t.p.maths Exact Differential Equatns.

TERM PAPER

TOPIC: APPLICATION OF EXACT DIFFERENTIAL EQUATION IN ENGINEERING. USING EQUATIONS OF YOUR OWN CHOICE.

SUBMITTED TO THE FACULTY OF TECHNOLOGY AND SCIENCES FOR THE PARTIAL FULFILMENT OF COURSE MTH 102

DATE OF SUBMISSION: 5/07/2010

SUBMITTED TO: SUBMITTED BY:

MR. SHOBIK SONI MS. SAHIBA

DEPTT. OF MATHS ROLL No.: RE1901B40

REG. No.: 10902639

SEC.: E-1901

ACKNOWLEDGEMENT

I convey my sincere thanks to Mr. SHOBIK SONI our MATHS TEACHER who have given his immense help and guidance to do my term paper in simple and lucid style. This term paper is submitted as a part of the curriculum included in the B.Tech – M.Tech (CSE) examination during the academic 2nd semester.The topic of the term paper is APPLICATION OF EXACT DIFFERENTIAL EQUATION IN ENGINEERING. USING EQUATIONS OF YOUR OWN CHOICE. The term paper contains the information about the topic in the differential equations i.e. exact differential equations. It includes the theorem of exact differential equation, the process of solving these types of differential equations being exact or non exact differential equations , its applications in the field of engineering which are reliably discussed in the term paper with their examples.

SUBMITTED BY:

SAHIBA

REG. No.: 10902639

CONTENTS

1. INTRODUCTION TO DIFFERENTIAL EQUATIONS2. INTRODUCTION TO EXACT DIFFERENTIAL EQUATIONS3. APPLICATIONS OF EXACT DIFFERENTIAL EQUATIONS 4. SOME OTHER APPLICATIONS OF EXACT DIFFERENTIAL EQUATIONS IN THE FIELD OF ENGG.

DIFFERENTIAL EQUATIONS

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. Differential equations play a prominent role in engineering, physics, economics and other disciplines.

Differential equations arise in many areas of science and technology: whenever a deterministic relationship involving some continuously varying quantities (modelled by functions) and their rates of change in space and/or time (expressed as derivatives) is known or postulated. This is illustrated in classical mechanics, where the motion of a body is described by its position and velocity as the time varies. Newton's Laws allow one to relate the position, velocity, acceleration and various forces acting on the body and state this relation as a differential equation for the unknown position of the body as a function of time. In some cases, this differential equation (called an equation of motion) may be solved explicitly.

An example of modelling a real world problem using differential equations is determination of the velocity of a ball falling through the air, considering only gravity and air resistance. The ball's acceleration towards the ground is the acceleration due to gravity minus the deceleration due to air resistance. Gravity is constant but air resistance may be modelled as proportional to the ball's velocity. This means the ball's acceleration, which is the derivative of its velocity, depends on the velocity. Finding the velocity as a function of time requires solving a differential equation.

EXACT DIFFERENTIAL EQUATIONS

A differential equation of the form M(x, y) dx + N(x, y) dy = 0 is said to be exact if its left hand member is the exact differential of some function u (x,y) i.e. du = Mdx + Ndy =0. Its solution, therefore , is u(x,y)=c.The necessary and sufficient condition for the differential equation Mdx + Ndy = 0 to be exact is

My(x, y) = Nx(x,y)Condition is necessary:

The equation Mdx + Ndy 0 will be exact, if

Mdx + Ndy = duWhere u is some function of x and y.

But du = dx + dy

Therefore, equating coefficients of dx and dy in (1) and (2) , we get M = and N =

= and

But

Therefore which is the necessary conditions for exactness.

CONDITION IS SUFFICIENT: i.e. if , then Mdx + Ndy = 0 is exact.

Consider the equation

        f(x,y)  =  C

Taking the gradient we get

        fx(x,y)i + fy(x,y)j  =  0

We can write this equation in differential form as 

        fx(x,y)dx+ fy(x,y)dy  =  0

Now divide by dx to get

        fx(x,y)+ fy(x,y) dy/dx  =  0

Which is a first order differential equation. The goal of this section is to go backward.  That is if a differential equation if of the form above, we seek the original function f(x,y) (called a potential function).  A differential equation with a potential function is called exact. 

