This includes classification, properties, and biological importance of biomolecules. It will introduce the students to the concept of genetic code and concept of heredity. The key emphasis is placed on understanding the basic principles that govern the biological functions of biomolecules. Table of Contents. A special case is ordinary differential equations ODEs , which deal with functions of a single variable and their derivatives.
In mathematics, a partial differential equation PDE is a differential equation that contains unknown multivariable functions and their partial derivatives. SIZE — PAGES — S Chand Publishing is a leading publishing company in India. Their books have touched three generations of Indian educated community.
They have been publishing quality books and meet the right needs from primary level books to higher academy books in Engineering, Management, Commerce, Basic Sciences, and other subjects. They have been recipients of national awards every year. The tests are directed twice in a year for the most part in the months of June and December. Candidates will have to pick any one subject and apply for the same.
COM does no longer owns this book neither created nor scanned. We simply offer the hyperlink already to be had on the internet. In other words, integral trans- forms change problems into easier ones.
The transformed problem is then solved, and its inverse is obtained to find the solution to the original problem; this general strategy is illustrated in Figure Solutron Easy Solve the easv problem of problem! Step 21 easy problem Integral transform Inverse transform!
Step 11! In Figure The transform and its inverse together form what is called a transform pair. Table TABLE L inverse finite-sine transform 5. For instance, in the case of the Laplace transform, the notation! The current lesson does not attempt to study all of these transform pairs- only the sine and cosine transforms 1. Questions about the relationship between the transforms, when to apply them, advantages and disadvantages of each, will be answered as we go along.
However, before we begin the study of integral transforms, it will be useful to study what is called the spectrum of a function or the spectral resolution of a function. The Spectrum of a Function Integral transforms and the spectrum of a function are closely related; in fact, an integral transformation can be thought of as a resolution of a function into a certain spectrum of components.
How the transform actually resolves the function changes from transform to transform, but the function is being resolved into something nevertheless. With this intuitive explanation of the spectrum of a function, we now get to the nuts and bolts of integral transforms. The first step would be to list a few properties of these transforms that make them work. Sine and Cosine Transforms of Derivatives 1.
We now show how the sine transform can solve an important initial-boundary-value problem. STEP 1 To solve this, we break it into three simple steps. First our strategy is to transform the x-variable via the Fourier sine transform so that we get an ODE in time. We now have a slight modification, since u x,t depends on x and t.
You should use the formulas according to which variable is being transformed and treat the others as constants. In this case, the transform is with respect to x, and, hence, tis just carried along as a constant. The exact values of these well-known functions can be found in most tables for physics and chemistry. It should be noted that these integrals cannot be inte- grated by the usual elementary tricks of calculus.
For different values of time, we have the graph of a complementary-error function with different scale factors on the x-axis. As time increases, the error function gets pulled out; for a graph of the solution at different values of time, see Figure Prove the identities What assumptions do you need to assume about the function f?
Solve the ordinary-differential equation problem Operational Mathematics by R. An excellent text covering many of the integral transforms; good problems and many tables. Tables of Integral Transform by A. One of the most comprehensive tables of integral transform. Integral Transforms in Mathematical Physics by C. Chapman and Hall Science Paperbacks , A small, but concise paperback; easy to read with many examples.
These integrations are routine calculus evaluations. To find Euler's formulas for an and bn, respectively, we multiply each side of equation For the sawtooth wave, the Fourier represen- tation is given by As a matter of fact, we can represent a function inside an interval with many different types of Fourier series by considering different types of extensions outside the interval some converge faster than others.
The reader shouldn't get the idea that every periodic function can be rep- resented by a Fourier series expansion. What we do know is that if a function. Indeed, the differen- tiated series will not even converge for any x the reader can verify this himself or herself. The exact conditions that insure that a functionf x will have a Fourier series representation and that the representation can be differentiated term by term are found in the recommended reading for this lesson.
For our purposes, we are content to know the important result of P. Dirichlet's Deer-ish-lay theorem, which states Dirichlet's theorem sufficient conditions for a function to have a Fourier! For example, in Figure We are now almost ready to introduce the Fourier transform.
