Aufgaben:Problem 14

Problem 4a

Let \( T > 0,\eta \in C^{1}([0,T]), \phi \in C^{0}([0,T]) \), and assume that: $$ \frac{d}{dt}\eta(t) \leqslant \eta(t)\phi(t), \forall t \in [0,T] $$

Show that: $$\eta(t) \leqslant \eta(0)e^{\int_{0}^{t} \phi(s) ds}, \forall t \in [0,T] $$

Hint: consider \( \xi(t) := \eta(t) e^{-\int_{0}^{t} \phi(s) ds} \)

Solution

Using chain rule: $$ \frac{d}{dt} \xi(t) = \frac{d}{dt} (\eta(t) e^{-\int_{0}^{t} \phi(s) ds}) = (\frac{d}{dt} \eta(t)) e^{-\int_{0}^{t} \phi(s) ds} + \eta(t) (\frac{d}{dt} e^{-\int_{0}^{t} \phi(s) ds}) = (\frac{d}{dt} \eta(t)) e^{-\int_{0}^{t} \phi(s) ds} - \eta(t) \phi(t)e^{-\int_{0}^{t} \phi(s) ds}$$

Substituting \( \eta(t) = \xi(t) e^{\int_{0}^{t} \phi(s) ds} \) in our assumption, then we get \( \forall t \in [0,T] \):

$$ \begin{align} \frac{d}{dt}\eta(t) &\leqslant \eta(t)\phi(t) \\ \frac{d}{dt}(\xi(t) e^{\int_{0}^{t} \phi(s) ds}) &\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \\ (\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds} + \xi(t) \frac{d}{dt}(e^{\int_{0}^{t} \phi(s) ds})&\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \\ (\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds} + \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t)&\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \\ (\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds}&\leqslant 0 \\ \frac{d}{dt}\xi(t) &\leqslant 0 \\ \end{align} $$ , where we used in the last step that \(\forall x: e^{x}>0 \).

Now integrating the result from 0 to t and since \(g(x) \leqslant h(x) \Rightarrow \int g(x) dx \leqslant \int h(x) dx \):

$$ \begin{align} \int_{0}^{t} \frac{d}{dt'}\xi(t') dt' = \xi(t) - \xi(0) &\leqslant 0 = \int_{0}^{t} 0 dx \\ \eta(t) e^{-\int_{0}^{t} \phi(s) ds}-\eta(0) e^{-\int_{0}^{0} \phi(s) ds} &\leqslant 0 \\ \eta(t) e^{-\int_{0}^{t} \phi(s) ds}-\eta(0) e^{0} &\leqslant 0 \\ \eta(t) e^{-\int_{0}^{t} \phi(s) ds}&\leqslant \eta(0) \\ \eta(t) &\leqslant \eta(0)e^{\int_{0}^{t} \phi(s) ds} \\ \end{align} $$

Prblem 4b

Consider the initial value problem (IVP) $$ \begin{align} \frac{\partial}{\partial t} u(x,t)-\frac{1}{2} \frac{\partial^2}{\partial x^2} u(x,t) &= (u(x,t))^2 \\ u(x,0) &= u_0(x) \in C^2(S^1) \\ \end{align} $$

i) Show that, if \(u,v \in C^2(S^1 \times [0,T]) \) are two solutions of (IVP) for the same \(T>0\), then \(u \equiv v \) on \( S^1 \times [0,T] \)

Hint: consider \(\eta(t):=\int_0^{2\pi} (u(x,t)-v(x,t))^2 dx \)