Example

Solve the differential equation 

        y + (2xy - e-2y)y'  =  0

Solution

We have 

        M(x,y)  =  y    and    N(x,y)  =  2xy - e-2y

Now calculate 

        My  =  1    and    Nx  =  2y

Since they are not equal, finding a potential function f is hopeless.  However there is a glimmer of hope if we remember how we solved first order linear differential equations.  We multiplied both sides by an integrating factor .  We do that here to get

        M + Ny'  =  0

For this to be exact we must have 

        ( M)y  =   ( N)x  

Using the product rule gives

        yM + My  =   xN + Nx 

We now have a new differential equation that is unfortunately more difficult to solve than the original differential equation.  We simplify the equation by assuming that either is a function of only x or only y.  If it is a function of only x, then y  =  0 and 

        My  =   xN + Nx         

Solving for x, we get

                    My - Nx

        x  =                                               N

If this is a function of y only, then we will be able to find an integrating factor that involves y only.

If it is a function of only y, then x  =  0 and 

        yM + My  =  Nx         

Solving for y, we get

                    Nx - My

        y  =                                               M

If this is a function of y only, then we will be able to find an integrating factor that involves y only.

For our example

                    Nx - My                  2y - 1        y  =                        =                         =  (2 - 1/y)                            M                          y 

Separating gives

        d/  =  (2 - 1/y) dy

Integrating gives

        ln   =  2y  -  ln y

          =  e2y - ln y  =  y -1e2y

Multiplying both sides of the original differential equation by gives

        y(y -1e2y) + (y -1e2y)(2xy - e-2y)y'  =  0

        e2y + (2xe2y - 1/y)y'  =  0

Now we see that 

        My  =  2e2y  =    Nx  

Which tells us that the differential equation is exact.  We therefore have

        fx (x,y)  =  e2y 

Integrating with respect to x gives

        f(x,y)  =  xe2y + C(y)

Now taking the partial derivative with respect to y gives

        fy(x,y)  =  2xe2y + C'(y)  =  2xe2y - 1/y

So that 

        C'(y)  =  1/y

Integrating gives

        C(y)  =  ln y

The final solution is 

        xe2y + ln y  =  0

EXAMPLE:

Q. Solve y / sinx + x cosy+ x =0

Solution: given equation can be written as

(y cos x + siny + y)dx + (sinx +x cos y+ x)dy = 0

Here M = y cosx + siny +y and N = sinx +x cosy + x.

Therefore ,

Thus the equation is exact and its solution is

i.e.

or y sinx + (sin y + y)x = c.

APPLICATIONS OF EXACT DIFFERENTIAL EQUATIONS

We present examples where differential equations are widely applied to model natural phenomena, engineering systems and many other situations.

1. Radioactive Decay

Many radioactive materials disintegrate at a rate proportional to the amount present. For example, if X is the radioactive material and Q(t) is the amount present at time t, then the rate of change of Q(t) with respect to time t is given by

where r is a positive constant (r>0). Let us call the initial quantity of the material X, then we have

Clearly, in order to determine Q(t) we need to find the constant r. This can be done using what is called the half-life T of the

material X. The half-life is the time span needed to disintegrate half of the material. So, we have . An easy

calculation gives . Therefore, if we know T, we can get r and vice-versa. Many chemistry text-books contain the

half-life of some important radioactive materials. For example, the half-life of Carbon-14 is . Therefore, the constant r associated with Carbon-14 is . As a side note, Carbon-14 is an important tool in the archeological research known as radiocarbon dating.

Example: A radioactive isotope has a half-life of 16 days. You wish to have 30 g at the end of 30 days. How much radioisotope should you start with?

Solution: Since the half-life is given in days we will measure time in days. Let Q(t) be the amount present at time t and the amount we are looking for (the initial amount). We know that

,

where r is a constant. We use the half-life T to determine r. Indeed, we have

Hence, since

,

we get

2. Simple harmonic motion :

Let us reexamine the problem of a mass on a spring .Consider a mass which slides over a horizontal frictionless surface. Suppose that the mass is attached to a light horizontal spring whose other end is anchored to an immovable object. Let be the extension of the spring: i.e., the difference between the spring's actual length and its unstretched length. Obviously, can also be used as a coordinate to determine the horizontal displacement of the mass.

