{"loaderData":{"0-0-56":{"bookData":{"_id":"63e5cec7467a0065b0205630","category_id":"66cecf01c3d3f3f05eac01b6","author_id":["630db4d2b7c550d2e2f5742b","630da3e25d2b6424645ccb69"],"book_name":"Applied Differential Equations","book_image":"/books/DifferentialEquationspng-1681127429563.png","book_description":"

A First Course in Differential Equations With Applications in Engineering
(Coming Soon)

","have_express_version":true,"is_published":false,"status":"ACTIVE","slug":"applied-differential-equations","regular_version_cost":0,"express_version_cost":0,"regular_hard_copy_cost":0,"express_hard_copy_cost":0,"created_at":"2023-01-30T06:04:06.237Z","__v":0,"updated_at":"2024-08-28T07:18:35.350Z","buy_express_link":"undefined","buy_regular_link":"undefined","chapter_name_type":1,"sequence":8,"chapter_data":[{"_id":"63eefe76d1c28b91680cd002","book_id":"63e5cec7467a0065b0205630","chapter_name":"Introduction to Differential Equations","sequence":1,"chapter_content":"

Mathematical modeling is an attempt to describe how the real world\nworks using mathematics and symbols. Models can help us understand and\npredict the behavior of a system, or investigate the impacts of various\ncomponents on its behavior. Consequently, mathematical modeling is\nincreasingly used in the natural sciences (physics, chemistry, biology,\netc.), engineering, the social sciences (sociology, economics,\npsychology, etc.), and even music and linguistics.

In many situations mathematical models involve the search for an\nunknown function whose derivatives satisfy an equation. Such equations\nare called differential equationss. We obviously want\nto find a solution of a given differential equation. However, you’ll\ndiscover as the course progresses that studying differential equations\nentails more than merely knowing techniques developed by others to solve\nthem.

Since learning the terminology of any specialized subject is\nessential for studying that subject, we start with definitions, and some\nbasic concepts of differential equations.

\n","public_visibility":false,"book_slug":"applied-differential-equations","slug":"ch-1-introduction","created_at":"2023-02-16T12:09:45.711Z","__v":0,"chapter_express_content":"","allow_part_see":false,"header_data":[{"sequence":5,"public_visibility":true,"header_type":"TEXT","chapter_slug":"ch-1-introduction","slug":"definitions-and-classifications","header_name":"Definitions and Classifications","type":"header-1","header_data":[],"_id":"696ddcb14f3840e41e39040f"}]},{"_id":"63eefe76d1c28b91680ccff2","book_id":"63e5cec7467a0065b0205630","chapter_name":"First-Order Differential Equations","sequence":2,"chapter_content":"

First-Order Equations

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Can I examples?

","public_visibility":false,"book_slug":"applied-differential-equations","slug":"ch-2-first-order-equations","created_at":"2023-02-16T12:09:45.711Z","__v":0,"chapter_express_content":" Ordinary Differential Equations\n\n

3 First-Order Equations

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","express_authors":"","regular_authors":"","header_data":[{"sequence":1,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"separable-equations","header_name":"Separable Equations","type":"header-1","header_data":[],"_id":null},{"sequence":2,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"bernoullis-equation","header_name":"Bernoullis Equation","type":"header-1","header_data":[],"_id":null},{"sequence":3,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"integrating-factor","header_name":"Integrating Factor","type":"header-1","header_data":[],"_id":null},{"sequence":4,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"2","header_name":"2","type":"header-1","header_data":[],"_id":null},{"sequence":5,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"linear-differential-equations-of-the-first-order","header_name":"Linear Differential Equations Of The First Order","type":"header-1","header_data":[],"_id":null},{"sequence":6,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"exact-first-order-equations","header_name":"Exact First Order Equations","type":"header-1","header_data":[],"_id":null},{"sequence":7,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"riccatis-equation","header_name":"Riccatis Equation","type":"header-1","header_data":[],"_id":null},{"sequence":8,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-2-first-order-equations","slug":"introduction-1","header_name":"Introduction 1","type":"header-1","header_data":[],"_id":null}]},{"_id":"63f469c8d1c28b91680cdbab","book_id":"63e5cec7467a0065b0205630","chapter_name":"Modeling With First-Order Differential Equations","sequence":3,"chapter_content":"

This is a test.

Link to doc

Figure 1.

","chapter_express_content":"","public_visibility":false,"book_slug":"applied-differential-equations","slug":"how-to-add-figure","created_at":"2023-02-16T12:09:45.711Z","__v":0,"header_data":[]},{"_id":"63eefe76d1c28b91680ccff6","book_id":"63e5cec7467a0065b0205630","chapter_name":"Higher-Order Differential Equations","sequence":4,"chapter_content":" Ordinary Differential Equations\n\n

2 Higher Order Differential\nEquations

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","public_visibility":false,"book_slug":"applied-differential-equations","slug":"ch-4-higher-order-differential-equations","created_at":"2023-02-16T12:09:45.711Z","__v":0,"chapter_express_content":" Ordinary Differential Equations\n\n

