Multiple positive solutions for a class of nonlinear four-point boundary value problem with p-La
公共交通好处 英语作文
公共交通好处英语作文全文共3篇示例，供读者参考篇1The Benefits of Public TransportationAs students, we are always looking for ways to save money and reduce our environmental impact. One solution that can help address both of these concerns is using public transportation. While it may not be the most convenient option for everyone, the advantages of taking the bus, train, or subway are numerous and significant.Cost SavingsLet's start with the financial benefits. For most students, money is tight. We are often living on a limited budget, juggling tuition fees, textbook costs, rent, and other expenses. Owning and operating a personal vehicle can be a substantial financial burden. Between gas, insurance, maintenance, and parking fees, the costs quickly add up.In contrast, using public transportation is much more affordable. Many cities offer discounted transit passes or fare options for students, making it an extremely cost-effective wayto get around. Even without a student discount, the cost of a monthly transit pass is usually a fraction of what it would cost to own and operate a car.Environmental ImpactIn addition to the cost savings, using public transportation is also a responsible choice for the environment. Transportation is a major contributor to greenhouse gas emissions and air pollution, and personal vehicles are a significant part of the problem.When we choose to take the bus, train, or subway instead of driving, we are reducing our carbon footprint and helping to mitigate the negative impacts of transportation on the environment. Public transit vehicles are generally morefuel-efficient than personal cars, and by combining multiple passengers into a single vehicle, we are reducing the overall emissions per person.Moreover, by reducing the number of cars on the road, we are also helping to alleviate traffic congestion, which can further reduce emissions and improve air quality in urban areas.Time ManagementAnother often-overlooked benefit of public transportation is the opportunity it provides for better time management. As students, our schedules are frequently packed with classes, assignments, extracurricular activities, and social commitments. Driving can be a stressful and time-consuming endeavor, especially in urban areas with heavy traffic.On the other hand, when we take public transit, we can use that commute time more productively. Whether it's catching up on reading, working on assignments, or simply taking awell-deserved break, the time spent on the bus or train can be a valuable opportunity to multitask and make the most of our day.SafetyAnother advantage of public transportation is increased safety. Driving can be a risky activity, with the potential for accidents, injuries, and traffic violations. By leaving the driving to professional operators, we can reduce our risk of being involved in a collision and avoid the hassle of navigating congested streets and finding parking.Additionally, many public transit systems have implemented safety measures such as security cameras and emergency call buttons, providing an extra layer of protection for passengers.Community BuildingBeyond the practical benefits, using public transportation can also contribute to a sense of community and social cohesion. When we share a ride with our fellow citizens, we are exposed to a diverse cross-section of society and have the opportunity to interact with people from different backgrounds and walks of life.This exposure can foster greater understanding, empathy, and a shared sense of belonging to a larger community. It can also help break down social barriers and promote a more inclusive and connected society.AccessibilityFinally, public transportation plays a crucial role in ensuring accessibility for individuals with disabilities or mobility challenges. Many transit systems are designed with features such as low-floor buses, wheelchair ramps, and priority seating to accommodate passengers with special needs.By supporting and utilizing public transportation, we are contributing to a more inclusive and equitable society, where everyone has the opportunity to participate fully and access essential services and amenities.ConclusionIn conclusion, the benefits of public transportation are numerous and compelling. From cost savings and environmental responsibility to time management, safety, community building, and accessibility, embracing public transit can have a positive impact on our lives as students and on society as a whole.While it may require some adjustments and sacrifices in terms of convenience, the advantages of using public transportation are worth considering. By making the conscious choice to take the bus, train, or subway whenever possible, we can not only save money and reduce our environmental footprint but also contribute to a more sustainable, inclusive, and connected community.As students, we have the opportunity to lead by example and advocate for sustainable transportation solutions. By embracing public transit, we can demonstrate our commitment to environmental stewardship, fiscal responsibility, and social responsibility, setting a positive example for others to follow.篇2The Advantages of Public Transportation: A Convenient and Eco-Friendly SolutionAs students, we often find ourselves juggling multiple responsibilities, from attending classes and completing assignments to participating in extracurricular activities and socializing with friends. Amidst this hectic schedule, the need for reliable and efficient transportation cannot be overstated. However, relying solely on private vehicles can be costly, both financially and environmentally. This is where public transportation emerges as a compelling solution, offering a plethora of advantages that cater to the needs of students like ourselves.One of the most significant benefits of public transportation is its cost-effectiveness. Let's face it, as students, we are often on a tight budget, and the expenses associated with owning and maintaining a private vehicle can be a significant financial burden. By opting for public transportation, we can save a substantial amount of money that would otherwise be spent on fuel, insurance, and maintenance costs. This financial relief allows us to allocate our limited resources towards other essential expenses, such as tuition fees, textbooks, or recreational activities.Furthermore, public transportation is a convenient and time-saving option for students. Imagine the frustration of beingstuck in traffic jams or navigating through crowded parking lots, only to arrive late for classes or appointments. By utilizing buses, trains, or subways, we can avoid these hassles and navigate through the city with ease. Many public transportation systems offer dedicated lanes or prioritized traffic signals, ensuring a smoother and more predictable commute. Additionally, the ability to multitask while on board, such as reading, studying, or even catching up on assignments, makes public transportation an efficient use of our valuable time.Beyond the practical advantages, public transportation plays a crucial role in promoting environmental sustainability, a cause that resonates strongly with many students today. By choosing to commute via public transportation, we actively contribute to reducing our carbon footprint and mitigating the negative impacts of greenhouse gas emissions. Unlike private vehicles, which contribute significantly to air pollution and climate change, public transportation options like buses and trains are more energy-efficient and produce fewer emissions per passenger. By making this eco-friendly choice, we can take pride in playing our part in creating a greener and more sustainable future for ourselves and generations to come.Moreover, public transportation fosters a sense of community and social interaction. As students, we often seek opportunities to connect with our peers and expand our social circles. Public transportation provides a shared space where we can engage with people from diverse backgrounds, strike up conversations, and potentially forge new friendships or relationships. This social aspect not only enriches our personal experiences but also contributes to building a more inclusive and connected society.However, it is important to acknowledge that public transportation systems are not without their challenges. Overcrowding during peak hours, delays, and occasional service disruptions can be frustrating experiences. Nevertheless, many cities and municipalities are actively working to improve and expand their public transportation networks, investing in infrastructure upgrades, increasing service frequencies, and implementing innovative solutions to address these issues.As students, our voices and actions can play a vital role in shaping the future of public transportation. By advocating for better services, providing feedback, and actively supporting sustainable initiatives, we can contribute to creating a more efficient, accessible, and environmentally friendly transportationsystem that meets the needs of our generation and those to come.In conclusion, public transportation offers a multitude of advantages for students like ourselves. From cost savings and convenience to environmental sustainability and social connectivity, embracing this mode of transportation is a wise choice that aligns with our values and practical needs. By making the conscious decision to utilize public transportation, we not only simplify our daily commutes but also contribute to creating a more sustainable and inclusive society. Let us embrace this opportunity and embark on a journey that transcends mere transportation, shaping a better future for ourselves and the world around us.篇3The Numerous Benefits of Public TransportationAs students, we are always looking for ways to save money and reduce our environmental impact. One solution that accomplishes both of these goals is using public transportation. While it may not be the most convenient option for everyone, the advantages of taking the bus, train, or subway are numerous and significant. In this essay, I will explore the financial,environmental, and social benefits of utilizing public transit systems.Financial BenefitsFor most students, money is tight. We are often living on a limited budget, trying to balance the costs of tuition, textbooks, housing, and living expenses. Owning and operating a personal vehicle can be a major financial burden, one that many of us simply cannot afford. The costs of gas, insurance, maintenance, and parking can quickly add up, putting a strain on our already stretched budgets.Public transportation, on the other hand, is a much more economical option. In many cities, students can purchase discounted transit passes or qualify for reduced fares. Even without a student discount, the cost of a monthly bus or train pass is typically far less than the expenses associated with owning a car. By utilizing public transit, we can save a significant amount of money that can be put towards other important expenses, such as tuition or rent.Environmental BenefitsIn addition to the financial advantages, public transportation is also a more environmentally friendly choice. As we becomeincreasingly aware of the impact of human activities on the planet, it is crucial that we take steps to reduce our carbon footprint and minimize our contribution to pollution and climate change.Personal vehicles are a major source of greenhouse gas emissions and air pollution. According to the Environmental Protection Agency (EPA), transportation accounts for approximately 27% of total greenhouse gas emissions in the United States. By choosing to take the bus, train, or subway instead of driving a personal vehicle, we can significantly lower our individual carbon footprint and reduce our contribution to air pollution.Furthermore, public transit systems are generally more efficient in terms of energy consumption and resource utilization.A single bus or train can transport dozens or even hundreds of passengers, removing the need for each individual to operate their own vehicle. This shared transportation model reduces the overall energy consumption and emissions associated with travel.Social BenefitsBeyond the financial and environmental advantages, public transportation also offers social benefits that are oftenoverlooked. Using public transit can foster a sense of community and provide opportunities for social interaction.When we take the bus or train, we are sharing a space with fellow commuters from various backgrounds and walks of life. This exposure to diverse individuals and perspectives can broaden our understanding and appreciation of different cultures and lifestyles. It can also encourage us to be more mindful and considerate of others, as we are sharing a common space and experience.Additionally, public transit systems can promote a more equitable and accessible society. For individuals without access to a personal vehicle, whether due to financial constraints, disability, or other reasons, public transportation can provide a vital means of mobility and independence. By supporting and utilizing these systems, we contribute to creating a more inclusive and connected community.Counterarguments and ResponsesWhile the benefits of public transportation are significant, there are also potential drawbacks and counterarguments that should be addressed.One common argument against using public transit is the issue of convenience and flexibility. Personal vehicles offer the freedom to travel directly from point A to point B without the need to follow predetermined routes or schedules. This can be particularly appealing for those with busy or unpredictable schedules, or for those living in areas with limited public transit options.However, it is important to consider the trade-offs involved. While personal vehicles may offer more flexibility, they also come with higher costs, increased environmental impact, and the potential for traffic congestion and parking challenges. In many urban areas, public transit systems are becoming increasingly efficient and accessible, with expanded routes, more frequent service, and the integration of real-time information and mobile applications.Another concern often raised is the issue of safety and security on public transit. There is a perception that buses, trains, and subway stations may be more prone to crime or unsavory incidents. While it is true that crowded public spaces can present certain risks, it is important to note that transit authorities and law enforcement agencies work diligently to ensure the safety and security of passengers.Additionally, the presence of other commuters and the shared nature of public transit can actually enhance safety through increased visibility and accountability. By being aware of our surroundings and taking reasonable precautions, we can mitigate potential risks and enjoy the benefits of public transportation without undue concern for our well-being.ConclusionIn conclusion, the advantages of utilizing public transportation are numerous and compelling. From financial savings and environmental benefits to social connectivity and accessibility, public transit systems offer a multitude of advantages for students and society as a whole.While personal vehicles may provide convenience and flexibility, the costs and impacts associated with their use cannot be ignored. By embracing public transportation, we can reduce our individual expenses, minimize our carbon footprint, and contribute to the creation of more sustainable and equitable communities.As students, we have the opportunity to lead by example and shape the future of transportation. By making the choice to use public transit whenever possible, we can not only save money and reduce our environmental impact but also promotepositive social change and foster a more connected and inclusive society.。
支持新科技的原因英语作文
支持新科技的原因英语作文**Title: Embracing New Technologies: A Leap into the Future**In the relentless march of human progress, new technologies have consistently emerged as pivotal drivers of transformation. They not only reshape our daily lives but also redefine societal structures and global economies. The embrace of these innovations is crucial for several compelling reasons.Firstly, **enhanced efficiency** stands as a cornerstone benefit. Technological advancements streamline processes in industries ranging from healthcare to agriculture, enabling tasks to be completed faster and with greater precision. For instance, AI-driven diagnostic tools in medicine can swiftly identify diseases, while precision farming technologies optimize crop yields, ensuring food security for an ever-growing population.Secondly, **unprecedented connectivity** fosters global collaboration and understanding. The internet and social media platforms have shrunk the world, allowing ideas andknowledge to circulate freely across borders. This interconnectedness spurs innovation, cultural exchange, and the formation of a global community united in addressing shared challenges like climate change.Moreover, **sustainable solutions** are beingfacilitated by technology. Renewable energy technologies, smart grids, and electric vehicles are key to mitigating climate change and transitioning towards a greener future. These innovations not only reduce carbon footprints but also stimulate economic growth through the creation of new jobs and industries.Lastly, **personalized experiences** are now possible due to advancements in data analytics and machine learning. From tailored health treatments to personalized education, technology empowers individuals with customized services that cater to their unique needs, thereby enhancing overall quality of life.In summary, the adoption of new technologies is indispensable for harnessing progress, fostering global unity, ensuring environmental sustainability, and enriching individual experiences. As we navigate this digital age,embracing these innovations is not merely an option—it's a vital step forward into a brighter, more interconnected,and sustainable future.---**拥抱新技术：迈向未来的一跃**在人类进步的不懈进程中，新兴技术不断涌现，成为变革的关键驱动力。
如何成为一个好的管理者英语作文
如何成为一个好的管理者英语作文Becoming a Good ManagerBeing a manager is a challenging but rewarding role. A good manager is someone who can effectively lead a team, inspire others, make tough decisions, and achieve results. If you are looking to become a successful manager, here are some key qualities and skills you will need to develop:1. Communication Skills: One of the most important skills fora manager is the ability to communicate effectively. This includes being able to clearly convey instructions, provide feedback, listen to others, and resolve conflicts. Good communication helps build trust and collaboration within a team.2. Leadership Skills: A good manager is a strong leader who can set a clear vision, motivate others, and lead by example. Leadership involves inspiring and guiding team members to achieve common goals, providing support and direction when needed.3. Decision-Making Skills: Managers are often faced with tough decisions that can impact the success of a project or team. Good managers are able to make informed decisions underpressure, weighing the pros and cons of different options and considering the potential outcomes.4. Time Management: Time management is crucial for managers who juggle multiple tasks, projects, and deadlines. Good managers are able to prioritize their workload, delegate tasks effectively, and make the most of their time.5. Team Building: Building a strong and cohesive team is essential for a manager's success. Good managers know how to recruit the right talent, foster a positive work environment, and provide opportunities for professional development and growth.6. Problem-Solving Skills: Managers often encounter obstacles and challenges in their roles. Good managers are able to think critically, analyze problems, and come up with creative solutions to overcome them.7. Adaptability: The business world is constantly evolving, and managers need to be able to adapt to changes and new situations. Good managers are flexible, open to new ideas, and willing to learn from their experiences.8. Emotional Intelligence: Emotional intelligence is the ability to understand and manage your own emotions, as well as theemotions of others. Good managers are empathetic, self-aware, and able to build strong relationships with their team members.9. Accountability: Managers are responsible for the success or failure of their team's efforts. Good managers take accountability for their actions, admit mistakes, and learn from them to improve their performance.10. Continuous Learning: Becoming a good manager is an ongoing process of learning and self-improvement. Good managers are constantly seeking new opportunities to grow, develop their skills, and stay up-to-date with industry trends.In conclusion, becoming a good manager requires a combination of skills, qualities, and attitudes that can be developed over time. By investing in your personal and professional growth, honing your leadership abilities, and continuously learning and improving, you can become a successful and effective manager.。
NOTE ON MULTIPLE POSITIVE SOLUTIONS TO NON-HOMOGENOUS MULTI-POINT BVPS FOR SECOND ORDER p-LAPLAC
30Leabharlann No.1X. Y. Liu, etc., MULTIP LE PO SITIVE SOLUTIO NS TO BVP S
� [ φ ( x � (t ))] � + q ( t ) f (t, x ( t ) , x � ( t )) = 0 , 0 ≤ t ≤ 1 , � � m m X X � αi x( ξ i ) = 0 , x(1) − β i x ( ξi ) = 0 , � x (0) −
Xingy uan L iu, Dongqing Xia, Aimin Tang
( D ep t. o f Ma th., S ha o ya ng Un iv ersity , Sh a oy an g 4 2 20 0 0, Huna n ; E-m a il: liuxing yuan syx@2 1cn.co m ( X . Liu) )
Ann . o f Diff . Eqs. 26 :1(2010) , 30-44
NOTE ON MULTIPLE POSITIVE SOLUTIONS TO NON-HOMOGENOUS MULTI-POINT BVPS FOR SECOND ORDER p-LAPLACIAN EQUATIONS ∗ †
∗ Su ppo rt ed by N a t ura l Science Found at io n o f Ed ucat io na l Co m mitt ee of H un an Pro vince (N o .0 8 C7 94 ) . † M anuscript received M ay 4 , 2 0 08
A bs tra ct
In th is pa pe r, so me new suffi cient cond it io ns g uara nteeing t he e xistence o f a t least t hre e p ositive solut io ns t o no n-ho mo g enou s mult i-p oint bo unda ry va lue p ro ble m for seco nd o rder p-L a pla cian equa tions are est a blish ed. A n exam ple is presented t o illust rat e t he m ain re sult. Key wor ds p -L apla cia n diff erent ia l equa t io n; no n-ho m og eno us mu lt i-p oint bo unda ry value prob lem ; p ositive so lut ion 20 00 M athem atic s Sub ject C lassific ation 3 4 B1 0; 3 4B 15 ; 35 B1 0
amd cpu封测方法
amd cpu封测方法The process of sealing and testing AMD CPUs is acrucial step in ensuring the quality and reliability of these essential components for computer systems. This process involves several key steps and considerations, all of which are aimed at verifying the performance and functionality of the CPUs before they are released to the market. In this article, we will explore the AMD CPU sealing and testing methods from multiple perspectives, including the technical aspects, quality assurance considerations, and the importance of this process in delivering high-quality products to consumers.From a technical perspective, the sealing and testing of AMD CPUs involves a series of rigorous procedures and protocols to verify their performance and functionality. This includes the application of advanced testing equipment and software to assess various aspects of the CPU's operation, such as speed, power consumption, and thermal performance. Additionally, the sealing process is designedto protect the delicate components of the CPU from external contaminants and physical damage, ensuring that they remain in pristine condition until they are installed in a computer system.Quality assurance is another critical aspect of the AMD CPU sealing and testing process. This involves the implementation of strict quality control measures to ensure that every CPU meets the company's high standards for performance and reliability. This includes conducting thorough inspections and tests at various stages of the sealing and testing process to identify any potential issues or defects. By adhering to these stringent quality assurance procedures, AMD can confidently deliver CPUs that meet or exceed the expectations of consumers and industry professionals.The sealing and testing of AMD CPUs also play a vital role in ensuring the overall reliability and longevity of computer systems. By subjecting each CPU to comprehensive testing and validation, AMD can minimize the risk of premature failures or malfunctions, which can lead tocostly downtime and repairs for end-users. This is particularly important in mission-critical applications, such as enterprise servers and workstations, where system stability and performance are of utmost importance.Furthermore, the sealing and testing process is essential for maintaining the competitive edge of AMD in the highly competitive CPU market. By consistently delivering high-quality and reliable products, AMD canbuild trust and loyalty among consumers, leading to repeat business and positive word-of-mouth recommendations. This, in turn, can help AMD maintain its market share and reputation as a leading provider of CPU solutions for a wide range of applications and industries.In conclusion, the sealing and testing of AMD CPUs is a critical process that encompasses technical, quality assurance, and market competitiveness considerations. By employing advanced testing methods, stringent quality control measures, and a focus on delivering reliable and high-performance products, AMD can ensure that its CPUs meet the demands of today's computer systems andapplications. This ultimately benefits consumers, businesses, and the industry as a whole by providing access to cutting-edge CPU technology that can drive innovation and productivity in a variety of settings.。
THE EXISTENCE OF POSITIVE SOLUTIONS FOR A CLASS OF INDEFINITE WEIGHT SEMILINEAR ELLIPTIC BO
α Our method works provided that the linearized problem for (Iλ ), viz, −∆u = λg (x)u in Ω (Lα ) (1 − α) ∂u + αu = 0 on ∂ Ω. ∂n
Then, when α > 0, it is well known that positive solutions must satisfy 0 < u < 1 and are precisely those arising out of bifurcation from the zero solution; moreover the equation has no positive solutions if λ < λ1 where λ1 denotes the least eigenvalue of the Laplacian. We shall show, however, that, when g changes sign, the variational method proves the existence of a positive solution for all λ, 0 < λ < λ+ (α), where λ+ (α) denotes the positive principal eigenvalue of (Lα ) and that such solutions are not bounded above by 1. The plan of the paper is as follows. In section 2 we first recall the facts that we shall require about eigencurves and show how eigencurves can be used to generate an equivalent norm for W 1,2 (Ω); then using this equivalent norm we prove the existence of solutions by applying variational methods. In section 3 we discuss α the solutions of (Iλ ) which arise from bifurcations and compare these with the variational solutions obtained in section 2 for the case where α ∈ (0, 1], i.e., where we have Dirichlet or the standard Robin boundary condition. 2. Variational Solutions We first recall some facts about how the method of eigencurves can be used to prove the existence of principal eigenvalues of (Lα ) (see, e.g., [2]). For fixed λ we denote by µ(α, λ) the principal eigenvalue of the Schr¨ odinger problem (1.1). Clearly α λ is a principal eigenvalue of (L ) if and only if µ(α, λ) = 0. It can be shown that µ(α, λ) has the variational characterisation µ(α, λ) = inf {
MULTIPLE POSITIVE SOLUTIONS TO p-LAPLACIAN DYNAMIC EQUATION ON TIME SCALES
MULTIPLE POSITIVE SOLUTIONS TO p-LAPLACIAN DYNAMIC EQUATION ON TIME SCALES ∗†
Wu Chufen1 , Song Changxiu 2, Weng Peixuan1
1
Introduction
In th is pap e r, we assum e T to b e a tim e sc ale [1] su ch th at a, b ∈ T and [ a, b] T = [ a, b] ∩ T . For conve nien ce , we still let [ a, b] de note [ a, b ]T . We are c onc erne d with th e e xistenc e of p ositive solu tion s to p - Laplac ian d yn amic e qu ation on a time scale ( ( φ p ( u� � ( t ))) ∇ ( t ) + f ( t, u� ( t)) = 0 , t ∈ [a , b] , (1.1) u ( a ) = u � ( b) = u � � ( a ) = 0 , whe re φ p (u ) is th e p -L aplacian op e rator, that is, φ p ( u ) = | u| p − 2 u , p > 1, ( φ p ) − 1 ( u ) = φ q ( u ) , whe re 1 /p + 1 /q = 1 an d (A) Fun ction f : [ a, b] × R → R + := [0 , + ∞ ) is continu ous. For ord in ary d iffe rential equ ations an d fi nite d iffe ren ce equ ation s, p -Lap lac ian p roblem s with two-, thre e- , m- point b ou nd ary c ond ition s have be en stud ied exte nsively, see [2- 6] and the re fere nc es the re in. H owe ver, few stu dy h as be en con du cted on th e p -Lap lacian p roblem s on time sc ales. In this p ap e r, we use fixe d p oint the orems to obtain at least on e p ositive solu tion to bou nd ary value p roblem (BVP for short) (1.1). For conve nien ce , we in tro du ce the followin g d efi nition s, wh ich can also b e fou nd in [1]. De fini tion 1.1 For t < sup T an d r > in f T , d efi ne forward jum p ope rator σ and back ward ju mp op erator ρ as follows, σ ( t) = in f { τ ∈ T | τ > t } ∈ T , ρ ( t ) = sup { τ ∈ T | τ < t } ∈ T , for all t ∈ T .
自我介绍英文高级表达
自我介绍英文高级表达1Oh my goodness! Let me introduce myself to you. I am a person full of passion and drive. I have achieved remarkable success in academic competitions, winning several prestigious awards. This not only showcases my intelligence but also my determination and hard work. Can you imagine the joy and pride I felt at those moments?I am also proficient in multiple languages, which has opened up a vast world of communication and knowledge for me. It has allowed me to explore different cultures and perspectives, enriching my understanding of the world. How wonderful is that?When it comes to hobbies, reading is my constant companion. Through the pages of books, I have traveled to distant lands and experienced countless adventures. Isn't it amazing how a simple book can transport us to different realms? And my love for painting has allowed me to express my innermost thoughts and emotions on the canvas. It has been a source of self-expression and creativity.In conclusion, I am a unique individual with a diverse range of experiences and interests. I am constantly striving to grow and learn, and I am excited to see what the future holds!2I am a person full of unique charm and diverse qualities. When it comes to my personality, optimism is one of the most prominent traits. Time and again, I have found myself in difficult situations, but my unwavering optimism has always been my guiding light. For instance, during a challenging project at work, where everyone was stressed and pessimistic, I chose to see the silver lining. I believed that every problem had a solution, and with a positive mindset, we could overcome any obstacle. And guess what? We did it!Another remarkable quality of mine is tenacity. In a team collaboration for a community event, we faced numerous setbacks and challenges. However, my determination never wavered. I led the team with a firm belief in our ability to succeed. Despite the odds, we persevered and ultimately achieved our goal. Was it easy? No! But did we give up? Never!I firmly believe that these qualities not only define who I am but also shape my approach towards life. With optimism and tenacity, I am ready to embrace whatever comes my way. How wonderful is that?3Oh, dear friends! Let me introduce myself to you. I have a clear vision for my future and I'm passionate about achieving my goals. I dream of becoming a doctor one day! Why? Because I want to help people, to easetheir pain and bring them hope and health. How am I working towards this? I'm studying hard, learning all the medical knowledge I can. I spend hours poring over thick textbooks and attending various courses. And it's not just about theory. I actively participate in volunteer activities to gain practical experience. I assist in local clinics, helping doctors and interacting with patients. It's tough but rewarding! I know the road ahead is long and challenging, but I'm determined and never give up! I believe with my perseverance and hard work, I will make my dream come true. Won't I?4I come from a family where love and support abound! How fortunate I am to have grown up in such an environment. My parents have always been my guiding stars, shaping my values and influencing the way I deal with people and the world around me.My father, a man of wisdom and kindness, taught me the importance of honesty and responsibility. His words of advice still ring in my ears whenever I face a difficult decision. "Be true to yourself and others, and always take responsibility for your actions," he would say. And my mother, with her endless patience and warmth, showed me the power of compassion and understanding. She would listen to my problems and offer comfort and solutions, making me believe that kindness can heal any wound.The love and unity within my family have instilled in me a strongsense of belonging and security. I have learned to appreciate the simple joys of life, to care for others, and to never give up in the face of challenges. How could I not be grateful for such a nurturing upbringing? It has truly made me the person I am today! I am certain that the values I have acquired will continue to guide me through the journey of life.5I am an individual who firmly believes in the power of fairness and justice. This belief is not just a fleeting thought but a guiding light that shapes my every decision and action in life! How do I pursue this ideal? Well, in every interaction with others, I strive to listen attentively and understand their perspectives. I question unfairness whenever I encounter it, and I am not afraid to speak out against injustice. I believe that small acts of fairness can accumulate and bring about significant changes in our society. Why is this so important to me? Because I have witnessed the negative impact of injustice on people's lives, and I am determined to be a force for positive change. I constantly ask myself: How can I do more? How can I inspire others to join this pursuit? In my daily life, I treat everyone with respect and equality, regardless of their background or status.I believe that by doing so, I can create a ripple effect of fairness and justice that spreads far and wide. This is who I am, a passionate advocate for fairness and justice, constantly striving to make the world a better place!。
pohozaev流形方法的开题报告
假设 ,且 是 的临界点的一个必要条件是
这个条件定义了一个Nehari流形
Euler函数
首先找出方程的Nehari流形，定义Euler函数，讨论关于函数的一些引理，结合（PS）条件，山路引理和Sobolev嵌入定理确定合适的 ，使得 ，方程至少存在两个正解。
研究的意义：
方程在当今科学研究领域中扮演着重要的角色.近年来,很多源于物理学,工程学等科学领域,具有实际应用背景的微分方程边值问题引起了人们的极大兴趣.众多作者综合运用变分法,山路引理，临界点理论等多种方法分析研究方程解的存在性与多重性.方程在物理、化学、生物、金融数学等不同的学科领域已得到广泛的应用。具N-Laplace算子的微分方程边值问题也早已运用到工程、物理学等领域，随着分数阶微分方程在实际生产和生活中的不断应用，关于分数阶微分方程边值问题理论的研究已引起了国内外学者的广泛关注并逐渐成为研究热点。椭圆型偏微分方程在工程技术科学与自然科学中的应用很广泛,许多重要的物理,力学学科的基本方程本身就是偏微分方程,许多领域中的数学模型都可以用偏微分方程来描述.因此,求解偏微分方程就变得很重要.N-Laplace方程在晶体错位、守恒律、金融、火焰传播、极小平面、多元散射、材料学、最优化等方面都有广泛应用，近几年来得到了广泛关注。
[5]LilianeA.Maia Positive solutions of a symptotically linear equations via Pohozaev manifold.RaquelLehrera.
[6]A.Ambrosetti, G.Cerami, D.Ruiz, Solutions of linearly coupled systems of semilinear non-autonomous equations on R N, J.Funct.Anal.254 (2008)2816–2845.
科技的双面性英语作文
科技的双面性英语作文The Dual Nature of TechnologyTechnology, a double-edged sword, has brought about profound changes in our lives, both positive and negative. On one hand, it has greatly enhanced our lives, making them more convenient and efficient. On the other hand, however, it has also posed certain challenges and risks.Firstly, the positive impacts of technology are evident in various aspects of our daily lives. The advancement of information technology has made it possible for us to access information and communicate with others instantaneously, regardless of geographical distance. Smartphones, computers, and the internet have revolutionized the way we work, learn, and entertain ourselves. Additionally, technology has also improved our healthcare system, enabling doctors to diagnose and treat diseases more accurately and efficiently.However, the negative impacts of technology cannot be ignored. The excessive use of technology, especially among children and young adults, has led to issues such as addiction, decreased social skills, and poor mental health. Moreover, the widespread use of technology has also increased the risk ofcyberbullying, privacy breaches, and cybercrime.Furthermore, technology has also had an impact on our environment. The production and disposal of electronic waste, known as e-waste, has become a significant environmental problem. These devices often contain harmful materials that can pollute the soil and water if not disposed of properly.In conclusion, technology is a double-edged sword. While it has brought about numerous benefits, it has also posed certain challenges and risks. It is, therefore, essential for us to use technology responsibly and to mitigate its negative impacts while maximizing its positive potential.。
EXISTENCE AND MULTIPLE EXISTENCE OF POSITIVE SOLUTIONS TO SECOND-ORDER m-POINT BOUNDARY VALUE PR
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写作业就是学生的烦恼英语
写作业就是学生的烦恼英语Certainly! Here's a piece of content that fits the title "Doing Homework is a Student's Worry":The Perpetual Struggle: Homework as a Student's WorryHomework is an integral part of the educational system, designed to reinforce and deepen the understanding of concepts taught in class. However, for many students, it often becomes a source of worry and stress.Firstly, the sheer volume of homework can be overwhelming. Students are expected to juggle multiple subjects, each with its own set of assignments. This can lead to late nights and a significant reduction in leisure time, which is essential for a well-rounded development.Secondly, the difficulty level of some homework can be a major concern. When assignments are too challenging, students may feel inadequate or frustrated, leading to a decrease in self-confidence and motivation. It's crucial for educators to provide support and guidance to help students overcome these hurdles.Thirdly, the pressure to perform well on homework can be daunting. In many educational systems, homework gradescontribute to a student's overall academic performance. This can lead to anxiety, especially for those who areperfectionists or who fear disappointing their parents or teachers.Lastly, the lack of a conducive environment for studying can exacerbate the issue. Distractions at home, lack of resources, or even a crowded study space can make it difficult for students to focus and complete their homework efficiently.To alleviate the worries associated with homework, it's important for educators to communicate with students and parents, set reasonable expectations, and provide asupportive learning environment. Additionally, studentsshould be encouraged to develop effective time management and study skills to help them navigate through their academic responsibilities with less stress.In conclusion, while homework is a necessary tool for learning, it should not become a source of constant worry for students. A balanced approach that considers the well-beingof students is essential for a positive educational experience.This content is designed to address the concerns and challenges students face when it comes to doing homework, offering insights and potential solutions to mitigate the stress associated with it.。
含导函数Stieltjes积分边界条件下二阶问题的正解
应用泛函分析学报Vol.22, No.4Dec., 2020第22卷第4期2020年12月ACTA ANALYSIS FUNCTIONALIS APPLICATA DOI ： 10.12012/1009-1327(2020)04-0193-14文献标识码：A含导函数Stieltjes 积分边界条件下二阶问题的正解计倩,张国伟(东北大学理学院数学系，沈阳110819)摘 要 本文研究了一类含导函数Stieltjes 积分边值条件下二阶边值问题的正解.由 于边值条件中带有导数，导致讨论过程与已有文献不同，并且给出相应的格林函数.应 用不动点指数理论证明非线性项/(x,y,z)关于x , y 有超(次)线性增长情形下方程正 解的存在性.通过两个具体例子进行说明理论结果的有效性，例子中边值条件包含积分 型与多点型的形式.关键词 正解；不动点指数;锥中图分类号O175.14; O177.91Positive Solutions for Second Order Problems under Stieltjes Integral Boundary Conditions with DerivativeJI Qian, ZHANG Guowei(Department of Mathematics, College of Science, Northeastern University, Shenyang 110819, China)Abstract In this paper, we study positive solutions for a class of second order prob lems under Stieltjes integral boundary conditions with derivative. Due to the derivative in the boundary conditions, the procedure of discussing is different from one in previ ous literature, and Green's function corresponding to the problem is given. The fixed point index theory is applied to prove the existence of positive solutions when the non linear term /(x, y, z) has superlinear or sublinear growth on x and y. The validity of the theoretical results is illustrated by two concrete examples, in which the boundary conditions include the forms of integral and multi-point types.Keywords positive solution; fixed point index; cone收稿日期：2020-09-13作者简介：计倩(1994-),女辽宁本溪人硕士研究生，研究方向：非线性泛函分析，E-mail: ******************.194应用泛函分析学报第22卷Chinese Library Classification O175.14;O177.911引言近些年来，非线性微分方程边值问题在科学研究和工程技术等领域中都具有重要的应用，并受到诸多学者的广泛关注，取得了许多研究成果二阶非线性常微分方程边值问题的正解存在性及多解性成为一个重要研究领域，文献[3]利用锥上不动点指数方法讨论了边值条件中带有Stieltjes积分的方程—u〃(t)=/(t,u(t),G[0,1],au(0)—bu z(0)=a[u],cu(1)+du'(1)=0[u]正解的存在性，但是在Stieltjes积分中不含未知函数的导数Stieltjes积分中含有未知函数导数的边值问题也有一些研究结果文献[5]研究方程—u''(t)=g(t)f(t,u(t)),t G[0,1],u(0)=a[u],u'(1)=0[u]+入[u']正解的存在性，其中a[u]=/u(t)d A(t),0[u]=/u(t)d B(t),入[u']=/u'(t)dA(t),丿0Jo JoA,B和A为界变差函数但是非线性项函数不含有未知函数的导数受此启发，本文用锥上不动点指数方法讨论如下含导函数Stieltjes积分边界条件下二阶边值问题正解的存在性：—u''(t)=f(t,u(t),u'(t)),t G[0,1],(1)u'(0)=a[u],u(1)=0[u]—A[u'].关于非线性项和Stieltjes积分形式边值条件含未知函数导数的工作可见文献[9,10].本文讨论方程(1)的内容和所使用的方法与[9,10]不相同.2预备知识定义c”0,1]空间的范数为||u||ci=max{||训c,||u'||c}.首先我们假设，(C1)f:[0,1]x R+x R t R+是连续函数其中R+=[0,Q.(C2)A(1)-A(s)、0,V s G[0,1].由于方程⑴边值条件中含有入[u'],类似于[5]中Webb所使用的方法,我们需要给出相应的Green函数.弓|理1在(C i)的条件下，考虑当a[u]=0[u]=0,即—u''(t)=f(t,u(t),u'(t)),t G[0,1],(2)u'(0)=0,u(1)+入[u']=0第4期计倩等：含导函数Stieltjes 积分边界条件下二阶问题的正解195时,⑵在C 1[0,1]中的解由如下定义(Hu )(t ) = / (A(1) — A(s ))f (s,u (s ),u '(s ))d s + [ (t, s )f (s,u (s ),u '(s )) d s丿0丿0:=/ kH (t, s )f (s,u (s ),u '(s ))d s J0的算子H 不动点给出，其中⑶{1 — s, 0 < t < s < 1,1 — t, 0 < s < t < 1,⑷I A (1) — A(s ) + 1 — s, 0 < t < s < 1, k H (t, s )= <(A (1) — A(s ) + 1— t, 0 < s < t < 1.⑸证明 首先,对—u ''(t ) = f (t,u (t ),u '(t ))在[0,t 和[t, 1]求两次积分,并利用 ⑵中的边值条件就可得到(3)式•其次，(Hu )(t ) = / (A(1) — A (s ))f (s,u (s ),u '(s ))d s + [ k o(t, s )f (s,u (s ),u '(s )) d s丿o丿o =[(A(1) — A(s ))f (s,u (s ),u '(s ))d s + [ (1 — t )f (s,u (s ),u '(s ))d s +丿0丿0/ (1 — s)f(s,u(s),u '(s))ds,[f (s, u (s ), u '(s )) d s + (1 — t )f (t, u (t ), u '(t )) — (1 — t )f (t, u (t ), u '(t ))0(Hu)''(t) = —f (t,u(t),u '(t)),显然(Hu )(1) = J 0X (A(1) — A(s ))f (s,u (s ),u '(s ))d s ,而tf (s, u (s ), u '(s )) ds)dA(t )f (s, u (s ), u '(s )) dA(t )d s = f (A (1) — A(s ))f (s, u (s ), u '(s )) d s = (Hu )(1),丿0所以(Hu )''(t ) = —f (t, u (t ), u '(t )), (Hu )'(0) = 0, (Hu )(1) + 入[(Hu )'] = 0•由此可见，u 是 H 不 动点.弓I 理2如果(C 2)满足，则存在非负函数①h (s ) = A(1) — A(s ) + 1 — s,使得V t, s e [0,1]有(1 一 t )① H (s ) < k H (t, s ) < ① H (s ).容易证明在C 1 [0,1]中,BVP(1)有解当且仅当如下的积分方程u (t ) =(t 一 dA(t ) — 1)a [u ] + 0[u ] + (Hu )(t )I f(s,u(s),u '(s))ds,丿0dA(t ) — t) a [u ]-入[(Hu )']f (s, u (s ), u '(s )) d s dA(t )dA(t)) 0[u ] + (Hu )(t ) := (Tu)(t)⑹196应用泛函分析学报第22卷存在解，其中a [u ] = 0[u ] — a [u ]•记2是a [u ]对应的有界变差函数，并且假设(C 3) K a (s ) := / k n (t, s )d A (t ) > 0,K b (s ) ：= / k n (t, s )d B (t ) > 0, V s G [0,1].丿0 丿0令 y (t ) = 1 — t + fl dA(t ), d (t ) = t — fl dA(t ).再假设(C 4)0 < S [7] < 1,0[y ] > 0, 0 < 0[d ] < 1, a [d ] > 0, D := (1 — d [7])(1 — 0[d ]) — d [J ]0[7] > 0. 定义算子S 如下(Su)(t)：=丄 d - dA (t)—t + 人"A") (1 一 00]) / K A (s)f (s,u (s ),u z (s ))d s + 丘0] / K b (s )f (s,u (s ),u z (s ))d s_ 丿0 丿0 .0[Y ] / K A (s )f (s,u (s ),u '(s ))ds + (1 — a[Y ]) / k b (s)f (s,u(s ),u '(s ))ds 一 丿0 丿0 _t + - D+ k n (t, s )f (s,u (s ),u z (s ))d s丿0/ ks (t, s )f (s, u (s ), u z (s )) d s,丿0即(Su )(t ) = / k s (t, s )f (t,u (t ),u '(t ))d s.丿0⑺显然k s (t,s ) =1 — t +D 0 dA(t ) [(1 — 0[d ])K A (s ) + a [d ]K B (s )] +-~『叮人")[0[Y ]K A (s) + (1 — a[Y ])K B (s)] + k n (t, s).由上述条件可以很容易的看出(Su )(t ) > 0.并且dk s (t, s )~dt -=D [(1 — 00])K A (s ) + a[J]K B (s)] + D [0[Y ]K A (s) + (1 — a[Y ])k b (s)]—以仔 s )—1 1< D [(1 — 0[d ])K A(s ) + a [d ]K B(s )] + 万[0[y ]K a (s ) + (1 — S [y ])K b (s )] + 1 :=巫(s ).(8)⑼引理3假设满足(C 2)〜(C 4),则存在非负函数Q(s )和v(t) = min {t, 1—讣使得V t, s G [0,1],v (t )Q(s ) < k s (t, s ) < $(s ),其中 Q(s ) = D [(1 — 0Q ])K a (s ) + &[d ]K B (s )] + D [0[y ]K a (s ) + (1 —殆])恥(s )] + $n (s ).我们定义如下两个锥和三个线性算子：P = {u G C 1[0,1] : u(t) > 0, V t G [0,1]},(10)(L i u )(t )K = {u G P : u (t ) > v(t) ||u||C , V t G [0, 1]},[K s (t, s)(a 2u(s) + C 2)d s, @2,C 2为正常数),(11)(12)第4期计倩等：含导函数Stieltjes 积分边界条件下二阶问题的正解197(13)(14)(02u )(s ) = / K s (t, s )u (t )d t, u G C [0,1],Jo (L g u )(t ) = f K s (t, s )u (s )d s, u G C [0,1].JoV x,y G X ，若x — y G P ，记为x A y 或者y Y x ,称为由锥P 导出的半序.弓I 理4假设(C i )〜(4)都成立，那么S : P t K 和L i : C [0,1] t C [0,1]均是全连续算 子并且 L i (P ) U K, (i =1, 2, 3).证明 当(7), (8)和(C i )〜(C 4)都成立，则当u G P 时，有(Su )(t ) > 0.由(C i )条件可 以得到S : P T C i [0,1]是连续的算子取锥P 上的有界集合F ，则存在一个数M > 0,使得 ||训。
2006 Positive solutions for singular non-linear third-order periodic boundary value problems
Nonlinear Analysis64(2006)1528–1542/locate/na Positive solutions for singular non-linear third-order periodic boundary value problemsJifeng Chu a,∗,Zhongcheng Zhou ba Department of Applied Mathematics,College of Science,Hohai University,Nanjing210098,PR Chinab School of Mathematics and Finance,Southwest Normal University,Chongqing400715,PR ChinaReceived26May2005;accepted5July2005AbstractIn this paper,we are concerned with the problem of the existence of positive solutions for non-linear third-order periodic boundary value problemu + 3u=f(t,u),0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1,2.Here, ∈(0,1√3)is a positive constant and our non-linearity f(t,u)may be singular at u=0.The proof relies on a non-linear alternative of Leray–Schauder type and on Krasnoselskiifixed point theorem on compression and expansion of cones.᭧2005Elsevier Ltd.All rights reserved.MSC:34B15Keywords:Periodic boundary value problem;Positive solutions;Singular;Leray–Schauder alternative;Fixed point theorem in cones∗Corresponding author.E-mail address:jifengchu@(J.Chu).0362-546X/$-see front matter᭧2005Elsevier Ltd.All rights reserved.doi:10.1016/j.na.2005.07.005J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421529 1.IntroductionIn this paper,we establish the existence of positive solutions for the third-order periodic boundary value problemu + 3u=f(t,u),0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1,2,(1.1)where ∈(0,1√3)is a positive constant and the type of the non-linear term f(t,u)we aremainly interested in the case that f(t,u)has a singularity near u=0.In recent years,the non-linear periodic boundary value problems have been widely studied by many authors.For details,see[3–10,12–15,17–20]and the references therein.For the boundary value problem(1.1),we recall the following two results.In[10],by using Schauder fixed point theorem,together with perturbation technique,it is established the existence of at least one positive solution under the following conditions:(A1)f(t,u)is a non-negative function defined on[0,2 ]×(0,+∞)and f(t,u)is inte-grable on[0,2 ]for eachfixed u∈(0,+∞);(A2)f(t,u)is non-increasing in u>0for almost all t∈[0,2 ]andlim u→0+f(t,u)=+∞,limu→+∞f(t,u)=0hold uniformly for t∈[0,2 ].(A3)For eachfixed constant >0,the inequality 2f(s, )d s<+∞holds.In[17],the condition of monotonicity on f(t,u)in(A2)is removed and the existence of multiple positive solutions under suitable conditions on f(t,u)is obtained by using thefixed point index theory.Instead of Schauderfixed point theorem used in[10]andfixed point index theory in[17],our main tool in this paper is a non-linear alternative of Leray–Schauder type and afixedpoint theorem in cones due to Krasnoselskii.We remark that it is sufficient to prove that T:K∩(¯ 2\ 1)→K is continuous and completely continuous in Theorem2.2(Section 2).This point is essential and advantageous for singular problems.The method used inthis paper was applied successfully by Agarwal and O’Regan in[1,2]for the existence ofsingular Dirichlet,Conjugate,Focal and(n,p)problems and in[7,8]for the multiplicityof positive periodic solutions to superlinear repulsive singular equations.In this paper,wediscuss both positone case and semi-positone case.The remaining part of the paper is organized as follows.In Section2,some preliminaryresults will be given,including a famousfixed point theorem in cones due to Krasnoselskii.In Section3,we are devoted to the positone case,i.e.,f(t,u)is positive.In this case,weestablish the existence of at least two positive solutions.In Section4,the semi-positonecase,i.e.,f(t,u) −L for some L>0,is studied.In this case,we obtain the existence ofat least one positive solution.Some illustrating examples will be given.1530J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15422.PreliminariesIn this section,we present some preliminary results which will be needed in Sections3 and4.First,as in[17],we transform our problem into an integral equation.For any function u∈C[0,2 ],we define the operator(J u)(t)=2g(t,x)u(x)d x,(2.1)whereg(t,x)=⎧⎪⎨⎪⎩e− (t−x)1−e−2,0 x t 2 ,e− (2 +t−x)1−e−2,0 t x 2 .(2.2)By a direct calculation,we can easily obtain20g(t,x)d x=1.Now,we consider the problemu − u + 2u=f(t,J u),u(i)(0)=u(i)(2 ),i=0,1.(2.3) If u is a positive solution of problem(2.3),i.e.,u(t)>0for t∈[0,2 ],it is easy to verify that y(t)=(J u)(t)is a positive solution of problem(1.1).Therefore,we will concentrate our study on problem(2.3).Lemma2.1(Sun and Liu[17]).The boundary value problem(2.3)is equivalent to integral equationu(t)=2G(t,s)f(s,(J u)(s))d s,(2.4)whereG(t,s)=⎧⎪⎨⎪⎩2e( /2)(t−s)[sin√32(2 −t+s)+e− sin√32(t−s)]√3 (e +e− −2cos√3 ),0 s t 2 , 2e( /2)(2 +t−s)[sin√32(s−t)+e− sin√32(2 −s+t)]√3 (e +e− −2cos√3 ),0 t s 2 .(2.5)Moreover,for G(t,s),we have the estimates0<m=2sin(√3 )√3 (e +1)2G(t,s) 2√3 sin(√3 )=M∀s,t∈[0,2 ].(2.6)The technique we present is based on Krasnoselskii’sfixed point theorem in a cone, which we state here for the convenience of the reader.J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421531 Theorem2.2(Krasnosel’skii[11]).Let X be a Banach space and K(⊂X)be a cone. Assume that 1, 2are open subsets of X with0∈ 1,¯ 1⊂ 2,and letT:K∩(¯ 2\ 1)→Kbe a continuous and compact operator such that either(i) T u u ,u∈K∩j 1and T u u ,u∈K∩j 2;or(ii) T u u ,u∈K∩j 1and T u u ,u∈K∩j 2.Then T has afixed point in K∩(¯ 2\ 1).In applications below,we take X=C[0,2 ]with the supremum norm · and defineK=u∈X:u(t) 0for all t∈[0,2 ]and min0 t 2u(t) u,(2.7)where =m/M.One may readily verify that K is a cone in X.Now,suppose that f:[0,2 ]×R→[0,∞) is a continuous function.Define an operator T:X→X by(T u)(t)=2G(t,s)f(s,(J u)(s))d s(2.8) for u∈X and t∈[0,2 ].Lemma2.3.T is well defined and maps X into K.Moreover,T is continuous and completely continuous.Proof.It is easy to see that T is continuous and completely continuous since f:[0,2 ]×R→[0,∞)is a continuous function.Thus,we only need to show that T(X)⊂K.Let u∈X,then we havemin 0 t 2 (T u)(t)=min0 t 22G(t,s)F(s,(J u)(s))d s m2F(s,(J u)(s))d smax0 t 22G(t,s)F(s,(J u)(s))d s= T u ,i.e.,T u∈K.This implies that T(X)⊂K and the proof is completed.3.Positone caseIn this section,we establish the existence and multiplicity of positive solutions to(1.1) in the positone case.Recall that(1.1)is positone if f(t,u)>0for all t∈[0,2 ]and all u>0.In this paper,the notation 0means that (t) 0for all t∈[0,2 ]and (t)>0 for t in a subset of positive measure.1532J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–1542Theorem3.1.Suppose f(t,u)satisfies the following conditions:(F1)For each constant H>0,there exists a function H 0such that f(t,u) H(t) for all(t,u)∈[0,2 ]×(0,H].(F2)There exist continuous,non-negative functions g(u)and h(u)on(0,∞)such that f(t,u) g(u)+h(u)for all(t,u)∈[0,2 ]×(0,∞),and g(u)>0is non-increasing and h(u)/g(u)is non-decreasing in u∈(0,∞). (F3)There exists a positive number r such thatrg( r/ )(1+h(r/ )/g(r/ ))>1 2,where =m/M,m and M are given as in Section2.Then boundary value problem(1.1)has at least one positive solution with0< u <r/ . Proof.We only need to show problem(2.3)has at least one positive solution with0< u <r. If this is true,then(J u)(t)is a positive solution of(1.1)with0< J u <r/ since0< J u =max0 t 2 2g(t,x)u(x)d x u2g(t,x)d x=1u <r.To do so,we will use the Leray–Schauder alternative principle,together with a truncation technique.Let N0={n0,n0+1,...},where n0∈{1,2,...}is chosen such that1 2gr1+h(r/ )g(r/ )+1n0<r.(3.1)See(F3).Fix n∈N0,the idea isfirst to show that the“modified”problemu − u + 2u=f n(t,J u)+ 2n,0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1(3.2) has a solution u n∈C2[0,2 ]with0< u n r,wheref n(t,x)=f(t,1/n ),x 1/n ,f(t,x),x 1/n .(3.3)For ∈(0,1),consider the family of problemsu − u + 2u= f n(t,J u)+ 2n,0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1.(3.4)J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421533 Let u be a solution of problem(3.4),then it follows from Lemma2.1thatu(t)=20G(t,s)f n(s,(J u)(s))+2nd s= 2G(t,s)f n(s,(J u)(s))d s+1n,(3.5)where G(t,s)is the Green function given by(2.5).First,it is claimed that any solution u of problem(3.5)verifiesu =max0 t 2|u(t)|=r(3.6)independently of .Otherwise,assume that u is a solution of(3.5)for some ∈(0,1)such that u =r. Note that f n(t,(J u)(s)) 0.By Lemma2.3,for all t∈[0,2 ],u(t) 1/n andu(t)−1n=2G(t,s)f n(s,(J u)(s))d s m2f n(s,(J u)(s))d smMmax0 t 22G(t,s)f n(s,(J u)(s))d s=u−1n.Thus,u(t)u−1n+1nu −1n+1n> r.Therefore,we have from condition(F2),for all t∈[0,2 ],u(t)=20G(t,s)f n(s,(J u)(s))d s+1n= 2G(t,s)f(s,(J u)(s))d s+1n2G(t,s)f(s,(J u)(s))d s+12G(t,s)g((J u)(s))1+h((J u)(s))g((J u)(s))d s+1ngr1+hrgr2G(t,s)d s+1n0=1gr1+hrgr+1(3.7)since r/ (J u)(t) r/ . Therefore,r= u 12gr1+hrgr+1n0.This is a contradiction to the choice of n0and the claim is proved.1534J.Chu,Z.Zhou /Nonlinear Analysis 64(2006)1528–1542Define the operator T :X →X by(T u)(t)= 2 0G(t,s)f n (s,(J u)(s))d s +1n ,(3.8)then (3.5)is equivalent to the fixed point problem u =(1− )1n+ T u .(3.9)It is easy to see that T :X →X is continuous and completely continuous.Define the set U ={u ∈C [0,2 ]: u <r }.Since (3.6)holds,the non-linear alternative of Leray–Schauder [16]guarantees that T has a fixed point,denoted by u n ,in U ,i.e.,problem (3.2)has a solution u n with u n <r .Since u n satisfies u n =T (u n ),u n (t) 1/n for all t ∈[0,2 ]and u n is actually a positive solution of problem (3.2).Next,we claim that these solutions u n have a uniform positive lower bound,i.e.,there exists a constant >0,independent of n ∈N 0,such thatmin t ∈[0,2 ]u n (t) (3.10)for all n ∈N 0.To see this,we know from (F 1)that there exists a function r 0such that f (t,J u) r (t)for (t,u)∈[0,2 ]×(0,r ].Now,let u r (t)be the unique solution to the problem u − u + 2u = r (t),0 t 2 ,u (i)(0)=u (i)(2 ),i =0,1,(3.11)thenu r (t)= 20G(t,s) r (s)d s m r 1>0,where · 1denotes the usual L 1-norm over [0,2 ].So,we haveu n (t)= 2 0G(t,s)f n (s,(J u n )(s))d s +1/n= 2 0G(t,s)f (s,(J u n )(s))d s +1/n 2 0G(t,s) r (s)d s +1/n =u r (t)+1/n m r 1=: .In order to pass the solutions u n of the truncation problems (3.2)to that of the original problem (2.3),we need the following fact:u n K(3.12)for some constant K >0(independent of n ∈N 0)and for all n ∈N 0.J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421535 In fact,there exists t0∈[0,2 ]such that u n(t0)=0since u n(0)=u n(2 ).Integrate(3.2) in[0,2 ],it is obtained2 2u n(t)d t=2f n(t,(J u n)(t))+2nd t,(3.13)since u n(0)=u n(2 ). Thenu n =max0 t 2 |u n(t)|=max0 t 2tt0u n(s)d s=max0 t 2tt0[f n(s,(J u n)(s))+ 2/n− 2u n(s)+ u n(s)]d s2[f n(s,(J u n)(s))+ 2/n]d s+ 22u n(s)d s+ |u n(t)−u n(t0)|=2 2 2u n(s)d s+ |u n(t)−u n(t0)|<4 2r+2 r=:K.As u n <r,by(3.12),{u n}n∈N0is a bounded and equi-continuous family on[0,2 ].Bythe Arzela–Ascoli Theorem,{u n}n∈N0has a subsequence,{u nk}k∈N,converging uniformlyon[0,2 ]to a function u∈X.From the fact u n <r and(3.10),u satisfies u(t) rfor all t∈[0,2 ].Moreover,u nksatisfies the integral equationu nk (t)=2G(t,s)f(s,(J u nk)(s))d s+1/n k.Letting k→∞,we arrive atu(t)=2G(t,s)f(s,(J u)(s))d s,where the uniform continuity of f(t,J u)on[0,2 ]×[ ,r]is used.Therefore,u is a positive solution to problem(2.3).Finally,it is not difficult to show that u <r,by noting that if u =r,the argument similar to the proof of thefirst claim(3.6)will yield a contradiction.Corollary3.2.Let the non-linearity in(1.1)bef(t,u)=b(t)u− + c(t)u +e(t),0 t 2 ,(3.14)where >0, 0,b(t),c(t),e(t)∈X are non-negative functions and b(t)>0for all t∈[0,2 ],and >0is a positive parameter.Then(i)if <1,problem(1.1)has at least one positive solution for each >0;and(ii)if 1,problem(1.1)has at least one positive solution for each0< < ∗,where ∗is some positive constant.1536J.Chu,Z.Zhou /Nonlinear Analysis 64(2006)1528–1542Proof.Remark that assumption (F 1)is fulfilled by H (t)=H − min t b(t).To verify (F 2),one may simply takeg(u)=b 0u − ,h(u)= c 0u +e 0,whereb 0=max t b(t)>0,c 0=max t c(t) 0,e 0=max te(t) 0.Now,the existence condition (F 3)becomes< r +1 +2−b 0 + −e 0r c 0r +for some r >0.So,problem (1.1)has at least one positive solution for0< < ∗:=sup r>0r +1 +2−b 0 + −e 0r c 0r + .Note that ∗=∞if <1and ∗<∞if 1.We have the desired results.Next,we will find another positive solution to problem (1.1)by using Theorem 2.2for certain non-linearities.Theorem 3.3.Suppose that (F 1)–(F 3)are satisfied.Furthermore ,assume that(F 4)There exist continuous ,non-negative functions g 1(u)and h 1(u)on (0,∞)such thatf (t,u)g 1(u)+h 1(u)for all (t,x)∈[0,2 ]×(0,∞)and g 1(u)>0is non-increasing and h 1(u)/g 1(u)is non-decreasing in u ∈(0,∞);and(F 5)There exists a positive number R >r such that Rg 1(R/ ) 1+h 1( R/ )g 1( R/ ) 1,where is as in Section 2.Then problem (1.1)has another positive solution ˜u with r/ < ˜u R/ .Proof.Let X =C [0,2 ]and K be the cone in X defined by (2.7).Let1={u ∈X : u <r }, 2={u ∈X : u <R }and define the operator T :K ∩(¯2\ 1)→K by (T u)(t)= 2 0G(t,s)f (s,(J u)(s))d s,0 t 2 .(3.15)J.Chu,Z.Zhou /Nonlinear Analysis 64(2006)1528–15421537Since r u R for u ∈K ∩(¯ 2\ 1),thus 0< r/ (J u)(t) R/ .Since f :[0,2 ]×[ r/ ,R/ ]→[0,∞)is continuous,it follows from Lemma 2.3that the operator T :K ∩(¯ 2\ 1)→K is well defined and is continuous and completely continuous.First,we showT u < u for u ∈K ∩j 1.(3.16)In fact,if u ∈K ∩j 1,then u =r and r u(t) r for 0 t 2 .So we have(T u)(t)= 2 0G(t,s)f (s,(J u)(s))d s2 0G(t,s)g((J u)(s)) 1+h((J u)(s))g((J u)(s)) d s g r 1+h r g r 2 0G(t,s)d s =1 2g r 1+h r g r<r = u .This implies T u < u ,i.e.,(3.16)holds.Next,we show thatT u u for u ∈K ∩j 2.(3.17)To see this,let u ∈K ∩j 2.Then u =R and R u(t) R .As a result,it follows from (F 4)and (F 5)that,for 0 t 2 ,(T u)(t)= 2 0G(t,s)f (s,(J u)(s))d s2 0G(t,s)g 1((J u)(s)) 1+h 1((J u)(s))g 1((J u)(s)) d s 2 0G(t,s)g 1 R 1+h 1( R/ )g 1( R/ ) d s =1 2g 1 R 1+h 1( R/ )g 1( R/ )R = u .Now,(3.16),(3.17)and Theorem 2.2guarantee that T has a fixed point u ∈K ∩(¯ 2\ 1)with r u R .Clearly,˜u =J (u)is a positive solution of (1.1)and actually satisfiesr/ < ˜u R/ .Let us consider again the example (3.14)in Corollary 3.2for the superlinear case,i.e., >1.We assume also that c(t)>0for all t ∈[0,2 ].To verify (F 4),one may simply takeg 1(u)=b 1u − ,h 1(u)= c 1u +e 1,whereb 1=min t b(t)>0,c 1=min t c(t)>0,e 1=min t e(t) 0.Now,the existence condition (F 5)becomesR +1 +2−b 1 + −e 1 R c 1 + R + .(3.18)Since >1,the right-hand side goes to 0as R →+∞.Thus,for any given 0< < ∗,where ∗is as in Corollary 3.2,it is always possible to find such R ?r that (3.18)is satisfied.Thus,problem (1.1)has an additional positive solution ˜u .Corollary 3.4.Assume in (3.14)that >1and b(t)>0,c(t)>0for all t ∈[0,2 ].Then ,for each with 0< < ∗,with ∗given as in Corollary 3.2,the corresponding problem value problem (1.1)has at least two different positive solutions .4.Semi-positone caseIn this section,we establish the existence of at least one positive solutions of (1.1)in the semi-positone case.By the semi-positone case of (1.1),we mean that f (t,u)may change sign and satisfies(G 1)There exists a constant L >0such that F (t,u):=f (t,u)+L 0for all (t,u)∈[0,2 ]×(0,∞).Theorem 4.1.Suppose that f (t,u)satisfies (G 1).In addition ,we assume that the following conditions are satisfied :(G 2)There exist continuous non-negative functions g(u)and h(u)on (0,∞)such thatF (t,u) g(u)+h(u)for all (t,u)∈[0,2 ]×(0,∞)and g(u)>0is non-increasing and h(u)/g(u)is non-decreasing in u ∈(0,∞);(G 3)There exist continuous ,non-negative functions g 1(u)and h 1(u)on (0,∞)such thatF (t,u) g 1(u)+h 1(u)for all (t,u)∈[0,2 ]×(0,∞)and g 1(u)>0is non-increasing and h 1(u)/g 1(u)is non-decreasing in u ∈(0,∞);(G 4)There exists r > / such that r gr − 1+h(r/ − )g(r/ − ) >1 2,where =L/ 3, =m/M is the same as in Section 2.(G 5)There exists a positive number R >r such that R g 1 R − 1+h 1( R/ − )1 1 2.Then problem (1.1)has a solution u ∈C 2[0,2 ]with u(t)>0for t ∈[0,2 ]and r/ < u + <R/ .Proof.We only need to show that problem (2.3)has a positive solution y(t)with r < y + R .To do so,we will show that problemu − u + 2u =F (t,(J u)(t)− ),0 t 2 ,u(0)=u(2 ),u (0)=u (2 )(4.1)has a positive solution u ∈C 2[0,2 ]with (J u)(t)> for t ∈[0,2 ]and r < u R .Let X =C [0,2 ]and K be a cone in X defined by (2.7).Let1={u ∈X : u <r }, 2={u ∈X : u <R }and define the operator T :K ∩(¯2\ 1)→K by (T u)(t)=2 0G(t,s)F (s,(J u)(s)− )d s,0 t 2 ,(4.2)where G(t,s)is the Green function given by (2.5).Since r u R for u ∈K ∩(¯ 2\ 1),thus 0< r/ − (J u)(s)− R/ − .Since F :[0,2 ]×[ r/ − ,R/ − ]→[0,∞)is continuous,it follows from Lemma 2.3that the operator T :K ∩(¯ 2\ 1)→K is well defined and is continuous and completely continuous.First,we showT u < u for u ∈K ∩j 1.(4.3)In fact,if u ∈K ∩j 1,then u =r and (J u)(t) r/ > for 0 t 2 .So we have from conditions (G 2)and (G 4)that,for 0 t 2 ,(T u)(t)= 2 0G(t,s)F (s,(J u)(s)− )d s2 0G(t,s)g((J u)(s)− ) 1+h((J u)(s)− )g((J u)(s)− ) d s 2 0G(t,s)g r − 1+h(r/ − )g(r/ − ) d s =1 2g r − 1+h(r/ − )g(r/ − )<r = usince r/ − (J u)(s)− r/ − .This implies T u < u ,i.e.,(4.3)holds.Next,we showT u u for u ∈K ∩j 2.(4.4)To see this,let u ∈K ∩j 2,then u =R and (J u)(t) R/ > for 0 t 2 .As a result,it follows from (G 3)and (G 5)that,for 0 t 2 ,(T u)(t)= 20G(t,s)F (s,(J u)(s)− )d s2 0G(t,s)g 1((J u)(s)− ) 1+h 1((J u)(s)− )g 1((J u)(s)− ) d s 2 0G(t,s)g 1 R − 1+h 1( R/ − )g 1( R/ − ) d s =1g 1 R − 1+h 1( R/ − )1 R = u .This implies T u u ,i.e.,(4.4)holds.Now,(4.3),(4.4)and Theorem 2.2guarantee that T has a fixed point u ∈K ∩(¯ 2\ 1)with r u R .Note u =r by (4.3).Clearly,this u is a positive solution of (4.1).Let y(t)=u(t)− ,then y(t)is a positive solution of (2.3)and r < y + R sincey (t)− y (t)+ 2y(t)=u (t)− u (t)+ 2(u(t)− )=F (t,(J u)(t)− )− 3=F (t,(J u)(t)− )−L=f (t,(J u)(t)− )=f (t,(Jy)(t))for all t ∈[0,2 ].Corollary 4.2.Let the non-linearity in (1.1)be (3.14),where 1, >1,b(t),c(t):[0,2 ]→(0,∞)are continuous functions ,e(t)∈C [0,2 ]and >0is a positive pa-rameter.Then (1.1)has at least one positive solution for each 0< < ∗,where ∗is some positive constant .Proof.We will apply Theorem 4.1with L =e 0=max 0 t 2 |e(t)|andg(u)=b 0u − ,h(u)= c 0u +2e 0,g 1(u)=b 1u − ,h 1(u)= c 1u .Here,b 1=min t b(t)>0,c 1=min t c(t)>0,b 0=max t b(t)>0,c 0=max tc(t)>0.Then conditions (G 1)–(G 3)are satisfied.The existence condition (G 4)becomes< 2r( r/ − ) −b 0−2e 0(r/ − )c 0(r/ − ) +for some r> / .Let∗=supr> / 2r( r/ − ) −b0−2e0(r/ − )c0(r/ − ) +.Note that ∗<∞since >1.Another existence condition(G5)becomes2R(R/ − ) −b1c1( R/ − ) +.(4.5)Since >1,the right-hand side of(4.5)goes to0as R→+∞.Thus,for any given 0< < ∗,it is always possible tofind such R?r that(4.5)is satisfied.Thus,problem (1.1)has at least one positive solution.5.AcknowledgmentThe authors express their thanks to the referee for his valuable suggestions. 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用差分方法求解一类二阶两点边值问题
用差分方法求解一类二阶两点边值问题邹序焱【摘要】Presents a difference method to solve a second order two-point boundary value problem.The method has second-order accuracy,the coefficient matrix is tridiagonal matrix,and Thomas method is used to get the solutions.Through an example verifies the accuracy of the algorithm.%对一类二阶两点边值问题给出一种差分算法。
该算法具有二阶精度,差分格式的系数矩阵为三对角矩阵,可用追赶法求解,并通过实例验证了算法的精度。
【期刊名称】《湖南工业大学学报》【年(卷),期】2012(026)003【总页数】3页(P13-15)【关键词】常微分方程;边值问题;差分算法;追赶法【作者】邹序焱【作者单位】宜宾学院数学学院,四川宜宾644000【正文语种】中文【中图分类】O241.81近年来，许多学者对非线性微分方程边值问题正解的存在性进行了研究[1-5]，如文献[5]利用不动点定理研究了边值问题得到了在一定条件下存在解的结论。
大多数文献都只证明了解的存在性，并未给出解的求法以及解的表达式，这是因为求边值问题的解析解比较困难。
本文将采用差分方法讨论边值问题（1）的数值解法。
式中：p(x)∈ C1[a,b]，p(x)≥ 0 ；r(x)，f(x)∈C[a,b]；,,λ1,λ2为已知常数。
1 用差分方法求解边值问题用有限差分方法解两点边值问题，一般分两步进行：第一步将求解区间进行网格剖分，取为网格步长，以xm=a+mh(m=0,1,…,N)为网格结点，将区间[a,b]分成N等分。
第二步用差商代替微商把原方程变成等价的离散方程。
MULTIPLE POSITIVE SOLUTIONS TO BOUNDARY VALUE PROBLEMS FOR SECOND-ORDER SYSTEM OF QUASILINEAR DI
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上大学的优点英语作文
上大学的优点英语作文Attending university offers a multitude of benefits that can significantly impact an individuals personal and professional life. Here are some of the key advantages of pursuing higher education1. Academic Growth University education provides a comprehensive and indepth understanding of a chosen field. It allows students to specialize in their area of interest and gain expertise through rigorous academic training.2. Critical Thinking and Problem Solving Higher education encourages students to think critically and analytically. It equips them with the skills to approach problems from multiple perspectives and develop innovative solutions.3. Access to Resources Universities are hubs of knowledge offering access to extensive libraries research facilities and technological resources that can enhance learning and research capabilities.4. Networking Opportunities Attending university provides the chance to meet and interact with a diverse group of individuals including fellow students professors and industry professionals. These connections can be invaluable for future collaborations and career opportunities.5. Personal Development Living away from home and being part of a university community can foster independence and personal growth. Students learn to manage their time finances and responsibilities which are crucial life skills.6. Career Advancement A university degree is often a prerequisite for many professional positions. Higher education can open doors to better job prospects and higher earning potential.7. Cultural Exposure Universities often host cultural events guest lectures and international exchange programs providing students with a broader understanding of the world and its diverse cultures.8. Research Opportunities For those interested in academia or research universities offer the chance to participate in cuttingedge research projects which can lead to significant contributions to their field.9. Skill Development Beyond the classroom universities offer various extracurricular activities and clubs that can help students develop a range of skills such as leadership teamwork and public speaking.10. Lifelong Learning The pursuit of higher education instills a habit of lifelong learning. Graduates are more likely to continue seeking knowledge and adapting to new information throughout their careers.In conclusion attending university is not just about obtaining a degree its an investment in ones intellectual and personal development. It provides a solid foundation for a successful career and a fulfilling life.。
UKMT-Senior-Maths-Challenge-Solutions-2012
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According to one astronomer, there are one hundred thousand million galaxies in the universe, each containing one hundred thousand million stars. How many stars is that altogether? A 1013 B 10 22 C 10100 D 10120 E 10121
University of Leeds, Leeds LS2 9JT.
or by post to: SMC Solutions, UKMT Maths Challenges Office, School of Mathematics, Quick Marking Guide 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 E D E B D A A B A E E D B C B B
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Solution: B The interior angles of an equilateral triangle, a square and a regular pentagon are 60 , 90 and
108 , respectively. Hence θ = 360 − (60 + 90 + 108) = 360 − 258 = 102 .
Solution: B One hundred thousand is 10 5 and one million is 10 6 . So one hundred thousand million is 10 5 × 10 6 = 10 5+ 6 = 1011 . Therefore the total number of stars is 1011 × 1011 = 1011+11 = 10 22 .