The equilibrium state of the system corresponds to the situation where the mass is at rest, and the spring is unextended (i.e., ). In this state, zero net force acts on the mass, so there is no reason for it to start to move. If the system is perturbed

from this equilibrium state (i.e., if the mass is moved, so that the spring becomes extended) then the mass experiences a restoring force given by Hooke's law:

Here, is the force constant of the spring. The negative sign indicates that is indeed a restoring force. Note that the

magnitude of the restoring force is directly proportional to the displacement of the system from equilibrium (i.e., ). Of course, Hooke's law only holds for small spring extensions. Hence, the displacement from equilibrium cannot be made too large. The motion of this system is representative of the motion of a wide range of systems when they are slightly disturbed from a stable equilibrium state.

Newton's second law gives following equation of motion for the system:

This differential equation is known as the simple harmonic equation, and its solution has been known for centuries. In fact, the solution is

where , , and are constants. We can demonstrate that Eq. is indeed a solution of Eq. by direct substitution. Substituting

Eq. into Eq., and recalling from calculus that and , we obtain

It follows that Eq. is the correct solution provided .

Figure 1 shows a graph of versus obtained from Eq.1. The type of motion shown here is called simple harmonic motion. It

can be seen that the displacement oscillates between and . Here, is termed the amplitude of the oscillation. Moreover, the motion is periodic in time (i.e., it repeats exactly after a certain time period has elapsed). In fact, the period is

This result is easily obtained from Eq.1by noting that is a periodic function of with period . The frequency of the motion (i.e., the number of oscillations completed per second) is

It can be seen that is the motion's angular frequency (i.e., the frequency converted into radians per second). Finally, the

phase angle determines the times at which the oscillation attains its maximum amplitude, : in fact,

Here, is an arbitrary integer.

Table 1: Simple harmonic motion.

0 0

0 0

0 0

Table 1 lists the displacement, velocity, and acceleration of the mass at various phases of the simple harmonic cycle. The information contained in this table can easily be derived from the simple harmonic equation, Eq.1. Note that all of the non-zero values shown in this table represent either the maximum or the minimum value taken by the quantity in question during the oscillation cycle.

We have seen that when a mass on a spring is disturbed from equilibrium it executes simple harmonic motion about its

equilibrium state. In physical terms, if the initial displacement is positive ( ) then the restoring force overcompensates,

and sends the system past the equilibrium state ( ) to negative displacement states ( ). The restoring force again overcompensates, and sends the system back through to positive displacement states. The motion then repeats itself ad infinitum. The frequency of the oscillation is determined by the spring stiffness, , and the system inertia, , via Eq.. In contrast, the amplitude and phase angle of the oscillation are determined by the initial conditions. Suppose that the

instantaneous displacement and velocity of the mass at are and , respectively. It follows from Eq. that

Here, use has been made of the well-known identities and . Hence, we obtain

and

since and .

The kinetic energy of the system is written

Recall, from Sect. 5.6, that the potential energy takes the form

Hence, the total energy can be written

since and . Note that the total energy is a constant of the motion, as expected for an isolated system. Moreover, the energy is proportional to the amplitude squared of the motion. It is clear, from the above expressions, that simple harmonic motion is characterized by a constant backward and forward flow of energy between kinetic and potential components. The kinetic energy attains its maximum value, and the potential energy attains it minimum value, when the displacement is zero (i.e., when ). Likewise, the potential energy attains its maximum value, and the kinetic energy

attains its minimum value, when the displacement is maximal (i.e., when ). Note that the minimum value of is zero, since the system is instantaneously at rest when the displacement is maximal.

3. Orthogonal Trajectories

We have seen before that the solutions of a differential equation may be given by an implicit equation with a parameter something like

This is an equation describing a family of curves. Whenever we fix the parameter C we get one curve and vice-versa. For example, consider the families of curves

where m and C are parameters. Clearly, we may change the names of the variables and still have the same geometric curves. For example, the above families define the same geometric object as

Note that the first family describes all the lines passing by the origin (0,0) while the second the family describes all the circles centered at the origin (including the limit case when the radius 0 which reduces to the single point (0,0)) (see the pictures below).

and

In this page, we will only use the variables x and y. Any family of curves will be written as

One may ask whether any family of curves may be generated from a differential equation? In general, the answer is no. Let us see how to proceed if the answer were to be yes. First differentiate with respect to x, and get a new equation involving in

general x, y, , and C. Using the original equation, we may able to eliminate the parameter C from the new equation.