Before we do, however, it will be useful to introduce the idea of the frequency spectrum of a periodic function. We now introduce the Fourier transform. The Fourier Transform The major difficulty with Fourier series representation is that nonperiodic func- tions defined on - oo,oo cannot be represented.
It is possible, however, to find an analogous representation for some of these functions. L ] changes to the Fourier integral representation continuous frequency resolution Some examples of functions f x and their frequency spectrums are given in Figure Properties of this trans- form pair will be discussed in the next lesson along with problems using these transforms.
The absolute value of the Fourier transform F E is the frequency spectrum of f x. Not all functions have Fourier transforms [the integral Only functions that go to zero sufficiently fast as lxl - oo have transforms. In applications, we apply the Fourier transform to temperature functions, wave functions, and other physical phenomena that go to zero as lxl - oo.
Also graph the frequency spectrum of f x. Show that if we differentiate the Fourier series expansion HINT First multiply the numerator and denominator by 1 - iE to get rid of the complex number i in the denominator.
L cos mrx! A well-written text with a fairly extensive section on the Fourier series and transform Chapters 5, Most of the important questions dealing with whether a function actually has a Fourier series or integral representation, whether the representation can be dif- ferentiated term by term or under the integral to get the derivative of the function, and so forth, are answered in these chapters.
In particular, it is shown how the Fourier transform changes dif- ferentiation to multiplication, so differential equations change into algebraic equations. Also, the idea of the infinite convolution is introduced.
For example, Table We can see from the examples that the transformed function F E may or may not be a complex-valued function of E. In the first example, the transformed function F E contains the complex number i, so we call it a complex-valued function of the real variable E E ranges from -oo to oo. In other words, the argument Eis real, but the value of the function is complex. The usefulness of the Fourier transform as with most integral transforms comes from the fact that it changes the operation of differentiation into multi- plication; that is, differential equations are changed into algebraic equations.
There are also a host of other properties that make the Fourier transform a useful operational tool; we list a few of the more important ones. J - I e-lxl See Figure The reader can spend a few minutes to verify this property, which is used over and over again. It's given by the formula The importance of the convolution We are now in a position to work an important problem in PDE theory. In other words, we look for the solution to the initial-value problem IVP , sometimes called a Cauchy problem STEP 1 Transforming the problem Since the space variable x ranges from -co to co, we take tile Fourier transform of the PDE and IC with respect to this variable x the variable of integration in the transform is x.
The reader should note here that the function U actually depends on both t and the new transformed variable E. STEP 2 Solving the transformed problem Remember the new variable Eis nothing more than a constant in this differential equation, so the solution to this problem is Using this property, we can write Before stopping, however, let's analyze this result.
Note that the integrand is made up of two terms 1. These resulting temperatures are then added integrated to obtain solution Later, we will see that this general idea is known as superposition. From a practical point of view, solution If this integration cannot be carried out analytically, the solution can be found at any point x,t by numerically integrating the integral.
Only functions that damp to zero sufficiently fast as lxl - oo have transforms. Verify that the Fourier and inverse Fourier transforms are linear transforms. Allyn and Bacon, ; Dover, An excellent book gives the reader an intuitive as well as rigorous viewpoint of Fourier series and transforms. The Laplace transform has an advantage over the Fourier transform because it contains the damping factor e-"that allow us to transform a wider class of functions. Inasmuch as the transform operates on functions defined on [O,oo , it is mostly applied to the time variable t.
After discussing some useful properties of the Laplace transform, we will solve an important problem in PDE theory.
Of all the integral transforms we will study in this book, the Laplace transform In particular, we will attempt to apply the Laplace transform to any variable x, y, z, t, We must then decide how to solve this new problem maybe by another transform, by separation of variables, and so on.
Before actually solving a very interesting problem, we enumerate some useful properties of this transform. In fact, the exact conditions that insure that a function f t has a Laplace transform are given by the following theorem: Sufficient Conditions to Insure the Existence of a Laplace Transform If 1. F s e" ds '1 c-ioo We must often resort to contour integration in the complex plane and the theory of residues. We won't bother ourselves with this topic here but will use the tables in the appendix for finding inverse transforms.
