Finding a continous solution to an integral

[l888l888l888]

Homework Statement

find a continuous solution to the integral equation:
f(x)=(x^3)+(1/2)*integral from 0 to 1 of ((x*y)/(y+1))*f(y)dy, by finding a fixed point of the function v:(C([0,1]), d)-->((C([0,1]), d) defined by
v(f)(x)=(x^3)+(1/2)*integral from 0 to 1 of ((x*y)/(y+1))*f(y)dy

Homework Equations

d is the supremum norm.
d(v(f1),v(f2))=sup(x in [0,1]) |(v(f1)(x)-v(f2)(x)|<=(1/4)*sup(y in [0,1]) |f1(y)-f2(x)|

The Attempt at a Solution

it says to use maple to do calculatins. tried and failed

 

[Susanne217]

Homework Statement

find a continuous solution to the integral equation:
f(x)=(x^3)+(1/2)*integral from 0 to 1 of ((x*y)/(y+1))*f(y)dy, by finding a fixed point of the function v:(C([0,1]), d)-->((C([0,1]), d) defined by
v(f)(x)=(x^3)+(1/2)*integral from 0 to 1 of ((x*y)/(y+1))*f(y)dy

Homework Equations

d is the supremum norm.
d(v(f1),v(f2))=sup(x in [0,1]) |(v(f1)(x)-v(f2)(x)|<=(1/4)*sup(y in [0,1]) |f1(y)-f2(x)|

The Attempt at a Solution

it says to use maple to do calculatins. tried and failed

Am I right this is complex analysis?

 

[Gib Z]

Susanne217- This would be categorized as the theory of Integral equations. Here, specifically, the theory borrows tools from the theory of Metric spaces, in particular the Banach Contraction Mapping theorem.

OP - I don't know how to use Maple, so I'm not sure how this question was intended to be solved. Though I will complain, the question could have been better worded, since saying "Find the fixed point of v(f)(x) " is exactly the same as saying "solve the integral equation".

An idea though: We know that the fixed point of v(f)(x) is the limit of the recursion f_{n+1} = v(f_n) with any initial f_0 \in C([0,1], \mathbb{C} ). I started with f_0(x) = 1. Then f_1(x) = x^3 + x(1-\log 2)/2. Repeat this a few more times, and you'll notice that the same integrals appear over and over, with only some constants changing, and we always end up having f_n (x) = x^3 + C_n x where C_n is a sequence of real constants (it's an n-th degree polynomial in log 2, but we don't need that). So then one could well conjecture that our fixed point has the form f(x) = x^3 + Cx where C is some constant. So plug this into our fixed point relation v(f) = f, and solve for the value of C which satisfies this. This gives you a fixed point, which you know is also unique, and completes the problem.

 

[l888l888l888]

f(x)=x^3+Cx where C is some constant. So plug this into our fixed point relation v(f) = f, and solve for the value of C which satisfies this. This gives you a fixed point, which you know is also unique, and completes the problem.

Is this what you mean?...

1/2*integral (xy/y+1)(y^3 +Cy)dy +x^3=x^3+CX

 

[Gib Z]

Yes, it is.