Example. Find the differential equation satisfied by the family

Answer. We differentiate with respect to x, to get

Since we have

then we get

You may want to do some algebra to make the new equation easy to read. The next step is to rewrite this equation in the explicit form

this is the desired differential equation.

Example. Find the differential equation (in the explicit form) satisfied by the family

Answer. We have already found the differential equation in the implicit form

Algebraic manipulations give

Let us reconsider the example of the two families

If we draw the two families together on the same graph we get

As we see here something amazing happened. Indeed, it is clear that whenever one line intersects one circle, the tangent line to the circle (at the point of intersection) and the line are perpendicular or orthogonal. We say the two curves are orthogonal at the point of intersection.

Definition. Consider two families of curves and . We say that and are orthogonal whenever any curve from

intersects any curve from , the two curves are orthogonal at the point of intersection.

For example, we have seen that the families y = m x and are orthogonal. One may then ask the following natural question:

Given a family of curves , is it possible to find a family of curves which is orthogonal to ?

The answer to this question has many implications in many areas such as physics, fluid-dynamics, etc... In general this question is very difficult. But in some cases, we may be able to carry on the calculations and find the orthogonal family. Let us show how.

Consider the family of curves . We assume that an associated differential equation may be found, say

We know that for any curve from the family passing by the point (x,y), the slope of the tangent at this point is f(x,y). Hence the

slope of the line perpendicular (or orthogonal) to this tangent is which happens to be the slope of the tangent line to the orthogonal curve passing by the point (x,y). In other words, the family of orthogonal curves are solutions to the differential equation

From this we see what we have to do. Indeed consider a family of curves . In order to find the orthogonal family, we use the following practical steps

Step 1. Find the associated differential equation. Step 2. Rewrite this differential equation in the explicit form

Step 3. Write down the differential equation associated to the orthogonal family

Step 4. Solve the new equation. The solutions are exactly the family of orthogonal curves. Step 5. You may be asked to give a geometric view of the two families. Also you may be asked to find a specific curve from the orthogonal family (something like an IVP).

Example. Find the orthogonal family to the family of circles

Answer. First, we look for the differential equation satisfied by the circles. We differentiate with respect to the variable x to get

We rewrite this equation in the explicit form

Next we write down the equation for the orthogonal family

This is a linear as well as a separable equation. If we use the technique of linear equations, we get the integrating factor

which gives

We recognize the family of lines and we confirm our earlier observation (that the two families are indeed orthogonal).

This example is somehow easy and was given here to illustrate the technique.

Example. Find the orthogonal family to the family of circles

Answer. We differentiate with respect to x, to get

Since we have

then we get

You may want to do some algebra to make the new equation easy to read. The next step is to rewrite this equation in the explicit form

this is the desired differential equation.

Example. Find the differential equation (in the explicit form) satisfied by the family

Answer. We have already found the differential equation in the implicit form

Algebraic manipulations give

Let us reconsider the example of the two families

If we draw the two families together on the same graph we get

As we see here something amazing happened. Indeed, it is clear that whenever one line intersects one circle, the tangent line to the circle (at the point of intersection) and the line are perpendicular or orthogonal. We say the two curves are orthogonal at the point of intersection.

Definition. Consider two families of curves and . We say that and are orthogonal whenever any curve from

intersects any curve from , the two curves are orthogonal at the point of intersection.

For example, we have seen that the families y = m x and are orthogonal. One may then ask the following natural question:

Given a family of curves , is it possible to find a family of curves which is orthogonal to ?

The answer to this question has many implications in many areas such as physics, fluid-dynamics, etc... In general this question is very difficult. But in some cases, we may be able to carry on the calculations and find the orthogonal family. Let us show how.

