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## Tuesday, March 23, 2010

### Solution: Happy Saint Math Trick's Day

Original problem: http://dharmath.blogspot.com/2010/03/happy-saint-math-tricks-day.html

This problem is inspired by this comic.

Prove that the trick in the comic always works. In other words, if $p$ is a prime number greater than 3, prove that $p^2 + 14 \equiv 3 \mod 12$

#### Solution

$p^2 + 14 \equiv 3 \mod 12$ means that $p^2+11$ is divisible by 12, which means that $p^2+11-12 = p^2-1$ is divisible by 12.

If $p$ is a prime number greater than 3, then $p$ is odd. This means that $p$ divided by 4 has remainder either 1 or 3. In both cases, $p^2$ divided by 4 has remainder 1, which means $p^2-1$ is divisible by 4.

If $p$ is a prime number greater than 3, then $p$ divided by 3 has remainder either 1 or 2. In both cases, $p^2$ divided by 3 has remainder 1, which again means $p^2-1$ is divisible by 3.

Since $p^2-1$ is divisible by both 3 and 4, it is divisible by 12.

### Happy Saint Math Trick's Day

This problem is inspired by this comic.

Prove that the trick in the comic always works. In other words, if $p$ is a prime number greater than 3, prove that $p^2 + 14 \equiv 3 \mod 12$

Solution: http://dharmath.blogspot.com/2010/03/solution-happy-saint-math-tricks-day.html

## Friday, March 19, 2010

### Second Solution: Osculating Circle, Ellipse, and Cone

Original Problem: http://dharmath.blogspot.com/2010/03/osculating-circle-ellipse-and-cone.html

An osculating circle of a point on a curve is defined as a circle that:
1. passes through that point
2. whose slope at that point is the same of the slope of the curve at that point
3. whose radius is the same as the radius of curvature of the curve at that point

In other words, it is a second-degree approximation circle of the curve at that point.

http://en.wikipedia.org/wiki/Osculating_circle

Given a cone whose half-angle is $\theta$, we take a cross section with a plane whose incident angle is $\theta$. That is, the plane is perpendicular to one of the cone rays. Naturally, the cross section forms an ellipse.

If O is the intersection of the main axis of the cone and the cross section, and A is the point on the ellipse's major axis that's closest to O, then prove that a circle with center O and radius OA is an osculating circle to the ellipse at A.

Hint: for people without any knowledge of calculus, the radius of osculating circle at A is $b^2/a$ where $b$ is half the length of minor axis and $a$ is half the length of major axis (standard ellipse notation).
The rest of the problem can be done without using calculus.

#### Second Solution

As given in the hint, the radius of the osculating circle is $b^2/a$. And clearly the circle in the problem passes through A and its tangent at A is perpendicular to the major axis, hence coincides with the ellipse's tangent. We are left to prove that $OA = b^2/a$. However, astute readers will note that $b^2/a$ is exactly the length of semi latus-rectum of the ellipse. So suppose $D$ is the focus that's closest to $A$, and $GG_1$ is the latus rectum passing through $D$, we will show that $DG = OA$.

Let $B$ be the point on the major axis that's farthest to $O$, and let $S$ be the vertex of the ellipse. Let $x = 2 \theta$ be the angle of the cone. We also note that $AO \perp AS$.

Let $\tau$ be the plane that passes through $GG_1$ (and hence through D) and perpendicular to the cone axis. Let $E$ be the point of intersection of the axis and this plane. The cross section of the cone with $\tau$ is a circle with center $E$. Let $F$ be the point on the circle such that $EF$ passes through $D$.

Since $SO$ is an angle bisector, we have $AO:OB = SA:SB = \cos x$, thus
$$AO = \frac{\cos x }{ \cos x+1} AB = \frac{\sin x \cos x }{ \cos x+1} SB$$

Now, $D$ is the point at which the smaller Dandelin Sphere touches the ellipse. If we consider the triangle $SAB$, then $D$ is where the incenter of that triangle touches $AB$. Since $SAB$ is a right angle at $A$, then $DA$ is the radius of that incenter.
$$DA = \frac{AS.AB}{AS+AB+SB} = \frac{\sin x \cos x}{1 + \sin x + \cos x} SB$$

Thus $AO : DA = (1+ \sin x + \cos x) : (1 + \cos x) = 1 + (\sin x) / (1 + \cos x)$
Which means $DO:AD:AO = \sin x : (1 + \cos x) : (1 + \sin x + \cos x)$

Now, looking back at the triangle $SAB$ and the plane that contains it,
$$DF = \frac{AD}{\cos \theta} = \frac{1 + \cos x}{(1 + \sin x + \cos x) \cos \theta} OA$$
and
$$DE = DO \cos \theta = \frac{\sin x \cos \theta}{1 + \sin x + \cos x} OA$$
So
$$DG^2 = EG^2 - DE^2 = EF^2 - DE^2 = (DE + DF)^2 - DE^2 = DF(DF + 2DE)$$
$$= \frac{OA^2}{(1 + \sin x + \cos x)^2} \frac{1 + \cos x}{\cos \theta} \left(\frac{1 + \cos x}{\cos \theta} + 2\sin x \cos \theta \right)$$
Since $x = 2\theta$, then $1 + \cos x = 2 \cos^2 \theta$
$$(1 + \sin x + \cos x)^2 = (2 \cos^2 \theta + 2 \sin \theta \cos \theta)^2 = 4 \cos^2 \theta (\sin \theta + \cos \theta)^2$$
and
$$\frac{1 + \cos x}{\cos \theta} + 2\sin x \cos \theta = 2 \cos \theta + 2 \sin x \cos \theta = 2 \cos \theta (1 + \sin x)$$
So
$$DG^2 = \frac{OA^2}{4 \cos^2 \theta (\sin \theta + \cos \theta)^2} \frac{2 \cos^2 \theta}{\cos \theta} 2 \cos \theta (1 + \sin x)$$
$$= OA^2 \frac{1 + \sin x}{(\sin \theta + \cos \theta)^2}$$

But
$$1 + \sin x = 1 + 2 \sin \theta \cos \theta = \cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta = (\sin \theta + \cos \theta)^2$$

So $DG^2 = OA^2$ which means $DG = OA$

### Which jug is poisoned?

You have 1000 jugs of wine, one of which has been poisoned. You have access to rats that would die within 23 hours of drinking that poison. You need to determine which jug is poisoned within 24 hours. How many rats do you need at minimum?

## Thursday, March 18, 2010

### Which stack has counterfeit coins?

Credit to this problem goes to MIT Technology Review Puzzle Corner, November/December 2009 Edition.

Given thirteen stacks each containing four coins, we are told that exactly one stack contains all counterfeit coins. A counterfeit coin weighs less than a good coin by an amount not exceeding 5 grams, and all good coins weigh an integral number of grams.

We are given a precision scale with a very wide area to put the coins on. We need to answer all these three questions all in two weighings:
1. What is the weight of a good coin?
2. What is the weight of a counterfeit coin?
3. Which stack has the counterfeit coins?

### Solution: Osculating Circle, Ellipse, and Cone

Original Problem: http://dharmath.blogspot.com/2010/03/osculating-circle-ellipse-and-cone.html

An osculating circle of a point on a curve is defined as a circle that:
1. passes through that point
2. whose slope at that point is the same of the slope of the curve at that point
3. whose radius is the same as the radius of curvature of the curve at that point

In other words, it is a second-degree approximation circle of the curve at that point.

http://en.wikipedia.org/wiki/Osculating_circle

Given a cone whose half-angle is $\theta$, we take a cross section with a plane whose incident angle is $\theta$. That is, the plane is perpendicular to one of the cone rays. Naturally, the cross section forms an ellipse.

If O is the intersection of the main axis of the cone and the cross section, and A is the point on the ellipse's major axis that's closest to O, then prove that a circle with center O and radius OA is an osculating circle to the ellipse at A.

Hint: for people without any knowledge of calculus, the radius of osculating circle at A is $b^2/a$ where $b$ is half the length of minor axis and $a$ is half the length of major axis (standard ellipse notation).
The rest of the problem can be done without using calculus.

#### Solution

As given in the hint, the radius of the osculating circle is $b^2/a$. And clearly the circle in the problem passes through A and its tangent at A is perpendicular to the major axis, hence coincides with the ellipse's tangent. We are left to prove that $OA = b^2/a$.

Let B be the point on the major axis that's farthest to O, and let C be the vertex of the ellipse. Let $x = 2 \theta$ be the angle of the cone. We also note that $AO \perp AC$.

Now the length of major axis $2a = AB = BC \sin x \iff \frac{a}{BC} = \frac{\sin x}{2}$.

Since CO is an angle bisector, then $OA/OB = CA/CB = \cos x$
$$\frac{OA}{AB-OA} = \cos x \iff OA = \frac{AB \cos x}{1 + \cos x} = \frac{2a \cos x}{1 + \cos x}$$

Now, let D be the point where the smaller Dandelin Sphere touches the cutting plane.
http://en.wikipedia.org/wiki/Dandelin_spheres

D is the focus that's closest to A. Therefore $DA = a-c$ where $c = \sqrt{a^2-b^2}$ (using the standard ellipse notation). But the center of the Dandelin sphere is also the incenter of the triangle ABC, so DA is the same as inradius of ABC. Using the inradius formula, and since $\angle BAC = \pi/2$

$$DA = \frac{AC.AB}{AC+AB+BC} = \frac{BC \cos x . BC \sin x}{BC \cos x + BC \sin x + BC} = BC \frac{\cos x \sin x}{1+\sin x + \cos x}$$

So
$$a-c = BC \frac{\cos x \sin x}{1+\sin x + \cos x}$$
$$c/BC = a/BC - \frac{\cos x \sin x}{1+\sin x + \cos x} = \frac{\sin x}{2} - \frac{\cos x \sin x}{1+\sin x + \cos x} = \frac{\sin x}{2} . \frac{1 + \sin x - \cos x}{1 + \sin x + \cos x}$$
Thus $$\frac{c}{a} = \frac{1 + \sin x - \cos x}{1 + \sin x + \cos x}$$

Because $a^2 = b^2+c^2$,
$$\left( \frac{b}{a} \right)^2 = 1 - \left( \frac{c}{a} \right)^2 = 1 - \frac{(1 + \sin x - \cos x)^2}{(1 + \sin x + \cos x)^2} = \frac{4(1+\sin x)(\cos x)}{(1 + \sin x + \cos x)^2}$$
But
$$(1 + \sin x + \cos x)^2 = (1 + \sin^2 x + \cos^2 x + 2 \sin x + 2 \cos x + 2 \sin x \cos x)$$
$$= 2(1+\sin x)(1+\cos x)$$
So
$$\frac{b^2}{a^2} = \frac{2 \cos x}{1 + \cos x} = \frac{OA}{a}$$

Which means $OA = b^2/a$

Alternative solution available here: Second Solution

## Friday, March 12, 2010

### Osculating Circle, Ellipse, and Cone

An osculating circle of a point on a curve is defined as a circle that:
1. passes through that point
2. whose slope at that point is the same of the slope of the curve at that point
3. whose radius is the same as the radius of curvature of the curve at that point

In other words, it is a second-degree approximation circle of the curve at that point.

http://en.wikipedia.org/wiki/Osculating_circle

Given a cone whose half-angle is $\theta$, we take a cross section with a plane whose incident angle is $\theta$. That is, the plane is perpendicular to one of the cone rays. Naturally, the cross section forms an ellipse.

