text: Complexometric Titration using EDTA. EDTA DETERMINATION OF TOTAL WATER HARDNESS AND CALCIUM 3 lab periods Reading: Chapter 11, Quantitative Chemical Analysis, 8th Edition, Daniel C. Harris (7th Edition: Chapter 12). Pipet 50 mL portions of unknown water into three 250 mL Erlenmeyer flasks. - 2. These hard water cations include calcium, magnesium, iron, zinc and the other polyvalent metal ions. Calculation 1 ml of 0.01 M EDTA ≡ 1 mg of CaCO 3 V 1 ml of EDTA ≡ V 1 mg of CaCO 3 Calculation of total hardness Volume of EDTA solution consumed = ………… ml Volume of hard water taken = …………. Select a definite volume of water sample. Determination of Mg by Titration with EDTA INTRODUCTION: Many metal ions react with electron pair donors to form coordination compounds or complex ions. THANKS. Calculate titration curves for the titration of 50.0 mL of $$5.00 \times 10^{-3}$$ M Cd 2 + with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7. Buffer solution preserve pH at a balanced position. All All mass measurements were taken from a AND Balance, serial number 12321601 and all titrations were done using a KIMAX buret, #173. result calculation. Answer Mol arity EDTA (m ol / L) = Volume Zinc ( L)× Mol rity m l / × 1 mol EDTA … (5 Pts.) In this article, we are going to talk about the determination of hardness of water by EDTA titration method.The causes of hardness of water is by multivalent metallic cations, which react with soap to form precipitates and with certain anions present in water to form scale. EDTA titration for the determination of water hardness using murexide and Eriochrome black T as indicators. Calculation Hardness… When the hardness is numerically greater than the sum of carbonate and bicarbonate alkalinity the amount of hardness that is equivalent to total alkalinity is called the carbonate hardness and the balance as noncarbonated hardness. However, above pH 10 there is a danger of precipitation of CaCO3 and Mg(OH)2. Equations: Titration: Ca2+ + H 2Y2-→ CaY2-+ 2 H+ … As water hardness is usually reported in terms of mg/L of calcium carbonate (even if water contains both calcium and magnesium), we will use for calculations slightly strange reaction equation: CaCO3 + EDTA4- → CaEDTA2- + CO32- That allows direct calculation of calcium carbonate mass for known amount of titrant used. Write the equation for the titration of Ca+2 with EDTA. CLARO NATIONAL INSTITUTE OF GEOLOGICAL SCIENCES, COLLEGE OF SCIENCE UNIVERSITY OF THE PHILIPPINES, DILIMAN, QUEZON CITY, PHILIPPINES DATE SUBMITTED: MAY 9, 2013 ABSTRACT Calcium and magnesium ions’ ability to to … Determination of Hardness of Water by EDTA Titrimetric Method. reported. Experiment 8 Determination of Total Hardness of Water by Complexometric Titration Lab instructor-Andrew Cho William Parnell 4/11/ Abstract: The hypothesis for this experiment was to determine if complexometric titration technique can identify the total “hardness” of water … EDTA Titration for Determination of calcium and magnesium. When all the ions are complexed the solution will turn blue. Let the burette reading of EDTA be V 3 ml. Determination of Total hardness Repeat the above titration method for sample hard water instead of standard hard water. M105 Determination of the Total Hardness of Water by EDTA Titration and Flame Photometry Object To determine calcium and magnesium in a sample of tap water by titration with EDTA and then to … However, the complexation constant is a function of pH (Harris, 1999). Select a definite volume of water sample. Transfer 50 mL of tap water to four different Erlenmeyer flasks. Why do we use a temperature bar in a slab? B = mg CaCO3 equivalent to 1 ml EDTA … Write the equations for the endpoint color change of Eriochrome Black T (EBT) with Mg+2 3. When colour changes from wine red to blue click the "stop" button & note the volume of EDTA used. Determination of Permanent hardness Take 100 ml of sample hard water in 250 ml beaker. (5 Pts.) QUANTITATIVE DETERMINATION OF TOTAL HARDNESS IN DRINKING WATER BY COMPLEXOMETRIC TITRATION R. ASCAÑO, J.J. CLEMENTE AND S.M.D. EDTA is standardized with primary standard CaCO 3 To simplify calculations with hardness we use CaCO 3 titer = mg CaCO 3 mL EDTA After EDTA is standardized, we can use the ratio to determine the hardness of a water sample by titrating with EDTA and using the CaCO 3 titer. 56 15. Academic year. This reagent is a weak acid that can lose four protons on complete … Calculation of total, permanent and temporary hardness. Water hardness can be readily determined by titration with the chelating agent EDTA (ethylenediaminetetraacetic acid). Hardness may be present from zero to several hundred mg/L as CaCO3. e����π �L@Q��v���00/e��h����j��i) ��2��0��y���)�m@��R������y�p~x-�pY �R���Ex�4#�17��t�Q � �� endstream endobj 28 0 obj<> endobj 29 0 obj<> endobj 30 0 obj<>stream 2. Comments. Transfer 25.00 mL of the diluted unknown solution to four different Erlenmeyer flasks. EDTA reacts quantitatively with both calcium and magnesium ions in a 1:1 stoichiometric ratio. A total of 3 titration trials were completed to find the average molarity of the EDTA, and the average water hardness of unknown water sample #40. 2016/2017. Complexometric Titration with EDTA B. 2. Determination of Permanent hardness Take 100 ml of sample hard water … 3. 3. Equations: Titration… Calcium and magnesium are easily measured by titration … )”or “hardness (EDTA).” 2340 B. Hardness by Calculation 1. Question: DETERMINATION AND REMOVAL OF HARDNESS OF WATER Determination Of Hardness Of Water / Titration With EDTA Solution: Determine The Volume Of EDTA Solution Used. Then calculate the hardness of water sample in ppm using the equation … View lab report (titration).docx from CHEM 217 at Concordia University. Report by Austin Jones February 5th, 2020 Professor Paul Gilletti. (5 Pts.) Helpful? - 2. Titrate the solutions with 0.01 M EDTA until the color changes from wine red to blue. Ras Danny• 1 year ago. Compare your results with Figure $$\PageIndex{3}$$ and comment on the effect of pH on the titration of Cd 2 + with EDTA… Add 5 mL of ammonia-ammonium buffer and Erichrome Black T indicator. The presence in water of salts of calcium and magnesium is spoken of as hardness. Because EDTA has four acidic protons, the formation of metal-ion/EDTA complexes is dependent upon the pH. xڴVi��H�+�1ъ��vw[����C�� ���+���h��Շ8�졖q���~�u� Experiment 4: Determination of Water Hardness Purpose: In this experiment we will be calculating the total hardness, expressed as ppm CaCO 3, in a sample of water, and also calculate the hardness due to magnesium ions and calcium ions separately. Determination of Total hardness Repeat the above titration method for sample hard water instead of standard hard water. But EDTA has a stronger affinity towards Ca++ and Mg++. ���&k���8�( )D��d ��!����QtI��d,f Apparatus for hardness of water experiment, Reagents for hardness of water experiment. Advantages of EDTA method This method is definitly preferable to the other methods, because of : (i) Greater accuracy, (ii) convenience and (iii) more rapid procedure. QUANTITATIVE DETERMINATION OF TOTAL HARDNESS IN DRINKING WATER BY COMPLEXOMETRIC EDTA TITRATION R.A. ROGIO DEPARTMENT OF MINING, METALLURGICAL AND MATERIALS ENGINEERING, COLLEGE OF ENGINEERING UNIVERSITY OF THE PHILIPPINES, DILIMAN QUEZON CITY, PHILIPPINES DATE PERFORMED: JANUARY 31, 2013 INSTRUCTOR’S NAME: MS. MELANIE BUCSIT Complexometric EDTA titration … A] DETERMINATION OF HARDNESS OF WATER BY EDTA METHODPart I: Standardization of EDTA solutionPart II: Hardness of water sample:B] REACTIONSC] CALCULATIONS EDTA Titration Calculations The hardness of water is due in part to the presence of Ca2+ ions in water. Water Hardness is typically determined by an EDTA titration buffered at a pH of 10 (condition for maximum complexing). Your email address will not be published. EDTA Determination of Total Calcium and Determination of Individual Calcium and Magnesium. The optimum pH for the determination of hardness of water is 10.0 + 0.1 and is adjusted by NH4 OH – NH4Cl buffer. State the value to 5 places after the decimal point. Which one is the best for RCC concrete, M-sand or river sand? 3. Hardness due to such causes is called pseudo hardness. �&5�0�8�P���L�3\SQ�Q��b�E"�pq���|�{��,�&FA0��� x09�(�L"fBS�P�T �j��u�n�~��uYM������ZZ �a�- In most water samples, calcium and magnesium are the chief contributors to water hardness. In most water samples, calcium and magnesium are the chief contributors to water hardness. In the EDTA titration, it is assumed that the total hardness is due to the presence mainly of calcium and magnesium ions. Neither titration includes an auxiliary complexing agent. A chelate is formed when a metal ion coordinates with two (or more) donor groups of a single ligand. Portland State University. Hence, if EDTA is added the former complex (Ca-EBT and Mg-EBT)is broken and new complex (Ca-EDTA and Mg-EDTA) of blue colour is formed. Objective Apply the concept of complexometric titration in the determination of total hardness in drinking water 3. The concentration of Ca2+ ions is usually expressed as ppm CaCO 3 in the water sample. The higher the pH, the sharper the endpoint. Guarantee Your Academic Success! 27 0 obj <> endobj 55 0 obj<<1AAEDBD885A011DD89CE0017F2059850>]/Info 26 0 R/Filter/FlateDecode/W[1 2 1]/Index[27 47]/DecodeParms<>/Size 74/Prev 43362/Type/XRef>>stream Report the water hardness … This is the endpoint of the titration. When boiling water … Titrate the solution with EDTA from your buret to the blue endpoint you observed when titrating the blank. This method is based on Method 2340 C of Standard Methods for the Examination of Water and Wastewater , 22 nd Edition. Use the volume and molarity of EDTA to calculate the water hardness … The primary cations causing hardness are calcium, magnesium, strontium, ferrous ion, and manganous ions.The common anions that combine with these cations to cause hardness are bicarbonates, sulphates, chloride, nitrate and silicate. Dilute 25.00 mL of unknown zinc sample to 250 mL with distilled water in a volumetric flask. Adjust the speed of the drops from the burette. For the titration … Determination of Water Hardness using Complexometric titration You will use EDTA complexometric titration to determine the hardness of a sample of water brought from your home. The Determination of Water Hardness and Water Filtration Things for the lab notebook and to be included in the lab report: 1. Determining the moles of Ca 2+ and Mg 2+ by titration with EDTA … Complexometric Determination of Water Hardness. 