# If $\lVert u(t) \rVert \leq \lVert u_0 \rVert$ for solution of PDE, is $\lVert u(t) \rVert \leq \lVert u(s) \rVert$ for all $t \leq s$?

Suppose I have some nonlinear PDE on some time interval $[0,T]$ $$u_t + A(u) = 0$$ $$u(0)=u_0$$ and I have managed to show existence and uniqueness of solution with $\lVert u(t) \rVert \leq \lVert u_0 \rVert$ for some norm for all $t \in [0,T]$. Does it follow that for all $t < s$ in $[0,T]$, $$\lVert u(t) \rVert \leq \lVert u(s) \rVert?$$ I think so because I can split the equation into $[0,s]$ and $[s,T]$ where the initial value of the equation on $[s,T]$ is the final value of the solution of the first equation. Am I right?

• it doesn't seem to work for $y''=-y$, solution is $\cos(x)$ which is always below 1 but you can't say anything about monotonicity of $y$. Jul 17, 2014 at 11:35
• @MattB. Your example is not of the form $u_t + A(u) = 0$. Jul 17, 2014 at 11:37
• What kind of operator is $A$? What norm are you using? Jul 17, 2014 at 11:48
• @JuliánAguirre Let $A(u) = -\Delta (f(u))$ where $f$ is some polynomial. The norm is some $L^p$ norm (assume $u(t) \in L^p$ for each $t$). Just curious to know whether my technique is Ok. I have a feeling it is only OK for linear operators $A$ but maybe not... Jul 17, 2014 at 12:06

Yes, this works for any autonomous first-order equation $u_t = A(u)$, for the reason you stated. A more formal way to express this is to introduce the flow maps $\Phi_t$ which are defined so that $\Phi_t(u_0)=u_t$. These maps form a semigroup: $\Phi_{t+s}=\Phi_t\circ \Phi_s$. So, if you are interested in a property that is preserved under composition (such as being norm-decreasing), it suffices to establish the property for $0< t \le \epsilon$ for some $\epsilon>0$. The rest follows by composing flow maps.