Partial differentiation pdf

Link:Unit 3 Notes. Link:Unit 4 Notes. Link:Unit 5 Notes. Periodic Function,Fouries Series,Uses of Fouries Series,Dirichlet Condition,Even function,odd function,partial dirrentiation and its applications,functions of two or more variables partial derivatives,total differential and differntiability,derivaties of composite and implict functions,change of variables.

Multiple integrals and their multiplications,double integral,change of order of integration double integral in polar coordinates.

Applications of double integral to find area enclosed by lane curves and volume of solids of revolution. Follow us on Facebook and Support us with your Like. 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 A well-written book that contains many more aspects of dimensional analysis than does this lesson. However, it is true, too, that most students do not get very excited about studying something they know nothing about, and for that reason, we have waited until now to introduce the topic of classifying PDEs.

The purpose here is to classify the PDE In order for the reader to understand the classification scheme, we first give four examples of hyperbolic, parabolic, and elliptic equa- tions.

Note that we have called the time variable y in the general equation. In fact, we would have gotten the same result if we called the x in the heat equation the y in the general equation and then called the time variable t the x in the general equation.

The reader should note, too, that whether equation We now come to the major portion of this lesson: rewriting hyperbolic equa- tions in their canonical form. It turns out that if an equation is hyperbolic in a given region of space , then it is possible to introduce new coordinates Eand 11 characteristic coordinates in place of the original x and y, so that the equation takes on the simple form The reader will get a chance to carry them out in the problem set.

The next step in our process is to set the coefficients A and C in equation Form of the HyperboHc Equation! The only restriction is that we don't pick the same roots, or else we will end up with the two coordinates the same. Finding func- tions satisfying these conditions is really quite easy once we look for a few moments at Figure These particular new coordinates are drawn in Figure This com- pletes the discussion on how to find the new coordinates; the last step is to find the new equation.

Before we complete this lesson, let's apply the general procedure to see how it works in a specific example. We consider the problem of finding new coordinates that will change the original equation to canonical form for x and y in the first quadrant. One question the reader may ask is why someone would be interested in classifying and transforming a PDE into canonical form. The mathematical solutions of these three types of equations are quite different.

If we have an equation and want to study properties of the solution, we must convert it to canonical form and use existing results. After finding the solution in terms of the new coordinates, we can always convert back to the original coordinates.

Verify equations Verify that the equation is hyperbolic for all x and y and find the new characteristic coordinates. Continue with problem 4 by finding the alternative canonical form 6. Find the new characteristic coordinates for Solve the transformed equation in the new coordinate system and then transform back to the original coordinates to find the solution to the original problem.

J x,y,z in three dimensions and show how this solution satisfies Huygen's principle. J x,y It is then shown that the two-dimensional solution does not satisfy Huygen's principle. Finally, the method of descent is used once more to show that the one-dimensional version of this problem has the D'Alembert solution which we have seen before.

Earlier, we discussed the infinite vibrating string with I Cs and showed how it gave rise to the D' Alembert solution. The reader should realize that another application of the one-dimensional wave equation would be in describing plane waves in three dimensions. For instance, sound waves that are reasonably far from their source are essentially longitudinally vibrating plane waves and, hence, are described by this equation. The general situation is 1. One-dimensional waves are called plane waves 2.

Two-dimensional waves are called cylindrical waves 3. The problem of this lesson is to generalize the D' Alembert solution to two and three dimensions. The interpretation of this solution is that the initial disturbance iV radiates out- ward spherically velocity c at each point, so that after so many seconds, the point x,y,z will be influenced by those initial disturbances on a sphere of radius ct around that point Figure The actual value of the solution It might be interesting for the reader to try to evaluate this solution for a few simple functions lfl.

Now, to finish the problem. What about the other half; that is, This gives us the solution to problem Knowing this, we now have the solution to our general three-dimensional problem PDE u,.

This is known as Poisson's formula for the free-wave equation in three di- mensions, and it is the three-dimensional generalization of the D' Alembert formula. The most important aspect of the Poisson formula is the fact that the two integrals in Cf and iji are integrated over the surface of a sphere, which enables us to make the following important interpretation of the solution.

