Integral equations -- Picard method of succesive approximation

[LagrangeEuler]

Equation
[tex]\varphi(x)=x+1-\int^{x}_0 \varphi(y)dy[/tex]
If I start from ##\varphi_0(x)=1## or ##\varphi_0(x)=x+1## I will get solution of this equation using Picard method in following way
[tex]\varphi_1(x)=x+1-\int^{x}_0 \varphi_0(y)dy[/tex]
[tex]\varphi_2(x)=x+1-\int^{x}_0 \varphi_1(y)dy[/tex]
[tex]\varphi_3(x)=x+1-\int^{x}_0 \varphi_2(y)dy[/tex]
...
Then solution is given by
[tex]\varphi(x)=\lim_{n \to \infty}\varphi_n(x)[/tex].
When I could say that this sequence will converge to solution of integral equation. How to see if there is some fixed point? I know how to use this method, but I am not sure from the form of equation, when I can use this method. Thanks for the answer.

 

[mathman]

[tex]\phi_0(x)=1\ results \ in \ \phi_1(x)=1[/tex], you're done!

 

[LagrangeEuler]

[tex]\phi_0(x)=1\ results \ in \ \phi_1(x)=1[/tex], you're done!

This is not my question. I know how to solve this. I am not sure when I can use this method. When sequence of functions ##\varphi_0(x)##, ##\varphi_1(x)##... will converge to ##\varphi(x)##.

 

[mathman]

Since all [tex]\phi_n(x)=1[/tex] are the same, the sequence trivially converges to [tex]\phi(x)=1.[/tex] I am not sure what you are looking for.

 

[WWGD]

Since all [tex]\phi_n(x)=1[/tex] are the same, the sequence trivially converges to [tex]\phi(x)=1.[/tex] I am not sure what you are looking for.

I think s/he is looking for general conditions for convergence, not just for this particular problem.

 

[LagrangeEuler]

I think s/he is looking for general conditions for convergence, not just for this particular problem.

Yes. Thanks.

 

[MathematicalPhysicist]

You can use this method when you have: ##\int \lim_{n\to\infty} \varphi_n(y)dy = \lim_{n\to \infty} \int \varphi_n(y)dy##.

 

[WWGD]

You can use this method when you have: ##\int \lim_{n\to\infty} \varphi_n(y)dy = \lim_{n\to \infty} \int \varphi_n(y)dy##.

Isn't this equivalent to dominated or monotone convergence?

 

[mathman]

Isn't this equivalent to dominated or monotone convergence?

Dominated convergence is a sufficient condition, but not necessary.

 

[WWGD]

Dominated convergence is a sufficient condition, but not necessary.

Ah, yes, Dominated, no reason for Monotone here. Need some caffeine.

 

[LagrangeEuler]

Yes but if I have for example equation in the form
[tex]\varphi(x)=f(x)+\lambda \int^x_0K(x,t)\varphi(t)dt[/tex]
could I see this just for looking in kernel ##K(x,t)## and parameter ##\lambda##?

 

[MathematicalPhysicist]

@LagrangeEuler in your last post this is an eigenvalue problem: if we denote by: ##K\varphi(x) = \int_0^x K(x,t)\varphi(t)dt##

Then you want to solve the equation: ##(I-\lambda K)\varphi = f##; you need to solve the equation ##\det |I-\lambda K| \ne 0 ## and then you have a solution: ##\varphi(x) = (I-\lambda K)^{-1}f(x)##; how to find the inverse, check any functional analysis textbook or Courant's and Hilbert's first volume.