4 Higher Order Differential\nEquations

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","header_data":[{"sequence":1,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-4-higher-order-differential-equations","slug":"the-use-of-one-known-solution-to-find-another","header_name":"The Use Of One Known Solution To Find Another","type":"header-1","header_data":[],"_id":null},{"sequence":2,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-4-higher-order-differential-equations","slug":"particular-solutions-by-the-method-of-undetermined-coefficients","header_name":"Particular Solutions By The Method Of Undetermined Coefficients","type":"header-1","header_data":[],"_id":null},{"sequence":3,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-4-higher-order-differential-equations","slug":"the-cauchyeuler-equation","header_name":"The Cauchyeuler Equation","type":"header-1","header_data":[],"_id":null},{"sequence":4,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-4-higher-order-differential-equations","slug":"variation-of-parameters","header_name":"Variation Of Parameters","type":"header-1","header_data":[],"_id":null},{"sequence":5,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-4-higher-order-differential-equations","slug":"homogeneous-linear-differential-equations-with-constant-coefficients","header_name":"Homogeneous Linear Differential Equations With Constant Coefficients","type":"header-1","header_data":[],"_id":null}]},{"_id":"640107f9d1c28b91680ce7d0","book_id":"63e5cec7467a0065b0205630","chapter_name":"Modeling With Higher-Order Differential Equations","sequence":5,"chapter_content":"

This is a link:

\n\n

This is a test!

\n\n

Can we increase the font? Adding figure: Does MathJax Work \\[y=f(x).\\]

\n\n

\\[\\mathbf{v}=\\frac{d\\mathbf{x}}{dt}\\]

\n\n

hfiwef ufi 4u4

\n\n

\\(\\boldsymbol{\\sigma}\\).

\n\n

Here is the regular \\(\\dfrac{dy}{dx}\\).

\n\n

J integral

\n\n

Zhigang Suo @zhigangsuo

\n\n

Notes prepared for a course on Fracture

\n\n

You are invited to discuss these notes on Twitter

\n\n

The J integral can be developed for both linear and nonlinear theory of elasticity. The nonlinear theory will be used in class, and the linear theory of elasticity will be used in a homework problem.

\n\n

Fracture mechanics, with or without a field theory

\n\n

For a crack in an elastic body subject to a load, the elastic energy stored in the body is a function of two independent variables: the displacement of the load, and the area of the crack, . This function of two independent variables can be determined by alternative methods.

\n\n

Experimental method. Measure the load-displacement curve for a body containing a pre-cut crack. During the measurement, the crack does not extend. Integrate the load-displacement curve to obtain the elastic energy for the body with the fixed area of crack. Repeat the procedure for the body with a pre-cut crack of a different area. This method invokes no field theory of elasticity.

\n\n

Computational method. Solve the boundary-value problem for a body containing a pre-cut crack of a given area. Calculate the energy density at every material particle in the body. Integrate the energy density over all material particles to obtain the elastic energy for the body with the specific area of crack. Repeat the procedure for the body with a pre-cut crack of a different length. This method requires that the body be modeled by a field theory of elasticity.

\n\n

Energy release rate is defined as

\n\n

This definition of the energy release rate assumes that the body is elastic, but invokes no field theory of elasticity. Indeed, the energy release rate can be determined experimentally by measuring the load-displacement curves of identically loaded bodies with cracks of different areas. No field needs to be measured.

\n\n

J integral. Many materials, however, can be modeled with a field theory of elasticity. When a material is modeled by such a field theory, we can represent the energy release rate in terms of the field in the body:

\n\n

The integral is called the J integral (J for James R. Rice). This lecture describes the J integral, along with examples of calculation. Uses of the J integral are often better appreciated in the context of individual applications, which we will describe in later lectures. In current practice, the field in an elastic body is commonly determined by using the finite element method. Once the field is determined, the finite element code calculates the J integral, which gives the energy release rate.

\n\n

Express energy release rate by J integral

\n\n

Express energy release rate in terms of field variables. Model a body with the nonlinear theory of elasticity. Under the plane-strain or plane-stress conditions, we represent the body by a region in the plane, and the boundary of the body by a curve in the plane. Consider a sample of unit thickness. Each point on the boundary is subject to constant nominal traction . The potential energy per unit thickness of the body is

\n\n

The first integral extends over the body, and the second integral extends over the boundary of the body.

\n\n

Now compare two specimens, one having a crack of length in the undeformed state, and other having a crack of length in the undeformed state. The cracks are in the direction of coordinate . The two specimens are of the same general shape. In the deformed states, the cracks in the two specimens are traction-free, and external boundaries of the specimens are subject to the same tractions. The two specimens have different potential energies, and .

\n\n

\nIn the undeformed state, we translate one specimen relative to the other by a distance , so that the tips of the cracks in the two specimens coincide. Consider an element of the boundary dL. Let N be the unit vector normal to the element. As one specimen translates relative to the other by a distance , the element dL sweeps across a region in the shape of a parallelogram. The area of the parallelogram is . In the deformed states of the two specimens, the parallelogram contributes a difference in the elastic energies in the two specimens, .

\n\n

Consider a set of material particles. In the undeformed state, the set of particles forms a straight segment . In the deformed state, the set of material particles forms a segment represented by vector . The applied force acting on the element is . Consequently, the applied force acting on the element dL gives a difference in the potential energies of the two specimens, .