If O is the intersection of the main axis of the cone and the cross section, and A is the point on the ellipse's major axis that's closest to O, then prove that a circle with center O and radius OA is an osculating circle to the ellipse at A.

Hint: for people without any knowledge of calculus, the radius of osculating circle at A is $b^2/a$ where $b$ is half the length of minor axis and $a$ is half the length of major axis (standard ellipse notation).
The rest of the problem can be done without using calculus.

Solution:
http://dharmath.blogspot.com/2010/03/osculating-circle-ellipse-and-cone.html

Second Solution:
http://dharmath.blogspot.com/2010/03/second-solution-osculating-circle.html

### Solution: Tennis Tournament

Original problem: http://dharmath.blogspot.com/2010/03/tennis-tournament.html

In a tennis tournament of 100 people, each player plays every other players exactly once. There is no draw in a tennis game, one side always wins.

Given that no player loses all of his games, prove that there is a cycle of exactly 3 players. That is, there are players A,B, and C such that A defeats B, B defeats C, and C defeats A.

First Solution

First we prove that there exists a cycle of any length in the tournament. Indeed, take player A, "mark" him, and take any player that A defeats, say B, and also mark that player.. And take any player that B defeats, say C, mark him, and so on. At each turn, we are guaranteed to find a player that loses to that player. If we ever find a player that we have marked before, we obtain a cycle. But the number of marked players increases steadily while there is only a finite number of players in the tournament. Sooner or later, we will run out of players and we are guaranteed to return to a previously marked player.

If this cycle that we found has length 3, then we are done. Now we establish the following statement via induction:

A cycle of length at least 3 contains a cycle of length 3.

The base case is trivial. Now suppose we have a cycle of length $k: A_1, A_2, \cdots, A_k$ where $A_i$ beats $A_{i+1}$ and $A_k$ beats $A_1$.

If $A_1$ beats $A_{k-1}$ then we have a cycle of length 3: $A_1$ beats $A_{k-1}$ beats $A_k$ beats $A_1$.

If $A_{k-1}$ beats $A_1$ then we have a cycle of length $k-1: A_1, \cdots, A_{k-1}$ which also contains a cycle of length 3 by induction hypothesis.

## Tuesday, March 9, 2010

### Tennis tournament

In a tennis tournament of 100 people, each player plays every other players exactly once. There is no draw in a tennis game, one side always wins.

Given that no player loses all of his games, prove that there is a cycle of exactly 3 players. That is, there are players A,B, and C such that A defeats B, B defeats C, and C defeats A.

### Migrated here from wordpress

Due to some mix-ups with the hosting and domain company, and since I'm not really willing to pay for it anymore, I have decided to move my math problem repository here to blogspot. This past mix-up also explains my brief hiatus for about a month and a half. Thus, when reading any post older than March 10, 2010, please keep in mind that those posts were written for wordpress and were migrated automatically, thus you might lose some layout and formatting. I'll do my best to clean them up but I might miss some.

From now on, problems and solutions will still be two separate posts. This is mainly because blogspot does not allow partial hiding or spoiler tags. Thus to prevent solutions from spoiling the problem posts, I will continue to post them separately.


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# Tag Info

One example you might want to start with is the Poincaré homology sphere. It is a Seifert fibred space over $S^2$ with three singular fibres $M(\frac{1}{2}, - \frac{1}{3}, -\frac{1}{5})$ supporting exactly one Stein fillable contact structure. Note that the the Poincaré homology sphere with reversed orientation $\overline{\Sigma}(2,3,5)$ was the first ...

format: plain


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# Question #e788e

Jul 6, 2017

A salt solution is not a pure substance because it can be broken down into salt and water by a physical change.

#### Explanation:

Pure substances can only be separated by a chemical change, and elements not even by a chemical change.

Mixtures and solutions are not pure substances. Mixtures and solutions can be separated by physical changes.

Jul 6, 2017

So common salt itself is a pure substance (it contains only sodium chloride, NaCl) and pure water (as opposed to sea water, or water with additives in it) is a pure substance, it contains only ${H}_{2} O$ molecules.
But common salt solution is a mixture of water molecules and solvated $N {a}^{+}$ and $C {l}^{-}$ ions, more than one distinct chemical species, and therefore not a pure substance.


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diffbind binding affinity matrix
1
0
Entering edit mode
@akankshabafna-23275
Last seen 6 weeks ago
United Kingdom

I am using Diffbind to compare H3K27ac peaks between two time points. Recently, we added 4 more time points to the existing 2 to do time-series comparison. From my understanding, Diffbind could conventionally compare and provide differential bound sites between 2 samples .

Q1: Is there a way to compare and contrast 6 time-points, considering each has 2 biological replicates. Q2 If the answer to Q1 is no, when I compare A to B and separately A to C , why do I get different counts for A in the binding affinity matrix arising from A vs B to A vs C for the same peak interval.

diffbind DiffBind • 132 views
0
Entering edit mode
Rory Stark ★ 4.2k
@rory-stark-5741
Last seen 1 day ago
CRUK, Cambridge, UK

DiffBind supports the use of any GLM design formula accepted by the underlying analysis packages DESeq2 and edgeR. I try to avoid getting involved on telling people how to set up their GLM design.

Regarding the different numbers of counts, that depends on how you are setting things up, Are you talking about the values you see when you call dba.peakset() with bRetrieve=TRUE? In that case there is only one matrix to see. Are you talking about normalized read counts returned by dba.report() when bCounts=TRUE? In that case it depends on how the contrasts are set up. For example, if design=FALSE, the normalization will be different for each contrast.

If you can show some more of the script that you are using where you are seeing different counts for the same sample, I may be able to see what is happening in your case.


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# What does $x_{1:n}$ notation mean? [closed]

Notation in my book, Introduction to Probability and Mathematical Statistics by Bain and Engelhardt, is $x_{1:n}$ and $x_{n:n}$. For example, see Example 10.2.3 (page 340). Here is a screenshot.

Can anybody tell me what this denotes?

## closed as unclear what you're asking by Shailesh, hardmath, coldnumber, Claude Leibovici, JMPFeb 5 '16 at 7:42

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• You might need to include more detail. Usually, notations don't have a unique meaning. – Em. Feb 5 '16 at 3:00
• Most likely it stands for $x_1, x_2, \ldots , x_n$ – Shailesh Feb 5 '16 at 3:02
• Notation in which book? – David G. Stork Feb 5 '16 at 3:04
• Sorry if I wasn't very specific. I believe @Shailesh is correct. Thanks – Bbb Feb 5 '16 at 3:05
• Tell us what book? Help people help you. – Thomas Andrews Feb 5 '16 at 3:07

In my experience this might denote a sequence of numbers $x_{1}$, $x_{2}$, ..., $x_{n}$


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# Updating Ghost

I will admit, I have not been very good about keeping Ghost up to date. Back in October I received an email about the big October update, I believe this was 0.5.3. I just updated to the now current version (0.5.6). I have to look up the process each time I update, so I figure I’ll write about it, that way I’ll have a solid footing when I update to 0.5.7, perhaps motivating me to keep things up to date.

# Finding Support

I love Ghost as a blogging platform. I can run my own installation of it and writing posts in it is easy. (Not to mention it looks great.) I, however, find looking for information on their site a less intuitive experience. There is something strange to me about the layout, for instance, I know they have a forum, but there is no link on the main page for it, instead it is under support.

Anyway, my usual approach is to just turn to Google, but Google often turns up results from http://www.howtoinstallghost.com, which I’m not sure is actually affiliated with the Ghost team. While I like their goal of providing easy to use instructions for those looking to just use the platform without any of the technical details, it lacks some of potentially crucial info, such as this note about backing up your blog’s data before performing an upgrade on the actual Ghost support page:

Note: You can, if you like, take a copy of your database from content/data but be warned you should not do this whilst Ghost is running. Please stop it first.

I actually ended up doing this while Ghost was still running, thinking it was a great idea, before I found that page. My data survived, but next time I might not be so lucky.

That being said, the first thing we should is back up our precious data.

# Backups

As noted here, to backup all we have to do is export a .json file from the administrative interface of Ghost. This previously was found under /ghost/debug/, but now is located under /ghost/settings/labs/. Simply click export and you are all set!

Something I’d like to see in a future release (heck it might exist now), is a way to automate this. I’d like to just have a cronjob setup to backup my blog every week (day, hour, minute, whatever, depending on how often you write new content - not often if you’re me - and how paranoid you are).

The last thing to backup is themes. Casper is the default theme, and lives in your ghost directory under /content/themes/. They’ve been updating Casper fairly frequently. And they suggest just getting rid of the old versions and installing the latest. I’ve modified the first version I used though, and the design influenced other parts of my website, so I like to keep old versions. If you simply move /content/themes/casper/ to something like /content/themes/casper_0.5.2/ then you’ll see the previous versions when you go to change your themes in the admin panel.

Now that we’ve backed up our data let’s get to the actual upgrade, shall we?

• We’ll need to download the latest release, which I find most convenient to get from their github.
• I usually make a tmp directory in /var/www/ghost/, where I have ghost installed, and extract this .zip file there.

• Next, in /var/www/ghost/ we’ll delete index.js, and package.json, then copy the new ones in.

• We’ll do the same with core/.
• Then we’ll copy over the new Casper version.

You’ll want to replace the 0.0.0 with a version number most likely. You can also just delete the old version if you’d like.

• Finally, we need to run sudo npm install --production. This will take a few minutes, go grab some coffee.

# Testing and Running with forever

When it’s done I like to test things out with a sudo npm start --production. This will start Ghost. To keep ghost running I had initially used forever. If you don’t have forever installed, run npm install -g forever within your ghost directory. I like to then start Ghost with forever with the following options.

The NODE_ENV definition should be changed to suite your needs, but I’m running the production environment. If you too are in an environment that isn’t the development one, and you forget this bit, you’ll probably panic when you see your blog come up empty. It startled me a bit until I realized my error. Finally, the -l /var/www/ghost/forever.log logs the output, which is always useful.

# Wrapping Up

We’ve successfully upgraded Ghost! If everything went smoothly you’ll be on the latest version. So quit reading this and go write a new post! Before you do though, let me just remind you to clean up a bit, mostly just get rid of that ./tmp/ directory.

Happy ghosting!

Updated:


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# How To Find Variance

Statistics normally means taking the values of some quantities and by plotting it or by arranging it in a meaningful manner. It gives us the process by which we can collect, analyze, interpret, present and organize data. Variance is an important topic in statistics. Variance is a measure of central dispersion. It is the average of the squared difference from the mean. Mean is the average of a given set of numbers. In this article, we will discuss the steps to find variance.