When ,. As time increases, the radius of the sphere around x,y,z increases with velocity c, and so after t2 seconds, it will finally intersect the initial disturbance region, and, hence, u x,y,z,t becomes nonzero. In other words, the wave disturbance originating from the initial-disturbance region has a sharp trailing edge. This general principle is known as Huygen's principle for three dimensions, and it is the reason why sound waves in three dimensions stimulate our ears but die off instanteously when the wave has passed.

It turns out that waves always have sharp leading edges, but the trailing edges are sharp only in dimensions 3, 5, 7, We will now show that the Huygen's principle does not apply to cylindrical waves.

This situation occurs when a water wave originates from a point where the trailing edge of the wave is not sharp but gradually damps to zero. In other words, if we analyze what this means in a manner similar to the three-dimensional case, we see that initial disturbances give rise to sharp leading waves, but not to sharp trailing waves. Thus, Huygen's principle doesn't ho!

We generally say here that Huygen's principle does not hold in one dimension. The actual calculus can be found in reference 2 of Other Reading. The general idea is that the solutions of problems in higher-dimensional spaces can be used to find the solution to problems in lower-dimensional ones by assuming certain bouQ. The reader should realize that this isn't the only problem to which the method of descent applies. To be worked with problem 3.

What is the physical interpretation of why Huygen's principle does not hold in two dimensions? Partial Differential Equations by P. An ex- cellent account of Huygen's principle; the book is considerably more advanced than this lesson. An excellent reference for problems in mathematical physics. L Sn sin mrx! Earlier, we learned about the regular Fourier and Laplace transforms and how problems are solved by transforming partial differential equations into ODEs.

The usual Fourier transform requires the variable being transformed to range from - oo to oo; hence, it is used to solve problems in free space no boundaries. In this lesson, we show how to solve boundary value problems with bound- aries by transforming the bounded variables first time we've done this.

First, let's forget the motivation; let's just define the transforms, their inverses, and use them. We will talk about why they work later. In short, however, transform methods can be thought of as resolving the functions of the problem into their various frequencies-solving an entire spectrum of problems for each frequency and then adding up the results.

We first start with a function f x defined on an interval [O, L. The reader will note that these transforms do nothing more than transform a function into the Fourier sine and cosine coefficients. Jo n even n Think about it. Note, too, that the sine transform of f x is a function defined only at the positive integers that is, it is just a sequence of numbers. In other words, the finite sine and cosine transforms transform functions into sequences. Properties of the Transforms Before solving problems, we must derive some of the useful properties of these transforms.

What about derivatives? Determine the transform Since the x-variable ranges from 0 to 1, we use a finite transform. Also, you will see why, in this case, we use the sine transform.

We could solve this problem with the Laplace transform by transforming t it would involve about the same level of difficulty as the finite sine transform. STEP 2. There are other transforms for BCs like these; refer to the generalized sine and cosine transforms in reference 2 of Other Reading.

In order to apply the finite sine and cosine transforms, the equation shouldn't contain first-order derivatives in x since the sine transform of the first derivative involves the cosine transform, and vice versa. The finite sine- and cosine-transform method essentially resolves all func- tions in the original problem like u,,, """' the ICs, BCs into a Fourier sine or cosine series, solves a sequence of problems ODE for the Fourier coefficients, and then adds up the results.

Derive the basic laws for S[u.. Graph the sine transform. What would the graph of the inverse transform look like for all values of x. Several examples are worked by means of the finite sine and cosine transforms. This book has good problems and is clearly written. An excellent book covering the topic of integral transforms and their applications; more advanced than this book, but useful to anyone seriously interested in the technique.

It is also shown that the two basic methods for solving linear equations, separation of variables and integral transforms, use the principle of superposition. For an engineer who wishes to find the response u to a linear system from an input f, a common approach is 1.

It turns out if the system is linear, then the sum u is the response we get if the function f were imputted directly; this is the principle of superposition Figure In separation of variables, we generally break down the initial conditions into an infinite number of simple parts and find the response to each part. We then sum these individual responses to find the solution to the problem. On the other hand, integral transforms also use superposition, for instance, let's show how the finite sine transform uses this principle.