\n\n

Combining the above contributions, we obtain the difference in the potential energies of the two specimens:

\n\n

The first term is due to the elastic energy of the body, and the second term is due to the potential energy of the applied traction.

\n\n

By definition, the energy release rate is

\n\n

The partial derivative means that we vary the length of the crack, but keep the loading conditions fixed. Comparing the above two expressions, we obtain that

\n\n

This equation expresses the energy release rate in terms of field variables.

\n\n

J Integral is path-independent

\n\n

J integral. Let us focus on the integral:

\n\n

The path of integration is a curve, drawn in the body in the undeformed state, connecting two material particles. The unit vector N is normal to the path of integration.

\n\n

Recall the relations in the theory of elasticity: , and . The J integral can also be written as

\n\n

The integral is a functional of , the field that describes the deformation of the body. The dummy variable of integration is X, the coordinates of material particles in the undeformed body.

\n\n

Yet another alternative expression of the J integral. A material particle occupies a place of coordinate X in the undeformed state, and occupies a place of coordinate x in the deformed state. The displacement of the material particle is

\n\n

Thus

\n\n

Inserting this expression into the definition of the J integral, we obtain that

\n\n

In reaching this expression, we have used the following relation:

\n\n

This relation results from the balance of forces.

\n\n

Divergence theorem. In the following development, we will need a result in calculus: the divergence theorem. Consider a region in a plane. Let be a smooth function defined in the region, and be the unit vector normal to the curve around the region. The divergence theorem is

\n\n

The integral on the left is over the area of a region, and integral on the right is over the curve surrounding the region.

\n\n

J integral is path-independent. The two material particles A and B can be connected by numerous paths. For any two paths, so long as no singularity exists between them, the J integrals along the two paths are identical.

\n\n

To prove this statement, reverse the direction of one path, so that the two paths form a closed contour. This closed contour is used as a path of integration. We need to prove that the J integral vanishes if the closed contour encloses no singularity. The proof invokes the divergence theorem, as well as the field theory of elasticity. We show that the second part of the J integral equals the first part:

\n\n

J integral associated with a traction-free crack. Consider a crack in a body subject to a load. The faces of the crack are traction-free. In the undeformed body, a path of integration starts from a material particle on the lower face of the crack, surrounds the tip of the crack, and ends at a material particle on the upper face of the crack. Because the path encloses the singular tip of the crack, the J integral does not vanish. We can make two statements:

\n\n

Given the two material particles on the faces of the crack, the J integral is independent of the path.
The J integral is the same independent of the choice of the material particles on the two faces of the crack.

\n\n

Use the J integral to calculate energy release rate in the finite-element method. For a crack in an elastic body subject to a load, the energy release rate is defined as the decrease in the elastic energy of the body associated with extension of the crack by unit area, while the load is held fixed. We have used this definition directly to calculate the energy release rate for a few cases. In each case, we somehow obtain the elastic energy in the body as a function of the area of the crack, and then take the partial derivative of this function with respect to the area of the crack.

\n\n

This procedure, however, is difficult to apply when we solve the boundary-value problem by using the finite-element method. Let us imagine how we might go about it. For a given area of the crack, we use the finite-element method to determine the field in the body, and then integrate to obtain the elastic energy stored in the body. We then change the area of the crack slightly, solve a new boundary-value problem, and then obtain the elastic energy in the body. We then take the ratio between the difference in the elastic energy and the difference in the area of crack. Here are two obvious difficulties:

\n\n

● In each boundary-value problem, the field is singular around the front of the crack.

\n\n

● The elastic energies in the two bodies are large quantities compared to their difference. To calculate the energy release rate this way would require us to determine the field very accurately.

\n\n

Both of the above difficulties are avoided if we use the J integral to calculate the energy release rate. The integrand involves the field in the body of a fixed crack. The path of integration needs to enclose the tip of crack, but can be chosen to be far away from the tip of the crack.

\n\n

Examples

\n\n

A long crack in a strip of a material pulled by two rigid grips. Rivlin and Thomas (1953) described this experimental setup. They obtained the energy release rate directly from its definition. The result is

\n\n

where H is the width of the undeformed strip, and is the width of the deformed stripe. The quantity can be determined from experimentally measured load-displacement curve of a strip with no crack, pulled by two rigid grips.

\n\n

The same result can be obtained by using the J integral along the dotted lines indicated in the undeformed strip (Rice, 1968). Use the J integral in the form

\n\n

The only nonzero contribution comes from the vertical line in the strip ahead of the crack.

\n\n

Layered materials. In the above, the material is taken to be homogeneous. When the two material particles are of the same deformation gradient, they have the same free energy function—that is, the nominal density of free energy is a function the deformation gradient:

\n\n

The function does not depend on X explicitly.

\n\n

For a layered material, which is homogenous in the -direction, but inhomogeneous in the -direction, the energy density takes the form

\n\n

Going through the same steps, you can confirm that the J integral is still path-independent.

\n\n

Finally consider a generally inhomogeneous material, for which the energy density takes the form

\n\n

Going through the same steps, you will find that the J integral is path-dependent.

