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As many of you will know, Peano's theorem states that if $f(x,y)$ is continuous and bounded in the strip $T: |x-x_0| \le a, |y|\le\infty $. Then the intitial value problem $y'=f(x,y), y(x_0)=y_0$, has at least one solution in $|x-x_0| \le a $.

Now, consider the following problem. Let $J= [0,a]$. Let $g(x)$ be a continuous function. Let $k(x,t,z)$ be a function which is continuous and bounded in {${(x,t,z)\in J\times J\times \mathbb{R}: t\lt x}$}. How can I prove that there is (at least) a solution of the equation

$$y(x)= g(x)+ \int_0 ^x k(x,t,y(t))dt, x \in J $$

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Do not need edit? –  user52188 Jan 8 '13 at 0:57
2  
What is the variable in the second slot in $k(x,y,y(t))$? –  timur Jan 8 '13 at 0:58
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@timur I think the 2nd variable should be $t$; the OP should clarify. // Introducing $z=y-g$ we get $z(x)=\int_0^x \tilde k(x,t,z(t))\,dt$ and Peano applies. (The function $\tilde k$ differs from $k$ to account for the change we made, but is still continuous.) –  user53153 Jan 8 '13 at 3:19
    
hmm... thats all? –  MSKfdaswplwq Jan 8 '13 at 13:47
    
Yes that variable should be a t, i've edited it –  MSKfdaswplwq Jan 9 '13 at 13:38

1 Answer 1

up vote 2 down vote accepted

Let $\tilde k (x,t,z) = k(z,t,z+g(t))$. This is a continuous function. By the Peano theorem, the equation $z(x)=\int_0^x \tilde k(x,t,z(t))\,dt$ has a solution. Then $y=z+g$ satisfies $$y(x)=g(x)+\int_0^x \tilde k(x,t,y(t)-g(t))\,dt = g(x)+\int_0^x k(x,t,y(t))\,dt $$

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How can I see that this function is bounded? –  MSKfdaswplwq Jan 9 '13 at 13:26
    
Can I say that g(x) is bounded, because every function on a compact set is bounded. Also z is bounded? I don't see how this will finish the proof... –  MSKfdaswplwq Jan 9 '13 at 13:40
    
You assumed that $k$ is bounded. Therefore, if I plug another function into $k$, they composition is also bounded. –  user53153 Jan 9 '13 at 13:53

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