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I have an ODE $$u' = f(t,u) = \int_0^t k(u(s)) \text{ d}s$$ with an initial data $u_0$. Here, $k \in C_b(\mathbb{R})$ is a continuous and bounded function such that $|k(x)| \leq C$ for all $x\in \mathbb{R}$.

I want to apply the Cauchy-Peano theorem to this ODE to show that there exists a $C^1$ solution $u$.

Let $u$ be fix. Then $t \mapsto f(t,u)$ is continuous, see this question.

Let $t$ be fix. And let $u_n(x) \to u(x)$ for every $x \in \mathbb{R}$ (almost every $x$ would be enough). Then, since $k$ is continuous and bounded, we also have $k(u_n(x)) \to k(u(x))$ for almost every $x$ and $|k(u_n(x))| \leq C$. Therefore, by the Lebesgue dominated convergence theorem, we have

$$f(t,u_n)=\int_0^t k(u_n(s)) \text{ d}s \longrightarrow \int_0^t k(u(s)) \text{ d}s=f(t,u),$$

and $u \mapsto f(t,u)$ is continuous for fixed $t$.

Therefore, Cauchy-Peano gives me the existence of a $C^1$ solution to the above ODE.

Is this correct?

Cahn
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1 Answers1

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Your argument does not give you the existence of a solution, only that if an approximating and converging sequence exists then the limit satisfies (parts of) the original equation.

But you can easily transform your equation into an ordinary ODE by differentiating once more, $$ u''=k(u), ~~ u(0)=u_0,~ u'(0)=0. $$ To that or its first order version $$ u'=v\\ v'=k(u) $$ you can now apply the usual Peano theorem and conclude the existence of local solutions.

Lutz Lehmann
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  • Was my continuity argument incorrect? And if I take two functions $u$, $v$, then $$|f(t,u)-f(t,v)| \leq \int_0^t |k(u(s))-k(v(s))| ds \leq L t |u-v|_\infty,$$ if I assume $k$ to be Lipschitz continuous. Does this work? Thank you for the alternative argument. – Cahn Jun 11 '19 at 17:12
  • What is the argument $u$ in $f(t,u)$? In your calculation it is a point in a function space. You are thus far outside the usual frame for the Cauchy-Peano theorem, where $f$ is a function of time and a finite dimensional state vector. – Lutz Lehmann Jun 11 '19 at 17:30
  • Do you think that it would make more sense if I look at $u'=f(t,u)=\int_0^t k(s,u) ds$? Then I am probably more in the setting of a classical ODE. I am just thinking what I would need for $k$ to apply Cauchy-Peano. Maybe $k(s,\cdot) \in \text{Lip}(\mathbb{R})$ and $k(\cdot,u) \in L^1(\mathbb{R})$? – Cahn Jun 12 '19 at 08:30