Finding $u_0 \in H_0^1(0,1)$ given $u'_0$

Question

For $\varepsilon > 0$ consider the function $$u'_0 := \frac{3\varepsilon}{8\pi}\cos(2\pi x \varepsilon^{-1}) \qquad x \in (0,1)$$ Now I want to find $u_0 \in H_0^1(0,1)$. So I have to integrate the above function and by definition of $H_0^1(0,1)$, we should have that $u_0(0) = u_0(1) = 0$. Somehow, if we set $$u_0(x) = \int_0^x u'_0(t) dt + C$$ we get the requirement $C = 0$ and $$C = -\frac{3\varepsilon^2}{16\pi^2}\sin(2\pi\varepsilon^{-1})$$ What am I doing wrong?

Answer 1

This problem seems overdetermined: if you have a solution $u_0$, then $$ \int_0^1 \frac{3 \varepsilon}{8 \pi} \cos (2 \pi x \varepsilon^{-1})\,\mathrm{d}x = u_0(1) - u_0 (0) = 0. $$ This will only hold if $\varepsilon^{-1}$ is an integer.