## What is Variance?

The average of the squared differences from the mean is known as variance.

Variance is denoted by σ2.

## How to calculate variance step by step?

You can use the following steps to calculate variance.
Step 1. Calculate the mean of the given numbers.

Step 2. Then for each number, subtract the mean and find the square of the difference.

Step 3. Calculate the average of those squared differences.

## Formula

x1, x2, …., xare the n observations.

### Solved Examples

Example 1: The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two are

(A) 2 and 9

(B) 3 and 8

(C) 4 and 7

(D) 5 and 6

Solution:

Given no. of observations = 5

Mean = 4

Variance = 5.2

Let unknown values be a and b.

[(1-4)2 + (2-4)2 + (6-4)2 + (a-4)2 + (b-4)2 ]/5= 5.2

(-3)2 + 22 + 22 + (a-4)2 + (b-4)2 = 26

9 + 4 + 4 + (a-4)2 + (b-4)2 = 26

(a-4)2 + (b-4)2 = 26 – 17

(a-4)2 + (b-4)2 = 9

Substitute the given options in the above equation.

We find that 4 and 7 satisfy the above equation.

Hence option C is the answer.

Example 2:

The variance of the first 50 even natural numbers is:

(A) 833/4

(B) 833

(C) 437

(D) 437/4

Solution:

The first 50 even natural numbers are 2, 4, 6,…100.

Variance is given by

= [(22+42+…1002)/50 ] – [(2+4+6+…100)/50]2

We will split this and solve it.

(22+42+…1002)/50 = 22(12 + 22+32+…502)/50

Use the formula n(n+1)(2n+1)/6 to find the sum of squares of natural numbers.

22(12 + 22+32+…502)/50 = (22/50)(50×51×101)/6

= 3434

[(2+4+6+…100)/50]2 = [2(1+2+3+..+50)/50]2 {use the formula (n/2)(2a+(n-1)d) to find sum of n natural numbers.}

= [(2/50)(50/2)(2+49)]2

= 512

= 2601

σ 2 = 3434-2601 = 833

Hence option (B) is the answer.

Example 3:

In an experiment with 15 observations on x, the following results were available

$\sum x^2 = 2830, \sum x = 170$

On observation that was 20, was found to be wrong and was replaced by the correct value 30. Then the corrected variance is

(A) 78.00

(B) 188.66

(C) 177.33

(D) 8.33

Solution:

Given Σx = 170

Σx2 = 2830

n = 15

20 was replaced by 30. So change in Σx = 10

New Σx = 170+10 = 180

New Σx2 = 2830+302-202

= 2830 + 900 – 400

= 2830 + 500

= 3330

New variance = (3330/15) – (180/15)2

= 222 -122

= 222 – 144

= 78

Hence option (A) is the answer.


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# Why do aluminum, magnesium, and calcium form white precipitate when mix with sodium hydroxide?

Why do aluminum, magnesium, and calcium form white precipitate when mix with sodium hydroxide? Shouldn't there be no reaction? Because sodium is more reactive than all of those element, it should be impossible for sodium to be displace by those three elements.

YouTube video - "NaOH + Al - sodium hydroxide and aluminum"

When adding aluminium, calcium or magnesium, you are most likely adding them as chlorides. They dissociate almost fully, similarly to sodium hydroxide.

Then you have a solution with these four elements:

1. Sodium cations
2. Calcium (for example) cations
3. Hydroxide anions
4. Chloride anions

The precipitate that forms depends on the solubility of the different combinations of anions and cations. Sodium chloride and hydroxide are soluble so they remain in solution. Metal chlorides are also soluble. However calcium, magnesium, and aluminium hydroxides are insoluble so they form the precipitate.

You’re basically neutralising the basic solution, formed a solution of sodium chloride and a “basic” solid precipitate.

• Thanks for the answer! And why does the precipitation disappear when we add an excess amount of Sodium Hydroxide to solution containing Aluminum as the cation. – user28603 May 9 '18 at 11:20


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# Winesap's Blog

## Plaid CTF 2015 PlaidDB Writeup

My exploit source code: https://gist.github.com/33d6e03d0c9a8c13f3ba

This challange is a ELF64 binary pwnable, which the libc.so library is also provided. The service provides some key-value operations, GET, PUT, DUMP and DEL, by a red-black tree (we spent lots of time on checking the implementation of the tree but nothing there). First we found an input sequence that may cause heap corruption:

The bug is in the function at 0x1040. The length of the buffer for the row key is not enough for the trailing '\0'. The last row key in the input above, which is exactly 24 bytes, causes an off-by-one null byte overwriting into the size entry of the next chunk in heap memory. This makes the next chunk shrink and clears the inuse bit so that the current chunk seems already freed. Since the row key buffer will not be freed when the row is not found in DEL operation, we use DEL to trigger this bug to avoid unwanted memory corruption and also overwrite the prevsize entry of the next chunk. After then, free the next chunk which the prevsize is overwrited and its forward chunk is also unlinked. The result is to put a large chunk which crosses over other chunks into the unsorted bin and we can get it back by malloc a data buffer with matching size later.

Since the data buffer is overlapping with other node structures, we can use GET to leak information on these structures or use PUT to overwrite them. Note that the PUT operation creates a new data buffer. So we have to call PUT twice to swap the current buffer back and fill it up. By overwriting the data pointer in the node struture, we are able to dump memory at arbitrary address by apply GET to the node (see the leak() function in my exploit). First we can leak heap base address by reading the pointer in node strutures.

There is a trick to defeat ASLR for the libc base address. In the libc implementation, malloc() call mmap instead for large size (0x21000 is enough). In general, these pages will be placed at the address just before the .tls section. There are some useful information on .tls, such as the address of main_arena, the canary value of stackguard and a pointer which points to somewhere on stack with fixed offset.

However, the data buffer pointer can not be used as arbitrary memory writing, since the PUT operation always create a new buffer. Thus we have to make malloc() return the address which we want to overwrite and fill it by node data. There are two approaches: free a fake chunk with the target address or corrupt the fastbin. Since free() check both head and foot of the chunk, corrupt fastbin is easier and the only sanity checking is the chunk size in malloc(). We can free a chunk in fastbin range and then overwrite its fd pointer with the target address. Then after malloc() called twice, the fake chunk at the target address should be return, if the chunk size has been set correctly. (See the fastbin path in libc source for more details, which start from line 3331 to 3358)

We decided to overwrite the stack start from a return address entry with ROP chain. Thus we have to setup a valid chunk size on stack somewhere, which is before but close enough to a return address so that a tiny fastbin chunk can successfully cover it. We found that there is a buffer on stack in the PUT function at 0x1240. The buffer stores the data size and the rest space (total 16 bytes) is large enough for a 8 bytes size value. Also the address (start from [rbp-0x38]) is close to the return address entry. The valid length of ROP chain is only 16 bytes, which is not enough for a system('/bin/sh') call, but its enough for jump to a leave; ret gadget and migrate stack to second ROP chain which placed on heap in advance. Note that the stackgaurd canary should be overwrite with correct value.

Cat the flag by shell: flag{one_null_byte_t0_rul3_them_all_4ecd68f0}


text:
# Diode in parallel with two DC sources

Determine the voltage across the diode in the following figure, using the complete diode model with r'd = 10 Ohms and r'R = 100 Ohms.

Since the lower point is grounded, voltage at upper node is -10 V. Hence diode is forward biased. Then in complete diode model we can consider the diode as a closed switch with a DC source and resister inside.

Is this explanation is right and if so how may I proceed from here?

• I suggest you combine the two voltage sources and two resistors into a single voltage source and a single resistor (Thévenin equivalent) and proceed from there. – Spehro Pefhany Jul 15 '14 at 4:22

As Spehro said, I would prefer combining the 2 sources and resistors into a single one. This can be done by noting that the two sources + resistances are in parallel branches.

The formula for equivalent Voltage in parallel branches is

V_eq = ( V1/R1 + V2/R2 + .....Vn/Rn) / ( 1/R1 + 1/R2 + .......1/Rn)

Be careful about the polarities of the batteries though. In your case the two batteries will have opposite polarities and so V2 will have negative sign.

The equivalent resistance is then given by

1/R_eq = (1/R1 + 1/R2 + .....1/Rn)

Using the above formulas, your circuit would be a single voltage + resistance branch and a diode.

Alternatively ,if I am correct, in your question r'd denotes the forward resistance of the diode and r'R denotes the reverse bias resistance. Then you can also use Superposition theorem taking one source at a time. In that case, when considering the 10V source, you would replace the diode with reverse bias resistance ( since it is reverse biased) and when considering the 20V source, you would replace the diode with a forward bias resistance ( since it is forward biased).

I don't know what the complete diode model you refer to is, but for an ideal diode:

Consider the circuit without the diode. What is the voltage between the resistors? 15V falls across each resistor, so it will be 10-15 = -5 volts. So when we add the diode back into the circuit it will be forward biased (or "on") and the cathode will move to just below zero volts: about -0.6V or -0.7V in practise.

Of course you probably still need to do it mathematically using your diode model, for an exact answer.

I wanted to comment to Spehro, (But I lack in reputation.)

As a first pass guess, assume Vdiode ~ zero.
Then there is 1 mA into the diode node from the 10 V supply and 2mA drawn from the 20V supply.
So there's something like 1mA going through the diode. In practice that would be enough. For your class you'll have to work out the "exact" answer. (for whatever diode model you are using.)


text:
Transport of power in random waveguides with turning points

# Transport of power in random waveguides with turning points

Liliana Borcea111Department of Mathematics, University of Michigan, Ann Arbor, MI 48109. borcea@umich.edu and derekmw@umich.edu    Josselin Garnier222Centre de Mathématiques Appliquées, Ecole Polytechnique, 91128 Palaiseau Cedex, France. josselin.garnier@polytechnique.edu    Derek Wood111Department of Mathematics, University of Michigan, Ann Arbor, MI 48109. borcea@umich.edu and derekmw@umich.edu
###### Abstract

We present a mathematical theory of time-harmonic wave propagation and reflection in a two-dimensional random acoustic waveguide with sound soft boundary and turning points. The boundary has small fluctuations on the scale of the wavelength, modeled as random. The waveguide supports multiple propagating modes. The number of these modes changes due to slow variations of the waveguide cross-section. The changes occur at turning points, where waves transition from propagating to evanescent or the other way around. We consider a regime where scattering at the random boundary has significant effect on the wave traveling from one turning point to another. This effect is described by the coupling of its components, the modes. We derive the mode coupling theory from first principles, and quantify the randomization of the wave and the transport and reflection of power in the waveguide. We show in particular that scattering at the random boundary may increase or decrease the net power transmitted through the waveguide depending on the source.

Key words. mode coupling, turning waves, scattering, random waveguide.

