## 5.3.1 Forced vibrating string.

Figure 5.3: Vibrating string.

The problem is governed by the equations

[ y_{tt}=a^2y_{xx} y(0,t)=0,~~~~~~~y(L,t)=0, y(x,0)=f(x),~~y_t(x,0)=g(x). ]

We saw previously that the solution is of the form

[ y= sum_{n=1}^{infty} left( A_ncos left( frac{npi a}{L}t ight) + B_nsin left( frac{npi a}{L}t ight) ight) sin left( frac{npi }{L}x ight),]

where (A_n) and (B_n) were determined by the initial conditions. The natural frequencies of the system are the (circular) frequencies (frac{npi a}{L}) for integers (n geq 1).

But these are free vibrations. What if there is an external force acting on the string. Let us assume say air vibrations (noise), for example a second string. Or perhaps a jet engine. For simplicity, assume nice pure sound and assume the force is uniform at every position on the string. Let us say (F(t)=F_0 cos(omega t)) as force per unit mass. Then our wave equation becomes (remember force is mass times acceleration)

[y_{tt}=a^2y_{xx}+F_0cos(omega t),]

with the same boundary conditions of course.

We want to find the solution here that satisfies the above equation and

[y(0,t)=0,~~~~~y(L,t)=0,~~~~~y(x,0)=0,~~~~~y_t(x,0)=0.]

That is, the string is initially at rest. First we find a particular solution (y_p) of (5.3.3) that satisfies (y(0,t)=y(L,t)=0). We define the functions (f) and (g) as

[f(x)=-y_p(x,0),~~~~~g(x)=- frac{partial y_p}{partial t}(x,0).]

We then find solution (y_c) of (5.3.1). If we add the two solutions, we find that (y=y_c+y_p) solves (5.3.3) with the initial conditions.

Exercise (PageIndex{1}):

Check that (y=y_c+y_p) solves (5.3.3) and the side conditions (5.3.4).

So the big issue here is to find the particular solution (y_p). We look at the equation and we make an educated guess

[y_p(x,t)=X(x)cos(omega t).]

We plug in to get

[ - omega^2Xcos(omega t)=a^2X''cos(omega t),]

or (- omega X=a^2X''+F_0) after canceling the cosine. We know how to find a general solution to this equation (it is a nonhomogeneous constant coefficient equation). The general solution is

[ X(x)=Acos left( frac{omega}{a}x ight)+Bsin left( frac{omega}{a}x ight)- frac{F_0}{omega^2}.]

The endpoint conditions imply (X(0)=X(L)=0). So

[ 0=X(0)=A- frac{F_0}{omega^2},]

or (A=frac{F_0}{omega^2}), and also

[ 0=X(L)= frac{F_0}{omega^2} cos left( frac{omega L}{a} ight)+Bsin left( frac{omega L}{a} ight)- frac{F_0}{omega^2}.]

Assuming that (sin left( frac{omega L}{a} ight) ) is not zero we can solve for (B) to get

[ B=frac{-F_0 left( cos left( frac{omega L}{a} ight)-1 ight)}{- omega^2 sin left( frac{omega L}{a} ight)}.]

Therefore,

[ X(x)= frac{F_0}{omega^2} left( cos left( frac{omega}{a}x ight)- frac{ cos left( frac{omega L}{a} ight)-1 }{ sin left( frac{omega L}{a} ight)}sin left( frac{omega}{a}x ight)-1 ight).]

The particular solution (y_p) we are looking for is

[ y_p(x,t)= frac{F_0}{omega^2} left( cos left( frac{omega}{a}x ight)- frac{ cos left( frac{omega L}{a} ight)-1 }{ sin left( frac{omega L}{a} ight)}sin left( frac{omega}{a}x ight)-1 ight) cos(omega t).]

Exercise (PageIndex{2}):

Check that (y_p) works.