\n\n

Detaching highly stretchable materials. Jinda Tang and Jianyu Li are studying the delamination of highly stretchable materials, such as elastomers and gels, using a specific experimental setup. Attach a film to a substrate, and leave a detached region as a pre-existing crack. Both materials are elastic and highly stretchable. When a force pulls the substrate, the tip of the crack blunts. When the force reaches some critical value, the crack runs. The length of the crack is large compared to the thicknesses of the film and the substrate, so that the crack runs in a steady state, and the energy release rate is independent of the length of the crack. We wish to relate the energy release rate of the crack to the applied force.

\n\n

First measure the stress-stretch relations of the film and the substrate separately. When a uniaxial force pulls the film, the nominal stress in the film relates to the stretch through a curve , and the area under this curve is the nominal density of the free energy in the film as a function of the stretch, . Similarly, when a uniaxial force pulls the substrate, the nominal stress in the substrate relates to the stretch through a curve , and the area under this curve is the nominal density of the free energy in the film as a function of the stretch, .

\n\n

When the laminate is in the undeformed state, the thickness of the film is , the thickness of the substrate is , and the width of the film and substrate is B. In the deformed state, a force P pulls the substrate. Far behind the tip of the crack, the film is stress-free, and the substrate is in a state of uniaxial stress, with the stretch determined by

\n\n

Far ahead the tip of the crack, the film and substrate are attached and both have the same stretch , determined by

\n\n

Now compare two specimens. Both specimens have the total length L in the reference state. One specimen has a crack of length in the undeformed state, and has potential energy

\n\n

in the deformed state.

\n\n

The other specimen has a crack of length in the undeformed state, and has potential energy

\n\n

in the deformed state.

\n\n

By definition, the energy release rate is the reduction in the potential energy associated with unit increase of the crack area:

\n\n

giving that

\n\n

In writing the potential energy of each specimen, we have assumed that the detached substrate is in a homogeneous state of stretch , and the attached laminate is in a homogeneous state of . In the region around the tip of the crack, however, the field is inhomogeneous. Because the crack is in a steady state, the inhomogeneous field in this region is identical in the two specimens. Consequently, this inhomogeneous field does not affect the calculation of the energy release rate.

\n\n

We can also use the J integral to obtain the above expression for the energy release rate.

\n\n

These relations together relate the energy release rate to the applied force P. We measure the stress-stretch curves of the two materials, and . The areas under these curves give the nominal densities of energy of the two materials, and . The applied force P determines the stretch in the detached substrate by , and determines the stretch in the attached laminate by .

\n\n

A thin, compliant film on a substrate. In a limiting case that the substrate is much stiffer than the substrate, the attached film does not constrain the substrate, so that the stretch in the detached substrate is nearly the same as the stretch in the attached laminate, . In this limiting case, the energy release rate reduces to

\n\n

When the stretch is of order unity, the free energy density is on the order of the elastic modulus, E. For a hydrogel, a representative value is E = 104 Pa. For the film of thickness = 0.1 mm, the energy release rate is on the order of 1 J/m2. For given materials, the laminate will not detach if the film is thin and stretch is small.

\n\n

A stiff film on a compliant substrate. In this limiting case, the attached laminate does not deform, = 1, so that the energy release rate is entirely due to the detached substrate:

\n\n

For experimental details, see Jingda Tang, Jianyu Li, Joost J. Vlassak, Zhigang Suo. Adhesion between highly stretchable materials. Soft Matter 12, 1093-1099 (2016). http://www.seas.harvard.edu/suo/papers/350.pdf

\n\n

Historical Notes

\n\n

Eshelby studied the change in the potential energy associated with the movement of a singularity. He called this change the force on the singularity, and expressed this force as an integral. He did not relate his idea to the Griffith theory of cracks. Rice and Cherepanov independently discovered the J integral within the context of fracture mechanics.

\n\n

● J.D. Eshelby, The force on an elastic singularity. Phil. Trans. Roy. Soc. London A 244, 87-111 (1951).

\n\n

● J.D. Eshelby, The continuum theory of lattice defects. Solid State Physics 3, 79-144 (1956).

\n\n

● J.R. Rice, A path independent integral and the approximate analysis of strain concentration by notches and cracks. J. Appl. Mech. 35, 379-386 (1968). http://esag.harvard.edu/rice/015_Rice_PathIndepInt_JAM68.pdf

\n\n

● J.R. Rice, Mathematical analysis in the mechanics of fracture, Chapter 3 of Fracture: An Advanced Treatise (Vol. 2, Mathematical Fundamentals) (ed. H. Liebowitz), Academic Press, N.Y., 1968, pp. 191-311. http://esag.harvard.edu/rice/018_Rice_MathAnalMechFract_68.pdf

\n\n

● G.P. Cherepanov, Crack propagation in continuous media. Journal of Applied Mathematics and Mechanics 31, 503 (1967).

\n\n

● G.P. Cherepanov, Cracks in solids. International Journal of Solids and Structures 4, 811-831 (1968).

\n\n

● S.J. Chang, Path-independent integral for rupture of perfectly elastic materials. Journal of Applied Mathematics and Physics 23, 149-152 (1972).

