(1) Some specific first and second-order liner differential equations:

(a)

(b)

(b)

(c)

Find the general solution using Laplace transforms (not by any other method) and plot solutions over a range of t values, for a given initial condition.

(2) Forced undamped motion: .

Ordinary Differential Equations, Tenenbaum & Pollard, 338-342.

(3) Kirchoff’s law for the current in a closed circuit: .

Ordinary Differential Equations, Tenenbaum & Pollard, 370-375.

(4) Extra credit: Discuss the solution of a system of 2 linear first-order differential equations by using Laplace transforms.

Ordinary Differential Equations, Tenenbaum & Pollard, Lesson 27, pages292-312

In this block we will apply the Laplace transform to the solution of linear differential equations involving higher order derivatives.

Laplace’s idea was to transform differential equations into algebraic equations, rearrange the algebra (sometimes easier said than done) and then look for the unique solution function that gave this algebraic answer. In other words, the differential equation gives us an algebraic gadget, and if we can find a function from a catalog that gives the same algebraic gadget then we have, thanks to the uniqueness of the algebraic gadget, found the solution to the differential equation.

The Laplace transform is an operator on functions, and typically those functions are solutions to linear differential equations.

What makes the Laplace transform work is that solutions to linear differential equations do not grow too fast. We will use the following basic fact, without going into any more details at this point:

if satisfies a linear differential equation (of any order) then the improper integral converges for some value of .

What it means for the improper integral to converge, is that exists, for some .

Here is an example where the improper integral does not converge for any value of :

Let . Then the improper integral

does not converge for any value of .

For convenience, let’s call a function for which the improper integral converges for some , exponentially-bounded. The name is meant to imply that, in some loose sense, these functions do not increase too much faster than an exponential function. The calculation above shows that the function is not exponentially-bounded. Solutions to linear differential equations are always exponentially-bounded.

For an exponentially-bounded function the Laplace transform of is .

Here is a very simple example:

the function is exponentially-bounded because it is constant, and its Laplace transform is for .

For more complicated functions (that is, just about everything else) the calculation of Laplace transforms becomes more complicated. Not so long ago people complied tables of Laplace transforms because calculation of the Laplace transform was relatively less easy, and because one wanted to read the answer backwards – knowing what Laplace transform we got from a solution to a linear differential equation, which known function also had this Laplace transform. The text for this course has short tables of Laplace transforms on pages 306 and 310.

Nowadays, there is some progress in this area. Because computer algebra systems have built-in versions fo the latest algorithms for calculating integals, it is no surprise that these systems can readily calculate Laplace transforms. So the hard work of calculating integrals is passed to the computer algebra system which does this not only as well as we can do it, but better than we can, and as well as it can possibly be done.

The MATLAB command for calculating the Laplace transform of a function is

>> syms s t; laplace(x(t) ,t,s)

Without the command “syms s t” MATLAB does not know how to handle the symbols “s” and “t”.

So, for example,

>> syms s t; laplace(1 ,t,s)

yields

1/s

The Mathematica command for the Laplace transform is

LaplaceTransform[x[t], t, s]

Laplace transforms and linear differential equations

It is a simple consequence of basic properties of integrals that for any two exponentially-bounded functions and any two constant we have . In other words, the Laplace transform is a linear operator on exponentially-bounded functions. It is this linearity property that allows us to usefully apply the Laplace transform to linear differential equations.

One more ingredient we will need is a formula to relate the Laplace transform of the derivative of a function to the Lapalce transform of .

The formula we want comes about by a straightforward use of integration by parts:

First-order homogeneous linear differential equations

These are differential equations of the form .

By applying a Laplace transfrom to both sides of this differential equation we see that we must have for a solution to the differential equation. That is, for .

So, if we can find a function with then , by the uniqueness of Laplace transforms for exponentially-bounded functions, we will have found, to within a constant, the solution to the differential equation.

Fortunately for us, both MATLAB and Mathematica have built-in commands for finding the inverse of the Laplace transform. For MATLAB the command is

>> syms A B s t; ilaplace(A/(B+s),s,t)

with the result

>> A*exp(-B*t)

For Mathematica the command is

Inverse LaplaceTransform[A/(B+s),s,t]

with the result

In our case, so the solution to the differential equation is .

First-order non-homogeneous linear differential equations

These are differential equations of the form where . A similar approach to the homogeneous case shows us that the Laplace transform of a solution must satisfy .