Property 3 Convolution Property Convolution plays the same role here as it did in the Fourier transform, but now the convolution is defined slightly differently.
In other words, in the finite convolution, we integrate from 0 to tinstead of from -oo to oo, as we did in the infinite convolution. For example :e-1 [! Suppose the liquid has an initial temperature of u0 and that the temperature of the air above the liquid is zero some reference temperature. Our goal is to find the temperature of the liquid at various depths of the container at different values of time.
To solve this problem, we transform the time variable t by means of the Laplace transform conceivably, we could also transform x by means of the Laplace transform, since x also ranges from 0 to co. Note that we have dropped the s-notation in U x,s in favor of the simpler notation U x , since the differential equation in To solve To find this inverse transform, we must resort to the tables of inverse Laplace transforms in the appendix; they will give us If we spend a little time analyzing this equation and graphing it by means of a computer with a plotter attachment, we will see that it looks like Figure NOTH 1.
The Laplace transform can also be applied to problems where the PDE is nonhomogeneous in separation of variables, the equation had to be ho- mogeneous , but the Laplace transform will generally work only if the equa- tion has constant coefficients in separation of variables, we could have variable coefficients. The following table lists the types of problems the two methods will handle.
The Laplace transform which transforms t can be interpreted as projecting the xt-plane onto the x-axis, and the original BCs, PDE, and IC are trans- formed into a new differential equation and BCs.
See the following diagram. What is the physical inter- pretation of this problem? Ginn and Co. This text contains an extensive section oil contour integration, which is the tool used for evaluating the inverse Laplace transform. Almost any beginning text in ODE will contain a chapter on the Laplace transform.
This principle has interpretations in ODE, but we will illustrate how it works in the context of a specific initial-boundary- value problem. In addition to providing a powerful tool for solving PD Es, the Laplace transform also provides insight into the nature of solutions to physical problems. With the help of the Laplace transform, we illustrate a very important and interesting concept known as Duhamel's principle in this lesson.
Before getting to this principle, however, let's discuss a problem that occurs frequently in engineering. Heat Flow within a Rod with Temperature Fixed on the Boundaries Quite often, it is important to find the temperature inside a medium due to time- varying boundary conditions.
So, solving Hard Probkm We can now take the so- lution w x,t to the constant BC problem 2. There is another aspect of Duhamel's formulas that makes them very useful. The Importance of Duhamel's Prlnclple In the problem just discussed, we were able to solve the easy problem with constant BCs, so we used Duhamel's formulas Quite often, however, even the easy problem constant BCs cannot be solved analytically.
What we can do, however, is observe the solution experimentally; in other words, we can rig a device that has constant BCs and experimentally measure the response.
We can then use Duhamel's principle to find the solution for any time-varying BC. In fact, we have only to observe the response w x,t to the constant BC problem once. When we have this data, we can then solve the problem with arbitrary BC f t by substituting into Duhamel's formulas Prove the Duhamel principle Suppose we have a metal rod laterally insulated and we supply an initial impulse of heat at the right-hand side the left-hand side is fixed at zero.
Differential Equations by C. Duhamel's principle is dis- cussed in conjunction with problems in ODE in Chapter 6 of this text. An excellent source of all kinds of applied problems; the Duhamel principle is discussed on page Phenomena described by this convection-diffusion equation exhibit both diffusion and convection prop- erties and are common in many situations.
How much diffusion and con- vection takes place depends on the relative size of the two coefficients D and V. Inasmuch as the convection of a substance represents material moving with the medium, it is possible to pick a moving coordinate system that moves with the medium. In this way, the convection term is eliminated and the equation can be solved in terms of the moving coordinate and then transformed back into the original variable x.
So far, we have been concerned with heat flow or diffusion of some kind in a one-dimensional domain. Suppose now we consider the problem of finding the concentration of a substance upwards from the surface of the earth where the substance both diffuses through the air and is carried upward convected by moving currents moving with velocity V.
