Integral Equation

[Camillo]

Solve the equation :

$ f(x) = Phi(x) +lambda* int_0^1 xy f(y)dy $ , where

$Phi(x)$ is a continuous known function, while $ f(x) $ is the unknown function (continuous).

[gugo82]

"Thomas":It's very nice that I knew a lot of objects u uses, but with a different name... for example I've met the regular values of the operators while studying quantum mecahnics (ok course we don't learn functional analysis at the level of a real mathematician ), but we defined the complmentary in C of that set and we called it the "spectrum" of an operator...

-) Anyway tell me if I'm right... when we have a (bounded?) operator on an Hilbert (at this point I think also Banach) space, we can define the spectrum (or the regular values) because the operators form a Banach Algebra...

Correct.

I'd like to say that there's a subtle distinction between regular values for an integral equation and for a bounded linear operator.
Actually we say that $mu in CC$ is a regular value for equation $u=f+mu\int_I K*u" d"m$ iff it's solvable (with unique solution) for every known term $f$; on the other hand, we call $lambda in CC$ a regular value for the bounded linear operator $T$ iff equation $(lambda*I-T)u=f$ is solvable (with unique solution) for every known term $f$...

Now, suppose $K$ be choosen in such a way that the operator defined by $Tu=\int_I K*u" d"m$ is linear and bounded: then by linearity we can state that for each non-zero regular value for the integral equation $(I-muT)u=f$ there is one and only one regular value for the operator $T$, namely $lambda=1/mu$; conversely, for each non-zero regular value for $T$ there is an unique regular value $mu$ for integral equation $(I-muT)u=f$, i.e. $mu=1/lambda$.

To avoid confusion, it's customary in some Functional Analysis book to use the name regular value only in Integral Equations theory and to prefer resolvent value instead of "regular value" in Bounded Linear Operators theory.

"Thomas":-)At this point the spectrum of the operator partitions into (is this trivial, actually I don't know): 1)point spectrum; 2)approximate spectrum; 3)residual spectrum; (and we can find the regular values this way only in the case of operators, not in any Banach Algebra).

-)A step farher is proving that if the operator is compact, only the point spectrum exists (and is discrete);

So point (3) permitted u to find the regular values of the operator (compact because of finite range) simply looking for its eigenvalues...

Right, but there is a little imperfection.

The spectum of a bounded linear operator is not always only punctual: infact, the theorem about the structure of the spectrum for compact operators guarantees that $sigma(T)-{0}=sigma_P(T)-{0}$ and that $sigma_P(T)-{0}$ is at most a sequence approaching to $0$ but says nothing about the spectral value $0$ (remember that $0 in sigma(T)$ for every compact operator $T$)... There are simple examples which show that each of the three alternatives $0 in sigma_P(T)$, $0 in sigma_A(T)$ and $0 in sigma_R(T)$ can hold.

[Thomas16]

Ok... now the solution has become almost clear to me... ....

It's very nice that I knew a lot of objects u uses, but with a different name... for example I've met the regular values o the operators while studying quantum mecahnics (ok course we don't learn functional analysis at the level of a real mathematician ), but we defined the complmentary in C of that set and we called it the "spectrum" of an operator...

-) Anyway tell me if I'm right... when we have a (bounded?) operator on an Hilbert (at this point I think also Banach) space, we can define the spectrum (or the regular values) because the operators form a Banach Algebra...

-)At this point the spectrum of the operator partitions into (is this trivial, actually I don't know): 1)point spectrum; 2)approximate spectrum; 3)residual spectrum; (and we can find the regular values this way only in the case of operators, not in any Banach Algebra).

-)A step farher is proving that if the operator is compact, only the point spectrum exists (and is discrete);

So point (3) permitted u to find the regular values of the operator (compact because of finite range) simply looking for its eigenvalues...


I wanted to ask u if what I wrote is correct... actually it's a mixing of personal rememberings and wikipedia


anyway now I realize that all this is quite useless, since u put in the hypothesis of fredholm theory the compactness of the operator and so "a priori" only point spectrum plays a role in the theory and the regular values are simply the "non eigenvalues".... mmm... anyway what I wrote is consistent?

