Revision as of 10:32, 29 December 2014

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

ii) Give an explicit solution of (IVP) for constant initial data \(u_0 \equiv w_0 \in \mathbb{R} \).

iii) Conclude that, then there is initial data for (IVP) for which no global solution \( u \in C^2(S^1 \times [0,\inf)) \) exists.

Solution i)

We want to show uniqueness of the IVP problem. First we show that \(\eta(t) \) satisfies the setting of 4a above for every fix \( x \in S^1 \).

Proof: We can change integral and differential in the second line, because \((u(x,t)-v(x,t))^2-(u(x,t_0)-v(x,t_0))^2)/(t-t_0) \leqslant g(x) \) is integrable by mean value theorem, compactness of \(S^1 \times [0,T] \) and dominated convergence theorem. (The details are not worth being discussed here. I think the task's focus does not lie on that step.) For the following denote: \(u(x,t)=u \), \(v(x,t)=v\). $$ \begin{align} \frac{d}{dt} \eta(t)&=\frac{d}{dt}\int_0^{2\pi} (u-v)^2 dx \\ &=\int_0^{2\pi}\frac{\partial}{\partial t} (u-v)^2 dx \\ &=\int_0^{2\pi} 2(u-v) (\frac{\partial}{\partial t}u -\frac{\partial}{\partial t}v) dx \\ &=\int_0^{2\pi} 2(u-v) ( u^2+\frac{1}{2} \frac{\partial^2}{\partial x^2}u - v^2-\frac{1}{2} \frac{\partial^2}{\partial x^2}v) dx \\ &=\int_0^{2\pi} 2(u-v) ( u^2- v^2)+2(u-v)(\frac{1}{2} \frac{\partial^2}{\partial x^2}u-\frac{1}{2} \frac{\partial^2}{\partial x^2}v) dx \\ &=\int_0^{2\pi} 2(u-v)^2 (u+v) dx +\int_0^{2\pi}(u-v)\frac{\partial^2}{\partial x^2}(u-v) dx \\ \end{align} $$

Since \(u,v\) are \(2\pi\)-periodic, the second integral using integrating by parts resolves to: $$ \int_0^{2\pi}(u-v)\frac{\partial^2}{\partial x^2}(u-v) dx=-\int_0^{2\pi}\frac{\partial}{\partial x}(u-v)\frac{\partial}{\partial x}(u-v) dx=-\int_0^{2\pi}(\frac{\partial}{\partial x}(u-v))^2 dx \leqslant 0 $$

So we get: $$ \frac{d}{dt}\eta(t) \leqslant 2\int_0^{2\pi}(u-v)^2 (u+v) dx \leqslant 2\int_0^{2\pi}(u-v)^2 (u_{max}+v_{max}) dx = 2 (u_{max}+v_{max})\int_0^{2\pi}(u-v)^2 dx= \phi(t)\eta(t) $$ , where \(\phi(t):=2(u_{max}+v_{max}) \) and \( u_{max}:=max\{|u(x,t)|: (x,t) \in [0,2\pi] \times [0,T]\} \). (Maximum exists, because \([0,2\pi] \times [0,T] \) is compact and \(u,v\) continuous)

Since now all contitions of problem 4a are satisfied, apply proposition 4a: $$ \begin{align} \eta(t) &\leqslant \eta(0)e^{\int_{0}^{t} \phi(s) ds} \\ 0 \leqslant \int_0^{2\pi}(u(x,t)-v(x,t))^2 dx &\leqslant \int_0^{2\pi}(u(x,0)-v(x,0))^2 dx e^{\int_{0}^{t} \phi(s) ds} = \int_0^{2\pi}(u_0(x)-v_0(x))^2 dx e^{\int_{0}^{t} \phi(s) ds}=0 \\ \end{align} $$ ,since \(u_0(x)=v_0(x) \). We can now conclude that \(u \equiv v\).