Clearly, it is possible for the convection of the substance to contribute more of a movement in the substance than the diffusion itself. It would depend on the relative size of the diffusion coefficient and the velocity of the air. To verify this equation for a concentration u x,t of a substance, we use two basic facts 1. Flux due to diffusion The flux of material from left to right across a point x due to diffusion is given by -Dux x,t , where D is the diffusion coefficient.
If we substitute these two expressions into the conservation equation in Lesson 4, we can prove that the basic PDE is To get an idea of what solutions look like or how they behave with the convection term included, let's first work a problem that is pure convection the diffusion term is zero. A typical problem would be dumping a substance into a clean river moving with velocity V and observing the concentration of the substance downstream.
The Convection Term u,. To see this mathematically, let's solve Since it is a linear PDE with linear BCs, we should think in terms of separation of variables and integral transforms; however, since the x-variable is unbounded, separation of variables is out. Let's use the Laplace transform on t. It does, however, become more interesting when the solute pollutant diffuses with the medium.
To see what happens when a moving wave diffuses, we solve the following problem: The initial concentration is shown in Figure Note that in the new problem In other. To begin, we make what is called a change of variables change of independent variables.
The diagram in Figure The variable T, on the other hand, depends only on t. So much for the transformation. We now substitute our computed u,, u.. In other words, depending on the relative size of D diffusion coefficient and V velocity of the stream , the solution moves to the right with velocity V while, at the same time, the leading edge is diffusing at a rate defined by D Figure By looking at physical systems with different coordinates, the equations are sometimes sim- plified.
What does the solution look like for various values of time? In this particular problem, it is possible to evaluate the integral This is the Fourier transform solution from Lesson It may be more convenient for the reader to rewrite this integrand and then look it up in a table of integrals.
The Convection Term u. It is also shown how this equation is derived as a result of Newton's equations of motion. So far, we have been concerned with physical phenomenon described by one- dimensional parabolic equations diffusion problems. We will now begin to study the second major class of PD Es, hyperbolic equations. We start by studying the one-dimensional wave equation, which describes among other things the transverse vibrations of a string.
Vibrating-String Problem ' We consider the small vibrations of a string that is fastened at each end. We assume the string is stretched tightly, made of a homogeneous material, unaf- fected by gravity, and that the vibrations take place in a plane Figure Essentially, the wave equation is nothii! Looking at Figure The most important forces are 1. Net force due to the tension of the string a 2u. Restoring force - -vu This is a force that is directed opposite to the displacement of the string.
We will now present an intuitive interpretation of the simple wave equation. To answer this, we must under- stand that the expression u11 represents the vertical acceleration of the string at a point x. Hence, equation See Figure Materials that have large Young's moduli vibrate rapidly; sound waves also constitute longitudinal waves.
If the vibrating string had a variable density p x , then the wave equation would be: In other words, the PDE would have variable coefficients. This is in contrast to the heat equation, where only one IC was required. Another situation that can be described by the one-dimensional wave equa- tion is an electric current along a wire. With the help of Kirchhoffs laws, we can arrive at a system of two first-order PDEs The voltage also has its own equation, which is Derive the transmission-line equation From your knowledge of the various terms of the wave equation, what would you expect the solution of the following problem to look like for various values of time?
What would the solution of the following problem look like for various values of time? Holden-Day, An excellent reference text. In addition, this solution has interesting interpretations in terms of moving wave motion. In the hyperbolic case wave problem , we will do the opposite. We start by solving the one-dimensional wave equation in free space. This technique is basically the same as the moving-coordinate method of Lesson In other words, we can say that the general solution all solutions of is This completes step 2.
STEP 3. Figure Carrying out this integration on the second equation of Before we complete the lesson, however, we present a few examples to show how the D'Alembert solution is applied to specific problems.
Examples of the D'Alembert Solution 1. The reader might try to imagine the resulting wave form. The reader should ask himself or herself if this solution is reasonable. This completes Lesson 17; in Lesson 18, we show how the D'Alembert so- lution can be used to draw useful interpretations in the xt-plane. Note that a second-order PDE has two arbitrary functions in its general solution, whereas the general solution of a second-order ODE has two arbitrary constants.