[gugo82]

"Camillo":I appreciate the completeness of Gugo solution (not always of immediate understanding for me ) and I'm following with deep interest the discussions between Gugo and Thomas.
Anyhow I want to take advantage of the special form of the kernel, providing a simple solution (also if not general : the case $ lambda = 3 $ is excluded).
Let $c = int_0^1 y f(y)dy $ and $ a = int_0^1 y Phi(y)dy $ .
Then the initial equation becomes :
$ f(x) = Phi(x)+ lambda xc $ .
We get $ c= int_0^1 x f(x) dx= int_0^1 xPhi(x)dx +int_0^1 xlambdaxcdx= a+(lambdac)/3 $
Solving for $ c$ , we have $ c= (3a)/(3-lambda) $and finally :

$f(x) = Phi(x)+lambdaxc = Phi(x)+(3alambdax)/(3-lambda) $ , which is valid for $ lambda ne 3 $.

That's exactly what I wrote in my first post...

[gugo82]

"elgiovo":[quote="Gugo82"] $C^**=NBV$

I didn't know about the dual of $C$: I searched on Google, but I couldn't find anything. Would you suggest me a reference? Thanks a lot.[/quote]
My main reference is an English translation of the classical book by F.Riesz & B. Sz.-Nagy Leçons d'Analyse Fonctionelle, cap. III, §§ 49-50; obviously, you can find some more theorems on the space $C$ on another classic treatise: N. Dunford-J. Schwartz, Linear Operators, vol. 1, cap. IV, § 6.
A more recent book is the one by Rudin, Real and Complex Analysis, cap. 2, theorem 2.4; while in the two above-mentioned books the authors approach the theory in an old-fashioned style (which is simpler but not fully satisfactory to modern reader's eyes), in the latter text there is a much more abstract way to present and proof the Riesz Representation Theorem.

Riesz Representation Theorem states that each function $alpha(x)$ of bounded variation on $[a,b]$ ($a
Now this theorem fails to give a unique representation of any functional in $C^**$, for each $A in C^**$ has actually infinitely many representants in $BV$. For example, the Dirac $delta in C([-1,1])^**$ is represented by means of the Heavyside function:

$H(x)=\{(0, " if " -1le x <0),(1, " if " 0le x le 1):}$

(which is of bounded variation cause it's increasing), but each function $h(x)=H(x)+y_0$ represents $delta$ as well; also each function of the type:

$bar(alpha)(x)=\{(0, " if "-1le x<0), (bary, " if " x=0), (1, " if " 0
where $bary$ is any number fixed in $[0,1[$, represents $delta$ as well as each $alpha(x)=bar(alpha)(x)+y_0$...
To provide uniqueness of the Riesz reprentating function we have to restrict the space in which we search it to a subclass of $BV$, namely $NBV([a,b])$.
We say that a function of bounded variation $alpha$ is normalized (and write $alpha in NBV$) iff

1) $alpha$ is continuous from the right of every point in $[a,b[$;

2) $alpha(a)=lim_(x to a^+) alpha(x)=0$.

The Riesz Theorem can finally be stated as follows:

For every bounded linear functional $A$ defined on $C([a,b])$ there exists one and only one normalized function of bounded variation $alpha(x)$ s.t.:

i) $quad AA f in C, quad =\int_a^b f(x)" d"alpha(x)$ and

ii) $quad ||A||_(C^**)=V_a^b(alpha)$

(the r.h.s. in the last equality being the total variation of $alpha(x)$ over $[a,b]$).

Hope you find this post useful.
Byes!

[elgiovo]

"Gugo82": $C^**=NBV$

I didn't know about the dual of $C$: I searched on Google, but I couldn't find anything. Would you suggest me a reference? Thanks a lot.

[Camillo]

I appreciate the completeness of Gugo solution (not always of immediate understanding for me ) and I'm following with deep interest the discussions between Gugo and Thomas.
Anyhow I want to take advantage of the special form of the kernel, providing a simple solution (also if not general : the case $ lambda = 3 $ is excluded).
Let $c = int_0^1 y f(y)dy $ and $ a = int_0^1 y Phi(y)dy $ .
Then the initial equation becomes :
$ f(x) = Phi(x)+ lambda xc $ .
We get $ c= int_0^1 x f(x) dx= int_0^1 xPhi(x)dx +int_0^1 xlambdaxcdx= a+(lambdac)/3 $
Solving for $ c$ , we have $ c= (3a)/(3-lambda) $and finally :

$f(x) = Phi(x)+lambdaxc = Phi(x)+(3alambdax)/(3-lambda) $ , which is valid for $ lambda ne 3 $.