Consider the family of curves . We assume that an associated differential equation may be found, say

We know that for any curve from the family passing by the point (x,y), the slope of the tangent at this point is f(x,y). Hence the

slope of the line perpendicular (or orthogonal) to this tangent is which happens to be the slope of the tangent line to the orthogonal curve passing by the point (x,y). In other words, the family of orthogonal curves are solutions to the differential equation

From this we see what we have to do. Indeed consider a family of curves . In order to find the orthogonal family, we use the following practical steps

Step 1. Find the associated differential equation. Step 2. Rewrite this differential equation in the explicit form

Step 3. Write down the differential equation associated to the orthogonal family

Step 4. Solve the new equation. The solutions are exactly the family of orthogonal curves. Step 5. You may be asked to give a geometric view of the two families. Also you may be asked to find a specific curve from the orthogonal family (something like an IVP).

Example. Find the orthogonal family to the family of circles

Answer. First, we look for the differential equation satisfied by the circles. We differentiate with respect to the variable x to get

We rewrite this equation in the explicit form

Next we write down the equation for the orthogonal family

This is a linear as well as a separable equation. If we use the technique of linear equations, we get the integrating factor

which gives

We recognize the family of lines and we confirm our earlier observation (that the two families are indeed orthogonal).

This example is somehow easy and was given here to illustrate the technique.

Example. Find the orthogonal family to the family of circles

Answer. We have seen before that the explicit differential equation associated to the family of circles is

Hence the differential equation for the orthogonal family is

We recognize an homogeneous equation. Let us use the technique developed to solve this kind of equations. Consider the new

variable (or equivalently y = x z). Then we have

and

Hence we have

Algebraic manipulations imply

This is a separable equation. The constant solutions are given by

which gives z=0. The non-constant solutions are found once we separate the variables

and then we integrate

Before we perform the integration for the left-hand side, we need to use partial decomposition technique. We have

We will leave the details to you to show that A = 1, B=-2, and C=0. Hence we have

Hence

which is equivalent to

where . Putting all the solutions together we get

Going back to the variable y, we get

which is equivalent to

We recognize a family of circles centered on the y-axis and the line y=0 (the x-axis which was easy to guess, isn't it?)

If we put both families together, we appreciate better the orthogonality of the curves (see the picture below).

4. Population Dynamics:

Here are some natural questions related to population problems:

What will the population of a certain country be in ten years? How are we protecting the resources from extinction?

More can be said about the problem but, in this little review we will not discuss them in detail. In order to illustrate the use of differential equations with regard to this problem we consider the easiest mathematical model offered to govern the population dynamics of a certain species. It is commonly called the exponential model, that is, the rate of change of the population is proportional to the existing population. In other words, if P(t) measures the population, we have

,

where the rate k is constant. It is fairly easy to see that if k > 0, we have growth, and if k <0, we have decay. This is a linear equation which solves into

,

where is the initial population, i.e. . Therefore, we conclude the following:

if k>0, then the population grows and continues to expand to infinity, that is,

if k<0, then the population will shrink and tend to 0. In other words we are facing extinction.

Clearly, the first case, k>0, is not adequate and the model can be dropped. The main argument for this has to do with environmental limitations. The complication is that population growth is eventually limited by some factor, usually one from among many essential resources. When a population is far from its limits of growth it can grow exponentially. However, when

nearing its limits the population size can fluctuate, even chaotically. Another model was proposed to remedy this flaw in the exponential model. It is called the logistic model (also called Verhulst-Pearl model). The differential equation for this model is

,

where M is a limiting size for the population (also called the carrying capacity). Clearly, when P is small compared to M, the equation reduces to the exponential one. In order to solve this equation we recognize a nonlinear equation which is separable. The constant solutions are P=0 and P=M. The non-constant solutions may obtained by separating the variables

,

and integration

The partial fraction techniques gives

,

which gives

Easy algebraic manipulations give

where C is a constant. Solving for P, we get

If we consider the initial condition (assuming that is not equal to both 0 or M), we get

,

which, once substituted into the expression for P(t) and simplified, we find

It is easy to see that

However, this is still not satisfactory because this model does not tell us when a population is facing extinction since it never implies that. Even starting with a small population it will always tend to the carrying capacity M.

Some other Applications to Engineering and Sciences

Historically, it has been the needs of the physical sciences which have driven the development of many parts of mathematics, particularly analysis. The applications are sometimes difficult to classify mathematically, since tools from several areas of mathematics may be applied. We focus on these applications not by discussing the nature of their discipline but rather their interaction with mathematics.