AMS subject classifications. 60F05, 35Q99, 78M35.

## 1 Introduction

Guided waves have been studied extensively due to their numerous applications in electromagnetics [13], optics and communications [27, 29], quantum waveguides [17], ocean acoustics [32], etc. The classic waveguides, with straight boundaries and filled with media that are either homogeneous or do not vary along the direction of propagation of the waves, are well understood. The wave equation in such waveguides is separable and the solution is a superposition of independent modes, which are propagating and evanescent modes and, in the case of penetrable boundary, radiation modes. When the geometry of the waveguide varies, or the medium filling it is heterogeneous, the modes are coupled. Examples of mathematical studies that account for mode coupling and lead to numerical methods that model such waveguides can be found in [11, 12, 15, 14, 23]. We are interested in mode coupling due to small random perturbations of the waveguide, which can be quantified more explicitly using asymptotic analysis.

Random models are useful for waveguides with rough boundaries and filled with composite media that vary at small scale, comparable to the wavelength. Such variations are typically small and unknown, so they introduce uncertainty in the model of wave propagation. The random models of the boundary and the wave speed are used to quantify how this uncertainty propagates to the uncertainty of the solution of the wave equation. This is useful information in applications like imaging [10, 7, 9, 2]. The mode coupling theory in waveguides filled with random media has been developed in [24, 16, 19, 21, 20] for sound waves and in [27, 4] for electromagnetic waves. The theory has also been extended to waveguides with random perturbations of straight boundaries [5, 7, 22].

In this paper we develop a mode coupling theory in random waveguides with turning points. We consider for simplicity a two-dimensional acoustic waveguide with sound soft boundary, but the theory can be extended to three dimensions and to electromagnetics. The waveguide has a slowly bending axis, a slowly changing opening and a randomly perturbed boundary. The slow variations occur on a long scale with respect to the wavelength, whereas the random perturbations are at a scale similar to the wavelength.

In the absence of the random perturbations, waves in slowly changing waveguides can be analyzed with a local decomposition in modes that are approximately independent [3, 31]. This is known as the adiabatic approximation [29, Section 19-2], and the result differs from that in waveguides with straight boundaries in one important aspect: The change in the opening of the waveguide causes the number of propagating modes to increase or decrease by at locations called turning points. Modes transition there from propagating to evanescent or the other way around, and due to energy conservation, the impinging propagating mode is turned back i.e., is reflected. Turning waves in slowly changing waveguides are studied mathematically in [6]. A recent study of their interaction with a random boundary is given in [8], for the case of weak random fluctuations that affect only the turning modes. Here we extend the results in [8] to stronger fluctuations, that couple all the waveguide modes.

Starting from the wave equation, and using the separation of scales of variation of the waveguide, we derive an asymptotic model for the wave field that accounts for coupling of all the modes, propagating and evanescent. This coupling is described by a stochastic system of differential equations for the random mode amplitudes, endowed with initial conditions that model the source excitation and radiation conditions. The excitation is due to a point source, but due to the linearity of the equation, other source excitations can be handled by superposition. We obtain an extension of the diffusion approximation theorem proved in [28] to carry out the asymptotic analysis of the mode amplitudes. The result simplifies when the random fluctuations are smooth, because the forward and backward going components of the propagating modes become independent. This is known as the forward scattering approximation, and applies to propagation between the turning points. At the turning points there is strong coupling of the components of the turning waves, described by random reflection coefficients. With the diffusion approximation theorem, and in the forward scattering approximation regime, we quantify the net effect of the random boundary on the transmitted and reflected power in the waveguide. This is the main result of the paper, and shows that the random boundary can be useful for increasing the transmitted power.

The paper is organized as follows: We begin in section LABEL:sect:formulation with the formulation of the problem, and the derivation of the asymptotic model for the wave field. Then we give in section LABEL:sect:wavedec the mode decomposition and derive a closed system of stochastic differential equations for the random amplitudes of the propagating modes between turning points. These equations are complemented by source excitation conditions, radiation conditions, as well as continuity and reflection conditions at the turning points. The asymptotic limit of the solution of these equations and the forward scattering approximation are in section LABEL:sect:results. We use these results to quantify the transmitted and reflected power in the waveguide in section LABEL:sect:netscat. The diffusion approximation theorem used to carry the asymptotic limit is stated and proved in section LABEL:app:adif. We end with a summary in section LABEL:sect:sum.

## 2 Formulation of the problem

In this section we give the mathematical model for time-harmonic waves in a random waveguide with variable cross-section and bending axis. We begin in section LABEL:sect:form1 with the setup, and describe the scaling in section LABEL:sect:form2 in terms of a small, dimensionless parameter . We use it in section LABEL:sect:form3 to write the wave problem in a form that can be analyzed in the asymptotic limit .

### 2.1 Setup

We consider a two-dimensional acoustic waveguide with sound-soft boundary. The waveguide occupies the semi-infinite domain , bounded above and below by two curves and , as shown in Figure LABEL:fig:setup. The top boundary is perturbed by small random fluctuations about the curve shown in the figure with the dotted line. The axis of the waveguide is at half the distance between and . It is a smooth curve parametrized by the arc length , that bends slowly, meaning that its tangent and curvature vary on a scale which is large with respect to the waveguide width . The width function has bounded first two derivatives, and to avoid complications in the analysis of scattering of the waves at the random boundary, we also assume that it is monotonically increasing.

Because of the changing geometry, it is convenient to use orthogonal curvilinear coordinates with axes along and , where is the unit vector orthogonal to , pointing toward the upper boundary. For any , written henceforth as , we have

 \itbfx=\itbfx∥(z)+r\itbfn(zL), \hb@xt@.01(2.1)

where is along the waveguide axis at arc length , satisfying

 ∂z\itbfx∥(z)=τ(zL), \hb@xt@.01(2.2)

and is the coordinate in the normal direction. The domain is the set

 Ω={(r,z): z∈R, r∈(r−(z),r+(z))}, \hb@xt@.01(2.3)

where

 r−(z)=−D(z/L)2, \hb@xt@.01(2.4)

is at the bottom boundary and

 r+(z)=D(z/L)2[1+1(−ZM,ZM)(z)σν(zℓ)], \hb@xt@.01(2.5)

is at the randomly perturbed top boundary . The perturbation is modeled by the random process and it extends over the interval , the support of the indicator function , where is a long scale needed to impose outgoing boundary conditions on the waves.***In practice may be chosen based on the duration of the observation time of the wave, using the hyperbolicity of the wave equation in the time domain. The single frequency wave analyzed in this paper is the Fourier transform of the time dependent wave field. We let the boundaries of the waveguide be straight and parallel for .

The random process is stationary with zero mean

 E[ν(ζ)]=0, \hb@xt@.01(2.6)

and auto-correlation function

 R(ζ)=E[ν(0)ν(ζ)]. \hb@xt@.01(2.7)

We assume that is mixing, with rapidly decaying mixing rate, as defined for example in [28, section 2], and it is bounded, with bounded first two derivatives, almost surely. This implies in particular that is integrable and has at least four bounded derivatives. We normalize by

 R(0)=1,∫∞−∞dζR(ζ)=O(1), \hb@xt@.01(2.8)

so that in (LABEL:eq:coord3) is the standard deviation of the fluctuations of , and quantifies their correlation length.

The waves are generated by a point source at , which emits a complex signal at frequency . We take the origin of at the source, so that . The waveguide is filled with a homogeneous medium with wave speed , and the wave field satisfies the Helmholtz equation

 Δp(ω,\itbfx)+k2p(ω,\itbfx)=f(ω)δ(\itbfx−\itbfx⋆),\itbfx=(r,z)∈Ω, \hb@xt@.01(2.9)

where is the wavenumber. In curvilinear coordinates this takes the form

 ⎡⎢ ⎢ ⎢⎣∂2r−1Lκ(zL)∂r1−rLκ(zL)+∂2z[1−rLκ(zL)]2+rL2κ′(zL)∂z[1−rLκ(zL)]3+k2⎤⎥ ⎥ ⎥⎦p(ω,r,z) =∣∣∣1−r⋆Lκ(0)∣∣∣−1f(ω)δ(z)δ(r−r⋆), \hb@xt@.01(2.10)

as shown in appendix LABEL:ap:CurvCoord, where is the derivative of the curvature . The sound soft boundary is modeled by the homogeneous Dirichlet boundary conditions

 p(ω,r+(z),z)=p(ω,r−(z),z)=0, \hb@xt@.01(2.11)

and at points with we have radiation conditions that state that is outgoing and bounded.

### 2.2 Scaling

There are four length scales in the problem: The wavelength , the width of the waveguide , the scale of the slow variations of the waveguide, and the correlation length of the random fluctuations of the boundary . They satisfy

 L≫D∼λ∼ℓ, \hb@xt@.01(2.12)

where denotes “of the same order as”, and we model the separation of scales using the dimensionless parameter

 ε=ℓL,0<ε≪1. \hb@xt@.01(2.13)

Our analysis of the wave field is in the asymptotic limit .

As shown in section LABEL:sect:wavedec, the ratio of and defines , the number of propagating components of the wave, called modes, where denotes the integer part. The assumption in (LABEL:eq:sc1) means that

 Nmin≤N(z)≤Nmax, \hb@xt@.01(2.14)

for all , where and are natural numbers, independent of .

The scales and are of the same order in (LABEL:eq:sc1) so that the waves interact efficiently with the random fluctuations of the boundary. This interaction, called cumulative scattering, randomizes the wave field as it propagates in the waveguide. The distance from the source at which the randomization occurs depends on the standard deviation of the fluctuations. We scale as

 σ=√ε~σ,~σ=O(1), \hb@xt@.01(2.15)

so that we observe the randomization at distances .

The scaled variables are defined as follows: The arc length is scaled by ,

 ~z=zL, \hb@xt@.01(2.16)

and the similar lengths , and are scaled by , to obtain

 ~D(~z)=D(z/L)ℓ,  ~r=rℓ,  ~k=kℓ. \hb@xt@.01(2.17)

We also introduce the scaled bound of the support of the random fluctuations, which is a large number, independent of .

### 2.3 Asymptotic model

Let us multiply equation (LABEL:eq:we3) by and use the scaling relations (LABEL:eq:sc4)-(LABEL:eq:sc5p). Dropping the tilde to simplify notation, because all variables are scaled henceforth, we obtain

 [∂2z+(1−εrκ(z))2ε2(∂2r+k2)−κ(z)(1−εrκ(z))ε∂r+εrκ′(z)(1−εrκ(z))∂z]p(ω,r,z) =f(ω)[1−εr⋆κ(0)]εδ(r−r⋆)δ(z), \hb@xt@.01(2.18)

with homogeneous Dirichlet boundary conditions (LABEL:eq:we2) at

 r−(z)=−D(z)2,  r+(z)=D(z)2[1+1(−ZM,ZM)(z)√εσν(zε)], \hb@xt@.01(2.19)

and appropriate radiation conditions for . These equations define the asymptotic model for the wave field, and we wish to analyze it in the limit .