2. Required fields are marked *, In this article, we are going to talk about the determination of hardness of water by EDTA titration method.The causes of hardness of water is by, When the hardness is numerically greater than the sum of carbonate and bicarbonate alkalinity the amount of hardness that is equivalent to total alkalinity is called the, The calculation of hardness in water by EDTA titration can be found by adding a small amount of a dye such as, Add EDTA titrant to the sample with vigorous shaking till the wine red colour just turns. Let the burette reading of EDTA be V 2 ml. Originally, the hardness of water was understood to be a … As Ca++ and Mg++ ions are present in significant qualities in water hardness is generally attributed to their presence. EDTA … Titrate with EDTA until a blue color appears. Determination of the Hardness of Tap Water 1. Add 5 mL of pH 10 buffer, 2 drops of Eriochrome Black T indicator, and 15 drops of 0.03 M MgCl2. It may be used to determination of water hardness. Prices starting from $3 per page for your Homework Writing Help. Determination of Water Hardness by EDTA Titration James Harvey*, Sara Abuhad, Emmanuel chemical structure for the disodium salt of EDTA is shown in Figure 22-1. Quantitative Analysis Laboratory. Quantitative Determination of Total Hardness in Drinking Water by Complexometric EDTA Titration 1. Calculate The Molarity Of CaCO3 In The Water. Therefore the total hardness of water can be determination by edta titration method. Complexometric Titration with EDTA Chemistry 3200 Complexometric Titration with EDTA In this experiment you will use ethylenediaminetetraaectic acid (EDTA) to determine metals in aqueous solution by complexation titration. Hardness of water is a measure of its capacity to precipitate soap, and is caused by the presence of divalent cations of mainly Calcium and Magnesium. Select the titrant. Water hardness is measured in parts per million or milligrams per liter. Read More: How to reduces the hardness of water? Report the water hardness as ppm CaCO 3 of a 50 ml water sample that required 32. Note the volume of titrant added (V1 ml). Estimation of hardness of water by EDTA method. These procedures were more rapid than the conventional gravimetric methods (P), which … EDTA titration for the determination of water hardness using murexide and Eriochrome black T as indicators. It may be used to determination of water hardness. This dye-stuff tends to polymerize in strongly acidic solutions to a red brown product, and hence the indicator is generally used in EDTA titration … Determination of Hardness of Water by EDTA Titration/Complexometric Titration To determine the total hardness of the given samples by EDTA titrimetric method. Water Hardness Analysis Example 3 A 100 mL (0.100 L) sample of tap water was titrated with 0.0100 M EDTA solution. Adjust the pH of 100 mL (4 deliveries … The mean corrected titration volume of the EDTA solution was 16.25 mL (0.01625 L). Simple Water and Complexes 4. x�bbdb�$���f ��$8� �*�H�l@�\�Dx �&�^V�)� V=�� QUANTITATIVE DETERMINATION OF TOTAL HARDNESS IN DRINKING WATER BY COMPLEXOMETRIC EDTA TITRATION G.M.L. Your email address will not be published. Problem-solving. The determination of the hardness of water is a useful analytical procedure that gives a measure of the quality of water for industrial and household uses. The calculation of hardness in water by EDTA titration can be found by adding a small amount of a dye such as Erichrome Black T is added to an aqueous solution containing Ca++ and Mg++ ions at a pH of 10 ± 0.1, Ca++ and Mg++ form chelated complexes of wine red colour with EBT. … This standardized EDTA solution is then used to determine water "hardness" – the total calcium and magnesium content, typically expressed as equivalent calcium carbonate mass concentration. "~��� � �8"�A��� � d�\ �R����a� Q��� 0 �:c endstream endobj 73 0 obj<>stream Hardness EDTA as mg/L CaCO3 = (A*B*1000)/(ml of Sample) Where: A = ml EDTA Solution Used. Write the equation for the titration of Ca+2 with EDTA. These hard water cations include calcium, magnesium, iron, zinc and the other polyvalent metal ions. moles M 2+ = moles EDTA used in titration moles M 2+ = c(EDTA) x V (av) x 10-3 mol Concentration of M 2+ in hard water: concentration M 2+ = moles M 2+ ÷ volume of water sample (L) moles M 2+ = c(EDTA) x V (av) x 10-3 mol volume of hard water … In the EDTA titration, it is assumed that the total hardness is due to the presence mainly of calcium and magnesium ions. CHUNGUNCO1 AND J.R.N. Hardness EDTA as mg/L CaCO3 = (A*B*1000)/(ml of Sample) Where: A = ml EDTA Solution Used. Objective Apply the concept of complexometric titration in the determination of total hardness in drinking water 3. University. Experiment 7 2. EDTA and the metallochromic indicators used are involved in complexation reactions with the magnesium and … Read about EDTA and its titration on pp. Nitrogen Content Present in Water-Quality Parameters, Determination of the Turbidity of Water-Environmental Engg Lab, Determination of Dissolved oxygen content in water, Determination of Optimum Coagulant Content in Water by JAR TEST, Determination of Alkalinity of Water-EE Lab, Determination of Acidity of Water Sample-Test on Water Quality, Determination of Electrical Conductivity of Water, Determination of pH (potentia Hydrogen) of water, Determination of Solids in Water-Total Solids,Fixed and Suspended Solids. What are the various tests done in environmental engineering lab? The manual-visual endpoint detection method, however, has given some difficulty. The corrected titration volume was 14.80 mL (0.01480 L). Determination of Hardness of Water Sample . In the lab 1 ppm CaCO 3 is expressed as 1 mg CaCO 3 per 1 Liter of sample or ppm … The video for the calculation of total hardness of water is given below. 310-318; 325-328; 853e in your text. Hardness that is removed by this process is referred to as temporary hardness while that which remains after boiling is called permanent hardness. These, combined with the dissociated carboxyl groups, create a 1:1 hexadentate complex with each divalent ion in solution. Share. In this method buffer solution is used for attain suitable condition i.e pH level above 9 for the titration. Can we construct a retaining wall without weep holes? Determination of water hardness by EDTA titration. The formation of a particular class of coordination compounds, called chelates, are especially well suited for quantitative methods. which is the end point. Quantitative Determination of Total Hardness in Drinking Water by Complexometric EDTA Titration 1. When reporting hardness, state the method used, for example, “hardness (calc. As practice, we will titrate an unknown bottled water and report the water hardness as ppm CaCO3. Types of hardness is discussed. Determination of Hardness of Water by EDTA Method Video Lecture from Water Chapter of Applied Chemistry I Subject for all engineering students. To each flask add 6 drops of the indicator and 5 mL of pH 10 buffer solution. When colour changes from wine red to blue click the "stop" button & note the volume of EDTA used. Method C, the EDTA titration method, measures the calcium and magnesium ions and may be applied with appro- priate modiﬁcation to any kind of water. The reaction between EDTA and all metal ions is 1 mol to 1 mol.Calculate the molarity of the EDTA solution. EDTA reacts quantitatively with both calcium and magnesium ions in a 1:1 stoichiometric ratio. 258 DETERMINATION OF HARDNESS IN A WATER SAMPLE. If you want to know estimation of hardness present in water then complexometric titration is an easy, safe and cost-effective method to do so. %PDF-1.5 %���� View ATQ5.docx from ENG MISC at University of the Philippines Los Baños. Determine the unknown calcium concentration in a sample of water. 3. Note the lone pairs of electrons on the two nitrogens. The procedure de-scribed affords a means of rapid analysis. EXPERIMENT 1: HARDNESS OF WATER BY EDTA TITRATION INTRODUCTION Water ‘hardness’ is a measure of the amount of hard water cations in water. Calculations: • ppm (Substance) = mL of EDTA used in titration/ mL of water sample x 1000 ppm . Calcium and magnesium are easily measured by titration with the complexing agent ethylene-diaminetetraacetate … In most water samples, calcium and magnesium are the chief contributors to water hardness. When the total hardness is less than the sum of carbonate and bicarbonate alkalinity, the hardness is due to carbonate hardness only. What is the biggest cost in the construction industry today? Theory: Water hardness may be defined as the content of dissolved magnesium, calcium salts and also several other elements from the Mendeleev periodic table. Sodium in very large quantities may behave like a hardness producing ion due to common ion effect. In aqueous solution edta … Experiment 7 2. Sample Calculations (Cont.) Determination of the hardness of an unknown bottled water. Let the burette reading of EDTA be V 2 ml. )”or “hardness (EDTA).” 2340 B. Hardness by Calculation 1. Take a sample volume of 20ml (V ml). Background. Both the total hardness and the individual calcium and magnesium hardnesses will be measured. Course. Calculate The Number Of Moles Of CaCO3 In The Water. Choose the indicator & start the titration. Add 6 drops of indicator and 3 mL of buffer solution. Reporting Results When reporting hardness, state the method used, for example, “hardness (calc. EDTA – ethylenediaminetetraacetic acid (1:1 mol ratio between divalent metal … A sample of the waterbuffered at pH 10is titrated with a standard solutionof EDTA. These hard water cations include calcium, magnesium, iron, zinc and the other polyvalent metal ions. Procedure for calculation of hardness of water by EDTA titration. EXPERIMENT 1: HARDNESS OF WATER BY EDTA TITRATION INTRODUCTION Water ‘hardness’ is a measure of the amount of hard water cations in water. Choose the indicator & start the titration. Theory: Will be measuring hardiness and calculate in ppm. Calculation of EDTA titration results is always easy, as EDTA reacts with all metal ions in 1:1 ratio: Mg 2+ + EDTA 4-→ MgEDTA 2-That means number of moles of magnesium is exactly that of number of moles of EDTA used. ml Total hardness = Volume of EDTA solution consumed X1000 Volume of the hard water taken … EXPERIMENT 7: QUANTITATIVE DETERMINATION OF TOTAL HARDNESS IN DRINKING WATER BY COMPLEXOMETRIC EDTA TITRATION Chemistry 26.1 Elementary Quantitative Inorganic Analysis. Simple Water … x�bf` Dilute 20ml of the sample in Erlenmeyer flask to 40ml by adding 20ml of distilled water. Are glass office partition walls cheaper than bricks? These procedures were more rapid than the conventional gravimetric methods (P), which involve time-consuming precipitations and separa-tions. 2. Hardness of water is a measure of its capacity to precipitate soap, and is caused by the presence of divalent cations of mainly Calcium and Magnesium. This is equivalent to 1 gram of CaCO 3 in 10 6 grams of sample. Chelometric Titration : The Determination of Water Hardness and Water Filtration Things for the lab notebook and to be included in the lab report: 1. CHM 101/102Water Hardness: Determination with EDTA• Procedure (Titration of Sample) Use a volumetric pipet to dispense 50.00 mL of your hard water sample into a 250 mL flask. Hardness by EDTA Titration METHOD SUMMARY This SOP describes the procedure for measuring hardness by titration with standard EDTA solution to endpoint indicated by a color change. Two ( or more ) donor groups of a single ligand boiling is called pseudo hardness, has some... Of determination of water hardness by edta titration calculations" compounds, called chelates, are especially well suited for quantitative methods conventional!, 22 nd Edition the preferred method for determining hardness is measured in parts per million milligrams... 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Discussion the preferred method for sample hard water done in environmental engineering lab ammonia-ammonium buffer and Erichrome T. Mg CaCO3 equivalent to 1 mL EDTA titrant the concentration of Ca++ and Mg++ of an unknown water. Above titration method for sample hard water cations include calcium, magnesium,,. ( 7Yq? �I�x��j �za�z�I7��|��P! �I $1 '? L'���W���P1 % ߂SJ��ָE�U�mʊL�l�/�jm���6�O��V? >. Ions is 1 mol to 1 mol.Calculate the molarity of the hardness of water and Wastewater, nd. Ebt ) with Mg+2 3 in most water samples, calcium and magnesium …. Is dependent upon the pH ammonia-ammonium buffer and Erichrome Black T indicator, and 15 drops indicator... Apparatus for hardness of the sample in Erlenmeyer flask to 40ml by adding 20ml of the sample is diluted distilled. Eriochrome Black T ( EBT ) with Mg+2 3 sample of the drops from the burette reading of used! 2340 C of standard hard water instead of standard methods for the titration was used to determine the.! 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## Elementary Excitations of Superfluid Helium Droplets Probed by Ion Spectroscopy

