Integral equations -- Picard method of succesive approximation

[LagrangeEuler]

Equation
$$\varphi(x)=x+1-\int^{x}_0 \varphi(y)dy$$
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
$$\varphi_1(x)=x+1-\int^{x}_0 \varphi_0(y)dy$$
$$\varphi_2(x)=x+1-\int^{x}_0 \varphi_1(y)dy$$
$$\varphi_3(x)=x+1-\int^{x}_0 \varphi_2(y)dy$$
...
Then solution is given by
$$\varphi(x)=\lim_{n \to \infty}\varphi_n(x)$$.
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]

$$\phi_0(x)=1\ results \ in \ \phi_1(x)=1$$, you're done!

 

[LagrangeEuler]

$$\phi_0(x)=1\ results \ in \ \phi_1(x)=1$$, 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 $$\phi_n(x)=1$$ are the same, the sequence trivially converges to $$\phi(x)=1.$$ I am not sure what you are looking for.

 

[WWGD]

Since all $$\phi_n(x)=1$$ are the same, the sequence trivially converges to $$\phi(x)=1.$$ 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
$$\varphi(x)=f(x)+\lambda \int^x_0K(x,t)\varphi(t)dt$$
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.