The boundary has dependent fluctuations, so to ensure that the boundary conditions are satisfied at all orders of , we change variables to

 r=ρ+[2ρ+D(z)]4√εσν(zε), \hb@xt@.01(2.20)

for , and denote the transformed wave field by

 pε(ω,ρ,z)=p(ω,ρ+(2ρ+D(z))4√εσν(zε),z). \hb@xt@.01(2.21)

At there are no fluctuations so the transformation is the identity . We use the same notation for the wave field at all , and analyze it separately in the regions with the random fluctuations and without. The results are connected by continuity at .

The change of variables (LABEL:eq:AM3) makes the boundary conditions independent of ,

 pε(ω,±D(z)2,z)=0, \hb@xt@.01(2.22)

and maps the random fluctuations to the differential operator in the wave equation. Explicitly, we show in appendix LABEL:ap:DerAs that the wave equation becomes

 ∞∑j=0εj/2−1Lεjpε(ω,ρ,z)=ˆf(ω)[1+O(√ε)]δ(ρ−ρ⋆)δ(z), \hb@xt@.01(2.23)

for and , with the leading term in the expansion of the operator

 Lε0=(ε∂z)2+∂2ρ+k2. \hb@xt@.01(2.24)

This is the Helmholtz operator in a perfect waveguide, with straight and parallel boundaries. The random fluctuations appear in the first perturbation operator,

 \hb@xt@.01(2.25)

The second perturbation operator has a deterministic part, due to the curvature of the axis of the waveguide, and a random part, quadratic in the random fluctuations,

 Lε2= −κ(z)[2ρ(∂2ρ+k2)+∂ρ]+σ24{3ν2(zε)+[2ρ+D(z)]24ν′2(zε)}∂2ρ +[2ρ+D(z)]σ24{ν(zε)ν′(zε)ε∂2ρz+[ν′2(zε)+12ν′′(zε)ν(zε)]∂ρ}. \hb@xt@.01(2.26)

The remaining operators in the asymptotic series in (LABEL:eq:AM5) depend on higher powers of the fluctuations , but play no role in the limit .

By assumption, there are no variations of the waveguide at , so the operator in the left hand side of (LABEL:eq:AM5) reduces to in this region.

## 3 Mode decomposition and coupling

To analyze the solution of the wave equation (LABEL:eq:AM5) with boundary conditions (LABEL:eq:AM6) in the limit , we begin in section LABEL:sect:dec1 with the mode decomposition of the wave field. The modes are special solutions of the wave equation with operator (LABEL:eq:AM7). They represent propagating and evanescent waves which are coupled by the perturbation operators (LABEL:eq:AM8)-(LABEL:eq:AM9), as explained in section LABEL:sect:dec2. We are interested in the propagating modes, which are left and right going waves with random amplitudes satisfying a stochastic system of equations derived in section LABEL:sect:dec3. It is this system that we analyze in the asymptotic limit to quantify the cumulative scattering effects in the waveguide.

### 3.1 Mode decomposition

The second term in (LABEL:eq:AM7) is the Sturm-Liouville operator acting on functions that vanish at , for any given . Its eigenvalues are real and distinct

 λj(z)=k2−μ2j(z),μj(z)=πjD(z),  j=1,2,… \hb@xt@.01(3.1)

and the eigenfunctions

 yj(ρ,z)=√2D(z)sin[(2ρ+D(z))2μj(z)], \hb@xt@.01(3.2)

form an orthonormal basis in . We use this basis to decompose the solution of (LABEL:eq:AM7) in one dimensional waves called modes, for each ,

 pε(ω,ρ,z)=∞∑j=1pεj(ω,z)yj(ρ,z). \hb@xt@.01(3.3)

As shown in appendix LABEL:ap:ModeCoup, the modes can be written as

 pεj(ω,z)=uεj(ω,z)[1+O(√ε)], \hb@xt@.01(3.4)

with satisfying a coupled system of one dimensional wave equations that we now describe:

In the perturbed section of the waveguide, satisfies

 1ε[(ε∂z)2+k2−μ2j(z)]uεj(ω,z)+σ√εμ2j(z)ν(zε)uεj(ω,z)+σ2gεj(ω,z)uεj(ω,z) ≈Cεj(ω,z)+f(ω)yj(ρ⋆,0)δ(z), \hb@xt@.01(3.5)

where the approximation indicates that we dropped lower order terms that have no contribution in the limit . The coefficient in the left hand side is

 gεj(ω,z)=−34μ2j(z)ν2(zε)−[(πj)212+116]ν′2(zε), \hb@xt@.01(3.6)

and

 Cεj(ω,z) =∞∑q=1,q≠j[σΓjq√εν′′(zε)+σ2γjq(zε)+γojq(z)]uεq(ω,z) \hb@xt@.01(3.7)

models the coupling between the modes. The leading coupling coefficients and are constants, given in equation (LABEL:eq:DC6) in appendix LABEL:ap:ModeCoup. The second order coefficients and are quadratic in , as described in equations (LABEL:eq:DC7)-(LABEL:eq:DC8), and the coefficients and , given in equations (LABEL:eq:DC9)-(LABEL:eq:DC10), are due to the slow changes in the waveguide.

In the region , where the waveguide has straight and parallel boundaries, the wave equation simplifies to

 1ε[(ε∂z)2+k2−μ2j(z)]uεj(ω,z)=0. \hb@xt@.01(3.8)

Depending on the index , its solution is either an outgoing propagating wave or a decaying evanescent wave. This wave is connected to the solution of (LABEL:eq:DC3) by the continuity of and at .

### 3.2 Random mode amplitudes

Equations (LABEL:eq:DC3) are perturbations of the wave equation with operator , where the perturbation term models the coupling of the modes. This coupling is similar to that in waveguides with randomly perturbed parallel boundaries, studied in [5, 10], but the slow variation of the waveguide introduces two differences: The first is the presence of the extra terms and in (LABEL:eq:DC5), given by (LABEL:eq:DC9)-(LABEL:eq:DC10), which turn out to play no role in the limit . The second difference is important, as it gives a dependent number

 N(z)=⌊kD(z)π⌋ \hb@xt@.01(3.9)

of mode indexes for which . These modes are oscillatory functions in , and represent left and right going waves. For indexes the modes are decaying evanescent waves.

#### 3.2.1 Turning points

The function (LABEL:eq:DC12) that defines the number of propagating modes is piecewise constant. Starting from the origin, where we denote the number of propagating modes by , the function (LABEL:eq:DC12) increases by at arc lengths , for , and decreases by at , for . The jump locations , ordered as

 −ZM<…

are the zeroes of the eigenfunctions (LABEL:eq:eigval), and are called turning points [26, 6]. We assume henceforth that the monotonically increasing satisfies

 D′(z(t)±)>0,∀t≥1, \hb@xt@.01(3.10)

so that the turning points are simple and isolated. Consistent with our scaling, they are spaced at order one scaled distances.

Between any two consecutive turning points and , where we set by convention the number of propagating modes equals the constant

 N(t−1)±=N(0)±(t−1). \hb@xt@.01(3.11)

This number is bounded above and below as in (LABEL:eq:ASN), with and so the bounds and on the indexes are

 \hb@xt@.01(3.12)

Beginning from the source location , which we assume is not a turning point, is defined as the unique, negative arc-length satisfying

 k=πN(t−1)−D(z(t)−),t=1,…,t−M, \hb@xt@.01(3.13)

where the uniqueness is due to the monotonicity of . Similarly, the jump location is defined as the unique, positive arc length satisfying

 k=π(N(t−1)++1)D(z(t)+),t=1,…,t+M. \hb@xt@.01(3.14)

The analysis of the modes is similar on the left and right of the source, so we focus attention in this section on a sector of the waveguide, for some , and simplify the notation for the number (LABEL:eq:DC13pp) of propagating modes

 N=N(t−1)−. \hb@xt@.01(3.15)

These modes are a superposition of right and left going waves, with random amplitudes that model cumulative scattering in the waveguide, as we explain in the next section.

#### 3.2.2 The left and right going waves

We decompose the propagating modes in left and right going waves, using a flow of smooth and invertible matrices ,

 (aεj(ω,z)bεj(ω,z))=Mε,−1j(ω,z)(uεj(ω,z)vεj(ω,z)), \hb@xt@.01(3.16)

where denotes the inverse of and

 vεj(ω,z)=−iε∂zuεj(ω,z),j=1,…,N. \hb@xt@.01(3.17)

We obtain from (LABEL:eq:DC3) that

 ∂z(aεj(ω,z)bεj(ω,z)) =Mε,−1j(ω,z){iε(01k2(ω)−μ2j(z)0)Mεj(ω,z)−∂zMεj(ω,z) −iCεj(ω,z)Mε,−1j(ω,z)(01), \hb@xt@.01(3.18)

and the decomposition is achieved by a flow that removes to leading order the large deterministic term in (LABEL:eq:FM6), the first line in the right hand side.

The matrix must have the structure

 Mεj(ω,z)=⎛⎝Mεj,11(ω,z)−¯¯¯¯¯¯¯¯¯¯¯¯Mεj,11(ω,z)Mεj,21(ω,z)¯¯¯¯¯¯¯¯¯¯¯¯Mεj,21(ω,z)⎞⎠, \hb@xt@.01(3.19)

where the bar denotes complex conjugate, so that the decomposition (LABEL:eq:FM5) conserves energy. The expression of the components in (LABEL:eq:res6) depends on the mode index, more precisely on the mode wave number denoted by

 βj(ω,z)=√k2−μ2j(z). \hb@xt@.01(3.20)

Note that is bounded away from zero for all , and it approaches zero as , for . This last mode is a turning wave which transitions from a propagating wave at to an evanescent wave at , as described in section LABEL:sect:TP2. Here we give the decomposition of the modes indexed by .