Now we get to the point that we skipped. Suppose that (sin left( frac{omega L}{a} ight)=0). What this means is that (omega) is equal to one of the natural frequencies of the system, i.e. a multiple of ( frac{pi a}{L}). We notice that if (omega) is not equal to a multiple of the base frequency, but is very close, then the coefficient (B) in (5.3.11) seems to become very large. But let us not jump to conclusions just yet. When (omega = frac{npi a}{L}) for (n) even, then (cos left( frac{omega L}{a} ight)=1) and hence we really get that (B=0). So resonance occurs only when both (cos left( frac{omega L}{a} ight)=-1) and (sin left( frac{omega L}{a} ight)=0). That is when (omega = frac{npi a}{L}) for odd (n).

We could again solve for the resonance solution if we wanted to, but it is, in the right sense, the limit of the solutions as (omega) gets close to a resonance frequency. In real life, pure resonance never occurs anyway.

The above calculation explains why a string will begin to vibrate if the identical string is plucked close by. In the absence of friction this vibration would get louder and louder as time goes on. On the other hand, you are unlikely to get large vibration if the forcing frequency is not close to a resonance frequency even if you have a jet engine running close to the string. That is, the amplitude will not keep increasing unless you tune to just the right frequency.

Similar resonance phenomena occur when you break a wine glass using human voice (yes this is possible, but not easy^{2}) if you happen to hit just the right frequency. Remember a glass has much purer sound, i.e. it is more like a vibraphone, so there are far fewer resonance frequencies to hit.

When the forcing function is more complicated, you decompose it in terms of the Fourier series and apply the above result. You may also need to solve the above problem if the forcing function is a sine rather than a cosine, but if you think about it, the solution is almost the same.

Example (PageIndex{1}):

Let us do the computation for specific values. Suppose (F_0=1) and (omega =1) and (L=1) and (a=1). Then

[ y_p(x,t)= left( cos(x)- frac{ cos(1)-1 }{ sin(1)}sin(x)-1 ight) cos(t).]

Write (B= frac{ cos(1)-1 }{ sin(1)} ) for simplicity.

Then plug in (t=0) to get

[f(x)=-y_p(x,0)=- cos x+B sin x+1,]

and after differentiating in ( t) we see that (g(x)=- frac{partial y_P}{partial t}(x,0)=0).

Hence to find (y_c) we need to solve the problem

[ y_{yy}=y_{xx}, y(0,t)=0,~~~~y(1,t)=0, y(x,0)=- cos x+B sin x+1, y_t(x,0)=0.]

Note that the formula that we use to define (y(x,0)) is not odd, hence it is not a simple matter of plugging in to apply the D’Alembert formula directly! You must define (F) to be the odd, 2-periodic extension of (y(x,0)). Then our solution would look like

[ y(x,t)= frac{F(x+t)+F(x-t)}{2}+ left( cos(x) - frac{cos(1)-1}{sin(1)}sin(x)-1 ight) cos(t).]

**Figure 5.4**: Plot of (y(x,t)= frac{F(x+t)+F(x-t)}{2}+ left( cos(x) - frac{cos(1)-1}{sin(1)}sin(x)-1
ight) cos(t).).

It is not hard to compute specific values for an odd extension of a function and hence (5.3.17) is a wonderful solution to the problem. For example it is very easy to have a computer do it, unlike a series solution. A plot is given in Figure 5.4

## 5.3.2 Underground temperature oscillations

Let (u(x,t)) be the temperature at a certain location at depth (x) underground at time (t). See Figure 5.5.

The temperature (u) satisfies the heat equation (u_t=ku_{xx}), where (k) is the diffusivity of the soil. We know the temperature at the surface (u(0,t)) from weather records. Let us assume for simplicity that

**Figure 5.5**: Underground temperature.

[ u(0,t)=T_0+A_0 cos(omega t),]

where (T_0) is the yearly mean temperature, and (t=0) is midsummer (you can put negative sign above to make it midwinter if you wish). (A_0) gives the typical variation for the year. That is, the hottest temperature is (T_0+A_0) and the coldest is (T_0-A_0). For simplicity, we will assume that (T_0=0). The frequency (omega) is picked depending on the units of (t), such that when (t=1), then (omega t=2pi). For example if (t) is in years, then (omega=2pi).

