How to show an inequality of integral

Question

I’m reading a paper in which there is an inequality as $$ 2k\phi (L,t)\int_{0}^{L}{{{e}^{-\frac{\mu x}{2}}}\phi (x,t)dx}\le \left( \frac{r{{e}^{-\mu L}}}{2} \right){{\phi }^{2}}(L,t)+\left( \frac{2L{{k}^{2}}{{e}^{\mu L}}}{r} \right)\int_{0}^{L}{{{e}^{-\mu x}}{{\phi }^{2}}(x,t)dx}$$ where $k,r,μ,L$ are three independent positive constants and $ϕ$ is a continuous function of $x,t.$

In the paper, the authors hinted that they used the Young and Cauchy–Schwarz inequalities to get this result, but they haven’t given any further details. Can anyone help me to show this inequality?

Answer 1

To simplify the notations, let

• $\psi(x,t):=e^{-\mu x/2}\phi(x,t)$
• $\alpha:=\frac r2$
• $\beta:=kLe^{\mu L/2}.$

Your inequality rewrites $$\frac{2\beta}L\psi(L,t)\int_0^L\psi(x,t)dx\le\alpha\psi^2(L,t)+\frac{\beta^2}{L\alpha}\int_0^L\psi^2(x,t)dx$$ i.e. $$ 2\alpha\beta\psi(L,t)\int_0^L\psi(x,t)dx\le L\alpha^2\psi^2(L,t)+\beta^2\int_0^L\psi^2(x,t)dx,$$ which is equivalent to the following obvious one: $$\int_0^L\left(\beta\psi(x,t)-\alpha\psi(L,T)\right)^2dx\ge0.$$