions using a piece-wise definition

$$ y(t) = \left\{\begin{array}{lcl} 0 & \text{ when } & t < 0 \\ \left(\dfrac{2}{3}t\right)^{3/2} & \text{ when } & t \ge 0 \end{array}\right.. $$
and

$$ y(t) = \left\{\begin{array}{lcl} 0 & \text{ when } & t < 0 \\ -\left(\dfrac{2}{3}t\right)^{3/2} & \text{ when } & t \ge 0 \end{array}\right.. $$
We can also shift our solution to the right $t_0$ and obtain solutions of the form

$$ y(t) = \left\{\begin{array}{lcl} 0 & \text{ when } & t < 0 \\ \pm\left(\dfrac{2}{3}(t-t_0)\right)^{3/2} & \text{ when } & t \ge 0 \end{array}\right.. $$
The graphs of these solutions are in your textbook and the integral curves all coincide on the interval $(-\infty,0]$. Here we use the square bracket to indicate that we include the endpoint $t=0$ in our interval.

Example 2.4.4

Determine without solving the differential equation a rectangle in the $ty$-plane where the hypotheses of Theorem 2.4.2 are satisfied.
$$ y' = \left(t^2 + y^2\right)^{3/2},\qquad y(2) = 3. $$

Video Solution

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