Solving Linear Integral Equation: Norm of Resolvent

  • MHB
  • Thread starter sarrah1
  • Start date
  • Tags
    Norm
In summary, the conversation discusses the use of a linear integral operator (related to integral equations) and its properties. The conclusion is that if a bounded linear operator $T$ with $\|T\|<1$ is used, then $\|(I-T)^{-1}\| \leqslant \frac1{1-\|T\|}$. This is helpful in solving problems involving integral equations and the operator.
  • #1
sarrah1
66
0
I have a linear integral operator (related to integral equations)
$(Ky)(x)=\int_{a}^{b} \,k(x,s)y(s)ds$
If $|b|. ||K||<1$ (b is a scalar)
can I say
$||(I-bK)-1||< 1 / (1-|b|.||K||)$
I think it's correct
is it?
thanks
 
Physics news on Phys.org
  • #2
sarrah said:
I have a linear integral operator (related to integral equations)
$(Ky)(x)=\int_{a}^{b} \,k(x,s)y(s)ds$
If $|b|. ||K||<1$ (b is a scalar)
can I say
$||(I-bK)-1||< 1 / (1-|b|.||K||)$
I think it's correct
is it?
thanks
Yes, that is correct.

If $T$ is a bounded linear operator with $\|T\|<1$ and $x = (I-T)y$ then $\|x\| = \|y - T(y)\| \geqslant \|y\| - \|T(y)\| \geqslant \|y\| - \|T\|\|y\| = \|y\|(1-\|T\|).$ Therefore $\|(I-T)^{-1}(x)\| = \|y\| \leqslant \frac1{1-\|T\|}\|x\|.$ It follows that $\|(I-T)^{-1}\| \leqslant \frac1{1-\|T\|}$. Now all you have to do is to put $T = bK.$
 
  • #3
extremely grateful
 

FAQ: Solving Linear Integral Equation: Norm of Resolvent

What is a linear integral equation?

A linear integral equation is an equation that involves an unknown function and an integral of that function. It can be represented in the form of f(x) = g(x) + λ ∫K(x,t)f(t)dt, where f(x) is the unknown function, g(x) is a known function, λ is a constant, and K(x,t) is a known integral kernel.

What is the norm of a resolvent?

The norm of a resolvent is a mathematical concept used to measure the size or magnitude of a resolvent, which is a specific type of operator in functional analysis. It represents the inverse of the distance between a given operator and the set of invertible operators.

What is the significance of solving linear integral equations?

Solving linear integral equations has many practical applications in various fields such as engineering, physics, and economics. It allows for the modeling and analysis of complex systems and processes, and can provide insight into the behavior of these systems.

How is the norm of a resolvent related to solving linear integral equations?

The norm of a resolvent is used to determine the convergence of a numerical method for solving linear integral equations. It is also used to analyze the stability and accuracy of these methods.

What are some common methods for solving linear integral equations?

Some common methods for solving linear integral equations include the Fredholm method, the method of successive approximations, and the collocation method. Other methods such as the Galerkin method and the boundary integral equation method may also be used depending on the specific problem at hand.

Similar threads

MHB Does the Norm of a Linear Integral Operator Equal Its Spectral Radius?
Replies
2
Views
2K
MHB On the spectral radius of bounded linear operators
Replies
3
Views
2K
MHB How Can I Calculate the Norm of the Operator \(I-L^{-1}K\)?
Replies
4
Views
2K
MHB Can Linear Integral Operators Be Combined to Prove a Trivial Inequality?
Replies
2
Views
2K
MHB Is the Triangle Inequality Applicable to Norms of Integral Operators?
Replies
1
Views
2K
MHB Proof of Fredholm-Volterra Equation Convergence
Replies
2
Views
2K
MHB Do Commuting Linear Integral Operators Share Eigenfunctions and Eigenvalues?
Replies
2
Views
1K
MHB Checking Proof of Theorem 6.2.8 Part (ii)
Replies
2
Views
2K
MHB Operator Form for Volterra Linear Integral Equations
Replies
1
Views
1K
MHB Norm of a linear integral operator
Replies
12
Views
4K

Hot Threads

Recent Insights

Back
Top