周次: 10 日期: April 27, 2022 节次: 1
In this lecture, we talk briefly on the numerical method for solving an initial value problem for a general first order differential equation. The method is called Euler’s Method, it is the most simplest numerical method for solving a differential equation. Secondly, we will look at how to sketch graphical solution curves for a first-order autonomous differential equations and give some examples.
Outline
§ 9.3 Euler’s Method
In this section, we would like to take a glance at numerical method for solving the general first-order differential equation with the initial condition being proposed. The method is the simplest one, called the Euler’s method.
As a numerical solution to the differential equation, what we really need is not finding the solution as a function, but providing a table containing pairs of $x$’s and $y$’s, i.e. find the correspondence relation on some sampling points.
Suppose a first-order initial value problem has the form
$$ \begin{align}&\dfrac{d y}{dx} = f(x,y),\ &y(x_0) = y_0.\end{align} $$
with the initial condition (2). Geometrically, one can trace the slope field starting from the initial position $(x_0,y_0)$ to get a curve roughly demonstrating a solution. Euler’s method will be a mathematical realization of this idea.
Euler’s Method
Suppose we are starting from point $(x_0,y_0)$. Given a very small increment $\Delta x$ in variable $x$, we would like to predict a $y$ from the equation (10). Denoting $x_1 = x_0+\Delta x,$ the linearization for unknown function $y$ provides a very good approximation for $y(x_1)$. Indeed,
$$ y(x_1)\approx y(x_0) + y'(x_0) (x_1-x_0). $$
But according to the DE, $y'(x_0)$ is equal to $f(x_0,y_0)$, which can be computed directly. Therefore, we would like to think that, roughly speaking, $y(x_1)$ can be replaced with the linear approximation of it.
$$ y(x_1) \approx y(x_0) + f(x_0,y_0) (x_1-x_0):=L(x_1). $$
This completed the first Euler step.
For the second Euler step, we start from $(x_1,y_1),$ use linear approximation of $y$ at $x_1$ to predict a $y_2$
$$ y_2 \approx y_1 + f(x_1,y_1)(x_2-x_1) $$
By repeating this process again and again, we get an approximated solution to the DE.
Example 1. Using Euler’s Method
Find the first three approximations $y_1$, $y_2$, $y_3$ using Euler’s method for the initial value problem
$$ y' = 1+y,\quad y(0)=1, $$
starting at $x_0 = 0$ with $\Delta x = 0.1.$
Solution.
- Send $x_0 = 0,$ $y_0=1$, then $y_1 = y_0 + (1+y_0) \Delta x = 1.2$
- Send $x_1 = 0.1$, $y_1=1.2$, then $y_2 = y_1 + (1+y_1)\Delta x = 1.42$
- Send $x_2 = 0.2$, $y_2 = 1.42$, then $y_3 = y_2 + (1+y_2)\Delta x = 1.662$.
Demonstrations On Euler’s Method
- For $\frac{dy}{dx} = (1-x)y+\frac{x}{2}$, with initial value indicated by cross point. Blue curve shows the real solution, red curve shows the line segment obtained by Euler’s method.
- Second Example $\frac{dy}{dx} = y-x^2$
- Third Example $y' = - \frac{2xy}{1+x^2}$.
§ 9.4 Graphical Solutions of Autonomous Differential Equations
Autonomous Differential Equation
For example, consider an implicit equation
$$ \frac15 \ln(5y-15) = x+1. $$
By differentiating this equation, we obtain
$$ \frac15 \left(\frac{1}{y-3}\right)\frac{dy}{dx} = 1. $$
Therefore the derivative $\frac{dy}{dx}$ is a function of $y$ only. (Not explicitly depends on $x$).
Definition.
A differential equation for which $dy/dx$ is a function of $y$ only is called an autonomous differential equation.
Examples
- $\frac{dy}{dx} = 3 (y-1) \sin (y)$ is an autonomous differential equation
- $\frac{dy}{dx} = 3(y-x) sin(y)$ is not an autonomous DE.
Equilibrium Values
Think of a differential equation is defining how the derivative $\frac{dy}{dx}$ behaves.
$$ \frac{dy}{dx} = 5(y-3)(1-y) $$
Then for an autonomous differential equation, at what value $y$ the the $\frac{dy}{dx}$ vanishes is called an Equilibrium value or a
rest point.
Thus, equilibrium values are those at which no change occurs in the dependent variable, so $y$ is at rest.
Example 1. Finding Equilibrium Values
The equilibrium values for autonomous differential equation
$$ \frac{dy}{dx} = (y+1)(y-2). $$
Example. Finding Equilibrium Values
The equilibrium values for autonomous differential equation
$$ \frac{dy}{dx} = k y. $$
Sketch the Graphical Solutions to an Autonomous DE
To sketch graphical solutions to the autonomous differential equation, we need to analyze the signs of $y'$ and $y''$, these informations can be read from the differential equation.
Example 2. Sketching Solution Curves
Sketch solutions to the equation
$$ \frac{dy}{dx} = (y+1)(y-2). $$
Step 1. Mark the equilibrium values on the $xy$-plane.
The equilibrium values are $y=-1$ and $y=2$, where $dy/dx = 0.$
Step 2. Identify and label the intervals (regions) where $y'>0$ and $y'<0.$
Step 3. Calculate $y''$ and mark the intervals where $y''>0$ and $y''<0$.
Since
$$ y' = (y+1)(y-2) = y^2 - y -2. $$
Differentiate both sides with respect to $x$, using implicit differentiation
$$ y'' = 2yy'-y' = (2y-1)y' = (2y-1)(y+1)(y-2). $$
Thus we add the sign information to the phase line.
Step 4. Sketch an assortment of solution curves in the $xy$-plane
Example 3. Cooling Soup
What happens to the temperature of the soup when a cup of hot soup is placed on a table in a room? What does a typical temperature curve look like as a function of time?
By Newton’s law of cooling, there is a constant $k$ such that
$$ \frac{dH}{dt} = -k(H-15), $$
where $15$ is the ambient temperature in Celsius.
Since $H'' = -k H' = k^2(H-15)$, we can sketch the phase line and general solutions on the $xy$-plane below.
Example 4. Analyzing the Fall of a Body Encountering a Resistive Force (Omitted)
Example 5. Analyzing Population Growth in a Limiting Environment
If $P$ represents the number of individuals and we neglect departures and arrivals, then
$$ \frac{dP}{dt} = kP, $$
where $k>0$ is the birthrate minus the death rate per individual per unit time.
As the population approaches the limiting population or carrying capacity, resources become less abundant and the growth rate $k$ decreases. Suppose the limiting population is denoting by $M$. Then a simple relationship exhibiting this behavior is
$$ k=r(M-P), $$
where $r>0$ is a constant. Notice that $k$ is negative if $P$ is greater than $M$. Then the model is
$$ \begin{equation}\frac{dP}{dt} = r(M-P) P = rMP-rP^2\end{equation} $$
Equation (3) is referred to as logistic growth.
Since
$$ \frac{d^2 P}{dt^2} = rM P' - 2r P P' = r(M-2P)P' = r^2(M-2P)(M-P)P, $$
we can analyze the phase line and sketch a graph on it.