Mechanics of particles and systems studies dynamics of sets of particles or solid bodies, including rotating and vibrating bodies. Uses variational principles (energy-minimization) as well as differential equations.

Mechanics of deformable solids considers questions of elasticity and plasticity, wave propagation, engineering, and topics in specific solids such as soils and crystals.

Fluid mechanics studies air, water, and other fluids in motion: compression, turbulence, diffusion, wave propagation, and so on. Mathematically this includes study of solutions of differential equations, including large-scale numerical methods (e.g the finite-element method).

Optics, electromagnetic theory is the study of the propagation and evolution of electromagnetic waves, including topics of interference and diffraction. Besides the usual branches of analysis, this area includes geometric topics such as the paths of light rays.

Classical thermodynamics, heat transfer is the study of the flow of heat through matter, including phase change and combustion. Historically, the source of Fourier series.

Quantum Theory studies the solutions of the Schrödinger (differential) equation. Also includes a good deal of Lie group theory and quantum group theory, theory of distributions and topics from Functional analysis, Yang-Mills problems, Feynman diagrams, and so on.

Statistical mechanics, structure of matter is the study of large-scale systems of particles, including stochastic systems and moving or evolving systems. Specific types of matter studied include fluids, crystals, metals, and other solids.

Relativity and gravitational theory is differential geometry, analysis, and group theory applied to physics on a grand scale or in extreme situations (e.g. black holes and cosmology).

Astronomy and astrophysics : as celestial mechanics is, mathematically, part of Mechanics of Particles (!), the principal applications in this area appear to be concerning the structure, evolution, and interaction of stars and galaxies.

Geophysics applications typically involve material in Mechanics and Fluid mechanics, as above, but for large-scale problems (this subject deals with a very big solid and a large pool of fluid!)

Systems theory; control study the evolution over time of complex systems such as those in engineering. In particular, one may try to identify the system -- to determine the equations or parameters which govern its development -- or to control the system -- to select the parameters (e.g. via feedback loops) to achieve a desired state. Of particular interest are issues in stability (steady-state configurations) and the effects of random changes and noise (stochastic systems). While popularly the domain of "cybernetics" or "robotics", perhaps, this is in practice a field of application of differential (or difference) equations, functional analysis, numerical analysis, and global analysis (or differential geometry).

Biology and other natural sciences whose connections merit explicit connection in the MSC scheme include Chemistry, Biology, Genetics, and Medicine, In chemistry and biochemistry, it is clear that graph theory, differential geometry, and differential equations play a role. Medical technology uses techniques of information transfer and visualization. Biology (including taxonomy and archaeobiology) use statistical inference and other tools.

Game theory, economics, social and behavioral sciences including Psychology, Sociology, and other social sciences as a group. The more behavioural sciences (including Linguistics!) use a medley of statistical techniques, including experimental design and other rather combinatorial topics. Economics and finance also make use of statistical tools, especially time-series analysis; some topics, such as voting theory, are more combinatorial. This category also includes game theory, which is actually not about games at all but rather about optimization; which combination of strategies leads to an optimal outcome.

Observe that the branches of mathematics most closely allied with the fields of mathematical physics are the parts of analysis, particularly those parts related to differential equations. The other sciences draw on these as well as probability and statistics and, increasingly, numerical methods.

REFRENCES

1. http://farside.ph.utexas.edu/teaching/301/lectures/node138.html 2. en.wikipedia.org/wiki/Exact_differential_equation

3. www.ltcconline.net/greenl/courses/204/.../ exact DiffEQs.htm

4. www.math.fsu.edu/~fusaro/EngMath/Ch1/SEDE.htm

5. ^ a b c d e f g Çengel, Yunus A.; Boles, Michael A. (1998) [1989]. "Thermodynamics Property Relations". Thermodynamics - An Engineering Approach. McGraw-Hill Series in Mechanical Engineering (3rd ed.). Boston, MA.: McGraw-Hill. ISBN 0-07-011927-9.

Perrot, P. (1998). A to Z of Thermodynamics. New York: Oxford University Press. Zill, D. (1993). A First Course in Differential Equations, 5th Ed. Boston: PWS-Kent Publishing Company.

6. B. s. garewal higher engg. mathematics