\n\n

\n","chapter_express_content":"","public_visibility":false,"book_slug":"applied-differential-equations","slug":"adding-images","created_at":"2023-02-16T12:09:45.711Z","__v":0,"header_data":[]},{"_id":"63eefe76d1c28b91680cd012","book_id":"63e5cec7467a0065b0205630","chapter_name":"Systems of Linear Differential Equations","sequence":6,"chapter_express_content":" Ordinary Differential Equations\n\n

5 Systems of Differential\nEquations

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","public_visibility":false,"book_slug":"applied-differential-equations","slug":"ch-6-systems-of-differential-equations","created_at":"2023-02-16T12:09:45.711Z","__v":0,"chapter_content":" Ordinary Differential Equations\n\n

3 Systems of Differential\nEquations

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","header_data":[{"sequence":1,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-6-systems-of-differential-equations","slug":"constant-coefficient-systems","header_name":"Constant Coefficient Systems","type":"header-1","header_data":[],"_id":null},{"sequence":2,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-6-systems-of-differential-equations","slug":"nonhomogeneous-linear-systems","header_name":"Nonhomogeneous Linear Systems","type":"header-1","header_data":[],"_id":null}]},{"_id":"6663621df3472b54c5ec871c","book_id":"63e5cec7467a0065b0205630","chapter_name":"Laplace Transform","sequence":7,"chapter_content":"

The Laplace Transform

What is an integral transform?

A powerful method for solving certain types of ordinary or partial differential equations is the use of integral transforms. In general, a linear integral transform of a function \\(f(t)\\) is an expression of the form \\[\\mathcal{T}\\{f(t)\\}=\\int_a^b K(s,t) f(t) dt \\tag{i}\\] where \\(K(s,t)\\) is called the kernel of the transform. The result of the above integral is a function of \\(s\\), say \\(F(s)\\). Note that (i) is called integral transform because it transforms a function in \\(t\\)-space into another space in \\(s\\)-space by integration. The kernel function \\(K(s,t)\\) is a function of two spaces.

Depending on what \\(K(s,t)\\), \\(a\\), and \\(b\\) are, we have various integral transforms. For instance,

If \\(K(s,t)=e^{-st}\\), \\(a=0\\), and \\(b=+\\infty\\), the integral transform is called the Laplace transform.⁴

If \\(K(s,t)=\\frac{1}{\\sqrt{2\\pi}}e^{-ist}\\), \\(a=-\\infty\\), and \\(b=+\\infty\\), the integral transform is called the Fourier transform.

Similarly we can define other integral transforms.

In this chapter, we will learn about the Laplace transform and some of its applications in solving differential equations. We will see that how the application of the Laplace transform simplify solving certain differential equations.

This is sometimes called the unilateral Laplace transform. If \\(a=-\\infty\\), the integral transform is called the bilateral Laplace transform. In this book, we only consider the unilateral Laplace transform.↩︎

","chapter_express_content":"","public_visibility":false,"book_slug":"applied-differential-equations","slug":"laplace-transform","regular_authors":"","express_authors":"","created_at":"2024-06-03T13:02:24.363Z","__v":0,"header_data":[{"sequence":1,"public_visibility":false,"header_type":"TEXT","book_slug":"applied-differential-equations","chapter_slug":"laplace-transform","slug":"definition-of-laplace-transform","header_name":"Definition of Laplace Transform","type":"header-1","header_data":[],"_id":null}]},{"_id":"63eefe76d1c28b91680cd00a","book_id":"63e5cec7467a0065b0205630","chapter_name":"Series Solutions of Differential Equations","sequence":8,"chapter_express_content":" Ordinary Differential Equations\n\n

6 Series Solutions of Ordinary\nDifferential Equations

So far, we have been able to express the solution of a differential\nequation as a combination of the functions that we know from calculus.\nThese functions include \\(x^{n}\\),\n\\(e^{x}\\), \\(\\ln x\\), trigonometric functions, inverse\ntrigonometric functions, etc. —there is not a very long list. Such a\nlist of functions may be enough for linear differential equations with\nconstant coefficients, but as soon as we explore linear differential\nequations with variable coefficients, we will find this list too\nrestrictive. In fact, many important linear differential equations that\nappear in various branches of science and engineering—if we put aside\nthose with constant coefficients—have no solutions that can be expressed\nin terms of elementary functions. Some of those important equations are\nLegendre’s, Bessel’s, and the hypergeometric equations. In this chapter,\nwe introduce two methods to solve linear differential equations with\nvariable coefficients.

In the first method, we write the solution in the form of a power\nseries. A power series in \\(x-x_{0}\\)\nis an infinite series of the form \\[\\sum_{k=0}^{\\infty}a_{k}(x-x_{0})^{k},\\]\nwhere \\(x\\) is a variable and \\(a_{0},a_{1},a_{2},\\dots,x_{0}\\) are\nconstants. The second method is the generalization of the first one. In\nthis method, we multiply a power series by a logarithmic function or a\nfractional power.

The series solutions that have extensive applications are frequently\nused to define new functions. The properties of these new functions are\ninvestigated by means of their series. All these new functions and\nvarious other functions not appearing in calculus are known as\nspecial functions.

