Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
Do not need edit? – user52188 Jan 8 '13 at 0:57
What is the variable in the second slot in $k(x,y,y(t))$? – timur Jan 8 '13 at 0:58
@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? – Applied mathematician Jan 8 '13 at 13:47
Yes that variable should be a t, i've edited it – Applied mathematician Jan 9 '13 at 13:38
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 $$y(x)=g(x)+\int_0^x \tilde k(x,t,y(t)-g(t))\,dt = g(x)+\int_0^x k(x,t,y(t))\,dt $$

share|cite|improve this answer
How can I see that this function is bounded? – Applied mathematician 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... – Applied mathematician 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.