[gugo82]

"Thomas":Ok... you were so nice in explaining me all these things! I had never seen the definition of the adjoint outside og Hilbert spaces...

Anyway, tell me if I've understood well:

1- u wrote that the dual of $C([0,1])$ is $L_1([0,1])$... can u check it? actually it sound strange to me...

Right, I was wrong.
Actually $C^**=NBV$ but, since $P(x)=x^2/2 in NBV$ and $P'(x)=p(x)=x$, nothing changes in the following computations.
Anyway I have to correct my statement, thanks a lot.

P.S.: Now I remember why I made that mistake... I wrote my first post while I was also studying the sequences spaces $c_0$ and $l^p$; now it happens to be $(c_0)^**=l^1$ and with this equality in mind I wrote $C^**=L^1$...

"Thomas":2- is it right to say that if $\lambda$ is a regular value alternative one holds, otherwise if it isn't alternative 2 holds? u didn't say that but I suppose this is the case

Yes.
This conclusion holds for compact operator: infact if $lambda$ is regular then $(I-lambdaT)^(-1)$ exists and a solution of $u=f+lambdaTu$ is of course unique for all $f in X$ (actually $u=(I-lambdaT)^(-1)f$); on the other hand, since $T$ is compact, all the singular values $lambda$ are the inverse of some eigenvalue of $T$ therefore $I-lambdaT=lambda*(1/lambdaI-T)$ is not invertible; moreover, in the latter case, Fredholm Theorem provide the rapresentation formula in II).

"Thomas":3- your way to showing that $\lambda$ not 3 is a regular value is simply to find the inverse in an explict way or not? and the way to show that $\lambda=3$ is not a regular value is to show that $I-T_3$ has a non-banal kernel?

Yes.
Actually it's a very easy way because $T$ is a finite range operator: for operators of this kind the problem to find regular and singular values reduces to solve a problem of Linear Algebra.

"Thomas":4- how did u demonstrate that the kernel $N(1-T_3)$ is simply $span{p}$ and not bigger?

$N(I-3T)$ coincide with $"span"{p}$: here's an easy proof.
For $p=3 *p$, we have $p in N(I-3T)$ and $"span"{p} subseteq N(I-3T)$ by homogeneity; on the other hand, if $f$ is a solution of the homogeneous equation $(I-3T)f=omega$ then we have $f=3*p in "span"{p}$ so $N(I-3T) subseteq "span"{p}$.

[Thomas16]

Ok... you were so nice in explaining me all these things! I had never seen the definition of the adjoint outside og Hilbert spaces...

Anyway, tell me if I've understood well:

1- u wrote that the dual of $C([0,1])$ is $L_1([0,1])$... can u check it? actually it sound strange to me...

2- is it right to say that if $\lambda$ is a regular value alternative one holds, otherwise if it isn't alternative 2 holds? u didn't say that but I suppose this is the case

3- your way to showing that $\lambda$ not 3 is a regular value is simply to find the inverse in an explict way or not? and the way to show that $\lambda=3$ is not a regular value is to show that $I-T_3$ has a non-banal kernel?

4- how did u demonstrate that the kernel $N(1-T_3)$ is simly $span{p}$ and not bigger?

[gugo82]

"Thomas":sorry gugo82, I don't know much about the argoment... I wanted to ask you:

- what do u mean for $C^**$... often you put $^**$ over elements with two stars... what does it precisely mean?

If $V$ is a normed vector space (in particular a Banch space), $V^**$ denotes the normed dual space of $V$ and $u^**$ is the generic element of $V^**$: that is a functional $u^** in V^**$ iff $u^**:V to CC$ is linear and bounded (or continuous, which is the same).
If $u^** in V^**$ and $u in V$, the pairing $$ denotes the value of the functional $u^**$ in $u$; it hasn't to be confused with the inner product of an Hilbert space.

These are Anglosaxon notations which I prefer to the standard European notation (for example, in European books you usually find the symbol $V'$ to denote the normed dual of $V$ and $f,g,h,\ldots$ to denote functionals in $V'$, instead of $V^**$ and $u^**,v^**,\ldots$).

"Thomas":- can you tell me what is your definition of the adjoint of an operator in an Hilbert space?