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","public_visibility":false,"book_slug":"applied-differential-equations","slug":"ch-8-series-solutions-of-ordinary-differential-equations","created_at":"2023-02-16T12:09:45.711Z","__v":0,"chapter_content":"

4 Series Solutions of Ordinary Differential Equations

So far, we have been able to express the solution of a differential equation as a combination of the functions that we know from calculus. These functions include \\(x^{n}\\), \\(e^{x}\\), \\(\\ln x\\), trigonometric functions, inverse trigonometric functions, etc. —there is not a very long list. Such a list of functions may be enough for linear differential equations with constant coefficients, but as soon as we explore linear differential equations with variable coefficients, we will find this list too restrictive. In fact, many important linear differential equations that appear in various branches of science and engineering—if we put aside those with constant coefficients—have no solutions that can be expressed in terms of elementary functions. Some of those important equations are Legendre’s, Bessel’s, and the hypergeometric equations. In this chapter, we introduce two methods to solve linear differential equations with variable coefficients.

In the first method, we write the solution in the form of a power series. A power series in \\(x-x_{0}\\) is an infinite series of the form \\[\\sum_{k=0}^{\\infty}a_{k}(x-x_{0})^{k},\\] where \\(x\\) is a variable and \\(a_{0},a_{1},a_{2},\\dots,x_{0}\\) are constants. The second method is the generalization of the first one. In this method, we multiply a power series by a logarithmic function or a fractional power.

The series solutions that have extensive applications are frequently used to define new functions. The properties of these new functions are investigated by means of their series. All these new functions and various other functions not appearing in calculus are known as special functions.

This is a test.

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","header_data":[{"sequence":1,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-8-series-solutions-of-ordinary-differential-equations","slug":"power-series","header_name":"Power Series","type":"header-1","header_data":[],"_id":null},{"sequence":2,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-8-series-solutions-of-ordinary-differential-equations","slug":"the-method-of-frobenius","header_name":"The Method Of Frobenius","type":"header-1","header_data":[],"_id":null}]},{"_id":"63eefe76d1c28b91680ccffa","book_id":"63e5cec7467a0065b0205630","chapter_name":"Prerequisite Math","sequence":9,"chapter_content":"

1 Prerequisite Math

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\\[f^2_x=x^2+1,\\quad \\xi_m\\]

Test 1:

Test1
Test2

","public_visibility":false,"book_slug":"applied-differential-equations","slug":"ch-A-prerequisite-math","created_at":"2023-02-16T12:09:45.711Z","__v":0,"chapter_express_content":" Ordinary Differential Equations\n\n

1 Prerequisite Math

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","header_data":[{"sequence":1,"public_visibility":false,"header_type":"TEXT","book_slug":"differential-equations","chapter_slug":"ch-A-prerequisite-math","slug":"review-of-complex-numbers","header_name":"Review Of Complex Numbers","type":"header-1","header_data":[],"_id":null}]},{"_id":"643b8305e30c2b41e7d4545b","book_id":"63e5cec7467a0065b0205630","chapter_name":"Numerical Solutions of Differential Equations","sequence":10,"chapter_content":"

How To Deal With Sines and Cosines

What if I make some changes here?

Greek letters being usual to denote angles, we will take as the usual letter for any variable angle the letter θ (“theta”). In this chapter, θ is measured in radians.²²

Derivative of Sine

Let us consider the function

What we have to investigate is the value of d(sinθ)dθ; or, in other words, if the angle θ varies, we have to find the relation between the increment of the sine and the increment of the angle, both increments being indefinitely small in themselves. Examine Fig. 15.1.1, wherein, if the radius of the circle is unity, the height of y is the sine, and θ is the angle. Now, if θ is supposed to increase by the addition to it of the small angle dθ—an element of angle—the height of y, the sine, will be increased by a small element dy. The new height y+dy will be the sine of the new angle θ+dθ, or, stating it as an equation,

y + d y = sin (θ + d θ);

and subtracting from this the first equation gives

d y = sin (θ + d θ) - sin θ .

The quantity on the right-hand side is the difference between two sines, and books on trigonometry tell us how to work this out. For they tell us that if M and N are two different angles,

sin M - sin N = 2 cos M + N 2 \cdot sin M - N 2 .

If, then, we put M=θ+dθ for one angle, and N=θ for the other, we may write

d y = 2 cos θ + d θ + θ 2 \cdot sin θ + d θ - θ 2,

or,

d y = 2 cos (θ + 12 d θ) \cdot sin d θ 2 .

But if we regard dθ as indefinitely small, then in the limit we may neglect 12dθ by comparison with θ, and may also take sindθ2 as being the same as 12dθ. The equation then becomes:

d y d y = 2 cos θ \times 12 d θ; = cos θ \cdot d θ,

and, finally,

d y d θ = cos θ .

[Notice that the approximation sindθ2≈dθ2 is true only when dθ is measured in radians.]

The accompanying curves, Fig. 15.1.2 and Fig. 15.1.3, show, plotted to scale, the values of y=sinθ, and dydθ=cosθ, for the corresponding values of θ.

Figure 15.1.2

Figure 15.1.3

Derivative of Cosine

Take next the cosine.

Let y=cosθ.

Now cosθ=sin(π2−θ).