Using inverse Laplace transforms we get where .

Second-order linear differential equations

These are linear differential equations of the form .

We apply the Laplace transform to to get

Applying this to the second-order differential equation above gives

.

This is an algebraic expression which we solve for . We then apply the inverse Laplace transform as before to find the solution of the linear differential equation.

Details for doing this are given on pages 185-187 of differential Equations with MATLAB, Hunt and others, or in the “Applications” section for “LaplaceTransform” in the Mathematica help system.

Note that in the second-order differential equation above, we have assumed , but there is no reason to do this: we can replace D by any function for which we can find a Laplace transform, that is, by any exponentially-bounded function.

Second-order equations and systems of first order equations

The second-order linear equation can be written as a system of two first-order linear equations by the simple device of writing . By this device, we can re-write the second-order linear differential equation as the following system of first-order linear equations:

$layex \frac{dx}{dt}=y(t)$.

In the other direction we can, in general, write a system of linear-first-order equations as a single second-order linear differential equation. The exception is when the two first-order equations are completely decoupled.

called a matrix.

This square matrix A encodes the essential information for the linear function T.  The matrix operates on vectors to produce the exact same result as does T:

.

The constants associated with finding the invariant lines of T are the eigenvalues of  , and they are the roots of the quadratic polynomial , which is known as the characteristic polynomial of A  (if you have studied some linear algebra, this is the determinant of ).

The characteristic polynomial, being a quadratic in can have two distinct real roots, one real root, and two complex roots which are complex conjugates.

The behavior of the associated differential equations

(which describe the smooth flow of the vector field corresponding to the linear function T)

is qualitatively different, according to the character of the solutions to the characteristic equation.

Case 1: Two distinct roots

The characteristic equation is which has roots so there are two distinct real roots exactly when .

This happens, for example, for the linear function with matrix , where the roots of the characteristic equation are . Let’s call these two roots and . There are two invariant lines for the linear function defined by the matrix A: the lines and .

The corresponding differential equations (for the flow of the vector field) for this linear function are:

On the line the differential equations reduce to a single differential equation:

.

This simple linear differential equation has an exponential solution , where is an arbitrary constant,  BUT ONLY ON THE LINE .

On the other invariant line, , the differential equations reduce to a single differential equation:

which has the exponential solution , where is an arbitrary constant,  BUT ONLY ON THE LINE .

What about solutions to the differential equations

that are not restricted to one or the other of these two invariant lines, but describe a flow through other parts of the plane?

The idea is to project such solutions onto both of the invariant lines, and add together the projected solutions we find on  the invariant lines.

In this example we see that one of the eigenvalues is negative and the other is positive. This gives us a flow for which the origin is a saddle point, with the flow coming in toward the origin on one of the invariant lines, and away from the origin along the other invariant line.

A very nice picture of this can be obtained using TEMATH(Tools for Exploring Mathematics).

TEMATH is installed on the computers in the lab. Open it and run this example following the instructions here. You can also download TEMATH and run it on your own computer.

In the TEMATH instructions, an eigenvector is any non-zero vector on the invariant line for an eigenvalue.

Tasks for this 3-week block

Using TEMATH and/or MATLAB, go through all the cases for the roots of the characteristic equation, describing the geometry of the solutions of the differential equations

for the linear function corresponding to the matrix . Use examples different to those on the TEMATH page.

In each case, also describe the exact solutions to the differential equations obtained from MATLAB, and relate the form of those solutions to the geometry of the flow.

In this 3-week block we will be looking at the behavior of systems of linear differential equations.

A system of two linear differential equations is a pair of linked differential equations of the form

where are constants.

There is a more general form for linear systems in which the vary with but in this block we will only look at the case in which are constant, and do not vary with .

The analysis of systems of linked linear differential equations revolves around the behavior of linear functions, so will first spend a little time looking at linear functions on the plane. Trying to get a picture of a linear function will lead us very naturally into systems of linear differential equations.

Linear transformations of vectors

The Euclidean plane consists of all ordered pairs where are real numbers. We think of the point as the endpoint of a vector segment that starts at the origin :

Sometimes we will refer to “the vector ” meaning the directed vector segment from .

A linear function on the Euclidean plane is a function of the form . A linear function is exactly a function that preserves vector addition and scalar multiplication: and for all vectors and all real numbers .

As a result, once we know what a linear function T does to the vectors we know what T does to every vector. The reason is that .