The general technique of changing coordinate systems in a PDE in order to find a simpler equation is common in PDE theory. The strategy of finding the general solution to a PDE and then substituting it into the boundary and initial conditions is not a common technique in solving PDEs.
The solution discussed in this lesson is the only one that utilizes this strategy. Usually, we cannot find the general solution to the PDE, and even if we could, it is generally too complicated to substitute it into the side conditions. Verify that the general D'Alembert solution Substitute A readable pres- entation of some of the elementary ideas on PDEs. This lesson will show the reader some interesting interpretations of this equation in the xt-plane space-time plane and how the equation can be modified to find the solution of the vibrating semi-infinite string.
We start with our interpretation of We now interpret the D'Alembert solution when the initial position is zero, but the velocity is arbitrary. Problem This completes our interpretation of the D'Alembcrt solution in the xt-plane. Solutlon of the Seml-lnflntte String via the D1Alembert Formula The object now is to find the wave motion of the vibrating string whose left end is fixed at zero and has given initial conditions.
To find the solution of As long as x - ct ;;i: 0, we have no problem, since we can substitute This completes our lesson; we will examine the interpretation of equation Equation The sign of the wave is changed when it's reflected. Solution Characteristics are generally associated with hyperbolic equations.
Note that in this problem, the IC u x,O is discontinuous. Chapter 2 discusses many variations of D' Alembert's equation, including the semi-infinite problem discussed in this lesson. So far, the only kind of wave motion we have discussed is the one-dimensional transverse vibrations of a string. The reader should realize that this is only the tip of the iceberg as far as wave motion is concerned.
A few other types of important vibrations are: l. Sound waves longitudinal waves 2. Electromagnetic waves of light and electricity 3. Vibrations in solids longitudinal, transverse, and torsional 4. Probability waves in quantum mechanics 5.
Vibrating string transverse waves The purpose of this lesson is to discuss some of the various types of BCs that are associated with physical problems of this kind. Here, we will stick to one- dimensional problems where the BCs linear ones are generally grouped into one of three kinds: 1. In the area of mathematical control theory, an important problem involves determining the boundary function g2 t , so that a vibrating string can be shaken to zero in minimum time.
Boundary conditions similar to these are presented in the following two ex- amples: a Free end of a longitudinally vibrating spring Consider a vibrating spring with the bottom end unfastened Figure Physical problems like these come about in physics when an electric field a force is applied to vibrating electrons.
Elastic Attachment on the Boundaries Consider finally a violin string whose ends are attached to an elastic arrangement like the one shown in Figure This completes our discussion of the most common types of BCs associated with hyperbolic problems.
In the next few lessons, we will solve problems having BCs similar to these. Another BC not discussed in this lesson occurs when the vibrating string experiences a force at the ends proportional to the string velocity and in the opposite direction. Can you guess the solution to this problem? What is the general nature of the BC h u.. An excellent text with an extensive chapter Chapter 2 on initial and boundary conditions.
What happens is that the traveling-wave solution to the PDE and IC keeps reflecting from the boundaries in such a way that the wave motion does not appear to be moving, but, in fact, appears to be vibrating in one position; for example, a few typical standing waves are shown in Figure We will now solve the guitar-string problem by the method of separation of variables.
It will be left as an exercise for the reader to verify that only negative values of X. Hence, our goal is to find the constants A, B, C, and D and the negative separation constant A so that the expression Plugging Hence, we have now found a sequence of simple vibrations which we subscript with n The reader should be able to see that this sequence of functions constitutes a family of standing waves which have the property that each point on the wave vibrates with the same frequency whose shapes look like Figure This will then be the solution to our problem.
L sin mrx! This completes the problem, but before we close, we will make a few useful observations. If the initial velocity of the string is zero, then the solution If we now add each individual vibration of this type, we will get the solution to our problem.
The n-th term in the solution By using a trigono- metric identity, we can write this harmonic as where Rn and 8n are the new arbitrary constants amplitude and phase angle. This new form of the n-th mode is more useful for analyzing the vibrations. The property that all sound frequencies are multiples of a basic one is not shared by all types of vibrations.
