I've been working on a problem for a while now and can't seem to arrive at a solution. I have to find the optimal constant $C$ that satisfies the inequality: $$Cu(0)^2 \le \int_0^1u(x)^2\ dx+\int_0^1u'(x)^2\ dx$$ for real valued $u\in C^1(0,1)\cap C([0,1])$.
So far, I've noted that the inequality would hold trivially if $u(0)=0$, so under the assumption that $u(0)\neq0$ I have formulated the Lagrangian: $$F(x,u(x),u'(x))=\frac{u(x)^2+u'(x)^2}{u(0)^2},$$ which would need to be minimized to find the optimal constant $C$.
I've tried using Euler's equation: $$F_u-\frac{d}{dx}F_{u'}=0$$ $$\Rightarrow \frac{2u(x)u(0)^2-2u(0)\big(u(x)^2+u'(x)^2\big)}{u(0)^4} - \frac{d}{dx}\bigg\{\frac{2u'(x)}{u(0)^2}\bigg\}=0,$$ $$\Rightarrow \frac{2\big(u(x)u(0)-(u(x)^2+u'(x)^2)\big)}{u(0)^3}-\frac{2u''(x)u(0)}{u(0)^3}=0,$$ $$\Rightarrow \frac{2}{u(0)^3}\bigg[u(0)\big(u(x)-u''(x)\big)-\big(u(x)^2+u'(x)^2\big)\bigg]=0.$$
But I can't work out what to do from here.
Does anybody know if this is even along the right lines? Any help would be really appreciated. Thanks!
Say you wish to find $$ C=\inf\left\{\frac{\int_0^1\left[u^2(x)+u'^2(x)\right]{\rm d}x}{u^2(0)}:u\in C(\left[0,1\right])\cap C^1(\left(0,1\right))\right\}. $$ Note that $$ \frac{\int_0^1\left[u^2(x)+u'^2(x)\right]{\rm d}x}{u^2(0)}=\int_0^1\left[\left(\frac{u(x)}{u(0)}\right)^2+\left(\frac{u(x)}{u(0)}\right)'^2\right]{\rm d}x. $$ Thus your target is equivalent to figure out $$ C=\inf\left\{\int_0^1\left[v^2(x)+v'^2(x)\right]{\rm d}x:v\in C(\left[0,1\right])\cap C^1(\left(0,1\right)),v(0)=1\right\}. $$ The functional derivative yields the governing equation $$ v''(x)=v(x), $$ with the variational boundary condition $$ v'(1)=0. $$ In addition, we also have the imposed boundary condition $$ v(0)=1. $$ These lead to a unique solution for $v$, i.e., $$ w(x)=\frac{e^x+e^{2-x}}{1+e^2}. $$ Thanks to this solution, the desired optimal $C$ follows immediately, i.e., $$ C=\int_0^1\left[w^2(x)+w'^2(x)\right]{\rm d}x=1-\frac{2}{1+e^2}\approx 0.7616. $$