As an example, we will look at how the linear function transforms vectors. One way to begin to get a geometric picture of how T transforms vectors is to look at what T does to the unit square with corners :

First, so .

The line segment joining consists of vectors and for these vectors we have which, as is a line segment running from .

Similarly, T maps the line segment from to the line segment from . So, T maps the unit square to a parallelogram as follows:

Vector fields for linear transformations

In principle, we could just draw, for each vector a directed line segment from to . However, a moment’s thought tells us that all these directed line segments would overlap in a mess, and give us no sensible picture at all. One way around this is to draw a very short directed line segment from to , so that these small directed line segments do not overlap:

This was produced with the following MATLAB code:

>> [x,y]=meshgrid(-1:1/10:1,-1:1/10:1);
>> u=2*x+y;
>> v=x+y;
>> quiver(x,y,u,v,1)

We see a rather blurred line running from top left to bottom right of the picture, and the vectors near that line seem to curve over.

Generally, Mathematica® produces better graphics than MATLAB. Below is a graphic produced with the following Mathematica® code:

Needs["VectorFieldPlots`"];

VectorFieldPlot[{2*x + y, x + y}, {x, -1, 1}, {y, -1, 1}, PlotPoints -> 20, ScaleFactor -> 0.2, Axes->True]

Here we can see a little more clearly the curving of the vectors near a line running from top left to bottom right.

Finding invariant lines

How could we find that line that seems to run from top left to bottom right of the picture?

The line seems to go through the origin, so if we could find any non-zero vector on that line, we would know what the line is, because all other vectors on the line would just be multiples of that vector.

The characteristic property that looks like it will help us find the line is that if is a vector on the line then is also on the line – just look at how the vectors seem to point on the line. So, we are looking for a non-zero vector   – that is, either – such that for some real number . In other words, for some and for some with either (or both).

This gives us the two equations:

Substituting gives us , and substituting   gives us .

Since one of is non-zero this gives us .

This is a quadratic for with roots .

Substituting in the equation gives us the equation for the line we see running from the top left to the bottom right of the picture.

Of course, the other root, , gives us another invariant line with equation . This line is a little harder to see from the vector field picture. Both lines are shown below:

This graphic was produced using the following Mathematica® code:

Needs["VectorFieldPlots`"];

A1 = VectorFieldPlot[{2*x + y, x + y}, {x, -1, 1}, {y, -1, 1}, PlotPoints -> 20, ScaleFactor -> 0.2, Axes -> True];
A2 = Plot[-x*(1 + Sqrt[5])/2, {x, -1, 1}];
A3 = Plot[x*(-1 + Sqrt[5])/2, {x, -1.5, 1.5}];
Show[A1, A2, A3]

Connection with a system of linear differential equations

The curl of the directed line segments away from the line indicates a flow on the plane, such that if we placed a leaf in the flow it would follow the directed line segments as time progressed.

The introduction of time as a parameter means that we are thinking of the coordinates changing with time as the leaf follows the flow. The leaf’s path will be such that the instantaneous rate of change of the leaf’s coordinates will be given by the directed line segment at that point. This gives us a linked pair of differential equations for the coordinates, as functions of time, along a flow line:

Some examples to try

1.    (a) Plot the vector field for the linear function .

(b) Find any invariant lines, and write down the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow.

(c) Try to describe the flow geometrically, in words.

2. Repeat this for the linear functions for several choices of integers chosen randomly in the range .

In this 3 week block we are going to address a major issue in using differential equations to model physical or biological phenomena: the nature of the solution curves to a system of coupled non-linear differential equations.

These types of systems of differential equations arise in most serious modeling of physical and biological phenomena. There are very few known exact methods of solution for these systems of differential equations. As a result, people working with these types of systems of differential equations have to resort to numerical methods for obtaining approximations to solution curves.

You have already used one such method in the first week: Euler’s method. We will see, in more detail in this 3-week block, that Euler’s method does not produce an accurate enough solution for long enough values of the variables, to be a serious method for obtaining approximate numerical solutions.

More powerful methods – with smaller error -  are needed and we will focus on Runge-Kutta methods.

Edward Lorenz’s equations and the Lorenz attractor

Edward Lorenz (born in New England – West Hartford, Connecticut in 1917, and died in April 2008 in Cambridge, Massachusetts, aged 90) set up a simplified model of convection rolls arising in the equations of the atmosphere, in 1963.