It seems reasonable that the temperature at depth (x) will also oscillate with the same frequency. This, in fact, will be the steady periodic solution, independent of the initial conditions. So we are looking for a solution of the form

[ u(x,t)=V(x)cos(omega t)+ W(x)sin(omega t).]

for the problem

[ u_t=ku_{xx,}~~~~~~u(0,t)=A_0cos(omega t).]

We will employ the complex exponential here to make calculations simpler. Suppose we have a complex valued function

[h(x,t)=X(x)e^{i omega t}.]

We will look for an (h) such that ({ m Re} h=u). To find an (h), whose real part satisfies (5.3.20), we look for an (h) such that

[ h_t=kh_{xx,}~~~~~~h(0,t)=A_0 e^{i omega t}.]

Exercise (PageIndex{3}):

Suppose (h) satisfies (5.3.22). Use Euler’s formula for the complex exponential to check that (u={ m Re} h)satisfies (5.3.20).

Substitute (h) into (5.3.22).

[ i omega Xe^{i omega t}=kX''e^{i omega t}.]

Hence,

[ kX''-i omega X=0,]

or

[ X''- alpha^2 X=0,]

where ( alpha = pm sqrt{frac{i omega }{k}}). Note that (pm sqrt{i}= pm frac{1=i}{sqrt{2}}) so you could simplify to ( alpha= pm (1+i) sqrt{frac{omega}{2k}}). Hence the general solution is

[ X(x)=Ae^{-(1+i)sqrt{frac{omega}{2k}x}}+Be^{(1+i)sqrt{frac{omega}{2k}x}}.]

We assume that an (X(x)) that solves the problem must be bounded as (x ightarrow infty) since (u(x,t)) should be bounded (we are not worrying about the earth core!). If you use Euler’s formula to expand the complex exponentials, you will note that the second term will be unbounded (if (B eq 0)), while the first term is always bounded. Hence (B=0).

Exercise (PageIndex{4}):

Use Euler’s formula to show that (e^{(1+i)sqrt{frac{omega}{2k}x}}) is unbounded as (x ightarrow infty), while (e^{-(1+i)sqrt{frac{omega}{2k}x}}) is bounded as (x ightarrow infty).

Furthermore, (X(0)=A_0) since (h(0,t)=A_0e^{i omega t}). Thus (A=A_0). This means that

[ h(x,t)=A_0e^{-(1+i)sqrt{frac{omega}{2k}x}}e^{i omega t}=A_0e^{-(1+i)sqrt{frac{omega}{2k}}x+i omega t}=A_0e^{- sqrt{frac{omega}{2k}}x}e^{i( omega t- sqrt{frac{omega}{2k}}x)}.]

We will need to get the real part of (h), so we apply Euler’s formula to get

[ h(x,t)=A_0e^{- sqrt{frac{omega}{2k}}x} left( cos left( omega t - sqrt{frac{omega}{2k}x} ight) +i sin left( omega t - sqrt{frac{omega}{2k}x} ight) ight). ]

Then finally

[u(x,t)={ m Re}h(x,t)=A_0e^{- sqrt{frac{omega}{2k}}x} cos left( omega t- sqrt{frac{omega}{2k}}x ight).]

Yay!

Notice the phase is different at different depths. At depth the phase is delayed by (x sqrt{frac{omega}{2k}}). For example in cgs units (centimeters-grams-seconds) we have (k=0.005) (typical value for soil), . Then if we compute where the phase shift (x sqrt{frac{omega}{2k}}=pi) we find the depth in centimeters where the seasons are reversed. That is, we get the depth at which summer is the coldest and winter is the warmest. We get approximately (700) centimeters, which is approximately (23) feet below ground.

Be careful not to jump to conclusions. The temperature swings decay rapidly as you dig deeper. The amplitude of the temperature swings is (A_0e^{- sqrt{frac{omega}{2k}}x}). This function decays very quickly as (x) (the depth) grows. Let us again take typical parameters as above. We will also assume that our surface temperature swing is (pm 15^{circ}) Celsius, that is, (A_0=15). Then the maximum temperature variation at (700) centimeters is only (pm 0.66^{circ}) Celsius.