The entries of (LABEL:eq:res6) are defined by

 Mεj,11(ω,z) =1√βj(ω,z)exp[iε∫z0dz′βj(ω,z′)], Mεj,21(ω,z) =βj(ω,z)Mεj,11(ω,z), \hb@xt@.01(3.21)

for . This definition looks the same as in perfect waveguides with straight and parallel boundary, except that the mode wave number varies with . We obtain from (LABEL:eq:res6)-(LABEL:eq:FM8) that the determinant of is constant

 \hb@xt@.01(3.22)

so the matrix is invertible, and the decomposition (LABEL:eq:FM5) can be rewritten as

 uεj(ω,z)=1√βj(ω,z)[aεj(ω,z)eiε∫z0dz′βj(ω,z′)−bεj(ω,z)e−iε∫z0dz′βj(ω,z′)], \hb@xt@.01(3.23)

and

 ε∂zuεj(ω,z)=i√βj(ω,z)[aεj(ω,z)eiε∫z0dz′βj(ω,z′)+bεj(ω,z)e−iε∫z0dz′βj(ω,z′)]. \hb@xt@.01(3.24)

Note that equations (LABEL:eq:FM10)-(LABEL:eq:FM11) are just the method of variation of parameters for the perturbed wave equation satisfied by the -th mode. They decompose the mode in a right going wave with amplitude and a left going wave with amplitude . In perfect waveguides these amplitudes would be constant, meaning physically that the waves are independent. In our case the amplitudes are random fields, satisfying the system of stochastic differential equations

 ∂z(aεj(ω,z)bεj(ω,z))=Hεj(ω,z)(aεj(ω,z)bεj(ω,z))−i2Cεj(ω,z)(¯¯¯¯¯¯¯¯¯¯¯¯Mεj,11(ω,z)Mεj,11(ω,z)), \hb@xt@.01(3.25)

obtained by substituting (LABEL:eq:res6) and (LABEL:eq:FM8) in (LABEL:eq:FM6). Here is the matrix valued random process

 \hb@xt@.01(3.26)

with entries satisfying the relations

 Hε(ba)j(ω,z)=¯¯¯¯¯¯¯¯¯¯¯¯¯¯Hε(ab)j(ω,z),Hε(bb)j(ω,z)=¯¯¯¯¯¯¯¯¯¯¯¯¯¯Hε(aa)j(ω,z), \hb@xt@.01(3.27)

and taking the values

 Hε(aa)j(ω,z)≈i2βj(ω,z)[σ√εμ2j(z)ν(zε)+σ2gεj(ω,z)], \hb@xt@.01(3.28)

and

 Hε(ab)j(ω,z)≈[¯¯¯¯¯¯¯¯¯¯¯¯¯¯Hε(aa)j(ω,z)−∂zβj(ω,z)2βj(ω,z)]exp[−2iε∫z0dz′βj(ω,z′)]. \hb@xt@.01(3.29)

As before, the approximation means up to negligible terms in the limit .

Equations (LABEL:eq:FM12) show that the amplitudes of the -th mode are coupled to each other by the process , and to the other modes by , defined by the series (LABEL:eq:DC5). The first terms in this series involve the propagating waves , for , decomposed as in (LABEL:eq:FM10)-(LABEL:eq:FM10). We describe in the next two sections the turning and the evanescent waves.

#### 3.2.3 The turning waves

The mode indexed by transitions at from propagating to evanescent. This transition is captured by the matrix , which has the same structure as in (LABEL:eq:res6), but its entries are defined in terms of Airy functions [1, chapter 10]. This is because near the simple turning point , equation (LABEL:eq:DC3) for is a perturbation of Airy’s equation. We refer to [6, 29] for classic studies of turning waves in waveguides, and to [8] for an analysis of their interaction with the random boundary. The setup in [8] is the same as here, with the exception that we consider a larger standard deviation of the random fluctuations, to observe mode coupling in the waveguide.

We use the same as in [8], with entries

 \hb@xt@.01(3.30)

and

 MεN,21(ω,z)=−iε∂zMεN,11(ω,z), \hb@xt@.01(3.31)

for where is a small, positive number, independent of . We go slightly beyond the turning point to capture the transition of the wave to an evanescent one. The phase in definition (LABEL:eq:TW1) is given by the function

 ϕN(ω,z)=∫zz(t)−dz′√|k2−μ2N(z′)|, \hb@xt@.01(3.32)

evaluated at the source location , and the amplitude factor

 QN(ω,z)=∣∣3ϕN(ω,z)/2∣∣1/6∣∣k2−μ2N(z)∣∣1/4, \hb@xt@.01(3.33)

is shown in [8, Section 3.1.1] to be bounded, and at least twice continuously differentiable. The Airy functions and are evaluated at

 \hb@xt@.01(3.34)

where is of order one in the vicinity of the turning point, and it is much larger than one in the rest of the domain .

We recall from [8, Lemma 3.1] that the matrix is invertible, with constant determinant

 detMεN(ω,z)=2,∀z∈(z(t)−−δ,z(t−1)−), \hb@xt@.01(3.35)

so the decomposition (LABEL:eq:FM5) is well defined. Moreover, [8, Lemma 3.2] shows that at the expressions (LABEL:eq:TW1)-(LABEL:eq:TW2) become like (LABEL:eq:FM8),

 MεN,11(ω,z) =1√βN(ω,z)exp[iε∫z0dz′βN(ω,z′)]+O(ε), MεN,21(ω,z) =βN(ω,z)MεN,11(ω,z)+O(ε), \hb@xt@.01(3.36)

so the turning wave behaves just like any other propagating wave until it reaches the vicinity of the turning point from the right. On the left side of the turning point, at , the entries of grow exponentially, as given in [8, Lemma 3.3]. The wave is evanescent in this region, and must be decaying in order to have energy conservation. This is ensured by the radiation condition

 \hb@xt@.01(3.37)

which sets to zero the coefficients of the growing Airy function and its derivative in the expression of and at the end of the domain. We refer to [8, Section 3.1] for more details, and for the proof that the result does not depend on the particular choice of which is small, but bounded away from in the limit .

The evolution equation of the turning mode amplitudes is of the same form as in (LABEL:eq:FM12), with the following entries of the matrix (LABEL:eq:res24)-(LABEL:eq:res24p) indexed by ,

 Hε(aa)N(ω,z)≈i∣∣MεN,11(ω,z)∣∣22[σ√εμ2j(z)ν(zε)+σ2gεj(ω,z)], \hb@xt@.01(3.38)

and

 Hε(ab)N(ω,z)≈−i[MεN,11(ω,z)]22[σ√εμ2j(z)ν(zε)+σ2gεj(ω,z)]. \hb@xt@.01(3.39)

These expressions reduce to those in (LABEL:eq:FM14)-(LABEL:eq:FM14p) at , with the extra term involving in (LABEL:eq:FM14p) coming from an correction of the amplitudes, induced by the residual in (LABEL:eq:FM8E).

#### 3.2.4 Coupling with the evanescent waves

The modes indexed by in equations (LABEL:eq:DC3) are evanescent waves, with wavenumber

 βj(ω,z)=√μ2j(z)−k2. \hb@xt@.01(3.40)

We analyze these waves in appendix LABEL:ap:evanesc, and show that they can be expressed in terms of the propagating ones. Explicitly, at arc length , satisfying , we obtain that

 uεj(ω,z)≈ −σ√ε2βj(ω,z)N∑q=1∫∞−∞dξ[γ(e)jq(ω,zε+ξ)aεq(ω,z)√βq(ω,z)e1ε∫z0dz′βq(ω,z′)+iξβq(ω,z) −¯¯¯¯¯¯¯¯γ(e)jq(ω,zε+ξ)bεq(ω,z)√βq(ω,z)e−1ε∫z0dz′βq(ω,z′)−iξβq(ω,z)]e−|ξ|βj(ω,z), \hb@xt@.01(3.41)

where the approximation means that we neglect the terms that are negligible in the limit . Here we introduced the notation

 γ(e)jq(ω,zε+ξ)=Γjqν′′(zε+ξ)+iβq(ω,z)Θjqν′(zε+ξ)), \hb@xt@.01(3.42)

with coefficients and defined in (LABEL:eq:DC6), and recall that the bar denotes complex conjugate. For the derivative we have

 ε∂zuεj(ω,z)≈ −σ√ε2βj(ω,z)N∑q=1∫∞−∞dξ[θ(e)jq(ω,zε+ξ)aεq(ω,z)√βq(ω,z)e1ε∫z0dz′βq(ω,z′)+iξβq(ω,z) −¯¯¯¯¯¯¯¯θ(e)jq(ω,zε


text:
## Exact solution to a multidimensional wave equation with delay.(English)Zbl 07425035

Summary: This paper deals with a mixed problem for the wave equation with discrete delay $$\tau > 0$$, $u_{t t} (t, \boldsymbol{x}) = a_1^2 \Delta_{\boldsymbol{x}} u (t,\boldsymbol{x}) + a_2^2 \Delta_{\boldsymbol{x}} u(t-\tau, \boldsymbol{x}) + b_1 u ( t , \boldsymbol{x}) + b_2 u (t-\tau,\boldsymbol{x}), \quad t > \tau ,\, 0 \leq \boldsymbol{x} \leq \boldsymbol{l} ,$ with Dirichlet boundary conditions. The exact infinite series solution is constructed by the method of separation of variables, where the time-dependent functions of the decomposition satisfy second-order delay differential equations. Our approach is based on and extends the work by F. Rodríguez et al. [ibid. 219, No. 6, 3178–3186 (2012; Zbl 1309.35045)].

### MSC:

 35R10 Partial functional-differential equations 35C05 Solutions to PDEs in closed form 35C10 Series solutions to PDEs

Zbl 1309.35045
Full Text:

### References:

 [1] Rodríguez, F.; Roales, M.; Martín, J. A., Exact solutions and numerical approximations of mixed problems for the wave equation with delay, Appl. Math. Comput., 219, 6, 3178-3186 (2012) · Zbl 1309.35045 [2] D. Y. Khusainov, M. Pokojovy, E. Azizbayov, Representation of classical solutions to a linear wave equation with pure delay, arXiv:1401.5667(2014). [3] Farlow, S. J., Partial Differential Equations for Scientists and Engineers (1993), Dover: Dover New York · Zbl 0851.35001 [4] Folland, G. B., Fourier Analysis and its Applications (1992), Wadsworth: Brooks: Wadsworth: Brooks Pacific Grove, CA · Zbl 0786.42001 [5] Gonzalez-Velasco, E. A., Fourier Analysis and Boundary Value Problems (1995), Academic Press: Academic Press New York · Zbl 0894.35001 [6] Wiener, J.; Debnath, L., A wave equation with discontinuous time delay, Int. J. Math. Math. Sci., 15, 4, 781-788 (1992) · Zbl 0766.35024 [7] Nicaise, S.; Pignotti, C., Stability and instability results of the wave equation with a delay term in the boundary or internal feedbacks, SIAM J. Control Optim., 45, 5, 1561-1585 (2006) · Zbl 1180.35095 [8] Fridman, E.; Orlov, Y., Exponential stability of linear distributed parameter systems with time-varying delays, Automatica, 45, 194-201 (2009) · Zbl 1154.93404 [9] Messaoudi, S. A.; Fareh, A.; Doudi, N., Well posedness and exponential stability in a wave equation with a strong damping and a strong delay, J. Math. Phys., 57, 11, 111501 (2016) · Zbl 1355.35025 [10] Zhang, L.; Stepan, G., Stability analysis of the wave equation with delayed boundary conditions, Procedia IUTAM, 22, 139-145 (2017) [11] Wiener, J.; Debnath, L., Boundary value problems for the diffusion equation with piecewise continuous time delay, Int. J. Math. Math. Sci., 20, 187-195 (1997) · Zbl 0879.35065 [12] Martín, J. A.; Rodríguez, F.; Company, R., Analytic solution of mixed problems for the generalized diffusion equation with delay, Math. Comput. Model., 40, 3-4, 361-369 (2004) · Zbl 1062.35157 [13] Reyes, E.; Rodríguez, F.; Martín, J. A., Analytic-numerical solutions of diffusion mathematical models with delays, Comput. Math. Appl., 56, 3, 743-753 (2008) · Zbl 1155.65387 [14] Reyes-Salguero, E., Soluciones Analítico-Numéricas de Ecuaciones en Derivadas Parciales con Retardo (2008), Department of Applied Mathematics, University of Alicante, Alicante, Spain, Ph.D.Thesis [15] Khusainov, D. Y.; Ivanov, A. F.; Kovarzh, I. V., Solution of one heat equation with delay, Nonlinear Oscil., 12, 2, 260-282 (2009) · Zbl 1277.35193 [16] Escolano, J.; Rodríguez, F.; Castro, M. A.; Vives, F.; Martín, J. A., Exact and analytic-numerical solutions of bidimensional lagging models of heat conduction, Math. Comput. Model., 54, 7-8, 1841-1845 (2011) · Zbl 1235.80046 [17] ID 219040, 2012 · Zbl 1250.35117 [18] D.Y. Khusainov, M. Pokojovy, E. Azizbayov, On classical solvability for a linear 1d heat equation with constant delay, arXiv:1401.5662(2014). [19] Reyes, E.; Castro, M.; Sirvent, A., Exact solutions and continuous numerical approximations of coupled systems of diffusion equations with delay, Symmetry, 12, 9, 1560 (2020) [20] Jódar, L.; Almenar, P., Accurate continuous numerical solutions of time dependent mixed partial differential problems, Comput. Math. Appl., 32, 5-19 (1996) · Zbl 0857.65097 [21] Almenar, P.; Jódar, L.; Martín, J. A., Mixed problems for the time-dependent telegraph equation: continuous numerical solutions with a priori error bounds, Math. Comput. Model., 25, 31-44 (1997) · Zbl 0881.35064 [22] Khusainov, D. Y.; Diblík, J.; Rŭžičková, M.; Lukáčová, J., Representation of a solution of the cauchy problem for an oscillating system with pure delay, Nonlinear Oscil., 11, 2, 276-285 (2008) · Zbl 1276.34055 [23] Khusainov, D. Y.; Pokojovy, M.; Azizbayov, E. I., On the cauchy problem for a linear harmonic oscillator with pure delay, Adv. Differ. Equ., 2015, 1, 1-20 (2015) · Zbl 1422.34188 [24] Cortés, J.-C.; Jornet, M., $$\operatorname{L}^p$$-solution to the random linear delay differential equation with a stochastic forcing term, Mathematics, 8, 6, 1013 (2020) [25] Cortés, J.-C.; Jornet, M., Analytic solution to the generalized delay diffusion equation with uncertain inputs in the random Lebesgue sense, Math. Methods Appl. Sci., 44, 2, 2265-2272 (2021) [26] Cortés, J.-C.; Sevilla-Peris, P.; Jódar, L., Analytic-numerical approximating processes of diffusion equation with data uncertainty, Comput. Math. Appl., 49, 7-8, 1255-1266 (2005) · Zbl 1078.65005 [27] Calatayud, J.; Cortés, J.-C.; Jornet, M., Uncertainty quantification for random parabolic equations with non-homogeneous boundary conditions on a bounded domain via the approximation of the probability density function, Math. Methods Appl. Sci., 42, 17, 5649-5667 (2019) · Zbl 1432.60062 [28] ID 397053, 2013 · Zbl 1291.80013 [29] García, M. A.; Castro, M. A.; Martín, J. A.; Rodríguez, F., Exact and nonstandard numerical schemes for linear delay differential models, Appl. Math. Comput., 338, 337-345 (2018) · Zbl 1427.65097 [30] García, M. A.; Castro, M. A.; Martín, J. A.; Rodríguez, F., Exact and nonstandard finite difference schemes for coupled linear delay differential systems, Mathematics, 7, 1038 (2019) [31] Calatayud, J.; Cortés, J.-C.; Jornet, M.; Rodríguez, F., Mean square convergent non-standard numerical schemes for linear random differential equations with delay, Mathematics, 8, 1417 (2020) [32] Castro, M. A.; Sirvent, A.; Rodríguez, F., Nonstandard finite difference schemes for general linear delay differential systems, Math. Methods Appl. Sci. (2021)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Plastic strain is a mixture of avalanches and quasi-reversible deformations: Study of various sizes

# Plastic strain is a mixture of avalanches and quasi-reversible deformations: Study of various sizes

## Abstract

Size-dependence of plastic flow is studied by discrete dislocation dynamical simulation of systems with various numbers of interacting linear edge dislocations while the stress is slowly increased. Regions between avalanches in the individual stress curves as functions of the plastic strain were found nearly linear and reversible, where the plastic deformation obeys an effective equation of motion with a nearly linear force. For small plastic deformation, the means of the stress-strain curves are power law over two decades. Here and for somewhat larger plastic deformations, the mean stress-strain curves converge for larger sizes, while their variances shrink, both indicating the existence of a thermodynamical limit. The converging averages decrease with increasing size, in accordance with size-effects from experiments. For large plastic deformations, where steady flow sets in, thermodynamical limit was not realized in this model system.

###### pacs:
61.72.Lk 81.40.Lm 83.50.-v

Introduction and overview: As known from research in the past decade, the stress-strain curve of micron-scale specimens contains random steps, where the plateaus mark avalanches, a behavior also observed in simulations. (1); (2); (3); (4); (5); (6) Recently there has been a series of studies about the statistics of avalanches, and some on probabilistic properties of the yield stress, (7); (4); (8) but the dependence of these features on the sample size was less investigated. (7); (9); (10) The stair like stress-strain response caused by the dislocation avalanches become significant if the system size is in the order of 1m. Nevertheless, the size dependence of the plastic deformation is observed already at much larger system sizes. In most cases the smaller sample requires larger stress level to get the same deformation. The effect is traditionally modeled by large scale discrete dislocation dynamical (DDD) (11); (12); (13); (14) simulations, phenomenological “nonlocal continuum theories” in which an appropriate gradient term is added to the stress-strain relation (15); (16), or by continuum theory of dislocations (17); (18); (19); (20); (21); (22); (23) . It is obvious, however, that a continuum theory is not applicable if the dislocation spacing is comparable to the system size.

In this paper the size-dependence of plastic deformations is studied by DDD simulation of systems with various numbers of interacting linear edge dislocations while the stress is slowly increased. Note that elastic deformations are not considered in this model, so strain and deformation are understood as purely plastic. Our main observations in this work are as follows. Between avalanches the stress grows close to linearly with the deformation. Tests of load cycles show that here the deformation is approximately reversible, so plasticity appears as a randomly alternating sequence of quasi-reversible deformations and avalanches. The linearity coefficient is random even for systems of the same size (number of dislocations ), with a sharpening distribution for larger . In each realization, assuming an effective equation of motion for the deformation, we find a close-to-linear effective force, with a random spring constant. The full individual staircase-like stress-strain curves of the same also have a mean, which is a smooth function, different for different ’s. Up to a certain threshold deformation the mean stress-strain curve seems to converge for large , while the variance goes to zero. So for large essentially the same sharp stress-strain curve emerges for each realization, indicating the existence of a thermodynamic limit. This refines the finding of Tsekenis et al. (9), where the natural scaling by let curves collapse even for smaller . Within the region of thermodynamical limit , for two decades in the strain, the mean stress-strain curve is a power law, with an exponent decreasing from for small to about for large . This is caused by the alternation of linear, quasi-reversible segments and plateaus of avalanches with random lengths in the stress-strain response function. Furthermore, for the model under study, there seems to be no thermodynamical limit for large deformations in the model under study.

Simulation method: Plastic deformation is mainly due to the motion of dislocations, (24) interacting via a long-range stress field. Here we study one of the simplest models of dislocation systems described as follows.(25); (1); (26); (27); (28) Straight and parallel edge dislocations with parallel slip axes are considered, essentially a D cross section of a D system. Periodic boundary conditions are used on a square of side , the slip axes are parallel to one edge of the square (the axis), and for each realization dislocations are initially randomly placed with a uniform distribution. In the beginning each realization contains a fixed number of dislocations, with equal number of positive and negative Burgers vectors , where is the lattice constant and . Only dislocation glide is taken into account so the dislocations’ vertical () coordinates are constant.

One positive dislocation induces the shear stress field

 τind(→r)=bDx(x2−y2)/r4, (1)

where is the radius vector from that dislocation, , , with being the shear modulus and the Poisson number. The ’th dislocation is exposed to the shear stress field of all the others’ and to the external field taken to be the uniform , wherein it performs overdamped motion with drag coefficient . Thus the equation of motion of the ’th dislocation is (6); (29)

 ˙xi=B−1bsi[N∑j=1j≠isjτind(→ri−→rj)+τext], (2)

where the is the sign of the ’th dislocation and its position. The plastic strain is calculated by , where is the change in the coordinate relative to the initial value. This equation is rendered periodic numerically by including sufficiently many mirror images of the ’th dislocation. The resulting equation of motion is solved with th order Runge-Kutta method. Adaptive step size is used to better treat narrow dipoles. For very narrow dipoles would demand excessive computation time, we annihilate (different sign) or merge (same sign) dislocations if they are closer than . Whereas annihilation decreases the dislocation number, in order to avoid ambiguity, in the conversion formulas we use the original dislocation number . Note that dislocations are not created in our model, corresponding to non-source-controlled plastic deformations. Throughout the paper simulations with were considered with ensembles numbering , respectively, and for the largest sizes our computational power allowed the scanning only of more restriced regions of simulated strain.

Equation (2) is represented in the computer by , yielding the density . The mapping to different sample sizes, while the physical dislocation density is kept constant, occurs by our introducing natural quantities , , , , used throughout this paper. (26) Then physical quantities are obtained from the physical density as , , , . To give an example, if on the natural scale then for the generic physical variables and we would obtain the physical deformation .

In the scenario presented here firstly we let the system relax without external stress to form the initial state with . Then we apply quasi-static stress loading, i.e., we increase the external stress by a small rate, similar to what was applied earlier.(7); (9); (30) Having experimented with various stress rates we finally chose . The stress is kept increasing so long as the mean absolute velocity of the dislocations remains under the threshold . If that threshold is surpassed (this is our definition of an avalanche) then the external stress is kept constant until the mean absolute velocity drops again below the threshold. The end state is a steady flow, that is, an infinite avalanche, because in the absence of dislocation creation no work hardening takes place for large deformations.

Individual stress-strain curves and effective motion between avalanches: We first show typical dislocation configurations for various plastic deformations along our scenario in Fig. 1.

Firstly, dislocations relax in the absence of external stress, forming a random-looking configuration of numerous smaller clusters, see Fig. 1a. Due to the increasing external stress the deformation generically increases, while clusters grow mainly in the direction (1b). Beyond some threshold steady flow emerges, marked usually by a single dipolar wall, spanning across the whole simulation area (1c), while one-or-two dislocations are circling quickly along their slip axes, in conformance to the periodic boundary conditions. Hence we must conclude that in the steady flow boundary effects are important, thus our model may not be realistic in this region, and so we concentrate our study to smaller deformations.