Edward Lorenz

His simplified model involves 3 linked differential equations of 3 variables. Specifically, the equations are:

In these equations, all parameters are positive.

The parameters and   are related to physical properties of a fluid in motion: kinematic viscosity and heat flow (the Prandtl number and Rayleigh number, respectively). In these equations it is common to take .

Different values of give qualitatively different solution curves for this system of equations. For example, for the solution is chaotic, meaning that nearby points diverge approximately exponentially rapidly (so prediction with time is difficult to impossible). The solution curves for these values form an object in 3-space that is known as the Lorentz attractor:

The Lorenz attractor

For other values of the system has solutions that are periodic, but knotted in space.  A classic example of a knotted solution occurs when

Euler approximation to solution curves

Excel

The method for setting up an Euler approximation to a system of first-order differential equations is scarcely more difficult than for a single first-order eqation. We just have to remember that in the Euler approximation formula the variables are linked.

Just as you did for a single differential equation, we can write Excel formulas to carry out Euler’s method for the 3 linked Lorenz equations. Below is such a spreadsheet, with the formulas shown (to enlarge the image, click on it):

Excel does not have a built-in feature that allows XYZ scatter plots, so we cannot easily plot the points in three-dimensional space, as varies.

However, there are some things we can do:

• We can plot each of as functions of :

• We can plot each of
  • y(t) versus x(t):

  

  • z(t) versus x(t):

• z(t) versus y(t):

MATLAB

MATLAB comes equipped with a 3D plot capability, so the first thing for us to do if we are to understand how a 3D plot arises from an Euler approximation to the Lorenz system of differential equations, is to implement Euler’s method for a system of differential equations in MATLAB .

To do this, we can either do it “inline” following the MATLAB prompt, or we can define an M-file to run. We will do the latter, as on pp. 94-95 of the text Differential Equations With MATLAB, modified for a system of differential equations rather than a single differential equation.

A very well laid out explanation of Euler’s method in MATLAB for a system of differential equations appears on the Website Simple ODE Methods. The code below is based on the code at that Web site.

Before we write code to numerically approximate solutions to a system of differential equations, we will first write the code for a single differential equation. Because MATLAB handles vectors so easily, we will then simply modify the code for a single differential equation to a system of equations by replacing a single solution function by a vector of solution functions.

We wiil first write a MATLAB M file called euler.m (saved as “euler”),  the code for which appears below:

% Euler’s method for a single differential equation.

% The differential equation is dy/dt = f(t,y).

function [ t, y ] = euler ( f, t_range, y_initial, nstep )

% We set a range of time values, from t(1) to t(2).

t(1) = t_range(1);

% We define dt by dividing the time range into an equal number of specified steps.

dt = ( t_range(2) – t_range(1) ) / nstep;

% We set the initial value of y at the beginning time t(1).

y(1) = y_initial;

% We use Euler’s method to update the value of y at new time steps.

%  “feval” is used instead of “eval” because we are passing the name of f to the program.

for i = 1 : nstep
t(i+1) = t(i) + dt;
y(i+1) = y(i) + dt * feval ( f, t(i), y(i) );
end

% The following command plots y as a function of t.

plot(t,y)

The M File euler.m returns two data vectors (= lists):

1. t, the value of the independent variable at nstep+1 equally spaced points;
2. y, the value of the dependent variable at each t.

and plots y as a function of t over the specified time interval.

Let’s implement this for the first order differential equation for and .

For this example, .

We set up a simple M file, let’s call it test_example.m (saved as “test_example) that describes this function:

function yprime = test_example ( t, y )
yprime = y*(2./t-1);

Now to run this in the command line we set up the initial condition, the time range, and the number of steps to take, and call “test_example” as an argument to euler.m:

>> y_init = 0.1;
>> [ t, y ] = euler ( ‘test_example’, [ 0.1, 9.0 ], y_init, 200 );

Note the following things:

• we specify the initial value of y at the beginning time;
• the name of the function test_example appears in single quotes;
• the time interval is specified;
• the number of steps is specified.

After running this program we get the following graphic output:

We now want to modify euler.m so that it can calculate a numerical approximation to the Lorenz equations. We will do this in a general way by writing the code for a system of first-order equations, in an M file called euler_system.m (saved as “euler_system”):

% Euler’s method for a system of differential equations.