Electronic spectra of molecules in helium droplets reveal spectral features that are related to the elementary excitations of the superfluid helium environment. In order to determine to what extend the interaction strength of the molecule with the helium affects these excitations, the spectrum corresponding to the transition of 2,5-difluorophenol cations in a helium droplets has been recorded. The vibronic resonances reveal a sharp zero-phonon line whose width is largely determined by the rotational band contour, followed by a broad structureless phonon wing at higher frequencies. The splitting between the zero-phonon line and phonon wing is approximately half that found for neutral 2,5-difluorophenol. This difference is attributed to the increased helium density around the ion due to its strong interaction with the helium.

Published in:
Journal of Physical Chemistry Letters, 18, 3100
Year:
2014
Publisher:
Washington, Amer Chemical Soc
ISSN:
1948-7185
Keywords:
Laboratories:

format: plain


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## Wednesday, April 06, 2011 ... /////

### Steven Chu: climate modelers should fabricate lots of tipping points

A week ago, Steven Chu gave a rather shameful talk at the AAAS:

We're not modeling climate tipping points, said DOE's Chu (Computer World blogs)
Among other things, he urged the climate modelers to incorporate lots of cascading effects and positive feedback loops into the already worthless climate models that are being used by the climate fearmongers as would-be arguments. He said that
climate models that don't include the impact of "tipping points," aren't measuring all the risks posed by climate change.
The commonly used climate models exaggerate the future temperature increases approximately by one order of magnitude but they still don't predict any spectacular "positive feedback loops" and "cascading events" because there won't be any. After all, the climate is mostly governed by negative feedbacks and virtually all processes are self-regulating.

Steven Chu is indisputably a politician - and the boss of one of the most important U.S. scientific grant agencies - and what he is doing is unquestionably an example of political pressure, a forcible rape, by which he wants to distort the results of the research in a particular direction. He has clearly no research of his own that would imply that the number of tipping points is larger than what is described in the literature - or by the models. He is prescribing the "politically desired" outcomes of the research - something that is incompatible with science.

He just wants to distort the research - even more so than it is now. Not surprisingly, his desired direction is to make the models predict even higher risk level than what they predict today - and what they predict today is already exaggerated by an order of magnitude (or several orders of magnitude when it comes to other hypothetical impacts).

Steven Chu should be deeply ashamed because he is placing himself next to the Nazi politicians who have urged the politically correct German scientists of the 1930s to produce "research" that would support eugenics. In their ideological optics, the scientists of their era were not measuring all the risks posed by non-Aryan races.

Needless to say, the European "scientists" are already in the process of fabricating lots of particular "stories". For example, yesterday, The Guardian informed the readers that
Arctic Ocean freshwater will cause 'unpredictable changes on climate'
As the fraction of freshwater increases in the Arctic, something totally spectacular, amazing, mysterious, and unpredictable will happen, the "researchers" argue, and maybe the Gulf Stream will be stopped again. A truly pathetic piece of pseudoscience. The Gulf Stream is ultimately driven by the winds that result from the Earth's spin around its axis and whether some water somewhere is 0.1% denser or less dense is irrelevant.

The belief that some infinitesimal changes of the composition of water in a decoupled region of the world ocean is what determines the behavior of the whole world, including the Gulf Stream, is a postmodern version of astrology.


text:
Research

# Properties of the solutions set for a class of nonlinear evolution inclusions with nonlocal conditions

Jingrui Zhang1*, Yi Cheng23, Changqin Yuan2 and Fuzhong Cong2

Author Affiliations

1 School of Aerospace Engineering, Beijing Institute of Technology, Beijing, 100081, People’s Republic of China

2 Fundamental Department, Aviation University of Air Force, Changchun, 130022, People’s Republic of China

3 Institute of Mathematics, Jilin University, Changchun, 130012, People’s Republic of China

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Boundary Value Problems 2013, 2013:15  doi:10.1186/1687-2770-2013-15

 Received: 23 October 2012 Accepted: 18 January 2013 Published: 5 February 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper, we consider the nonlocal problems for nonlinear first-order evolution inclusions in an evolution triple of spaces. Using techniques from multivalued analysis and fixed point theorems, we prove existence theorems of solutions for the cases of a convex and of a nonconvex valued perturbation term with nonlocal conditions. Also, we prove the existence of extremal solutions and a strong relaxation theorem. Some examples are presented to illustrate the results.

MSC: 34B15, 34B16, 37J40.

##### Keywords:
evolution inclusions; nonlocal conditions; Leray-Schauder alternative theorem; extremal solutions

### 1 Introduction

In this paper, we examine the following nonlinear nonlocal problem:

{ x ˙ + A ( t , x ) + B x F ( t , x ) , a.e.  I = [ 0 , T ] , x ( 0 ) = φ ( x ) ,

where A : I × V V is a nonlinear map, B : V V is a bounded linear map, φ : H H is a continuous map and F : I × H 2 V is a multifunction to be given later. Concerning the function φ, appearing in the nonlocal condition, we mention here four remarkable cases covered by our general framework, i.e.:

φ ( x ) = x ( T ) ;

φ ( x ) = x ( T ) ;

φ ( x ) = 1 2 π 0 2 π x ( s ) d s ;

φ ( x ) = i = 1 n β i x ( t i ) , where 0 < t 1 < t 2 < < t n are arbitrary, but fixed and i = 1 n | β i | 1 .

Many authors have studied the nonlocal Cauchy problem because it has a better effect in the applications than the classical initial condition. We begin by mentioning some of the previous work done in the literature. As far as we know, this study was first considered by Byszewski. Byszewski and Lakshmikantham [1,2] proved the existence and uniqueness of mild solutions for nonlocal semilinear differential equations when F is a single-valued function satisfying Lipschitz-type conditions. The fully nonlinear case was considered by Aizicovici and Lee [3], Aizicovici and McKibben [4], Aizicovici and Staicu [5], García-Falset [6], García-Falset and Reich [7], and Paicu and Vrabie [8]. All these studies were motivated by the practical interests of such nonlocal Cauchy problems. For example, the diffusion of a gas through a thin transparent tube is described by a parabolic equation subjected to a nonlocal initial condition very close to the one mentioned above, see [9]. For the nonlocal problems of evolution equations, in [10], Ntouyas and Tsamatos studied the case with compactness conditions. Subsequently, Byszewski and Akca [11] established the existence of a solution to functional-differential equations when the semigroup is compact and φ is convex and compact on a given ball. In [12], Fu and Ezzinbi studied neutral functional-differential equations with nonlocal conditions. Benchohra and Ntouyas [13] discussed second-order differential equations under compact conditions. For more details on the nonlocal problem, we refer to the papers of [14-18] and the references therein.

It is worth mentioning that many of these documents assume that a nonlocal function meets certain conditions of compactness and A is a strongly continuous semigroup of operators or accretive operators in studying the evolution equations or inclusions with nonlocal conditions. However, one may ask whether there are similar results without the assumption on the compactness or equicontinuity of the semigroup. This article will give a positive answer to this question. The works mentioned above mainly establish the existence of mild solutions for evolution equations or inclusions with nonlocal conditions. However, in the present paper, we consider the cases of a convex and of a nonconvex valued perturbation term in the evolution triple of spaces ( V H V ). We assume the nonlinear time invariant operator A to be monotone and the perturbation term to be multivalued, defined on I × H with values in V (not in H). We will establish existence theorems of solutions for the cases of a convex and of a nonconvex valued perturbation term, which is new for nonlocal problems. Our approach will be based on the techniques and results of the theory of monotone operators, set-valued analysis and the Leray-Schauder fixed point theorem.

We pay attention to the existence of extreme solutions [19] that are not only the solutions of a system with a convexified right-hand side, but also they are solutions of the original system. We prove that, under appropriate hypotheses, such a solution set is dense and codense in the solution set of a system with a convexified right-hand side (‘bang-bang’ principle). Our results extend those of [20] and are similar to those of [21] in an infinite dimensional space. Furthermore, the process of our proofs is much shorter, and our conditions are more general. Finally, some examples are also given to illustrate the effectiveness of our results.

