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I am interested in getting a good bound for solution of the following ODE: $$ f''(t) + n^2 f(t) = (\sin(\theta t))^n$$ with the boundary condition $f(0) = f'(0) = 0$ and $t \in [0,1]$, where $\theta$ might not be an integer multiple of $\pi$ (in fact you can take $\theta < \pi/2$ so that the left hand side is almost flat at $0$ with monotone derivatives etc) . Numerical computations show that the solution is bounded by a constant independent of $n$. But I have very little experience with ODEs to prove anything about it. The only thing I did was Fourier transform, but there would be terms of the form $f(1) - f(0)$ and $f'(1) - f'(0)$ in the estimate of the Fourier coefficients of $f$ which I don't know how to deal with.

I am also interested in a good source of lecture notes where bounds for ODE solutions are discussed. The question came up in analyzing a random walk on the special orthogonal group.

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up vote 6 down vote accepted

Multiplying by $f{\ }'$ you get $$(f{\ }'^2+n^2 f{\ }^2)'=2\sin(\theta t)^n f{\ }'$$ so that the quantity $E(t)=f{\ }'^2+n^2 f^2$ satisfies the inequality $$E'(t)\le 2 \sqrt{E(t)} \qquad \qquad(1)$$ i.e. $$\sqrt{E(t)}'\le1$$ i.e. $$\sqrt{E(t)}\le t$$ since $E(0)=0$. You can improve the factor 2 in eq. (1) of course since $\theta$ is small, and get sharper estimates.

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That's much easier than I expected. Thanks a lot! You saved my day. – John Jiang Nov 8 '11 at 18:40

It's a non-homogeneous linear equation with constant coefficients, that you can solve by the variation of constant formula. In complex notation, let $z(t):=f(\frac{t}{n})+ \frac{i}{n}f'(\frac{t}{n})$ for $t\in[0,n]$, and let $h(t):= \frac{1}{n^2}(\sin \frac{\theta t}{n} )^n$. So $z(0)=0$, and $\dot z(t)+iz(t)=ih(t)\, , $ whence $$z(t)=i \int_ 0 ^ t h(s) e^ {i (s-t)} ds \, ,$$ that one can re-write in terms of $f$ and $f'$. This already shows a row estimate $$\|f\|_{\infty,[0,1]}\leq 1/n$$ and $$\|f'\| _ {\infty,[0,1]}\leq 1\, ,$$ just because $\|h\| _ \infty\leq 1/n^2.$ As a better estimate, $$|f(t)| \le\frac{1}{n} \int_0^1 | \sin(\theta s)|^n ds = O(n^{-3/2})$$ and $$|f'(t)| \le \int_0^1 | \sin(\theta s)|^n ds = O(n^{-1/2})\, .$$

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This is also very instructive solution. – John Jiang Nov 8 '11 at 18:51
And of course in the case $|θ|<\pi/2$ the integral has even an exponential decay. – Pietro Majer Nov 8 '11 at 21:27

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