% The differential equations are dy/dt = f(t,y) where y is a vector of unknown functions.

function [ t, y ] = euler_system( f, t_range, y_initial, nstep )

% We set a range of time values, from t(1) to t(2).

t(1) = t_range(1);

% We define dt by dividing the time range into an equal number of specified steps.

dt = ( t_range(2) – t_range(1) ) / nstep;

% We set the initial value of the vector y at the beginning time t(1).

y(:,1) = y_initial;

% We use Euler’s method to update the value of y at new time steps.

%  “feval” is used instead of “eval” because we are passing the name of f to the program.

for i = 1 : nstep
t(i+1) = t(i) + dt;
y(:,i+1) = y(:,i) + dt * feval ( f, t(i), y(:,i) );
end

% The following command plots the components of y as functions of t.

plot(t,y)

% We also want a 3D plot of the vector components as they vary with t.

plot3(y(1,:),y(2,:),y(3,:))

There is no need to specify how many component differential equations there are -  MATLAB figures that from the form of the variables in the prescription for the function f, which is in a separate M file.

The colon  in y(:,1) = y_initial is used because y is a column vector (that’s what the “1″ tells us) with as many components (= rows) as there are unknown functions.

Now we write an M file, lorenz_system.m, which returns the derivative function as a column vector:

function yprime = lorenz_system ( t, y )
yprime = [ 10.0* (y(2)-y(1)); y(1)*(28.0-y(3))-y(2);y(1)*y(2)-8*y(3)/3 ];

To run this in the command line we set up the vector of initial condition, the time range, and the number of steps to take, and call “lorenz_system” as an argument to euler_system.m:

>> y_init = [ rand(); rand(); rand() ];

>> [ t, y ] = euler_system ( ‘lorenz_system’, [ 0.0, 20.0 ], y_init, 1000 );

After running this program we get the following graphic output:

MATLAB’s Numerical ODE solvers

MatLab has a number of built-in packages for obtaining approximate numerical solutions to differential equations and systems of differential equations. The one we use to get a plot of the Lorenz attractor is ode45.

To get background information about ode45 (and other ode solvers):

1. click on the MATLAB help link at the top of the command window;
2. enter “ode45″ in the search window and press “GO”;
3. in the Demo Results window click on the first line containing the phrase “ode45″;
4. a new window opens with the background information.

To use this for the Lorenz system we set up two M-files (ref. this Web site).

The first one, g.m (save it as “g”), is as follows:

function xdot = g(t,x)

xdot = zeros(3,1);

sig = 10.0;

rho = 28.0;

bet = 8.0/3.0;

xdot(1) = sig*(x(2)-x(1));

xdot(2) = rho*x(1)-x(2)-x(1)*x(3);

xdot(3) = x(1)*x(2)-bet*x(3);

The second, lorenz_demo.m (save it as “lorenz_demo”), is as follows:

function lorenz_demo(time)

% Usage: lorenz_demo(time)

% time=end point of time interval

% This function integrates the lorenz attractor

% from t=0 to t=time

[t,x] = ode45(‘g’,[0 time],[1;2;3]);

disp(‘press any key to continue …’)

pause

plot3(x(:,1),x(:,2),x(:,3))

print -deps lorenz.eps

Save these as M-files and then enter “lorenz_demo(200)” at the MATLAB prompt.

MATLAB’s built-in demonstration of the Lorenz attractor

MATLAB has a built-in demonstration of the dynamics of the Lorenz attractor in time. Just type “lorenz” after the MATLAB prompt.

To see the code, do the following:

1. click on the MATLAB help link at the top of the command window;
2. enter “lorenz” in the search window and press “GO”;
3. in the Demo Results window click on “Lorenz Attractor Animation”;
4. when the new window opens click on “Open lorenz.m in the Editor”.

MATLAB help

In addition to the book Differential Equations with MATLAB, you might find the following pages helpful:

• A Practical Introduction to MATLAB
• MATLAB help

]]> http://mth21202f08.wordpress.com/2008/09/22/%e2%80%a2-second-3-week-block-example-1/feed/ 0 Editor edward_lorenz lorenz_attractor lorenz_euler1 lorenz_euler2 lorenz_euler3 lorenz_euler4 lorenz_euler5 euler_method4 lorenz_euler6 lorenz_euler7 lorenz_euler8 http://mth21202f08.wordpress.com/2008/09/01/possible-differential-equations-for-the-first-3-week-block-task/ http://mth21202f08.wordpress.com/2008/09/01/possible-differential-equations-for-the-first-3-week-block-task/#comments Mon, 01 Sep 2008 13:56:30 +0000 Author http://mth21202f08.wordpress.com/?p=434 ]]> Tenenbaum & Pollard pp. 55-56, pp. 44-45.