You need not dig very deep to get an effective “refrigerator,” with nearly constant temperature. That is why wines are kept in a cellar; you need consistent temperature. The temperature differential could also be used for energy. A home could be heated or cooled by taking advantage of the above fact. Even without the earth core you could heat a home in the winter and cool it in the summer. The earth core makes the temperature higher the deeper you dig, although you need to dig somewhat deep to feel a difference. We did not take that into account above.

^{2}Mythbusters, episode 31, Discovery Channel, originally aired may 18th 2005.

## Stability of periodic steady-state solutions to a non-isentropic Euler–Maxwell system

This paper is concerned with a stability problem in a periodic domain for a non-isentropic Euler–Maxwell system without temperature diffusion term. This system is used to describe the dynamics of electrons in magnetized plasmas when the ion density is a given smooth function which can be large. When the initial data are close to the steady states of the system, we show the global existence of smooth solutions which converge toward the steady states as the time tends to infinity. We make a change of unknown variables and choose a non-diagonal symmetrizer of the full Euler equations to get the dissipation estimates. We also adopt an induction argument on the order of derivatives of solutions in energy estimates to get the stability result.

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## 1 Answer

[ Note : you should try to avoit fixing the arguments a=of a symbolic function. ]

The expressions of the shape $acos(u)+ bsin(u)$ can indeed be expressed as $ccos(u+d)$, but that is irrelevant to your question. More relevant is to consider your function as a polynomial of the dampening function $e^<-2x>$

This decomposition allows to decompose your function as the sum of a "steady" and a "dampened" functions :

The plot is therefore trivial :

**EDIT :** If you are interested in the trigonometric expressions manipulation, here's how to transform your expressions :

This relation must be true for any value of $sin x$ and $cos x$ it is therefore implicitly a system of (polynomial equations in $sin x$ and $cos x$, which we can get explicitly :

Neither Maxima nor Giac or Fricas can directly solve this system, which we will solve manually. Get a first expression of $c$ :

which we use to get (a function of) $d$ : sage: St=(((Sys[0].subs(Sc)==0)-b)/a).trig_reduce() St tan(d) == -b/a

Note that we have to manually introduce $ an d$. We can now get our candidate solutions :

Since we have used a solution of $c^2$ to get a solution for $c$, we may have introduced a spurious solution. Indeed, the first solution doesn't check :

but the second one does : sage: bool(Eq.subs(Sol[1]).simplify_full().canonicalize_radical()) True

also needs a manual explicitation of the implicit system,

expresses its results in an awkward form, not easily simplifiable :

Sympy can solve the same system, but its solution is . disputable. :

The solutions given by Sympy are therefore identical to those manually obtained the same limitations apply. However, their properties are way less obvious, and their transformation to "easier" forms quite unintuitive.

## NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3

NCERT Solutions for class 10 Maths Chapter 5 Exercise 5.3 (Samantar Shreni) AP PDF files in Hindi Medium and English Medium as well as Offline Apps are updated for new academic session 2021-22 free to download.

NCERT Solutions are in Video Format also for CBSE Board, MP Board, UP Board and the students following NCERT Books 2021-22 for the current session 2021-2022. Download CBSE Apps and updated NCERT Solutions in PDF format.

## NCERT Solutions for class 10 Maths Chapter 5 Exercise 5.3

### Class 10 Maths Exercise 5.3 Solutions

### 10 Maths Chapter 5 Exercise 5.3 Solutions

NCERT Solutions for class 10 Maths Chapter 5 Exercise 5.3 Arithmetic Progressions – AP in English Medium as well as Hindi Medium to use Online or View in Video Format free for new academic session 2021-22. Move Class 10 Maths Chapter 5 main page for other exercises whether download or online study online.