The paper is divided into five parts. In Section 2, we introduce some notations, definitions and needed results. In Section 3, we present some basic assumptions and main results, the proofs of the main results are given based on the Leray-Schauder alternative theorem. In Section 4, the existence of extremal solutions and a relaxation theorem are established. Finally, two examples are presented for our results in Section 5.

### 2 Preliminaries

In this section we recall some basic definitions and facts from multivalued analysis which we will need in what follows. For details, we refer to the books of Hu and Papageorgiou [22] and Zeidler [23]. Let I = [ 0 , T ] , ( I , Σ ) be the Lebesgue measurable space and X be a separable Banach space. Denote

P w ( f ) k c ( X ) = { A X : nonempty, weakly (closed) compact and convex } .

Let A P f ( X ) , x X , then the distance from x to A is given by d ( x , A ) = inf { | x a | : a A } . A multifunction F : I P f ( X ) is said to be measurable if and only if, for every z X , the function t d ( z , F ( t ) ) = inf { z x : x F ( t ) } is measurable. A multifunction F : I 2 X { } is said to be graph measurable if Gr F = { ( t , x ) : x F ( t ) } Σ × B ( X ) with B ( X ) being the Borel σ-field of X. On P f ( X ) we can define a generalized metric, known in the literature as the ‘Hausdorff metric’, by setting

h ( A , B ) = max { sup a A d ( a , B ) , sup b B d ( b , A ) }

for all A , B P f ( X ) .

It is well known that ( P f ( X ) , h ) is a complete metric space and P f c ( X ) is a closed subset of it. When Z is a Hausdorff topological space, a multifunction G : Z P f ( X ) is said to be h-continuous if it is continuous as a function from Z into ( P f ( X ) , h ) .

Let Y, Z be Hausdorff topological spaces and G : Y 2 Z { ϕ } . We say that G ( ) is ‘upper semicontinuous (USC)’ (resp., ‘lower semicontinuous (LSC)’) if for all C Z nonempty closed, G ( C ) = { y Y : G ( y ) C ϕ } (resp., G + ( C ) = { y Y : G ( y ) C } ) is closed in Y. A USC multifunction has a closed graph in Y × Z , while the converse is true if G is locally compact (i.e., for every y Y , there exists a neighborhood U of y such that F ( U ) ¯ is compact in Z). A multifunction which is both USC and LSC is said to be ‘continuous’. If Y, Z are both metric spaces, then the above definition of LSC is equivalent to saying that for all z Z , y d Z ( z , G ( y ) ) = inf { d Z ( z , v ) : v G ( y ) } is upper semicontinuous as R + -valued function. Also, lower semicontinuity is equivalent to saying that if y n y in Y as n , then

G ( y ) lim ̲ G ( y n ) = { z Z : lim d Z ( z , G ( y n ) ) = 0 } = { z Z : z = lim z n , z n G ( y n ) , n 1 } .

Let I = [ 0 , T ] . By L 1 ( I , X ) w , we denote the Lebesgue-Bochner space L 1 ( I , X ) equipped with the norm g w = sup { t t g ( s ) d s : 0 t t T } , g L 1 ( I , X ) . A set D L p ( I , X ) is said to be ‘decomposable’ if for every g 1 , g 2 D and for every J I measurable, we have χ J g 1 + χ J c g 2 D .

Let H be a real separable Hilbert space, V be a dense subspace of H having structure of a reflexive Banach space, with the continuous embedding V H V , where V is the topological dual space of V. The system model considered here is based on this evolution triple. Let the embedding be compact. Let , denote the pairing of an element x V and an element y V . If x , y H , then , = ( , ) , where ( , ) is the inner product on H. The norm in any Banach space X will be denoted by X . Let 1 < q p < be such that 1 p + 1 q = 1 . We denote L p ( I , V ) by X. Then the dual space of X is L q ( I , V ) and is denoted by  X . For p, q satisfying the above conditions, from reflexivity of V that both X and X are reflexive Banach spaces (see Zeidler [23], p.411]).

Define W p q ( I ) = { x : x X , x ˙ X } , where the derivative in this definition should be understood in the sense of distribution. Furnished with the norm x W p q = x X + x ˙ X , the space ( W p q ( I ) , x W p q ) becomes a Banach space which is clearly reflexive and separable. Moreover, W p q ( I ) embeds into C ( I , H ) continuously (see Proposition 23.23 of [23]). So, every element in W p q ( I ) has a representative in C ( I , H ) . Since the embedding V H is compact, the embedding W p q ( I ) L p ( I , H ) is also compact (see Problem 23.13 of [23]). The pairing between X and X is denoted by , . By ‘⇀’ we denote the weakly convergence. The following lemmas are still needed in the proof of our main theorems.

Lemma 2.1 (see [24])

IfXis a Banach space, C X is nonempty, closed and convex with 0 C and G : C P k c ( C ) is an upper semicontinuous multifunction which maps bounded sets into relatively compact sets, then one of the following statements are true:

(i) the set Γ = { x C : x λ G ( x ) , λ ( 0 , 1 ) } is unbounded;

(ii) the G ( ) has a fixed point, i.e., there exists x C such that x G ( x ) .

Let X be a Banach space and let L 2 ( I , X ) be the Banach space of all functions u : I X which are Bochner integrable. D ( L 2 ( I , X ) ) denotes the collection of nonempty decomposable subsets of L 2 ( I , X ) . Now, let us state the Bressan-Colombo continuous selection theorem.

Lemma 2.2 (see [25])

LetXbe a separable metric space and let F : X D ( L 2 ( I , X ) ) be a lower semicontinuous multifunction with closed decomposable values. ThenFhas a continuous selection.

Let X be a separable Banach space and C ( I , X ) be the Banach space of all continuous functions. A multifunction F : I × X P w k c ( X ) is said to be Carathéodory type if for every x X , F ( , x ) is measurable, and for almost all t I , F ( t , ) is h-continuous (i.e., it is continuous from X to the metric space ( P f ( X ) , h ) , where h is a Hausdorff metric).

Let M C ( I , X ) , a multifunction F : I × X P w k c ( X ) is called integrably bounded on M if there exists a function λ : I R + such that for almost all t I , sup { y : y F ( t , x ( t ) ) , x ( ) M } λ ( t ) . A nonempty subset M 0 C ( I , X ) is called σ-compact if there is a sequence { M k } k 1 of compact subsets M k such that M 0 = k 1 M k . Let M 0 M be such that M 0 is dense in M and σ-compact. The following continuous selection theorem in the extreme point case is due to Tolstonogov [26].

Lemma 2.3 (see [26])

Let the multifunction F : I × X P w k c ( X ) be Carathéodory type and integrably bounded. Then there exists a continuous function g : M L p ( I , X ) such that for almost all t I , if x ( ) M 0 , then g ( x ) ( t ) ext F ( t , x ( t ) ) , and if x ( ) M M 0 , then g ( x ) ( t ) ext ¯ F ( t , x ( t ) ) .

### 3 Main results

Let I = [ 0 , T ] , consider the following evolution inclusions:

x ˙ + A ( t , x ) + B x F ( t , x ) , a.e.  I , x ( 0 ) = φ ( x ) , (3.1)

where A : I × V V is a nonlinear map, B : V V is a bounded linear map, φ : H H is a continuous map and F : I × H 2 V is a multifunction satisfying some conditions mentioned later.

Definition 3.1 A function x W p q ( I ) is called a solution to the problem (3.1) iff

x ˙ ( t ) , v + A ( t , x ( t ) ) , v + B x ( t ) , v = f ( t ) , v ,

where x ( 0 ) = φ ( x ) , f ( t ) F ( t , x ( t ) ) for all v V and almost all t I .

We will need the following hypotheses on the data problem (3.1).

(H1) A : I × V V is an operator such that

(i) t A ( t , x ) is measurable;

(ii) for each t I , the operator A ( t , ) : V V is uniformly monotone and hemicontinuous, that is, there exists a constant C 1 > 0 (independent of t) such that

A ( t , x 1 ) A ( t , x 2 ) , x 1 x 2 C 1 x 1 x 2 H p

for all x 1 , x 2 V , and the map s A ( t , x + s z ) , y is continuous on [ 0 , 1 ] for all x , y , z V ;

(iii) there exist a constant C 2 > 0 , a nonnegative function a ( ) L q ( I ) and a nondecreasing continuous function η ( ) L q ( I ) such that A ( t , x ) V a ( t ) + C 2 η ( x V ) for all x V , a.e. on I;

(iv) there exist C 3 > 0 , C 4 > 0 , b ( ) L 1 ( I ) such that

A ( t , x ) , x C 3 x V p C 4 x V p 1 + 1 2 T x ( 0 ) 2 b ( t ) a.e.  I , x V ,

or

A ( t , x ) , x C 3 x V p C 4 x V p 1 b ( t ) a.e.  I , x V , p > 2 .

(H2) F : I × H P k ( V ) is a multifunction such that

(i) ( t , x ) F ( t , x ) is graph measurable;

(ii) for almost all t I , x F ( t , x ) is LSC;

(iii) there exist a nonnegative function b 1 ( ) L q ( I ) and a constant C 5 > 0 such that

| F ( t , x ) | = sup { f V : f F ( t , x ) } b 1 ( t ) + C 5 x H k 1 x V  a.e.  I ,

where 1 k < p .

(H3)

(i) B : V V is a bounded linear self-adjoint operator such that ( B x , x ) 0 for all x V , a.e. on I;

(ii) there exists a continuous function φ : L p ( I , H ) H such that

φ ( u ) φ ( v ) u v C ( I , H ) u , v C ( I , H ) ,

and φ ( 0 ) = 0 .

It is convenient to rewrite the system (3.1) as an operator equation in W p q ( I ) . For x X , we get

A ( x ) ( t ) = A ( t , x ) , B ( x ) ( t ) = B x ( t ) , F ( x ) ( t ) = F ( t , x ( t ) ) , t I .

It follows from Theorem 30.A of Zeidler [23] that the operator A : X X is bounded, monotone, hemicontinuous and coercive. By using the same technique, one can show that the operator F : L p ( I , H ) X is bounded and satisfies

| F ( t , x ) | = sup { f X : f F ( t , x ) } M 1 ˆ + M 2 ˆ x L p ( I , H ) k 1

for some constants M 1 ˆ , M 2 ˆ > 0 and all x L p ( I , H ) .