Therefore

\\begin{align}&#xA0;&amp;\\begin{align}&#xA0;dy&#xA0;=&#xA0;d\\left(\\sin\\left(\\frac{\\pi}{2}&#xA0;-&#xA0;\\theta\\right)\\right)&#xA0;&amp;=&#xA0;\\cos\\left(\\frac{\\pi}{2}&#xA0;-&#xA0;\\theta\\right)&#xA0;\\times&#xA0;d(-\\theta),&#xA0;\\\\&#xA0;&amp;=&#xA0;\\cos\\left(\\frac{\\pi}{2}&#xA0;-&#xA0;\\theta\\right)&#xA0;\\times&#xA0;(-d\\theta),&#xA0;\\end{align}&#xA0;\\\\&#xA0;&amp;\\frac{dy}{d\\theta}&#xA0;=&#xA0;-\\cos\\left(\\frac{\\pi}{2}&#xA0;-&#xA0;\\theta\\right).\\end{align}\" tabindex=\"0\" id=\"MathJax-Span-375\">$\\begin{aligned} dy = d\\left(\\sin\\left(\\frac{\\pi}{2} - \\theta\\right)\\right) &= \\cos\\left(\\frac{\\pi}{2} - \\theta\\right) \\times d(-\\theta), \\\\ &= \\cos\\left(\\frac{\\pi}{2} - \\theta\\right) \\times (-d\\theta), \\end{align} \\\\ &\\frac{dy}{d\\theta} = -\\cos\\left(\\frac{\\pi}{2} - \\theta\\right).\\end{aligned}$

And it follows that

d y d θ = - sin θ .

Derivative of Tangent

Lastly, take the tangent.

As tanθ=sinθcosθ, we can apply the Quotient Rule (see page ) to find d(tanθ)dθ:²³

d (tan θ) d θ = cos θ d (sin θ) d θ - sin θ d (cos θ) d θ cos 2 θ = cos θ \cdot cos θ - sin θ (- sin θ) cos 2 θ = cos 2 θ + sin 2 θ cos 2 θ

Since

cos 2 θ + sin 2 θ cos 2 θ = 1 + (sin θ cos θ) 2

we get

d (tan θ) d θ = 1 + tan 2 θ .

Also, since sin2θ+cos2θ=1, we get

d (tan θ) d θ = 1 cos 2 θ = sec 2 θ .

Therefore,

d (tan θ) d θ = sec 2 θ = 1 + tan 2 θ .

Summary of Results

Collecting these results, we have:

y sin θ cos θ tan θ d y d θ cos θ - sin θ sec 2 θ = 1 + tan 2 θ

To derive the above results, we replaced sin(dθ/2) by dθ/2. In general, sinx is approximately equal to x when (1) x is small (2) x is measured in radians

sin x \approx x (x is small and measured in radians) .

For example, 1∘ is the same as π180 radians, and we cannot approximate sin1∘ by 1

sin 1 \circ \neq 1

but

sin 1 \circ = sin π 180 \approx π 180 = 0.017.

Therefore, the results tabulated above are true only when θ is measured in radians.

Sometimes, in mechanical and physical questions, as, for example, in simple harmonic motion and in wave-motions, we have to deal with angles that increase in proportion to the time. Thus, if T be the time of one complete period, or movement round the circle, then, since the angle all round the circle is 2π radians, (equivalent to 360∘), the amount of angle moved through in time t, will be

θ = 2 π t T, in radians.

If the frequency, or number of periods per second, be denoted by n, then n=1T, and we may then write:

θ = 2 π n t .

Then we shall have

y = sin (2 π n t) .

If, now, we wish to know how the sine varies with respect to time, we must differentiate with respect, not to θ, but to t. For this we must resort to the chain rule explained in Chapter 9 (page ), and put

d y d t = d y d θ \cdot d θ d t .

Now dθdt will obviously be 2πn; so that

d y d t = cos θ \times 2 π n = 2 π n \cdot cos (2 π n t) .

Similarly, it follows that

d (cos (2 π n t)) d t = - 2 π n \cdot sin (2 π n t) .

Second Derivative of Sine or Cosine.

We have seen that when sinθ is differentiated with respect to θ it becomes cosθ; and that when cosθ is differentiated with respect to θ it becomes −sinθ; or, in symbols,

d 2 (sin θ) d θ 2 = - sin θ .

So we have this curious result that we have found a function such that if we differentiate it twice over, we get the same thing from which we started, but with the sign changed from + to −.

The same thing is true for the cosine; for differentiating cosθ gives us −sinθ, and differentiating −sinθ gives us −cosθ; or thus:

d 2 (cos θ) d θ 2 = - cos θ .

Sines and cosines are the only functions of which the second derivative is equal (and of opposite sign to) the original function.

Examples.

With what we have so far learned we can now differentiate expressions of a more complex nature.

If y=arcsinx, find dydx.

[In many modern calculus textbooks, the inverse sine is denoted by sin−1; that is, arcsinx=sin−1x. Note that sin−1x is NOT the same as 1sinx. To avoid confusion, the notation arcsinx may be preferred over sin−1x in some texts, including this one.]

Solution. If y is the arc whose sine is x, then x=siny.

d x d y = cos y .

