Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I ask my question in two parts: though the topic is similar, I would like to distinguish linear and general cases since methods may be too different while my questions are broad.

Consider a space $X$ which we assume to be Banach. We define a linear operator $A:X\to X$ which is bounded: $$ \|\mathcal A\| := \sup\limits_{x\in X}\frac{\|\mathcal Ax\|}{\|x\|}<\infty. $$ I would like to discuss an existence of solution for a fixpoint equation $$\mathcal Ax = x.\quad (1)$$

What do I know: this is an eigenvalue problem for $\lambda = 1$ or it is a problem of finding the kernel for $(\mathcal A-\mathcal I)$, $\mathcal I$ is an identity operator. Also, if $\dim X<\infty$ then $\mathcal A$ has a correspondent finite-dimensional matrix $A$ and all properties of $(1)$ can be studied through the rang of $A$. E.g. there exists a solution of $(1)$ iff $\det (A-I)=0$ for $I$ is an identity matrix of the same size as $A$.

For the general state-space my first question is:

1.If there is a method similar to calculating $\det (A-I)$ to verify the existence of solution for $(1)$?

If the dimension of $X$ is not necessary finite, one of the main methods is to use Banach theorem based on the contraction principle - so it is only valid if $|\lambda^*(\mathcal A)<1|$ for the maximum eigenvalue in the absolute sense.

2.What can we do if the spectrum is not bounded by $1$ but just does not contain it?

There are also procedures (usually in the discrete-time setting, e.g. $X = L^2$ and $\mathcal A$ is an integral operator) connected with the iterations of operator $\mathcal A$. There are examples (if one wants, I can put it here) when for some $x\neq 0$ the limit $$x' = \lim\limits_{n\to\infty}\mathcal A^nx\quad (2)$$ exists while $\mathcal A$ is not contractive.

3.Under which conditions $\mathcal Ax' = x'$?

Finally, there are some methods in the continuous time setting (e.g. $X = L^2$ again and we are talking about differential operators). If one would like to solve an equation $\mathcal Bx = 0$ then it's helpful to consider a function $f(t)$ such that $f(0) = x_0$ and $$ \frac{df}{dt}(t) = \mathcal B f(t).\quad (3) $$ Suppose, $\lim\limits_{t\to\infty}f(t) = x'\in X$.

4.Under which conditions $\mathcal Bx' = 0$?

Finally, we can put $\mathcal A = \mathcal B+\mathcal I$. Then if the conditions of 4. are satisfied, $\mathcal Ax' = x'$.

5. Why for the integral equations ("discrete-time") people commonly use $(2)$ while for, say PDEs they use $(3)$ rather then $(2)$?

So there are 5 questions, and I would appreciate if you can help me with answering at least one of them or referring me to a literature which covers these questions. I guess that 5. is can be an unclear question - so if it is, just tell, I will try to make it clear.

share|improve this question
1  
1. I think that if $A$ is a trace-class operator (assuming $X$ is Hilbert, I guess) then there is a sensible definition of $det(A-I)$ and that it can still be used determine whether $\ker(A-I)$ is trivial or not. I think that this is mentioned in Lax's functional analysis text. –  Mark Jul 25 '11 at 12:30
    
@Mark: thanks you. –  Ilya Jul 25 '11 at 12:32
    
I'm not sure why you refer to integral operators as the "discrete time setting" and to differential operators as the "continuous time setting". Wether or not there is a continuous time variable in your model or a discrete time variable is unrelated to the definitions of operators that you use in your model. –  Tim van Beek Jul 25 '11 at 12:52
    
@Tim: sure you're right, that's why I use words usually, e.g. and commonly. The question 5. in fact is unrelated to the time setting. Only from my impression as I've seen the technique (2) mostly used for the integral equation and in the discrete time problems, while the method (3) is widely used for solving PDEs and continuous time problems. –  Ilya Jul 25 '11 at 13:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.