#### Class 10 Maths Chapter 5 Exercise 5.3 Solution in Hindi Medium Video

#### Class 10 Maths Chapter 5 Exercise 5.3 Solution in Videos

Class 10 Maths Exercise 5.3 Explanation Video Class 10 Maths Chapter 5 Exercise 5.3 Solution#### Important Questions with Answer for Practice in AP

1. The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P. [Answer: 3, 5, 7, 9, …]

2. If Sn, the sum of first n terms of an A.P. is given by Sn = 5n² + 3n, then find its nth term and common difference. [Answer: an = 10n – 2, d = 10]

3. The sum of third and seventh terms of an A.P. is 6 and their product is 8. Find the sum of first 16th terms of the A.P. [Answer: 76, 20]

4. If the mth term of an A.P. is 1/n and the nth term is 1/m, show the sum of its first (mn) terms is 1/2(mn + 1).

5. If in an A.P. the sum of first m terms is equal to n and the sum of first n terms is m, prove that the sum of first (m + n) terms is – (m + n).

##### Questions from Board Papers

1. Determine the A.P. whose 4th term is 18 and the difference of 9th term from the 15th term is 30. [Answer: 3, 8, 13, …]

2. If the sum of first k terms of an A.P. is 1/2(3k² + 7k), write its kth term. Hence find its 20th term. [Answer: a20 = 62, ak = 3k + 2]

3. The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1:2. Find the first and fifteenth terms of the A.P. [Answer: 6, 48]

4. If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term. [Answer: 2n + 1]

5. The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 term is 161. Find the 28th term of this A.P. [Answer: 57]

Download NCERT Books and Offline Apps 2021-22 based on new CBSE Syllabus. Visit to discussion forum and ask your queries related to NIOS, UP Board or CBSE Board.

## Knots and links in steady solutions of the Euler equation

Given any possibly unbounded, locally finite link, we show that there exists a smooth diffeomorphism transforming this link into a set of stream (or vortex) lines of a vector field that solves the steady incompressible Euler equation in $mathbb

[EG00] J. B. Etnyre and R. W. Ghrist, "Contact topology and hydrodynamics. III. Knotted orbits," *Trans. Amer. Math. Soc.*, vol. 352, iss. 12, pp. 5781-5794, 2000. [FH91] M. H. Freedman and Z. He, "Divergence-free fields: energy and asymptotic crossing number," *Ann. of Math.*, vol. 134, iss. 1, pp. 189-229, 1991. [Acta] G. Koch, N. Nadirashvili, G. A. Seregin, and V. vSverák, "Liouville theorems for the Navier-Stokes equations and applications," *Acta Math.*, vol. 203, iss. 1, pp. 83-105, 2009. [KP81] S. G. Krantz and H. R. Parks, "Distance to $C^*J. Differential Equations*, vol. 40, iss. 1, pp. 116-120, 1981. [La56] P. D. Lax, "A stability theorem for solutions of abstract differential equations, and its application to the study of the local behavior of solutions of elliptic equations," *Comm. Pure Appl. Math.*, vol. 9, pp. 747-766, 1956. [LS00] 3.0.CO2-A’ title=’Go to document’> P. Laurence and E. W. Stredulinsky, "Two-dimensional magnetohydrodynamic equilibria with prescribed topology," *Comm. Pure Appl. Math.*, vol. 53, iss. 9, pp. 1177-1200, 2000. [Mo85] H. K. Moffatt, "Magnetostatic equilibria and analogous Euler flows of arbitrarily complex topology. I. Fundamentals," *J. Fluid Mech.*, vol. 159, pp. 359-378, 1985.

## Comprehensions

The solver interface also gives tools for using comprehensions over the solution. Using the tuples(sol) function, we can get a tuple for the output at each timestep. This allows one to do the following:

One can use the extra components of the solution object as well as using zip . For example, say the solution type holds du , the derivative at each timestep. One can comprehend over the values using:

Note that the solution object acts as a vector in time, and so its length is the number of saved timepoints.

## 5.3: Steady periodic solutions - Mathematics

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##### Mathematical Expression Editor

We define and discuss the complex trigonometric functions.

### The Complex Cosine

The series of interest are:

and the sum identity for the cosine is: We get the ball rolling by allowing an imaginary term in the sum identity: Next, we define the sine and cosine of a purely imaginary angle using their respective power series: and These power series can be simplified into hyperbolic functions (!) by noting that for all : and

Substituting these into the sum identity establishes We use this as our definition of the complex cosine.