Passing now from the inverse function to the original one, we get

d y d x = 1 d x d y = 1 cos y .

now since cos2y+sin2y=1,

cos y = \pm 1 - sin 2 y ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ \sqrt = \pm 1 - x 2 ‾ ‾ ‾ ‾ ‾ ‾ \sqrt;

hence

d y d x = 1 cos y = \pm 1 1 - x 2 ‾ ‾ ‾ ‾ ‾ ‾ \sqrt .

But which one is correct? plus or minus? If we look at the graph of y=arcsinx (Fig. 15.6.1), we realize that the slope of the curve is always positive, indicating that that we must take the positive square root. Hence,

d y d x = 1 1 - x 2 ‾ ‾ ‾ ‾ ‾ ‾ \sqrt .

Figure 15.6.1

\\mywboxtext$d(arcsinx)dx=11−x2‾‾‾‾‾‾√$

If y=cos3θ, find dydθ.

Solution. This is the same thing as y=(cosθ)3.

Let cosθ=v;then y=v3;dydv=3v2.

d v d θ d y d θ = - sin θ . = d y d v \times d v d θ = - 3 cos 2 θ sin θ .

If y=sin(x+a), find dydx.

Solution. Let x+a=v;then y=sinv.

d y d v = cos v; d v d x = 1 and d y d x = cos (x + a) .

If y=lnsinθ, find dydθ.

Solution. Let sinθ=v;y=lnv.

d y d v d y d θ = 1 v; d v d θ = cos θ; = 1 sin θ \times cos θ = cot θ .

If y=cotθ=cosθsinθ, find dydθ.

Solution.

d y d θ = - sin 2 θ - cos 2 θ sin 2 θ = - (1 + cot 2 θ) = - csc 2 θ .

If y=tan3θ, find dydθ.

Solution. Let 3θ=v;y=tanv;dydv=sec2v.

d v d θ = 3; d y d θ = 3 sec 2 3 θ .

If y=1+3tan2θ‾‾‾‾‾‾‾‾‾‾‾√, find dydθ.

Solution. y=(1+3tan2θ)12.

Let 3tan2θ=v.

\\begin{align}&#xA0;y&#xA0;&amp;=&#xA0;(1+v)^{\\frac{1}{2}};\\quad&#xA0;\\frac{dy}{dv}&#xA0;=&#xA0;\\frac{1}{2\\sqrt{1+v}}&#xA0;\\text{&#xA0;(see&#xA0;Example~\\ref{Ex:Ch9-ExNo1},&#xA0;page&#xA0;\\pageref{Ex:Ch9-ExNo1});}&#xA0;\\\\&#xA0;%&#xA0;\\frac{dv}{d\\theta}&#xA0;&amp;=&#xA0;6\\tan\\theta&#xA0;\\sec^2&#xA0;\\theta\\end{align}\" tabindex=\"0\" id=\"MathJax-Span-1924\">\\begin{align} y &= (1+v)^{\\frac{1}{2}};\\quad \\frac{dy}{dv} = \\frac{1}{2\\sqrt{1+v}} \\text{ (see Example~\\ref{Ex:Ch9-ExNo1}, page \\pageref{Ex:Ch9-ExNo1});} \\\\ % \\frac{dv}{d\\theta} &= 6\\tan\\theta \\sec^2 \\theta\\end{align}

(for, if tanθ=u,

v = 3 u 2; d v d u = 6 u; d u d θ = sec 2 θ;

hence dvdθ=6tanθsec2θ;)
hence

d y d θ = 6 tan θ sec 2 θ 2 1 + 3 tan 2 θ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ ‾ \sqrt .

If y=sinxcosx, find dydx .

Solution.

d y d x = sin x (- sin x) + cos x \times cos x = cos 2 x - sin 2 x .

Exercises 15

(See page for Answers.)

Differentiate the following:

(i) y (ii) y (iii) y = A sin (θ - π 2) . = sin 2 θ; and y = sin 2 θ . = sin 3 θ; and y = sin 3 θ .

Find the value of θ for which sinθ×cosθ is a maximum.

Differentiate y=12πcos(2πnt).

If y=sinax, find dydx.

Differentiate y=lncosx.

Differentiate y=18.2sin(x+26∘).

Plot the curve y=100sin(θ−π12); and show that the slope of the curve at θ=5π12 is half the maximum slope.

If y=sinθ⋅sin2θ, find dydθ.

If y=a⋅tanm(θn), find the derivative of y with respect to θ.

Differentiate y=exsin2x.

Differentiate the three equations of Exercises XIII. (page ), No. 4, and compare their derivatives, as to whether they are equal, or nearly equal, for very small values of x, or for very large values of x, or for values of x in the neighborhood of x=30.

Differentiate the following:

(i) y (iii) y (v) y = sec x . = arctan x . = tan x \times 3 sec x ‾ ‾ ‾ ‾ ‾ ‾ \sqrt . (ii) y (iv) y = arccos x . = arcsec x .

Differentiate y=sin(2θ+3)2.3.

Differentiate y=θ3+3sin(θ+3)−3sinθ−3θ.

Find the maximum or minimum of y=θcosθ.

","chapter_express_content":"

This is the express version.

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Applied Differential Equations

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