What we're really doing is resolving the input f x,t into components, finding the response Un due to each component, and adding these responses. Mathematically, it may not be quite so obvious, but watch carefully. Note, too, that we have resolved the input f x,t into simple components Fn t. What we would like to find is the response Un t to each F" t. Then we would add the Un t to get the solution u x,t. To find the simple responses U" t , we must take our resolved PDE We can solve each of these problems by using an integrating factor or Laplace transform, if we like ; in any case, we get We have now found the responses U,, t to the simple inputs F,, t.

The reader should note that each U,, t is weighted by sin nirx. These, of course, are the same weights used when f x,t was decomposed into F,, t. NOTES In the finite sine transform, the resolutions were infinite series, whereas in most other integral transforms, the resolutions are integrals continuous resolutions. This is essentially what was done in lesson 12 when we solved this problem with Green's function.

Ginn and Company, This text contains a good section on linearity and superposition; it also has a large section on complex variables. See Chapter 2 in particular. It turns out that if we change coordinates from x,t to appropriate new coordinates s,T characteristic coordinates , then our differential equation becomes an ordinary differential equation.

Hence, we solve the ODE to find u s,T , and, then, the last step is to plug in the values of s and Tin terms of x and t to get u x,t. Also, if the reader recalls, the initial-position disturbance of the violin string at a point x affected the solution along two lines in the tx-plane corresponding to two moving waves. First, consider the new coordinate s. We can make these ideas clear with an example. It would be better if we found u as a function of x and t.

This is done in Step 3. We now summarize what we have been saying. We must now solve for s and T in terms of x and t from the transformation we found in Step 1 and substitute these values into u s,T. From a mathematical point of view, we could have easily switched them around.

However, since the techniques for solving second-order equations don't rely heavily on first-order ideas and since second-order equations are more important, we decided to study second-order equations first.

In the study of ordinary differential equations, the situation is different. The method of solution, the general theory, and so forth, follow naturally from first- to second-order equations, so text books always cover the first-order case first.

What do the characteristics look like? Plot the solution for different values of time. And, of course, check the answer. In fact, the differential equation doesn't have to involve the time variable at all maybe u depends only on space variables. A clear, concise description of first-order PDEs; an application to gas dynamics is given in Chapter This example points out that PDEs can be used to describe physical phenomena other than the usual ones in physics, biology, and engineering.

In fluid dynamics, u x,t could stand for the density of a fluid at x, while f x,t could be the flux amount of liquid passing a point x at time t.

We first derive the conservation equa- tion. Setting these two integrals equal. So, the question is, how do we use this equation and what are we looking for?

In traffic control, the amount of cars passing a given point flux is generally found experimentally as a function of the car density u. It seems obvious that as the density u increases, the flux f increases to a point, anyway. Other flow rates could be 1. Both the general characteristics and this specific example can be seen in Figure It is now clear that by knowing these characteristic curves starting from each point and knowing the solution doesn't change along these curves , we can piece together the solution u x,t for all time t.

We won't actually find the explicit equation for the solution u x,t in terms of x and t but will use our knowledge of the characteristics of the equation to solve some interesting problems. When characteristics run together, we have the phenomenon of shock waves discontinuous solutions , and what we must ask is, how fast does the leading edge of the shock wave propagate along the road?

A more detailed discussion of shock waves and further references can be found in reference 2 of the recommended reading for this lesson. The shock wave in our example occurred because the flux grows very large as a function of the density u. What is your interpretation of this solution? What is the relationship between the flux and density in this problem? Would you expect the solution to behave as it does? Compare the solutions of problem 1 and problem 2. What is the interpretation of your solution?

Does your solution check? Does it make sense physically? An Introduction to Fluid Dynamics by G. Univesity Press, An excellent, comprehensive book on fluid dynamics.

Formation and Decay of Shock Waves by P. A readable account of how shock waves propagate in space, written by a leading mathematician. Zachmanoglou and D. Williams and Wilkins, ; Dover, Chapter 3 presents a nice description of first-order nonlinear equations with examples. Mathematical Theory of Traffic Control by F. A mathematical description of traffic control; one of the few books of this kind. In many areas of science, several unknown quantities and their derivatives are related by more than one equation.