We have a small hot body in large cooler environment. The small hot body loses heat to the environment and so drops in temperature. The environment is large enough that the small amount of heat gained from the small hot body does not significantly raise the temperature of the environment. However, as time progresses the environment is warming as result of the impact of the rising sun.

To get a mathematical form for a differential equation we need to know something about the rate of change of the small hot body. We get this information from Newton’s law of cooling, which states that the rate of change of temperature of a hot body with respect to time is proportional to the difference between the temperature of the body and the temperature of the environment. If the temperature of the environment were constant we could write Newton’s law of cooling as:

where is the temperature of the hot body at time , is the constant temperature of the environment, and is a positive constant.

However, we are assuming that the environment is warming up as time progresses, so let’s assume this warming is progressing at a constant rate, at least for a reasonable time period. In other words, we’re assuming that is changing as a linear function of .

So now our differential equation for the temperature of the hot body, as a function of time, is

…………………………(*)

The problem is this: given the differential equation (*), how does the temperature of the hot body change with time?

As it stands we cannot answer this question by giving a specific numerical function because we do not know the initial temperature of the hot body, or the constants . The problem of determining the behavior of the temperature of the hot body from the differential equation has, therefore, 4 free parameters. Let’s narrow this down to a problem with only 1 free parameter – the initial temperature of the hot body, by specifying the other parameters.

Let’s suppose that at time (which could be any clock time – e.g. 6 a.m.) the temperature of the environment is and that the temperature rises steadily so that 360 minutes (= 6 hours) later the temperature of the environment is . The rate of change of temperature (measured in degrees per minute) is so in the differential equation (*), .

We do not know the value for k, and can only reasonably obtain this from experiment. however, a reaonably typical value for a hot body such as a hot cup of coffee is which gives .

With these parameter values the differential equation (*) becomes;

…………………………(**)

At first glance it looks like there are no free parameters in (**), but there is one, lying hidden. It is the initial temperature of the hot body. The differential equation (**) can – and should – be thought of as a prescription for movement in a dynamical system – the transfer of heat from the hot body to the environment. As such, (**) is a prescription for how the heat flow goes with time: in particular, how the temperature changes with time. What (**) does not tell us is what the temperature of the hot body was at time . Without that knowledge we can only know the way in which the temperature changes -which is what (**) tells us – but not the actual value of the temperature at any given time.

So we know have a differential equation (**) with one free parameter – the value . This differential equation is a first-order equation because it involves only the first derivative of the variable .

The basic problem of differential equations, illustrated by the differential equation (**), is: how do we get some understanding of the behavior of the solutions of the differential equation from the differential equation itself? To put it more picturesquely, if less precisely: the differential equation is a local prescription for change. It tells us how the dependent variable changes at a particular value of the independent variable. What we seek is a global, overall, idea of the behavior of the solution curves.

We now turn to some very simple techniques for obtaining that global picture of the solution curves to (**).

Direction fields

(Also known as slope fields)

Tenenbaum & Pollard pp. 38-41.

Hunt, Lipsman, Osborn & Rosenberg pp. 65-71

At any given time, t, the differential equation tells us how the temperature is undergoing a change at that time. The differential equation tells us how the dynamics of heat exchange is changing, and points us from the current temperature to temperatures at close future times.  You all know by now that in physics, mathematics and engineering we indicate pointing by vectors. The idea of a direction field is to place, at each time t, a small vector indicating the magnitude and direction in the change of temperature, as governed by the differential equation. The great beauty of a direction field is that if we draw the direction vectors sufficiently well then we can literally see how a solution curve behaves with time. We get an overall picture of the solution of a differential equation. This is what we mean by a global solution. The differential equation tells us how things change at any given time. The direction field, drawn well, gives us a picture of the overall change over a range of time. This is a process of integration, and it is your eye that does the integration.

If we assume the hot body starts off at a higher temperature than the environment at time then (**) tells us that the rate of change of the temperature of the hot body is negative.

Direction field for the differential equation (**)

The direction field above was constructed using MatLab. You can use MatLab on the computers in the Mathematics Department Computer Lab, room 218, Liberal arts Building. For working at home, if you don’t have MatLab you can use the open-source software Octave, which is designed to be similar to MatLab (and, being open source software it is freely available).