The plastic stress-strain curves of individual realizations are like staircases, they exhibit a sequence of plateaus of constant stress corresponding to avalanches, in accordance with earlier results, (1); (2); (3); (4); (5); (7) see Fig. 2. A novel observation here, to our knowledge not noted earlier, is that between plateaus, where the stress increases, the functions are nearly linear.

For the sake of comparison we normalized each increasing segment to a function starting and ending at the opposite corners of the unit square as

 g(x)=τext(x(γ1−γ0)+γ0)−τ0τ1−τ0,  x∈[0,1], (3)

where , and , mark the borders of the chosen segment. Such normalized segments of individual runs with were separately averaged for fixed sizes , and the resulting curves fall onto each other and form in the inset of Fig. 2. Here, also the standard deviation of the normalized segments is plotted, which is again nearly the same for various ’s. So the segments between avalanches on the stress-strain curves, normalized according to Eq. (3), follow on the average a universal, nearly linear form, with a universal variance.

In order to test properties of the close-to-linear segments, we have run a few loading cycles on individual realizations with various stress rates. In most of the cases, apart from a small, smooth, transient due to the finiteness of the stress rate, and from tiny avalanches, reversibility was found. Motivated by this “quasi-reversible” response, we surmise that deformations obey an effective equation of motion

 ˙γ(t)≈−F(γ(t)−γ0)+τext(t)−τ0, (4)

where the endpoint of the last avalanche, where the system is assumed to be in equilibrium, and is the effective restoring force, depending only on the increment . Similarly as (3) associates with the stress, we normalize the force onto the unit square. Again, like the external stress in the inset of Fig. 2, the mean normalized force very weakly depends on , and this universal function is astonishingly close to , which is the consequence of the low stress rate resulting . Thus even the force is very close to linear. We emphasize, that near equilibrium of a dislocation configuration, for small displacements, the elastic energy of course grows quadratically, so there the response should be linear. The remarkable feature in our case is that linearity holds way up to near the onset of the next avalanche. This instability is indeed marked by the slight curving of the universal function close to one in Fig. 2. Note that its slope there is not zero, whereas it would be zero in the case of a force-activated escape from a 1D potential, because of rare negative avalanches with .

In order to visualize the statistical nature of the quasi-reversible regions, in Fig. 3 we plot the cumulative probability distribution function of the steepness for (chosen as a practical value). The curves visibly contract with increasing , showing convergence to a finite mean.

A nearly linear response here means that, due to the interaction of dislocations, an effective shear modulus arises. The latter can be interpreted as the plastic component of the total empirical shear modulus in real crystals. Note that in the present model elastic deformations are not included.

Mean stress as function of the strain: So far we concentrated on the quasi-reversible segments between avalanches, now we turn to the global statistical behavior of stress-strain curves. Firstly, we plot the average over ensembles with fixed in Fig. 4. A main feature is the power law behavior over decades up to a strain approximately , (see inset), with exponent close to for small sizes and decreasing with size to about . We can interpret this feature such that the nearly linear segments of the stress-strain curves are interrupted by the avalanche plateaus just in the way that an effective power function with a smaller-than-one exponent emerges. That is, avalanches soften the linearity of quasi-reversible segments and give rise statistically to a power law.

On physical scales, taking a lattice constant , a dislocation density , the corresponds to . It is important to note that this value is much smaller than the threshold value customarily considered to be the yield strain in engineering practice. The fact that the power law arises way below the empirical yield value indicates the determining role of avalanches for much lower plastic strains than expected earlier. On the other hand, this is in a convincing agreement with fatigue experiments on single crystals that showed irreversibility already below the physical range of . (31)

Another important property seen in Fig. 4 is that for the stress-strain curves seem to converge for large , a criterion for the existence of a thermodynamical limit. On the other hand, for large strains, the tendency is inconclusive from our simulation, the stress values may even diverge.

For intermediate strains the convergent bundle of the curves switches order for the sizes we considered. Note that for fixed strains the stresses decrease with . As discussed earlier in connection with physical units, increasing here can mean increasing size with constant dislocation density. Therefore, larger stresses for smaller ’s as in Fig. 4 can be interpreted as a version of the property “smaller is harder”. We emphasize, however, that our model has periodic boundary conditions, thus pileups, commonly held responsible for this phenomenon,(32) cannot develop. This tendency reverses for large strains, where the stress increases with . In this region, however, where the configuration resembling a single wall forms, as seen in Fig. 1c, we do not suggest the model bears general relevance to real materials.

Standard deviation of the stress as a function of strain: Given the fact that the stress-strain response for macroscopic crystals with a fixed orientation is a well-defined, sharp curve, it is expected that the variance vanishes with increasing size. Accordingly, a decreasing variance was observed in micropillar experiments by Uchic et al. (2) To study this effect, we plotted in Fig. 5 the standard deviation of the stress for ensembles with fixed for different strains. In the region , where the mean stress converged with (see Fig. 4), the standard deviation decreases. We tested a power law convergence by plotting in the inset of Fig. 5, and indeed the collapse demonstrates that the deviation vanishes like . Thus, recalling that we found a sharp limit for the average stress when , we can conclude that in this region there is a thermodynamical limit. On the contrary, for large strains Fig. 5 does not show convergence of the standard deviation, like there was no visible convergence of the mean in Fig. 4 either, so here thermodynamical limit is absent.

Conclusion and outlook: In this paper our aim was twofold. On the one hand, we uncovered quasi-reversible behavior with nearly linear response between avalanches in a 2D model of simulated dislocations. For large sizes, the effective plastic shear modulus appears to converge in average. As to the full stress-strain curves, the quasi-reversible segments conspire with the avalanche plateaus to yield on the average a power response curve. The second main question was about the thermodynamical limit, which is achieved by both a convergent mean and a vanishing variance of the stress for . In this region we observed also the analog of the size-effect, as found in micropillars.(2) For largest strains thermodynamical limit is not reached and we do not consider our simulations as conclusive there.

Our study opens a series of questions. The results on the quasi-reversible behavior between avalanches call for more detailed investigations. The statistical properties of the finite-size behavior is best characterized by distribution functions, among which here we only described that of the local effective plastic shear modulus, characterizing quasi-reversible regions. The distribution of the stresses is of obvious interest, and at avalanches are expected to be related to extreme statistics. A longstanding problem in this area is the transition to steady flow, wherein the absence of a single critical point, rather critical behavior for all strains, has been shown before,(29) but a detailed study is still overdue. Carrying forth experiments on micropillars(5) with various sizes and their comparison to the prediction from simulations would be of immediate interest.

Financial supports of the Hungarian Scientific Research Fund (OTKA) under contract numbers K-105335 and PD-105256, and of the European Commission under grant agreement No. CIG-321842 (StochPlast) are acknowledged.

### References

1. M.-C. Miguel, A. Vespignani, S. Zapperi, J. Weiss, and J.-R. Grasso, Nature 410, 667 (2001)
2. M. D. Uchic, D. M. Dimiduk, J. N. Florando, and W. D. Nix, Science 305, 986 (2004)
3. M. D. Uchic, P. A. Shade, and D. M. Dimiduk, Annu. Rev. Mater. Res. 39, 361 (2009)
4. F. F. Csikor, C. Motz, D. Weygand, M. Zaiser, and S. Zapperi, Science 318, 251 (2007)
5. P. D. Ispánovity, A. Hegyi, I. Groma, G. Györgyi, K. Ratter, and D. Weygand, Acta Mat. 61, 6234 (2013, Issue 16)
6. P. D. Ispánovity, I. Groma, G. Györgyi, F. F. Csikor, and D. Weygand, Phys. Rev. Lett. 105, 085503 (2010)
7. V. Beato, M. Zaiser, and S. Zapperi, arXiv, 1106.3444v1(2011)
8. M. LeBlanc, L. Angheluta, K. Dahmen, and N. Goldenfeld, Phys. Rev. Lett. 109, 105702 (2012)
9. G. Tsekenis, N. Goldenfeld, and K. A. Dahmen, Phys. Rev. Lett. 106, 105501 (2011)
10. P. M. Derlet and R. Maass, arXiv, 1401.3571(2014)
11. D. Balint, V. Deshpande, A. Needleman, and E. Van der Giessen, Mod. Sim. Mat. Sci. Eng. 14, 409 (2006)
12. P. Guruprasad and A. Benzerga, Phil. Mag. 88, 3585 (2008)
13. D. Weygand, M. Poignant, P. Gumbsch, and O. Kraft, Mat. Sci. Eng. 483-84, 188 (2008)
14. A. A. Benzerga, J. Mech. Phys. Solids 57, 1459 (2009)
15. N. Fleck and J. Hutchinson, J. Mech. Phys. Solids 49, 2245â2271 (2001)
16. M. E. Gurtin, J. Mech. Phys. Solids 50, 5 (2002)
17. I. Groma, F. Csikor, and M. Zaiser, Acta Mater. 51, 1271 (2003)
18. I. Groma, G. Györgyi, and B. Kocsis, Phys. Rev. Lett. 96, 165503 (2006)
19. A. Acharya, A. Roy, and A. Sawant, Scripta Mater. 54, 705 (2006)
20. J. Kratochvil and R. Sedlacek, Phys. Rev. B 77 (2008)
21. S. D. Mesarovic, R. Baskaran, and A. Panchenko, J. Mech. Phys. Solids 58, 311 (2010)
22. S. Sandfeld, T. Hochrainer, M. Zaiser, and P. Gumbsch, J. Mater. Res. 26, 623 (2011)
23. L. H. Poh, R. H. J. Peerlings, M. G. D. Geers, and S. Swaddiwudhipong, J. Mech. Phys. Solids 61, 913 (2013)
24. F. R. Nabarro, Theory of crystal dislocations, Vol. 1 (Clarendon, 1967)
25. I. Groma and G. S. Pawley, Phil. Mag. A 67, 1459 (1993)
26. F. F. Csikor, M. Zaiser, P. D. Ispánovity, and I. Groma, Journal of Statistical Mechanics: Theory and Experiment 2009, P03036 (2009)
27. L. Laurson, M. C. Miguel, and M. J. Alava, Phys. Rev. Lett. 105, 015501 (2010)
28. L. Laurson and M. Alava, Phys. Rev. Lett. 109, 155504 (2012)
29. P. D. Ispánovity, I. Groma, G. Györgyi, P. Szabó, and W. Hoffelner, Phys. Rev. Lett. 107, 085506 (2011)
30. P. M. Derlet and R. Maass, Modelling Simul. Mater. Sci. Eng. 21, 035007 (2013)
31. H. Mughrabi, Metall. Mater. Trans. B 40B, 431 (2009)
32. L. Kovács and L. Zsoldos, Dislocations and plastic deformation (Pergamon Press, 1973)
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