The problem we face here is to find the unknown functions p, u, v, and p simultaneously that satisfy the four equations along with initial and boundary conditions. There are other reasons for studying systems of equations. If the reader recalls ODE theory, a second-order ordinary equation can be written as a system of two first-order equations.

Although things are not quite so simple in PDE theory, it is often possible to write a single higher-order PDE as a system of first-order equations. For that reason, we present a short review of linear algebra.

We will now solve a simple system of two equations along with their ICs. We start by writing the two PDEs in matrix form or It turns out that v will satisfy a very easy system of equations the two equations in the new unknowns v1 and v2 are independent of each other , and, hence, we can find v1 and v2 easily.

We now substitute equations Many numerical methods have been developed to solve systems of equations, and, hence, computer programs are often written to solve a system of n first-order equations. For that reason, when using these programs, it will often be necessary to write your higher order equation as a system. The major difference is that now the matrix P of eigenvectors A x,t may be functions of x and t.

Write the system of PDEs in equations Verify that functions u 1 and u2 in equation A well-written text containing an excellent treatment of systems; more advanced than this text.

To find the vibrations of the drum- head that satisfy arbitrary initial conditions, we add the basic fundamental vibrations in such a way that the initial conditions are satisfied. The purpose of this lesson is straightforward: to find the vibrations of a circular drumhead with given boundary and initial conditions.

For simplicity, we let the radius of the circle be one and the. To solve this problem, we recall the violin-string problem from Lesson 20 whose solution involved the superposition of an infinite number of simple vi- brations.

For our next step, we want to solve the Helmholtz equation, but, first, it needs a boundary condition. The solutions U r,0 stand for the shapes of the fundamental modes of vibration of the drumhead, while the 'A's turn out to be the roots of certain Bessel functions and are proportional to the frequencies of these vibrations. So, the next step is to solve the Helmholtz eigenvalue problem this is an important problem in itself.

We solve it like most other linear, homogeneous PDEs with zero boundary conditions by separation of variables. The graphs of these functions are well-known and can be found in reference 2 of Other Reading. Also see Figure In order to find the functions Jn "-r and Yn Ar , we must resort to the method of Frobenius, which is to find solutions R r as power series.

It turns out that we can find two linearly independent power series J" Ar and Yn "-r. Tables of these roots are well-known, and computer programs are available to find them. A few of these roots are listed in Table We plot these functions for the different values of n and m in Figure The general shape of Unm r, 0 is the same for different values of the constants A and B.

Note, too, that the nodal circles where no vibration takes place have radii! Now that we have solved the Helmholtz equation for the basic shapes unm r,0 , the final step is to multiply by the time factor and add the products in such a way that the initial conditions are satisfied. Also note that we have lumped together the constants as A,,m and B,,m mere detail. Rather than going through the complicated process of finding A,,m and B,,m for the general case, we will find the solution for the situation where u is in- dependent of 0 very common.

In other words, we will assume that the initial position of the drumhead depends only on r. With these assumptions, the solution now becomes " Hence, we multiply each side of equation The solution to the vibrating membrane indepent of 0 is equation This solution is not"so complicated as the reader might think. It would be helpful for the reader to know the interpretation of J0 k 01 r , J0 k02r , We start by drawing J0 r Figure Now, in order to graph the functions J 0 k01 r , J0 k02r , The basic physical principle of the vibrating membrane is that each basic shape J 0 2.

Cs u, r,0,o -0 5. What is the highest frequency in the vibration? Graph the following functions for 0 :;;:; r :;;:; 1: a J 0 5. A comprehensive text with many examples; very good for engineers and physicists. The book solves the problem of the square drumhead; see Chapter 5.

Tables of Functions with Formulae and Curves by E. Jahnke and F. Stechert, ; Dover, One of the best known books of tables; roots of Bessel functions are tabulated here. Uae Spherical r r2 ,- r- sin "' Since the problem of transforming coordinates causes some students a great deal of difficulty, we will also discuss the chain rule that is used in making these transformations. The Laplacian operator a2 a2 a2 The question is, what does it mean and why should the sum of three second derivatives have anything to do with the laws of nature?

The answer to this lies in the fact that the Laplacian of a function allows us to compare the function at a point with the function at neighboring points. It does what the second derivative did in one dimension and might be thought of as a second derivative generalized to higher dimensions. We now give the basic interpretation of the Laplacian that makes it so useful. That is, the temperature at a point is increasing if the temperature at that point is less than the average of the temperatures on a circle around the point.

