Given the initial boundary value problem \begin{align*} &u_t = Du_{xx} + f(u), \quad 0<x<1, t>0 \\ &u(0,t) = u(1,t) = 0, \quad t>0 \\ &u(x,0) = u_{0}(x), \quad 0<x<1 \end{align*} We assume that $u_0$ is a continuous and non-negative function on $[0,1]$ with $u_{0}(0) = u_{0}(1) = 0$. Furthermore we assume that the reaction term $f$ is a bounded continuous function, $M = sup|f|$ and that $D$ is a positive diffusion constant

Show that $\displaystyle \int_{0}^{1} u f(u) \ dx \leq \dfrac{1}{2} \left(\frac{M}{\pi\sqrt{D}}\right)^{2} + \frac{D}{2} \int_{0}^{1} u_{x}^2 \ dx$ using Holder's, Young's and Poincare's ineqaulaity.
Given Information:

Holder's Inequality: If $p$ and $q$ $\in \mathbb{R}_{>0}$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$, then $$\int_{0}^{1} |u(x)v(x)| dx \leq \left[\int_{0}^{1} u(x)^p dx \right] ^{\frac{1}{p}} \left[\int_{0}^{1} v(x)^q dx \right] ^{\frac{1}{q}}$$ Young's Inequality: $$\int_{0}^{1} u(x)^2 dx \cdot \int_{0}^{1} v(x)^2 dx \leq \frac{1}{2} \left[\int_{0}^{1} u(x)^2 dx \right] ^{2} + \frac{1}{2} \left[\int_{0}^{1} v(x)^2 dx \right] ^{2}$$ Poincare inequality: Let u be a twice continuously differentiable function on $[0,1]$. If $u(0) = u(1) = 0$ then $$ \int_{0}^{1} u_{x}^2 \ dx \geq \pi^2 \int_{0}^{1} u(x)^2 \ dx$$

Attempted solution: $\int_{0}^{1} u f(u) \ dx \leq \left| \int_{0}^{1} u f(u) \ dx \right| \leq \int_{0}^{1} |u f(u)| \ dx \overset{Holder}\leq \left[\int_{0}^{1} u(x)^2 dx \right] ^{\frac{1}{2}} \left[\int_{0}^{1} f(u)^2 dx \right] ^{\frac{1}{2}} = \left(\left[\int_{0}^{1} u(x)^2 dx \right] \left[\int_{0}^{1} f(u)^2 dx \right] \right)^{\frac{1}{2}} \overset{Young}\leq \left(\frac{1}{2}\left[\int_{0}^{1} u(x)^2 dx \right]^2 +\frac{1}{2} \left[\int_{0}^{1} f(u)^2 dx \right]^2 \right)^{\frac{1}{2}}$

and then use Poincare but couldn't get any further.

I also tried to use Integration by parts as follows: $\int_{0}^{1} u f(u) \ dx \overset{IBVP}= \int_{0}^{1} u u_t \ dx - D \int_{0}^{1} uu_{xx}\ dx = \frac{\partial}{\partial t} \int_{0}^{1} (\frac{1}{2} u^2)\ dx + D \int_{0}^{1} (u_x)^2 \ dx$


1 Answer 1


I am not too sure about the Poincare inequality you give there. The classical Poincare Inequality reads $\Vert u \Vert_{L^p(\Omega)} \leq C(\Omega, p) \Vert \nabla u \Vert_{L^p(\Omega)}$. For the case when $\Omega$ is bounded and $u$ is zero on $\partial \Omega$, a sharper version holds:$\Vert u \Vert_{L^2(\Omega)} \leq \text{diam}(\Omega) \Vert \nabla u \Vert_{L^2(\Omega)}$ which translates in your case into $\int_0^1 u_x^2 \mathrm d x \geq \frac{1}{(1-0)^2} \int_0^1 u^2 \mathrm d x$. In particular, I am not too sure that the version you provide does hold (why should there be a $\pi$ appear suddenly, and why would you want it in your estimates?)

Assuming the Poincare inequality you give is true you can show the result by using "Young's Inequality with $\epsilon$" where here \begin{align} b&:= \sqrt{\int_0^1u^2 \mathrm dx} \\ a&:=\sqrt{\int_0^1f(u)^2 \mathrm dx} \\ \epsilon &:= \pi^2 D \end{align} Then, \begin{align} \left[\int_{0}^{1} u^2 \mathrm dx \right] ^{\frac{1}{2}} \left[\int_{0}^{1} f(u)^2 \mathrm dx \right] ^{\frac{1}{2}} \leq & \frac{ \pi^2D}{2} \int_{0}^{1} u^2 \mathrm dx + \frac{1}{2\pi^2 D} \int_{0}^{1} f(u)^2 \mathrm dx \\ \leq & \frac{ D}{2} \int_{0}^{1} u_x^2 \mathrm dx + \frac{1}{2\pi^2 D} \int_{0}^{1} M^2 \mathrm dx \\ =&\frac{ \pi^2D}{2} \int_{0}^{1} u_x^2 \mathrm dx + \frac{1}{2} \bigg( \frac{M}{\pi \sqrt{D}}\bigg)^2 \cdot (1-0) \end{align}

For the "classical" Poincare Friedrichs Inequality you would end up with $$ \frac{D}{2} \int_{0}^{1} u_x^2 \mathrm dx + \frac{1}{2} \bigg( \frac{M}{ \sqrt{D}}\bigg)^2, $$ so basically the same result. Also, I do not quite see the reason why you would want include $D$ in the result.


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