We are not dealing with an Hilbert space, but with a Banach space: infact $C=C([0,1])$ is a Banach space with the supremum norm; therefore the notion of adjoint operator, in all its generality, has to be given in Banach space as follows.
Let $X,Y$ be Banach spaces, $X^**,Y^**$ their normed duals, $x,y,x^**,y^**$ the generic elements of these four spaces and $A in L(X,Y)$ (that is $A:XtoY$ is a bounded linear operator); there exists one and only one operator $A^** in L(Y^**,X^**)$ s.t.:

$AA x in X, AA y^** in Y^**, quad = $ .

The operator $A^**$ is called adjoint operator of $A$.

"Thomas":- what do you mean for "regular values"?

Let $X$ be a Banach space, $I$ the identity operator on $X$, $A in L(X)$ (that is $A:Xto X$ is a bounded linear operator) and $mu in CC$. We say that $mu$ is a regular value for $A$ iff the bounded linear operator $I-muA$ is invertible, i.e. $(I-muA)^(-1)$ exists.

For any bounded linear operator $A$, the set of regular values is always non-empty for the whole open disc ${ mu in CC: |mu|<1/(||A||_(op))}$ is actually made of regular values (here $||A||_(op)$ denotes the norm of $A$ in $L(X)$)*.

"Thomas":- how do you see that the operator respects Fredholm alternative (I) or (II)?

This is a fundamental result in Fredholm-Riesz Theory:
If $T$ is a compact linear operator acting on a Banach space $X$, then for every $mu in CC$ Fredholm Alternative holds for $I-muT$ (an operator like $I-muT$ is said to be of Riesz type or a compact perturbation of $I$).

"Thomas":- how do you pass from the expression of $$ to the one of $f$?

It's very easy.
The function $f$ is a solution of our integral equation iff $f=Phi+ lambda* p$: note that this equality gives us also a representation of the solution in terms of $Phi$ and $$. Therefore, since we already know $Phi$ (this is the reason for it's called known term! ), our aim is to find $$: infact, once we found $ = alpha$, a simple substitution in our equation leads to $f=Phi + lambda*alpha p$.
This is the plan to solve the equation which is fully carried out in my previous post.


___________
* Obviously, if $A=O$ (here $O$ is the zero-operator on $X$ defined by: $AA x in X, Ox=0_X$), so that $||A||_(op)=0$, then the set of regular values coincides with the whole complex plane.

[Thomas16]

sorry gugo82, I don't know much about the argoment... I wanted to ask you:

- what do u mean for $C^**$... often u put $^**$ over elements with two stars... what does it precisely mean?

- can u tell me what is your definition of the adjoint of an operator in an Hilbert space?

- what do you mean for "regular values"?

- how do u see that the operator respects fredholm alternative (1) or (2)?

- how do u pass from the expression of $$ to the one of $f$?

[gugo82]

"Camillo":To my knowledge the integral equation is Fredholm type .
Volterra type should be like this $ f(x) = Phi(x) +lambda* int_0^x xy f(y) dy $.

I call:

- Volterra equation of the first kind an equation like $ f(x) = Phi(x) +lambda* int_a^b K(x,y) f(y)" d"y $ (with fixed extremes of integration);

- Volterra equation of the second kind an equation like $ f(x) = Phi(x) +lambda* int_a^x K(x,y) f(y)" d"y $ (with the upper extreme variable in $[a,b]$).

Of course the equation of the second kind is a special case of equation of the first kind: infact it suffices to set $hat(K)(x,y)= \{ (K(x,y) , " if " ale y le x le b), (0, " if " a le x le y le b) :}$ in order to obtain:

$int_a^x K(x,y) f(y)" d"y=int_a^b hat(K)(x,y) f(y)" d"y$.

A solution for a Volterra equation of the second kind always exists (and it is unique) for each $Phi$.

[Camillo]

To my knowledge the integral equation is Fredholm type .
Volterra type should be like this $ f(x) = Phi(x) +lambda* int_0^x xy f(y) dy $.

[gugo82]

"Camillo":Solve the equation :

(1) $quad f(x) = Phi(x) +lambda* int_0^1 xy f(y)dy $ , where

$Phi(x)$ is a continuous known function, while $ f(x) $ is the unknown function (continuous).

Equation (1) is of Volterra type, with kernel $K(x,y)=x*y in C([0,1]^2)$ and parameter $lambda in CC$ (in what follows all functions take their values in $CC$).