That is, the drumhead at a point is accelerating upward force is up if the drumhead at that point is less in height than the average of its neighbors. For example, a steady-state, stretched rubber membrane not moving satisfies Laplace's equation, hence, the height of the membrane at any point is equal to the average height of the membrane on a circle around the point.

Average value of II on a small circle Average value of Zt on a small circle around p is equal to 11 p around pis greater than u p Average value of 11 on a small circle around p is smaller than 11 p p is a point in two or three dimensions FIGURE What would be the nature of the potential field now that you know the meaning of the Laplacian? If g x,y is positive at a point, then heat is generated at that point. Negative g x,y means that heat is absorbed.

It's the equation we get when the time factor is separated from the wave equation or heat equation. So the question be- comes, how do we rewrite the Laplacian in different coordinate systems? As an illustration, we see how the two-dimensional Laplacian is transformed into polar coordinates.

So, we begin with To find the new Laplacian, we begin by drawing the diagram shown in Figure The purpose here is to illustrate exactly what depends on what. For example, the diagram in Figure We could also have drawn the diagram with rand 9 at the bottom, but since we want to derive expressions for """ and u,.

So, how do we find""" and u,. It's really very easy; we first compute u" and u,.. We find u" by adding up the paths in our diagram from u to x. By a similar argument, we have u,.

We'll find """ first u,. First, however, let's draw another diagram Figure The Laplaclan an Intuitive Description 2. The reader should realize at this point that if u depends on rand 0 which, in turn, depends on x and y , then u" will also depend in general on the same variables.

Adding u"" and uYY' we get the desired result This is the Laplacian in polar coordinates. It has the same intuitive meaning as the Laplacian in cartesian coordinates, but a different form.

Unfortunately, it has variable coefficients, so equations involving the polar Laplacian are more difficult to solve.

The Laplacian in cartesian coordinates is the only one with constant coef- ficients. This is one reason why problems in other coordinate systems are harder to solve. It is still possible to use the separation of variables method for these equations with variable coefficients; it's just that some of the resulting ordinary differential equations have variables coefficients.

Wear- rive at a lot of fairly complicated equations, such as Bessel's equation, Legendre's equation, and other so-called classical equations of physics. There are many more coordinate systems than the ones we've mentioned. For example, the vibration of an elliptical drumhead would require writing the Laplacian in an elliptical coordinate system. Rather than do this, how- ever, there is another approach: transform the ellipse into a region more to our liking like a circle.

We then solve the transformed problem into our new coordinates polar coordinates and then transform into the original coordinates. What are the solutions of this equation?

These are the circularly symmetric potentials in two dimensions. What is Laplace's equation in spherical coordinates if the solution u depends only on r? Can you find the solutions of this equation? These are the spher- ically symmetric potentials in three dimensions.

An excellent text, written from an intuitive viewpoint; gives the reader the between-the-lines descrip- tion of many concepts. These differential equations have no initial conditions like the hyperbolic-wave equation and the parabolic-heat equa- tion, but only boundary conditions.

For that reason, these problems are called boundary-value problems BVPs. The three most common types of boundary conditions BCs are: 1. BCs of the first kind Dirichlet BCs 2. BCs of the second kind Neumann BCs 3. BCs of the third kind Robin BCs These are explained and examples are shown to illustrate these ideas. Until now, the problems we've discussed involved phenomena that changed over space and time.

Tech branch to enhance more knowledge about the subject and to score better marks in the exam. The Cauchy Problem and Wave Equations: Mathematical modeling of vibrating string and vibrating membrane, Cauchy problem for second-order PDE, Homogeneous wave equation, Initial boundary value problems, Non-homogeneous boundary conditions, Finite strings with fixed ends, Non-homogeneous wave equation, Goursat problem. Method of Separation of Variables: Method of separation of variables for second-order PDE, Vibrating string problem, Existence and uniqueness of solution of vibrating string problem, Heat conduction problem, Existence and uniqueness of solution of heat conduction problem, Non-homogeneous problem.

Partial Differential Equations Notes Source: math.