LagrangeEuler
- 711
- 22
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.
\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.