I want to minimize $\int_1^2\frac{1}{x}\sqrt{1+u^{\prime}(x)^2}dx$ so that $u(1)=0$ and $u(2)=1$.
Using the Euler-Lagrange-equation I obtain $u(x)=2-\sqrt{5-x^2}$. Futhermore, the second variation of $u$ is given by $$\int_1^2\frac{1}{x}(1+u^{\prime}(x)^2)^{-\frac{3}{2}}(\varphi^{\prime}(x))^2dx=\int_1^2\frac{1}{x}(1-\frac{1}{5}x^2)^{\frac{3}{2}}(\varphi^{\prime}(x))^2dx$$ for any nonzero $\varphi\in C_c^{\infty}((a,b))$. How can I show that this critical point is a local $C^1$ minimizer? I know that $\int_1^2\frac{1}{x}(1-\frac{1}{5}x^2)^{\frac{3}{2}}(\varphi^{\prime}(x))^2dx\geq 2\lambda\int_1^2 |\varphi|^2+|\varphi^{\prime}|^2dx$ would impliy it, but I am not able to show that. Any help is greatly appreciated!
Let us denote by $h$ the function that appears under the last integral as a factor to $\varphi^2$, $$ h(x) = \frac 1x\left(1-\frac {x^2}5\right)^{3/2}\ , \qquad h:[1, 2]\to\Bbb R\ . $$ Then $h$ is minimal in $2$ and $h(2)=\frac 12\cdot\left(\frac 15\right)^{3/2}\ge \frac 1{25}$, since $\left(\frac 15\right)^3=\frac 1{125}=\frac 5{625}\ge \frac 4{625}= \left(\frac 2{25}\right)^2$, and let us write $m$ for the one or the other positive lower bound. Then we have for any test function $\varphi$ with compact support in $[1,2]$: $$ \begin{aligned} \int_1^2 \varphi'(x)^2\; dx &=\frac 1m\int_1^2 m\;\varphi'(x)^2\; dx \le \frac 1m\int_1^2 h(x)\;\varphi'(x)^2\; dx \ , \\[3mm] \varphi(x)^2 &=\left(\int_1^x \varphi'(x)\right)^2 \le \left(\int_1^2 \varphi'(x)\; dx\right)^2 \le \left(\int_1^2 1^2\right)\left(\int_1^2 \varphi'(x)^2\; dx\right)\ ,\text{ so} \\ \int_1^2 \varphi(x)^2\; dx&\le \int_1^2 \varphi'(x)^2\; dx\le \frac 1m\int_1^2 h(x)\;\varphi'(x)^2\; dx \end{aligned} $$ Add to see that the integral of $|\varphi|^2+|\varphi'|^2$ is controlled by $\int_1^2 h(x)\cdot |\varphi'(x)|^2\; dx$ with a convenient factor.