Let me replace from now on $NBV([0,1])$ and $C([0,1])$ respectively with $NBV, C$.
As long as the kernel separates the variables $x,y$, we can write the equation in the following manner:

(2) $quad f - *p =Phi quad$,

where $p(x)=x in C$, $P(x)=x^2/2 in NBV$ and $<\cdot, \cdot>$ is the duality product of the Banach spaces $NBV=C^**$ and $C$ equipped with their natural norm (in other words, the pairing $$ equals the Stiltjies integral $\int_0^1f" d"P$; since $P'=p$ we have $ = \int_0^1p*f" d"x$).
The operator $T_lambda: C to C$ which assigns to $f$ the function $ *p$ is linear with finite range, hence $T_lambda$ is a compact operator on $C$ for each $lambda in CC$. Then the operator $I-T_lambda$* is of Riesz type and the Fredholm Alternative holds:

I) either equation (2) (and (1) of course) has only one solution in $C$ for each choice of $Phi in C$,

II) or it has solution iff $Phi in N(T_lambda^**)^o$ (that is $ =0$ for each $g^** in N(T_lambda^**)$)** and the solutions of (2) are in the class $bar(f) + N(I-T_lambda)$, where $barf$ is a particular solution of (2).

The adjoint operator of $T_lambda$ is defined by $T_lambda^**g^**= < g^**, p> *lambdaP$, thus the adjoint homogeneous equation is $ *lambdaP=omega$*** or simply $ =0$.

After this prologue, we can try to solve (2).
Evaluating the functional $$ on both sides of (2) we got:

$*p > = quad => $

$quad => quad(1-lambda )* = $;

since $ =\int_0^1 x^2" d"x=1/3$, the last equality determines $$ iff $lambda != 3$ (hence all $lambda in CC-{3}$ are regular values for the compact operator $T=T_1$).
If $lambda != 3$, we have $ = 3/(3-lambda)$ and therefore from the Fredholm Alternative I) follows that the only solution of (2) is $f=Phi + (3lambda)/(3-lambda) *p$ and the corresponding solution of equation (1) is of course:

$f(x)=Phi(x)+(3lambda)/(3-lambda)*(\int_0^1y*Phi(y)" d"y)*x quad$.

If $lambda=3$, the function $p$ is an eigenfunction for $T_3$ with eigenvalue $1$ (infact $T_3p= <3P,p>*p=3*1/3*p=1*p$) so $N(I-T_3)="span"{p}$. For the Fredholm Alternative II) equation (2) has solution iff $ =0$; the function $bar(f)=Phi$ is a particular solution of (2) as long as we have:

$bar(f)=Phi=Phi+0=Phi+3*p=Phi+T_3bar(f) quad$;

we can write the rappresentation formula $f=Phi+alpha*x$ (with the constant $alpha in CC$ totally undetermined) which provides the whole of the solutions of (2) and the corresponding formula for the solutions of (1) is:

$f(x)=Phi(x)+alpha*x quad$.

Finally we can sum up the results in the following manner:

The Volterra equation of the first kind $f(x)=Phi(x)+lambda*\int_0^1xy*f(y)" d"y$ has:

a) one and only one solution for each $Phi in C$ iff $lambda!=3$: in this case the solution is exactly $f(x)=Phi(x)+(3lambda)/(3-lambda)(\int_0^1y*Phi(y)" d"y)*x$;

b) an $oo^1$ solution for each $Phi in C$ s.t. $\int_0^1y*Phi(y)" d"y=0$ if $lambda=3$: in this case the general integral is $f(x)=Phi(x)+alpha*x$ with the constant $alpha in CC$.

[size=75]EDIT: Thanks to Thomas who made me notice a bad error in the identification of $C^**$.[/size]
__________
* Here and in what follows $I$ is the identity operator in $C$: $AA f in C, If=f$.
** Here, with customary notations of Functional Analysis, $T_lambda^**$ denotes the adjoint operator of $T_lambda$; $N(T_lambda^**)$ is the null space of $T_lambda^**$ (and in general $N(\cdot)$ denotes the null space of any linear operator defined in $C$); the apex $o$ denotes the orthogonal (or annichilator) in the duality $L^1$-$C$ as specified in the brackets.
*** Here $omega$ is the function which takes the value zero over the whole of $[0,1]$: $AA x in [0,1], omega(x)=0$.