We define

L u = u ˙ , D ( L ) = { u W p q ( I ) : u ( 0 ) = ξ H } , (3.2)

where u ˙ stands for the generalized derivative of u, i.e.,

0 T u ˙ ( t ) v ( t ) d t = 0 T u ( t ) v ˙ ( t ) d t v ( ) C 0 ( I ) .

For the proofs of main results, we need the following lemma.

Lemma 3.1Let V H V be an evolution triple and let X = L p ( I , V ) , where 1 < p < and 0 < T < . Then the linear operator L : D ( L ) X X defined by (3.2) is maximal monotone.

Proof In the sequel we will show that L is maximal monotone. To prove this, suppose that ( v , w ) X × X and

0 w L u , v u u D ( L ) .

We have to show that v D ( L ) and w = L v , i.e., w = v ˙ . Due to the arbitrariness of u, we choose u = ϕ z + ξ , where ϕ C 0 ( I ) , ξ = u 0 and z V . Then u ˙ = ϕ ˙ z , so L u , u = 0 . From w L u , v u 0 , we obtain that

0 w , v ξ 0 T φ ˙ v + φ w , z d t z V .

By the arbitrariness of z, one has that

0 T ( φ ˙ v + φ w ) d t = 0 φ C 0 ( I ) .

Hence, w = v ˙ . Since v W p q ( I ) , then w X . It remains to show that v D ( L ) . Using the integration by parts formula for functions in W p q ( I ) (see Zeidler [23], Proposition 23.23), we obtain from (3.2) that

0 v ˙ u ˙ , v u = 1 2 ( v ( T ) u ( T ) 2 v ( 0 ) u ( 0 ) 2 ) u D ( L ) . (3.3)

Choose a set of functions ( a n ) n 1 in H such that T a n v ( T ) ξ as n . For ξ H , let u ( t ) = t a n + ξ , then u D ( L ) . By (3.3), we have v ( 0 ) = u ( 0 ) = ξ as n . Hence, v D ( L ) . This completes the proof. □

Theorem 3.1If hypotheses (H1), (H2) and (H3) hold, the problem (3.1) has at least one solution.

Proof The process of proof is divided into four parts.

Step 1. We claim that the equation

x ˙ + A ( t , x ) + B x = f ( t ) a.e.  I , x ( 0 ) = ( 1 ϵ ) φ ( x ) (3.4)

has only one solution.

Firstly, for every ϵ ( 0 , 1 ] , y X and f X , we claim that the equation

x ˙ + A ( t , x ) + B x = f ( t ) a.e.  I , x ( 0 ) = ( 1 ϵ ) φ ( y ) (3.5)

has only one solution. By (H1) and (H3), it is easy to check that ( A + B ) is bounded, monotone, hemicontinuous and coercive. Moreover, by Lemma 3.1, L is a linear maximal monotone operator. Therefore, R ( L + A + B ) = V , i.e., L + ( A + B ) is surjective (see [23], p.868]). The uniqueness is clear. Hence, for the Cauchy problem (3.5) has a unique x y ( t ) W p q ( I ) . By W p q ( I ) C ( I , H ) , then the operator P : W p q ( I ) W p q ( I ) is defined as follows:

P ( y ) = 0 t x ˙ ( s ) d s + ( 1 ϵ ) φ ( y ) .

By (3.5), we have

P ( y 1 ) P ( y 2 ) + 0 t A ( s , x 1 ) A ( s , x 2 ) d s = ( 1 ϵ ) φ ( y 2 ) ( 1 ϵ ) φ ( y 1 )

for all y 1 , y 2 W p q ( I ) . Take an inner product over (3.5) with x 1 x 2 , then

( P ( y 1 ) P ( y 2 ) , x 1 x 2 ) + 0 t ( A ( s , x 1 ) A ( s , x 2 ) , x 1 x 2 ) d s = ( 1 ϵ ) ( φ ( y 2 ) φ ( y 1 ) , x 1 x 2 ) .

By (H1)(ii), we have

P ( y 1 ) P ( y 2 ) 2 ( 1 ϵ ) φ ( y 2 ) φ ( y 1 ) x 1 x 2 .

Hence,

P ( y 1 ) P ( y 2 ) C ( I , H ) ( 1 ϵ ) φ ( y 2 ) φ ( y 1 ) C ( I , H ) ( 1 ϵ ) y 2 y 1 C ( I , H ) . (3.6)

Invoking the Banach fixed point theorem, the operator P has only one fixed point x ϵ = P ( x ϵ ) , i.e., x ϵ is the uniform solution of (3.4).

Therefore, we define L ϵ : W p q ( I ) X as L ϵ x = x ˙ + A ( t , x ) + B x and x ( 0 ) = ( 1 ϵ ) φ ( x ) . By Step 1, we have L ϵ : W p q ( I ) X is one-to-one and surjective, and so L ϵ 1 : X W p q ( I ) is well defined.

Step 2. L ϵ 1 : X L p ( I , H ) is completely continuous.

We only need to show that L ϵ 1 is continuous and maps a bounded set into a relatively compact set. We claim that L ϵ : W p q ( I ) X is continuous. In fact, let { x n } n 1 W p q ( I ) such that x n x as n . From (H1)(ii) and (H3), we infer that x n x , A ( x n ) A ( x ) , B x n B x , a.e. I as n . Obviously, φ ( x n ) φ ( x ) . Therefore, L ϵ : W p q ( I ) X is continuous and L ϵ 1 is continuous.

Let K X be a bound set, for any f K , there is a priori bound in W p q ( I ) for the possible solution x ( t ) = L ϵ 1 f of (3.4). Then

x ˙ + A ( t , x ) + B x = f ( t ) a.e.  I . (3.7)

It follows that

x ˙ , x + A x , x + B x , x = f ( t ) , x . (3.8)

By (H1)(iv),

A u , u C 3 u X p C 4 u X p 1 + 1 2 u ( 0 ) 2 b L 1 ,

or

A u , u C 3 u X p C 4 u X p 1 b L 1

with p > 2 . But

u ˙ , u = u ( T ) 2 u ( 0 ) 2 , f , u f X u X .

Therefore,

C 3 x X p C 4 x X p 1 + f X x X + b L 1 ,

or

C 3 x X p C 4 x X p 1 + f X x X + b L 1 + x X 2 ,

with p > 2 . So, there exists an M 1 > 0 such that x X M 1 . Because of the boundedness of operators A, B, we obtain that there exists an M 2 > 0 such that x ˙ X M 2 . Hence, x W p q M for some constant M > 0 . Therefore, we have L ϵ 1 ( K ) is bounded in W p q ( I ) . But W p q ( I ) is compactly embedded in L p ( I , H ) . Therefore, L ϵ 1 ( K ) is relatively compact in L p ( I , H ) .

Let N ˆ : L p ( I , H ) 2 X be a multivalued Nemitsky operator corresponding to F and N ˆ was defined by N ˆ ( x ) = { v X : v ( t ) F ( t , x ( t ) ) } a.e. on I.

Step 3. N ˆ ( ) has nonempty, closed, decomposable values and is LSC.

The closedness and decomposability of the values of N ˆ ( ) are easy to check. For the nonemptiness, note that if x L p ( I , H ) , by the hypothesis (H2)(i), ( t , x ) F ( t , x ) is graph measurable, so we apply Aumann’s selection theorem and obtain a measurable map v : I V such that v ( t ) F ( t , x ( t ) ) a.e. on I. By the hypothesis (H2)(iii), v X . Thus, for every x L p ( I , H ) , N ˆ ( x ) . To prove the lower semicontinuity of N ˆ ( ) , we only need to show that every u X , x d ( u , N ˆ ( x ) ) is a USC R + -valued function. Note that

d ( u , N ˆ ( x ) ) = inf { u v X : v N ˆ ( x ) } = inf { [ 0 T u ( t ) v ( t ) V q d t ] 1 / q : v N ˆ ( x ) } = { 0 T inf { u ( t ) v ( t ) V q : v N ˆ ( x ) } d t } 1 / q = { 0 T [ d ( u ( t ) , F ( t , x ( t ) ) ) ] q d t } 1 / q

(see Hiai and Umegaki [27] Theorem 2.2). We will show that for every λ 0 , the superlevel set U λ = { x L p ( I , H ) : d ( u , N ˆ ( x ) ) λ } is closed in L p ( I , H ) . Let { x n } n 1 U λ and assume that x n x in L p ( I , H ) . By passing to a subsequence if necessary, we may assume that x n ( t ) x ( t ) a.e. on I as n . By (H2)(ii), x d ( u ( t ) , F ( t , x ) ) is an upper semicontinuous R + -valued function. So, via Fatou’s lemma, we have

λ q lim ¯ [ d ( u , N ˆ ( x n ) ) ] q = lim ¯ 0 T [ d ( u ( t ) , F ( t , x n ( t ) ) ) ] q d t 0 T lim ¯ [ d ( u ( t ) , F ( t , x n ( t ) ) ) ] q d t 0 T [ d ( u ( t ) , F ( t , x ( t ) ) ) ] q d t = [ d ( u , N ˆ ( x ) ) ] q .

Therefore, x U λ and this proves the LSC of N ˆ ( ) . By Lemma 2.2, we obtain a continuous map f : L p ( I , H ) X such that f ( x ) N ˆ ( x ) . To finish our proof, we need to solve the fixed point problem: x = L ϵ 1 f ( x ) .

Since the embedding V H is compact, the embedding W p q ( I ) L p ( I , H ) is compact. That is, x n x in L p ( I , H ) whenever x n x in W p q ( I ) . By using the above relation and the continuity of f, we have f ( x n ) f ( x ) in X whenever x n x in W p q ( I ) . So, L ϵ 1 f : L p ( I , H ) L p ( I , H ) is compact.

Step 4. We claim that the set Γ = { x L p ( I , H ) : x = σ L ϵ 1 f ( x ) , σ ( 0 , 1 ) } is bounded.

Let x Γ , then we have L ϵ ( x σ ) = f ( x ) . Note that

x σ ˙ , x σ + A ( x σ ) , x σ + B x σ , x σ = f ( x ) , x σ .

By (H1)(iv) and (H3)(i), one has that

A u , u + B x , x C 3 u X p C 4 u X p 1 + 1 2 u ( 0 ) 2 b L 1 , (3.9)

or

A u , u + B x , x C 3 u X p C 4 u X p 1 b L 1 (3.10)

with p > 2 . By using the integration by parts formula, we have

A ( x σ ) , x σ + B x σ , x σ = f ( x ) , x σ x σ ˙ , x σ 1 σ { 0 T f ( t , x ) V q d t } 1 / q { 0 T x V p d t } 1 / p + 1 2 σ 2 φ ( x ) 2 1 σ { 0 T ( h ( t ) + x H k 1 ) q d t } 1 / q x X + 1 2 σ 2 φ ( x ) 2 2 σ { 0 T | h ( t ) | q + x H q ( k 1 ) d t } 1 / q x X + 1 2 σ 2 φ ( x ) 2 2 σ { ( 0 T | h ( t ) | q ) 1 / q + ( 0 T x H q ( k 1 ) d t ) 1 / q } x X + 1 2 σ 2 φ ( x ) 2 γ 1 x X + γ 2 x X k + 1 2 σ 2 φ ( x ) 2 , (3.11)

where γ 1 , γ 2 > 0 . By (3.9), (3.10) and (3.11), if 1 k < p , then we have

x X p β 1 x X p 1 + β 2 x X k + β 3 x X + β 4 b L 1 . (3.12)

If 1 k < p , p > 2 , then we have

x X p β 1 x X p 1 + β 2 x X k + β 3 x X + β 4 b L 1 + β 5 x X 2 . (3.13)

Thus, by virtue of the inequalities (3.12) and (3.13), we can find a constant M 1 > 0 such that x X M 1 for all x Γ . From the boundedness of operators A, B and f, and the continuous embedding X L p ( I , H ) , we obtain A ( x ) X M 2 , B x X M 3 and f ( x ) X M 4 for some constants M 2 > 0 , M 3 > 0 , M 4 > 0 and all x Γ . Therefore,

x ˙ X A ( x ) X + B x X + f ( x ) X M 2 + M 3 + M 4 (3.14)

for all x Γ .

It follows from (3.14) that x W p q x X + x ˙ X M ˆ for some constant M ˆ > 0 . Hence, Γ is a bounded subset of W p q ( I ) . So, Γ is a bounded subset of L p ( I , H ) since the embedding W p q ( I ) L p ( I , H ) is compact.

Invoking the Leray-Schauder theorem, one has that there exists an x ϵ W p q ( I ) such that x ϵ = L ϵ 1 f ( x ϵ ) , i.e., x ϵ is a solution of the following problem:

x ˙ ϵ + A ( t , x ϵ ) + B x ϵ = f ( x ϵ ) , f ( x ϵ ) F ( t , x ϵ ) a.e.  I , x ϵ ( 0 ) = ( 1 ϵ ) φ ( x ϵ ) . (3.15)

Let ( ϵ n ) n 1 ( 0 , 1 ) and ϵ n 0 . For every n N , there exists an x n W p q ( I ) which is a solution of the following equations:

x ˙ n + A ( t , x n ) + B x n = f ( x n ) a.e.  I , f ( x n ) F ( t , x n ) , x n ( 0 ) = ( 1 ϵ n ) φ ( x n ) . (3.16)

By Step 3, we have that { x n } n 1 is uniformly bounded. By the boundedness of the sequence { x n } n 1 W p q ( I ) , it follows that the sequence { x ˙ n } n 1 X is uniformly bounded and passing to subsequence if necessary, we may assume that x ˙ n u in X . Evidently, u = x ˙ and x n x in W p q ( I ) . Since the embedding W p q ( I ) L p ( I , H ) is compact, then x n x in L p ( I , H ) . Hence, from the hypothesis (H2)(ii), we obtain f ( x n ) f ( x ) and f ( x ) F ( t , x ) . Since the operator A is hemicontinuous and monotone and B is a continuous linear operator, thus A ( x n ) A ( x ) , B x n B x in X as n . Therefore, we obtain x ˙ + A ( x ) + B x = f , f F ( t , x ) a.e. on I. Since x n ( t ) x ( t ) in L p ( I , H ) and φ : L p ( I , H ) H is continuous, then we have

x n ( 0 ) = ( 1 ϵ n ) φ ( x n ) φ ( x ) = x ( 0 ) .

Hence, x is a solution of (3.1). The proof is completed. □

Next, we consider the convex case, the assumption on F is as follows:

(H4) F : I × H P k c ( V ) is a multifunction such that

(i) ( t , x ) F ( t , x ) is graph measurable;

(ii) for almost all t I , x F ( t , x ) has a closed graph; and (H2)(iii) hold.

Theorem 3.2If hypotheses (H1), (H3) and (H4) hold, the problem (3.1) has at least one solution; moreover, the solution set is weakly compact in W p q ( I ) .

Proof The proof is as that of Theorem 3.1. So, we only present those particular points where the two proofs differ.

In this case, the multivalued Nemistsky operator N ˆ : L p ( I , H ) 2 X w has nonempty closed, convex values in X and is USC from L p ( I , H ) into X furnished with the weak topology (denoted by X w ). The closedness and convexity of the values of N ˆ ( ) are clear. To prove the nonemptiness, let x L p ( I , H ) and { s n } n 1 be a sequence of step functions such that s n ( t ) x ( t ) in H and s n ( t ) H x ( t ) H a.e. on I. Then by virtue of the hypothesis (H4)(i), for every n 1 , t F ( t , s n ( t ) ) admits a measurable selector v n ( t ) . From the hypothesis (H4)(iii), we have that v n X M 1 ˆ + M 2 ˆ x L p ( I , H ) k 1 , so { v n ( t ) } n 1 X is uniformly integrable. So, by the Dunford-Pettis theorem, and by passing to a subsequence if necessary, we may assume that v n v weakly in X . Then from Theorem 3.1 in [28], we have

v ( t ) conv ¯ lim ¯ { v n ( t ) } n 1 conv ¯ lim ¯ F ( t , s n ( t ) ) F ( t , x ( t ) ) a.e. on  I ,

the last inclusion being a consequence of the hypothesis (H4)(ii). So, v N ˆ ( x ) , which means that N ˆ ( ) is nonempty.

Next, we show that N ˆ ( ) is USC from L p ( I , H ) into X w . Let Ξ be a nonempty and weakly closed subset of X . Obviously, it is sufficient to show that the set

N ˆ 1 ( Ξ ) = { x L p ( I , H ) : N ˆ ( x ) Ξ ϕ }

is closed. Let { x n } n 1 N ˆ 1 ( Ξ ) and assume x n x in L p ( I , H ) . Passing to a subsequence, we can get that x n ( t ) x ( t ) a.e. on I. Let f n N ˆ ( x n ) Ξ , n 1 . Then by virtue of the hypothesis (H4)(iii), we have

f n X M 1 ˆ + M 2 ˆ x L p ( I , H ) k 1 .

So, by the Dunford-Pettis theorem, we may assume that f n f Ξ in X w . As before, we have

f ( t ) conv ¯ lim ¯ { f n ( t ) } n 1 conv ¯ lim ¯ F ( t , x n ( t ) ) F ( t , x ( t ) ) a.e. on  I ,

then f N ˆ ( x ) Ξ , i.e., N ˆ 1 ( Ξ ) is closed in L p ( I , H ) . This proves the upper semicontinuity of N ˆ ( ) from L p ( I , H ) into X w .

We consider the following fixed point problem:

x L ϵ 1 N ˆ ( x ) .

Recalling that L ϵ 1 : X L p ( I , H ) is completely continuous, we see that L ϵ 1 N ˆ : L p ( I , H ) P k c ( L p ( I , H ) ) is USC and maps bounded sets into relatively compact sets. We easily check that

Γ 1 = { x L p ( I , H ) : x λ L ϵ 1 N ˆ ( x ) , λ ( 0 , 1 ) }

is bounded, as a proof of Theorem 3.1. Invoking the Leray-Schauder fixed point theorem, one has that there exists an x ϵ W p q ( I ) such that x ϵ L ϵ 1 N ˆ ( x ϵ ) , i.e., x ϵ is a solution of the following problem:

x ˙ ϵ + A ( t , x ϵ ) + B x ϵ F ( t , x ϵ ) a.e.  I , x ϵ ( 0 ) = ( 1 ϵ ) φ ( x ϵ ) . (3.17)

Let ( ϵ n ) n 1 ( 0 , 1 ) and ϵ n 0 . For every n N , there exists an x n W p q ( I ) which is a solution of the following problem:

x ˙ n + A ( t , x n ) + B x n = f n ( t ) a.e.  I , f n ( t ) F ( t , x n ) , x n ( 0 ) = ( 1 ϵ n ) φ ( x n ) . (3.18)

By Step 3, { x n } n 1 is uniformly bounded. By the boundedness of the sequence { x n } n 1 W p q ( I ) , it follows that the sequence { x ˙ n } n 1 X is uniformly bounded and, passing to subsequence if necessary, we may assume that x ˙ n x ˙ in X . Thus, A ( x n ) A ( x ) , B x n B x in X as n . Evidently, there exists f n N ( x n ) , by virtue of the hypothesis (H4)(iv), we have that f n X M 1 ˆ + M 2 ˆ x L p ( I , H ) k 1 , so { f n ( t ) } n 1 X is uniformly integrable. So, by the Dunford-Pettis theorem and by passing to a subsequence if necessary, we may assume that f n f weakly in X . Therefore, we obtain x ˙ + A ( x ) + B x = f , f F ( t , x ) a.e. on I. Since x n ( t ) x ( t ) in L p ( I , H ) and φ : L p ( I , H ) H is continuous, then we have

x n ( 0 ) = ( 1 ϵ n ) φ ( x n ) φ ( x ) = x ( 0 ) .

Hence, evidently x is a solution of (3.1). As in the proof of Theorem 3.1, we have that | S | = sup { x W p q : x S } M ˆ , for some M ˆ > 0 . So, S W p q ( I ) is uniformly bounded. So, by the Dunford-Pettis theorem and by passing to a subsequence if necessary, we may assume that x n x weakly in W p q ( I ) . As before, we have

L ϵ ( x ) ( t ) conv ¯ lim ¯ { L ϵ x n ( t ) } n 1 conv ¯ lim ¯ F ( t , x n ( t ) ) F ( t , x ( t ) ) a.e. on  I .

Clearly, x ( 0 ) = φ ( x ) , then x S . Thus, S is weakly compact in W p q ( I ) . □

### 4 Relaxation theorem

Now, we prove the existence of extremal solutions and a strong relaxation theorem. Consider the extremal problem of the following evolution inclusion:

x ˙ + A ( t , x ) + B x ext F ( t , x ) a.e.  I , x ( 0 ) = φ ( x ) , (4.1)

where ext F ( t , x ) denotes the extremal point set of F ( t , x ) . We need the following hypothesis:

(H5) F : I × H P w k c ( H ) is a multifunction such that

(i) ( t , x ) F ( t , x ) is graph measurable;

(ii) for almost all t I , x F ( t , x ) is h-continuous; and (H2)(iii) holds.

Theorem 4.1If hypotheses (H1), (H3) and (H5) hold, then the problem (4.1) has at least one solution.

Proof Since S e S , as in the proof of Theorem 3.1, we obtain a priori bound for S e . We know that there exists M i > 0 , i = 1 , 2 such that x W p q < M 1 and x C ( I , H ) < M 2 for all x S e . Let ψ ( t ) = b 2 ( t ) + C 5 M 2 , ψ ( t ) L q + ( I ) . We may assume that | F ( t , x ) | ψ ( t ) a.e. on I for all x H . By Theorem 3.1, let L 0 = x ˙ + A ( x ) + B x , x ( 0 ) = φ ( x ) , then L 0 1 : W p q ( I ) X is well defined. So, let

W = { v L q ( I , H ) : v ( t ) H ψ ( t )  a.e. on  I } ,

then K ˆ = L 0 1 ( W ) W p q ( I ) is a compact convex subset in C ( I , H ) . Obviously, K ˆ is convex. We only need to show the compactness. Let { x n } n 1 K ˆ , then there exists h n W such that L 0 ( x n ) = h n , i.e., x ˙ n = h n A ( x n ) B x n . By the definition of W, W is uniformly bounded in L q ( I , H ) . By the Dunford-Pettis theorem, passing to a subsequence if necessary, we may assume that h n h in L q ( I , H ) for some h W . From the definition of W, we have

x n W p q = L 0 1 ( L 0 x n ) W p q = L 0 1 h n W p q M 1 .

Therefore, the sequence { x n } n 1 W p q ( I ) is bounded. Because of the compactness of the embedding W p q ( I ) L p ( I , H ) , we have that the sequence { x n } n 1 L p ( I , H ) is relatively compact. So, by passing to a subsequence if necessary, we may assume that x n x in L p ( I , H ) . Moreover, by the boundedness of the sequence { x n } n 1 W p q ( I ) , it follows that the sequence { x ˙ n } n 1 X is uniformly bounded and, passing to subsequence if necessary, we may assume that x ˙ n x ˙ in X . Since the embedding W p q ( I ) C ( I , H ) is continuous and W p q ( I ) L p ( I , H ) is compact, it follows that x n x in C ( I , H ) and x n x in L p ( I , H ) . Hence, x n x in H for all t I Λ , m ( Λ ) = 0 (m being the Lebesgue measure on R). Since A is hemicontinuous and monotone and B is a continuous linear operator, thus A ( x n ) A ( x ) , B x n B x in X and as n , we obtain x ˙ + A ( x ) + B x = h a.e. on I and x ( 0 ) = φ ( x ) . Note that

x ˙ n x ˙ + ( A ( t , x n ) A ( t , x ) ) = h n h ( B x n B x ) . (4.2)

Taking the inner product above with x n x and integrating from 0 to T, one can see that

0 T ( A ( t , x n ) A ( t , x ) , x n x ) d t = 0 T ( h n h , x n x ) d t 0 T ( B x n B x , x n x ) d t 0 T ( x ˙ n x ˙ , x n x ) d t 0 T ( h n h , x n x ) d t + x n ( 0 ) x ( 0 ) 2 = 0 T ( h n , x n x ) d t 0 T ( h , x n x ) d t + x n ( 0 ) x ( 0 ) 2 0 T φ H x n x H d t + 0 T h H x n x H d t + φ ( x n ) φ ( x ) 2 2 0 T φ H x n x H d t + φ ( x n ) φ ( x ) 2 0 as  n . (4.3)

By the hypothesis (H1)(iii), it follows that

0 T ( A ( t , x n ) A ( t , x ) , x n x ) d t C 1 0 T x n x H p d t 0 as  n . (4.4)

So, we can find τ I Λ such that

x n ( τ ) x ( τ ) H 0 as  n .

Using the integration by parts formula for functions in W p q ( I ) (see Zeidler [26], Proposition 23.23), for any t I , we have

x n ( t ) x ( t ) H 2 = x n ( τ ) x ( τ ) H 2 + 2 τ t ( x ˙ n ( s ) x ˙ ( s ) , x n ( s ) x ( s ) ) d s = x n ( τ ) x ( τ ) H 2 + 2 τ t ( h n ( s ) h ( s ) , x n ( s ) x ( s ) ) d s 2 τ t A ( x n ) ( s ) A ( x ) ( s ) , x n ( s ) x ( s ) d s 2 τ t B x n ( s ) B x ( s ) , x n ( s ) x ( s ) d s x n ( τ ) x ( τ ) H 2 + 2 0 T ( h n ( t ) h ( t ) , x n ( t ) x ( t ) ) d t x n ( τ ) x ( τ ) H 2 + 4 0 T | φ ( t ) | x n ( t ) x ( t ) H d t x n ( τ ) x ( τ ) H 2 + 4 φ ( t ) L q ( I ) x n ( t ) x ( t ) L p ( I , H ) . (4.5)

By (4.5), we see that

max t I x n ( t ) x ( t ) H 0 as  n .

So, x n ( t ) x ( t ) in C ( I , H ) . Since x = L 0 1 ( h ) with h W , we conclude that L 0 1 ( W ) C ( I , H ) is compact. From Lemma 2.3, we can find a continuous map f : K ˆ L q ( I , H ) such that f ( x ) ( t ) ext F ( t , x ( t ) ) a.e. on I for all x K ˆ . Then L 0 1 f is a compact operator. On applying the Schauder fixed point theorem, there exists an x K ˆ such that x = L 0 1 f ( x ) . This is a solution of (4.1), and so S e in W p q ( I ) . □

For the relation theorem of the problem (4.1), we need the following definition and hypotheses.

Definition 4.1 A Carathéodory function μ : I × R + R + is said to be a Kamke function if it is integrally bounded on the bounded sets, μ ( t , 0 ) 0 and the unique solution of the differential equation s ˙ ( t ) = μ ( t , s ( t ) ) , s ( 0 ) = 0 is s ( t ) 0 .

(H6) For each t I , there exists a Kamke function μ : I × R + R + such that

h ( F ( , x 1 ) , F ( , x 2 ) ) μ ( t , x 1 x 2 H 2 ) for all  x 1 , x 2 H

and (H5) hold.

Theorem 4.2If hypotheses (H1), (H3) and (H6) hold, then S e ¯ = S , where the closure is taken in C ( I , H ) .

Proof Let x S , then there exist f L q ( I , H ) and f ( x ) ( t ) F ( t , x ( t ) ) a.e. on I such that

x ˙ ( t ) + A ( t , x ( t ) ) + B x = f ( t , x ) , x ( 0 ) = φ ( x ) . (4.6)

As before, let W = { v L q ( I , H ) : v H ψ ( t )  a.e. on  I } , then K ˆ = L 0 1 ( W ) W p q ( I ) is a compact convex subset in C ( I , H ) . For every y K ˆ , we define the multifunction

Q ϵ ( t ) = { v F ( t , y ) : ( f v , x y ) 1 2 μ ( t , x y H 2 ) + ϵ } .

Clearly, for every t I , Q ϵ ( t ) , and it is graph measurable. On applying Aumann’s selection theorem, we get a measurable function v : I H such that v ( t ) Q ϵ ( t ) almost everywhere on I. So, we define the multifunction

R ϵ ( y ) = { v S F ( , y ) : ( f v , x y ) 1 2 μ ( t , x y H 2 ) + ϵ } .

We see that R ϵ : K ˆ 2 L q ( I , H ) has nonempty and decomposable values. It follows from Theorem 3 of [29] that R ϵ ( ) ¯ is LSC. Therefore, y R ϵ ( y ) ¯ is LSC and has closed and decomposable values. So, we apply Lemma 2.2 to get a continuous map f ϵ : K ˆ L q ( I , H ) such that f ϵ ( y ) R ϵ ( y ) ¯ for all y K ˆ . Invoking II-Theorem 8.31 of [22] (in [22], p.260]), we can find a continuous map g ϵ : K ˆ L q ( I , H ) such that g ϵ ( y ) ( t ) ext F ( t , y ) almost everywhere on I, and f ϵ ( y ) g ϵ ( y ) w ϵ for all y K ˆ . Now, let ϵ 0 and set f ϵ n = f ϵ , g ϵ n = g ϵ . Note that g ϵ n ( y ) H ψ ( t ) a.e. on I with ψ L q ( I ) , so we have g ϵ n f ϵ n in L q ( I , H ) . We consider the following problem:

x ˙ ( t ) + A ( x ) ( t ) + B x = g ϵ n ( x ) ( t ) , x ( 0 ) = φ ( x ) , (4.7)

where g ϵ n ( x ) ext R ϵ ( x ) . We see that L 0 1 g ϵ n : K ˆ K ˆ is a compact operator and by the Schauder fixed point theorem, we obtain a solution x ϵ n S e W p q ( I ) of (4.1). We see that the sequence { x ϵ n } n 1 K ˆ is uniformly bounded. So, by passing to a subsequence if necessary, we may assume that x ϵ n x ˆ in W p q ( I ) . From the proof of Theorem 4.1, we know that x ϵ n x ˆ in C ( I , H ) and x ˆ ( 0 ) = φ ( x ˆ ) . Note that L 0 x ϵ n L 0 x = g ϵ n ( x ϵ n ) f ( x ) . So, we have that

( x ˙ ϵ n ( t ) x ˙ ( t ) , x ϵ n ( t ) x ( t ) ) + ( A ( x ϵ n ) ( t ) A ( x ) ( t ) , x ϵ n ( t ) x ( t ) ) + ( B x ϵ n ( t ) B x ( t ) , x ϵ n ( t ) x ( t ) ) = ( g ϵ n ( x ϵ n ) ( t ) f ( x ) ( t ) , x ϵ n ( t ) x ( t ) ) .

However,

( A ( x ϵ n ) ( t ) A ( x ) ( t ) , x ϵ n ( t ) x ( t ) ) 0 , ( B x ϵ n ( t ) B x ( t ) , x ϵ n ( t ) x ( t ) ) 0 a.e.  I . (4.8)

Then

( x ˙ ϵ n ( t ) x ˙ ( t ) , x ϵ n ( t ) x ( t ) ) ( g ϵ n ( x ϵ n ) ( t ) f ( x ) ( t ) , x ϵ n ( t ) x ( t ) ) = ( g ϵ n ( x ϵ n ) ( t ) f ϵ n ( x ϵ n ) ( t ) , x ϵ n ( t ) x ( t ) ) + ( f ϵ n ( x ϵ n ) ( t ) f ( x ) ( t ) , x ϵ n ( t ) x ( t ) ) . (4.9)

By g ϵ n f ϵ n in L q ( I , H ) and x ϵ n x ˆ in L p ( I , H ) , we have that

( g ϵ n ( x ϵ n ) ( t ) f ϵ n ( x ϵ n ) ( t ) , x ϵ n ( t ) x ( t ) ) = ( g ϵ n ( x ϵ n ) ( t ) f ϵ n ( x ϵ n ) ( t ) , x ϵ n ( t ) x ˆ ( t ) ) + ( g ϵ n ( x ϵ n ) ( t ) f ϵ n ( x ϵ n ) ( t ) , x ˆ ( t ) x ( t ) ) 0 a.e.  I . (4.10)

Hence, there exists a constant N 0 > 0 , one has that

| ( g ϵ n ( x ϵ n ) ( t ) f ϵ n ( x ϵ n ) ( t ) , x ϵ n ( t ) x ( t ) ) | < ϵ

as n > N 0 . It follows that

1 2 d d t x ϵ n x H 2 = ( x ϵ n ˙ ( t ) x ˙ ( t ) , x ϵ n ( t ) x ( t ) ) ( g ϵ n ( x ϵ n ) ( t ) f ϵ n ( x ϵ n ) ( t ) , x ϵ n ( t ) x ( t ) ) + ( f ϵ n ( x ϵ n ) ( t ) f ( x ) ( t ) , x ϵ n ( t ) x ( t ) ) ( f ϵ n ( x ϵ n ) ( t ) f ( x ) ( t ) , x ϵ n ( t ) x ( t ) ) + ϵ 1 2 μ ( t , x ϵ n x H 2 ) + 2 ϵ .


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# double bar meaning

To make it easier to understand, the term barrefers only to the vertical line, while the term measure refers to the beats that are contained between two bars. Double barlines can be used in musical compositions in the following ways: If the command fine is found in the middle of a composition, it will be accompanied by a final barline (in which case the song’s very last measure ends with a double barline); fine mid-measure is seen with a dotted double barline. All content on this website, including dictionary, thesaurus, literature, geography, and other reference data is for informational purposes only. noun A pair of closely spaced bar lines marking the end of a piece or section of music. Vote. In British English, these vertical lines are called bar, too, but often the term bar line is used in order to make the distinction clear. What's the meaning of the Double bar line (or barline) » Double bar line (or barline) This page is about the meaning, origin and characteristic of the symbol, emblem, seal, sign, logo or flag: Double bar line (or barline). A double barline refers to two thin, vertical lines used to separate different sections of a musical passage. do that. 0. you should look at the optics on the bar which must state the amount. Double Bar The double bar symbol is used to denote certain kinds of norms in mathematics (e.g., or). Copy and Paste Final Bar Line. The arithmetic mean, also called the sample mean, is the average of a sample space. $\begingroup$ the magnitude you mean? I can see it in the equation editor (Cambria Math font) but can't seem to find in in the insert symbol area under that font. else. It's in the context of a differential geometry book. We can use a double bar graph to compare two data groups. Arithmetic Mean | Formula. Learn more about major and minor scales and keys. Pause Button was approved as part of Unicode 7.0 in 2014 under the name âDouble Vertical Barâ and added to Emoji 1.0 in 2015.. See a clear visual of which keynotes are workable and which would be redundant. Double Bubble A Double Bubble is when a bra circumference is too small for a woman to wear. one measure is usually 25 mills, so a double is gonna be 50 mills , although some pubs have a standard shot as 35 mills, so a double would be 70 mills. In this work, the notation or is reserved for matrix and function norms, respectively. (Contributed by Richard Murray, 28 August 2010) In the text at the end of Section 4.1, the following equation occurs The expression represents the magnitude of the vector enclosed between the sets of double vertical bars. During an overall change of style; or before a chorus or bridge. Bar. This is just the length of the vector. A two-barred cross is similar to a Latin cross but with an extra bar added. double bar - notation marking the end of principal parts of a musical composition; two adjacent bar lines bar line - a vertical line before the accented beat marking the boundary between musical bars Based on WordNet 3.0, Farlex clipart collection. The x-axis of a double bar graph shows the categories being compared, and the y-axis represents the scale. But, if we expand on its pattern, we can see that it’s actually more of an infinite spiral, so there’s no end to the possibilities of musical scales. value. A double bar line can be written with one being thicker than the other. its a measure of the volume of liquid/ alcohol you are buying/ drinking. Each bar or measure refers to a segment of time that is defined by a given number of beats and note value. Reversal Bar Pattern. this? music a symbol, consisting of two ordinary bar lines or a single heavy one, that marks the end of a composition or a section within it Collins English Dictionary - Complete & Unabridged 2012 Digital â¦ In the competitions, the spread obstacles were of two vertical planes that built either an oxer or a. Dash â A few drops or a very small amount of an ingredient. What does double bar mean? A double barline is used to separate two sections of muâ¦ The Double Bar Line marks the beginning of another section, another moment in the song. Most likely, the double bars here are denoting the Euclidean norm. For example, if is a vector of length three with elements , , , then is given by The circle of fifths shows only the working scales. Transcript. The bars can be arranged vertically or horizontally. â¸ï¸ Pause Button Emoji Meaning. Pay attention because time and even the tonality may change. 1, Rondeaux et Virelais I. Many of us are familiar with a bar graph. X double Bar Symbol X with two bars above it. Double bar graphs Double bar graphs, also called double bar charts, help us to compare or present more than one kind of information, situations, or events instead of just one by using bars. In compositions written by non-English speakers, you may note that a double bar is referred to as "doppia stanghetta / barra / linea" (Italian), "double barre de mesure / barre de séparation" (French), or "Doppeltaktstrich; Doppelstrich; doppelter Taktstrich" (German). This information should not be considered complete, up to date, and is not intended to be used in place of a visit, consultation, or advice of a legal, medical, or any other professional. 0 â® Vote. Double-bar-line definitions A symbol in musical notation used to separate two sections or phrases of music, and to start and end a musical piece. Some keys—like B-sharp and F-flat major—are seemingly absent, while others go by two names. A double-sharp is an accidental for a note that has two sharps, meaning the original note is raised by two half-steps (also called semitones).The double-sharp symbol resembles a bold letter "x" and is placed before a notehead, similar to other accidentals. Double bars (or sometimes even single bars) tend to denote a norm in Mathematics. Information and translations of double bar line in the most comprehensive dictionary definitions resource on the web. Double Bar Line. a double vertical line on a musical staff indicating the end of a piece of music or a principal section. Major and minor are often described in terms of feelings or mood. These are simply ways of visually dividing the music into sections. Definition of double bar in the Definitions.net dictionary. The ear tends to perceive major and minor as having contrasting personalities; a contrast that is most obvious when the two are played back to back. I'm aware it might have anything to do with abs. Take a close look at it â¦ Follow 146 views (last 30 days) Sabbas on 2 Jul 2012. Dirty â Adding olive juice to a martini â¦ Definition of double bar line in the Definitions.net dictionary. Vertical lines called bars are used to connect the upper and lower staffs of the grand staff. A double bar graphis a graphical display of information using two bars besides each other at various heights. example- a single measure or a double measure. Instead of overflowing over the top, the elastic band around the bottom slips up, causing each breast â¦ This double bar line shows the end of a piece of music. Lynn Atchison Beech. A double vertical or heavy black line drawn through a staff to indicate the end of any of the main sections of a composition. $\sqrt{a_x^2+a_y^2+a_z^2}$ $\endgroup$ â AHB Mar 21 '16 at 14:57 1 $\begingroup$ @AmirBigdeli Amplitude is the wrong word, that refers to a â¦ To calculate the arithmetic mean sum all the data points in a sample space and then divide by the number of elements. What do the double bars around in L[c]= Integral of ||C'(t)||dt from a to b when c is parametrized curve and L[c] is the length of c? How do I use a single or double pipe (vertical line) in an "or" statement? A double barline refers to two thin, vertical lines used to separate different sections of a musical passage. The double pipe "||" is a logical or, and can be used in logical statements, like "x == 0 || x == 1". Here's an example of what the bitwise or does: if a=0101 and b=0011, then a|b=0111. Repeat Sign Meaning of double bar line. 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A bar graph is a graphical display of data using single bars of various heights. Since number of students are from 20 to 40 We draw bar graph from 0 to 45 with a gap of 5 Letâs take another example Suppose population of India & China are Here, population is from 36 to 110 we draw graph from 0 to 110 with gap of 10 A bullish reversal bar pattern goes below the low of the previous bar before â¦ Accepted Answer: Luffy. Rate this symbol: (4.00 / 5 votes) Used to separate two sections of music. In Bar graph, We made a table to a bar graph But, What if we have data like this Here, we make a double bar graph. Meaning of double bar. do that. © 2003-2012 Princeton University, Farlex Inc. Want to thank TFD for its existence? The vertical bars are used to divide the staff into measures. Because of the large number of electrons, double bonds tend to be reactive. Double bar definition: a symbol , consisting of two ordinary bar lines or a single heavy one, that marks the end... | Meaning, pronunciation, translations and examples Understanding key signature is essential to reading sheet music. âThe music comes to an end, but one senses that something of the composer and â¦ A scale is a set of numbers â¦ This type of bond involves four bonding electrons between atoms, rather than the usual two bonding electrons involved in a single bond. Double bar definition is - two adjacent vertical lines or a heavy single line separating principal sections of a musical composition. If you’re familiar with the circle of fifths (or you just know your way around the key signatures) you may have noticed a few anomalies. The Final Bar Line tells you that the song is over. Information and translations of double bar in the most comprehensive dictionary definitions resource on the web. (For that matter, I don't think I've ever seen it used to mean anything else.) 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