Euler’s method

Euler’s method consists in taking an approximation to the derivative, with small steps in the independent variable. In the example for this block the independent variable is time, t. So we take a small time step and we approximate the derivative at time t by . This means so that .

We know the form of from the differential equation (**) so we use the above approximation to estimate from knowledge of . To start this process we need an initial value for .

This simple approximation to a numerical solution to the differential equation (**) can be easily implemented in a spreadsheet, and the updating feature of a spreadsheet can easily allow us to see the effect of changing the initial condition :

Numerical solutions to (**) with different initial conditions, obtained using Euler's method

However, spreadsheets do not have good numerical accuracy, so we will use MatLab instead. This also gives you practice getting used to the MatLab environment.

Euler’s method in MatLab is described on pp. 91-95 of Hunt, Lipsman, Osborn & Rosenberg.

The basic problem with Euler’s method is that it has low accuracy over extended time periods. Basically we need to take very small to get good accuracy, but that gives us a trade-off in the overall time for which we get a numerical approximation to the solution.

Analytical techniques

(1) Substitution

We define a new function by and try to choose the constant so that – the reason being that this would give us a simpler differential equation for than we had previously for .

With the substitution we have so if we choose then we have .

The differential equation of is simpler than that for , and involves only one parameter, .

You might recall that so you might suspect that or perhaps, allowing for the fact that we have another parameter to account for: the temperature of the hot body at time 0. This guess is, in fact, correct, and we will see why by using separation of variables and integration.

Substitution is dealt with in Tennenbaum & Pollard pp. 101-103.

(2) Separation of variables

Here’s the intuitive idea:

it is very tempting to write the differential equation as , whatever that might mean

Following the formalism where it leads has a long and venerable history in mathematics: that is, for example, how complex numbers were discovered. However, blind formalism, where we accept formal manipulation as correct no matter what, can lead to absurd thinking. Here’s an example: let’s just imagine we can blindly make sense of the infinite sum in which each term in the sum is twice the one before. Let’s call this sum so . Now let’s formally suppose we can multiply all the terms in by 2: . Now operate formally and subtract to get , which is obviously not true!

So blind formal manipulation is not always a good idea. But inspired formal manipulation, as we have done with the differential equation, might be productive.

Her’s a little more formal manipulation of the differential equation that is very suggestive:

.

Thinking of an integral as an anti-derivative (why not?) this seems to give us where is constant, so as we had previously guessed.

This technique is called separation of variables because we manipulate the differential equation into a form where all the occurrences of the variable are on one side of the equation, and all the occurrences of the variable are on the other side of the equation.

Separation of variables is described in Tenebaum & Pollard pp. 47-55.

Computer-aided solutions

Almost any technique you know of, or anyone knows of, is encoded in standard computer packages that allow exact solutions of differential equations. such packages include Maple, Mathematica, and  MatLab, as well as their open source equivalents.

Here’s how we go about finding solutions to the differential equation (**) using MatLab:

(Hunt, Lipsman, Osborn & Rosenberg pp. 65-71)

>> dsolve(‘DT=-0.05*(T-(t/12+45))’,'t’)

The result is:

ans =130/3+1/12*t+exp(-1/20*t)*C1

The – unknown – constant comes about because we did not yet specify . Once we do that we get a solution with no free parameters:

>> dsolve(‘DT=-0.05*(T-(t/12+45))’, ‘T(0)=200′, ‘t’)

ans =

130/3+1/12*t+470/3*exp(-1/20*t)

So the computer does the hard work of finding an exact anyltic solution for us, when there is such an anlytic solution.

Of course, what we weould like ot know is how the oslution behaves as time increases. One way to do this is to look at a graph of versus .

In MatLab we can do this by naming the solution:

>> SOL = dsolve(‘DT=-0.05*(T-(t/12+45))’, ‘T(0)=200′, ‘t’)

and then plotting it:

>> ezplot(SOL,[0,360])

Temperature T versus time t,

Finally: does this make sense? Should the temperature-time curve dip like this then rise? Does this solution make sense in terms of the model we had?

]]> http://mth21202f08.wordpress.com/2008/08/10/example-first-3-week-block/feed/ 0 Editor Direction field for the differential equation (**) Numerical solutions to (**) with different initial conditions, obtained using Euler's method. Temperature T versus time t, $latex 0 \leq t \leq 360$