### The Complex Sine

Based on the results obtained in the method for defining the complex cosine, which of the following is the definition of the complex sine function ?

### Identities

#### Periodicity

Since the real sine and cosine functions are periodic, so are their complex extenstions.

#### Evenness and Oddness

Recall that and are odd functions and that and are even functions. As a result, is an odd function and is an even function.

The evenness of the complex cosine is demonstrated similarly (verify).

#### Shift by

#### Relation to the Complex Exponential

The result follows by dividing by . The proof of the second equation is similar (verify).

#### Co-functions

#### Sum and Difference Identities

The second equation follows from the first by replacing with and using evenness and oddness. The third and fourth equations are proved in the same manner as the first and second (verify).

### The Other Trig functions

The other four trigonometric functions are defined in terms of the sine and cosine.

## Numerical parameters

**Ntst**This is the number of mesh intervals for discretization of periodic orbits. If you are getting bad results or not converging, it helps to increase this. For following period doubling bifurcations, it is automatically doubled so you should reset it later.**Nmax**The maximum number of steps taken along any branch. If you max out, make this bigger.**Npr**Give complete info every Npr steps.**Ds**This is the initial step size for the bifurcation calculation.*The sign of Ds tells AUTO the direction to change the parameter.*Since stepsize is adaptive, (Ds) is just a ``suggestion.''**Dsmin**The minimum stepsize (positive).**Dsmax**The maximum step size. If this is too big, AUTO will sometimes miss important points.**Par Min**This is the left-hand limit of the diagram for the principle parameter. The calculation will stop if the parameter is less than this.**Par Max**This is the right-hand limit of the diagram for the principle parameter. The calculation will stop if the parameter is greater than this.**Norm Min**The lower bound for the*L_2*norm of the solution. If it is less than this the calculation will stop.**Norm Max**The upper bound for the*L_2*norm of the solution. If it is greater than this the calculation will stop.

## 5.3: Steady periodic solutions - Mathematics

Volatile viscous fluids on partially wetting solid substrates can exhibit interesting interfacial instabilities and pattern formation. We study the dynamics of vapor condensation and fluid evaporation governed by a one-sided model in a low-Reynolds-number lubrication approximation incorporating surface tension, intermolecular effects, and evaporative fluxes. Parameter ranges for evaporation-dominated and condensation-dominated regimes and a critical case are identified. Interfacial instabilities driven by the competition between the disjoining pressure and evaporative effects are studied via linear stability analysis. Transient pattern formation in nearly flat evolving films in the critical case is investigated. In the weak evaporation limit unstable modes of finite-amplitude nonuniform steady states lead to rich droplet dynamics, including flattening, symmetry breaking, and droplet merging. Numerical simulations show that long-time behaviors leading to evaporation or condensation are sensitive to transitions between filmwise and dropwise dynamics.

©2018 American Physical Society

#### Physics Subject Headings (PhySH)

#### Authors & Affiliations

- Department of Mathematics, University of California, Los Angeles, Los Angeles, California 90095, USA

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#### References (Subscription Required)

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## New Features of Anaconda 5.3

• **Compiled with Latest Python release:** Anaconda 5.3 is compiled with Python 3.7, taking advantage of Python’s speed and feature improvements.

• **Better Reliability:** The reliability of Anaconda has been improved in the latest release by capturing and storing the package metadata for installed packages.

• **Enhanced CPU Peformance:** The Intel Math Kernel Library 2019 for Deep Neural Networks(MKL 2019) has been introduced in Anaconda 5.3 distribution. Users deploying Tensorflow can make use of MKL 2019 for Deep Neural Networks. These Python binary packages are provided to achieve high CPU performance.

• **New packages are added:** There are over 230 packages which has been updated and added in the new release.

• **Work in Progress:** There is a casting bug in Numpy with Python 3.7 but the team is currently working on patching it until Numpy is updated.

You can download the Anaconda Distribution for your system using the following link: