Purchase Solution

Fredholm equation and Lipschitz condition

Not what you're looking for?

Ask Custom Question

Consider the nonlinear Fredholm equation where is continuous on [a,b] and is continuous and satisfies a Lipschitz condition: on the set . Show that the integral equation has a unique solution on [a,b] if .

Attachments
Purchase this Solution
Solution Summary

The Fredholm equation and Lipschitz condition are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

Purchase this Solution
Free BrainMass Quizzes
Multiplying Complex Numbers

This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.

Graphs and Functions

This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.

Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

Know Your Linear Equations

Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.

View More Free Quizzes
Related BrainMass Solutions

  • Local or Uniform Lipschitz Constants

    that it has locally Lipschitz condition since the Lipschitz constant here changes as the reagion changes?

  • Lipschitz Continuity and Initial Value Problems for ODE's

    Lipschitz continuity is a necessary condition to insure that the initial value problem for a first order ordinary differential equation is well-posed, i.e., that a solution exists and is unique.

  • Lipschitz functions

    (c) Any f∈C_[a,b] can be approximated by a straight-segments function like that in equation (a.6) with the only difference that f does not necessarily satisfy a Lipschitz condition.

  • Existence and uniqueness of solutions for IVPs

    For equation 2, Modulus[f(y1,t) - f(y2,t)] = Modulus[(y1-y2)/(1 - t^2)] which is less than or equal to Mod[1-t^2] * Mod[y1-y2] and so it is immediate that f satisfies the Lipschitz condition as long as t is not equal to plus or minus 1.

  • Real Analysis : Lipschitz Condition

    152221 Real Analysis using the Lipschitz Condition Please see the attached file for the fully formatted problems. Please see the attached file for the complete solution. A proof using the Lipschitz condition is provided in the solution.

  • Lipschitz functions

    But it does not satisfies the Lipschitz condition. Because when x approaches 1 or -1, the slope of the function goes to infinity. This is a proof regarding uniformly continuous and Lipschitz functions.

  • Differential Equations : Uniqueness Theorem and Initial Conditions

    51499 Differential Equations : Uniqueness Theorem and Initial Conditions Using uniqueness theorem, what can you conclude about the solution to the equation with the given inital conditions?

  • Advanced Calculus: The Existence Theorem for Nonlinear Differential Equations

    Proof: is Lipschitz continuous, then it satisfies the Lipschitz condition. There exists , such that is continuous, so we can suppose is continous in a rectangular region , .

  • Existence and Uniqueness of Solution to an ODE

    Useful information: Global existence and uniqueness Theorem: The ordinary differential equation d_t â-u=â-f(t,â-u (t)) â-u (0)=â-u_0 has a unique solution if â-fâ??C^0 (I)Ã-Lipschitz(L_â??

View More Related Solutions