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As an exercise I want to show that for

$$E(v)=\int_0^1 \frac 1 2 x^5 (v'(x))^2 - v(x) dx $$

and $H_0^1(0,1) = \{ v \in H^1(0,1) : v(0)=v(1)=0 \}$ there exists no solution to the problem $$\min_{v\in H_0^1} E(v).$$

Actually I have no clue how to find (or construct?) such a function $v$ with which I can prove this. Who can help?

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up vote 2 down vote accepted

Indirect answer:

The Euler Lagrange equation for $E$ is $$ \left(x^5 \nu'\right)' + 1 = 0 $$ which implies $$ x^5 \nu' = -x + C \implies \nu' = \frac{-x + C}{x^5} $$ which we can integrate and see that it is not compatible with the requirement that $\nu(0) = \nu(1) = 0$.

Direct answer:

We can construct explicitly a sequence $\nu_k$ such that $E_k = E(\nu_k)\searrow -\infty$. Let $\delta_k = \frac{1}{2^{5k}}$. Let $h_k = 2^{6k}$. Let $a_k = 2^{-4k} - \delta_k$ and $b_k = 2^{-4k}$. Note that for $k \geq 1$ we have $0 < a_k < b_k$. Let $\nu_k$ be the piecewise linear function defined by $$ \nu_k(x) = \begin{cases} 0 & x \not\in (a_k,b_k)\\ \frac{2h_k}{\delta_k}(x-a_k) & x\in (a_k, a_k + \frac12 \delta_k] \\ \frac{2h_k}{\delta_k}(b_k - x) & x \in (b_k - \frac12\delta_k, b_k)\end{cases} $$

The graph of $\nu_k$ is an isosceles triangle based on $(a_k,b_k)$ (so with width $\delta_k$) with height $h_k$. We see that

$$ \int_0^1 \nu_k(x) \mathrm{d}x = \frac12 \delta_k h_k = 2^{k-1} $$

Using that the slope of the two line segments are $\pm 2h_k /\delta_k$, we have that

$$ \int_0^1 x^5 (\nu')^2 \mathrm{d}x \leq b_k^5 \int_{a_k}^{b_k} \frac{4 h_k^2}{\delta_k^2} \mathrm{d}x = 4 b_k^5 h_k^2 \delta_k^{-1} = 2^{2 -20k + 12k + 5 k} = 2^{2 - 3k}$$


$$ E_k \leq 2^{1-3k} - 2^{k-1} \leq 2 - 2^{k-1} $$

And so we have that

$$ \lim_{k\to\infty} E_k = - \infty$$

and hence the functional is unbounded below.

Remark: Notice that replacing $\nu_k$ by $-\nu_k$ this also shows that the functional is unbounded above.

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