This has something to do with the pleasing sound of a violin or guitar string in contrast to a drumhead, where the higher-order frequencies are not multiple frequencies of the fundamental one.
Find the solution to the vibrating-string problem Is the solution periodic in time? What is the period? What is the solution of the vibrating-string problem Show that for A ;;;;. A very readable text that contains many interesting examples; see in particular Chapter 7. It is also pointed out how the vibrations of the beam compare with the vibrations of the violin string.
Without going into the mechanics of thin beams, we can show that this resistance is responsible for changing the wave equation to the fourth-order beam equation The derivation of this equation can be found in reference 1 of Other Reading. Since this is the first time the reader has seen an application of PDEs higher than second order in this text, it will be useful to solve a typical vibrating-beam problem.
Later, we will talk about other types of beam problems. By "simply fastened," we mean that the ends of the beam are held stationary, but the slopes at the end points can move the beam is held by a pin-type arrangement, Figure Using the theory of thin beams see reference 1 of Other Reading , we can show that the bending moment.
Hence, the vibrating beam in Figure We now substitute equation Order PDEJ Hence, all that remains to do is choose the constants a,, and b,, in such a way that the ICs are satisfied. Substituting equation In order for the reader to understand this problem, we present a simple example. Sample Vibrating Beam Consider the simply supported beam shown in Figure It would be interesting forthe reader to imagine just how each of these vibrations looks.
Note, however, that both higher frequencies are integer multiples of the fundamental frequencies. Beams are generally fastened in one of three ways a Free unfastened b Simply fastened c Rigidly fastened Some sketches are given in Figure Another important vibrating-beam problem is the cantilever-beam problem shown in Figure The solution to this problem can be found in reference 3 of Other Reading.
Solve the beam problem with these BCs and tell how to find the natural frequencies of vibration of this beam. Knowing the natural frequencies of the beam is important, since various kinds of inputs of the same frequency can give rise to resonance.
Chapter 5 contains a derivation of the vibrating-beam problem. Mathematical Methods in Physics and Engineering by J. McGraw-Hill, ; Dover, This text contains a large section on the Sturm-Liouville problem.
Advanced Mathematics for Engineers by C. This book contains the solution of the cantilever-beam problem. In this form, we replace the original variables of the problem by new dimensionless ones they have no units. By writing a problem in dimensionless form, specific equations from physical, chemistry, biology, and economics that originally look different become one and the same.
For this reason, the mathematical study of PDEs generally doesn't concern itself with the physical parameters in the equations. It is up to the chemist, physicist, or biologist to transform his or her equation into those in the textbook. The basic idea behind dimensional analysis is that by introducing new dimen- sionless variables in a problem, the problem becomes purely mathematical and contains none of the physical constants that originally characterized it.
To see how this process works, let's consider a simple example. Our goal is to change problem No physical parameters like a in the new equation 2. It's also clear why we chose U x,t. With a little effort, we can see that the original problem Next, we transform the space variable x. The final step is to introduce a dimensionless time 'T, so that the constant [al L ]2 disappears from the differential equation.
L 2t Using this transformation on our previous problem No parameters in the PDE 2. Simple BCs 3. IC hasn't essentially been changed still a known function 4. Problem is simpler and more compact than the original one The solution to this problem can be found once and for all, so if a scientist transformed his or her original problem There aren't any set rules on how the new variables are defined; we more or less have to use physical intuition and try various possibilities.
We finish this lesson with a simple example of how to transform into dimen- sionless form, solve the new problem, and transform back to the original lab- oratory coordinates.
Anyone using these pro- grams must transform the problem into the form accepted by the program, solve the transformed problem, and then transform the numerical results back to his or her own coordinates.
Dimensional analysis allows mathematicians to work with PDEs without bothering with a lot of parameters and constants that are not relevant to the mathematical analysis.
It's not always necessary to transform all the variables into dimensionless form; sometimes only one or two have to be transformed.
Transform the vibrating string problem Transform problem
0コメント