# Can we show this inequality? (PDE question)

I am attempting to show that $$\int^t_0 \left[e^{-\lambda \kappa (t-s)} a(s)\right] ds \leq \int^t_0 \frac{\lambda}{2\kappa} a(s)^2 ds \tag{1}\label{1}$$ for any $t$, where $\lambda, \kappa >0$, $a(s)$ is a function of $s$.

Can we prove $\eqref{1}$ with this much information? It is possible that I missed something.

Edit:

I guess I did something wrong, so here is the original question.

Let $D \subset \Bbb R^N$ be an open bounded domain and $v_k \in C^2(D) ∩ C^1(\bar{D})$ with $v_k(x) = 0$ on $\partial D$ and $\lVert v_k \rVert = 1$ is an eigenfunction of $−\Delta$ on $D$ with eigenvalue $λ_k$. Show that if $a_k \in C([0, \infty))$, then $$u_k(x, t) = \int ^t_0a_k(s)e^{−λkκ(t−s)}ds \, v(x)$$ solves $$(u_k)_t = \kappa \Delta u_k + a_k(t)v_k(x)$$on $\Omega = D × \Bbb R^+$. Also show using Schwarz inequality that for any $t > 0$ we have $$\lVert \Delta u_k(\cdot,t)\rVert^2 \leq \int^t_0 \frac{\lambda}{2\kappa} a(s)^2 ds$$

I have problem with the inequality part. Since $u(x,t) = A(t) v(x)$, then the space Laplacian $\Delta u = A(t) \Delta v$. I think I applied the Schwarz inequality correctly and get $\eqref{1}$. What is missing?

OK, I found I should not put a square (so I edited $\eqref{1}$). But still I don't know how to continue.

• You must have some typos in your $(1)$. Consider the limiting case $\lambda \to 0$, the LHS converges to a positive value (assume $a(s)$ not identically zero) while the RHS converges to $0$. i.e the inequality will break for sufficiently small $\lambda$. – achille hui May 9 '15 at 0:47
• Ah, OK, I think it is because I missed something. Please allow me to edit and put in the original question. – MonkeyKing May 9 '15 at 0:52

When you expand $\lVert\Delta u_k\rVert^2$ to get LHS of $(1)$, each copy of of $\Delta$ will give you an extra factor $\lambda$. The correct version of $(1)$ should be $$\left[ \lambda \int^t_0 e^{-\lambda \kappa (t-s)} a(s) ds \right]^2\le \int^t_0 \frac{\lambda}{2\kappa} a(s)^2 ds$$

Let $\displaystyle\;\mathcal{A} = \int^t_0 a(s)^2 ds.\;$ By Cauchy-Schwarz inequalities,

\begin{align} \text{LHS} &\le \lambda^2 \mathcal{A}\int_0^t e^{-2\lambda\kappa(t-s)}ds = \lambda^2 \mathcal{A}\int_0^t e^{-2\lambda\kappa s}ds\\ &\le \lambda^2 \mathcal{A}\int_0^\infty e^{-2\lambda\kappa s}ds = \frac{\lambda^2}{2\lambda\kappa}\mathcal{A} = \frac{\lambda}{2\kappa}\mathcal{A} = \text{RHS} \end{align}

• Sorry, I suddenly don't understand why we don't need the $v(x)$ part. I believe originally, I just put $\Delta$ on $v$ and misread $\lVert v_k \rVert$ as $\lVert \Delta v_k \rVert$ and which is wrong. Could you please expand $\lVert \Delta u_k \rVert^2$ explicitly? Thanks a lot. – MonkeyKing May 9 '15 at 14:34
• @MonkeyKing $\lVert v_k \rVert = 1$. – achille hui May 9 '15 at 14:44
• How does this tell us about $\Delta v_k$? – MonkeyKing May 9 '15 at 14:46
• @MonkeyKing by defn of $v_k$, $\Delta v_k = -\lambda_k v_k$. – achille hui May 9 '15 at 14:51
• Ahhhhhhhhhhhhhhh Why didn't I see that. Thank you. – MonkeyKing May 9 '15 at 14:53

Hint: $$e^{-2\lambda k (t-s)} \le e^{\ln \frac{\lambda}{2k}}$$

If and only if $$-2\lambda k (t-s) \le \ln \frac{\lambda}{2k}$$

The latter is a constant. So if you can show that $$-2\lambda k t \le \ln \frac{\lambda}{2k}$$, you have proved the inequality.