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

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

$$ \begin{align} \frac{d}{dt}\eta(t) &\leqslant \eta(t)\phi(t) \\ \frac{d}{dt}(\xi(t) e^{\int_{0}^{t} \phi(s) ds}) &\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \\ (\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds} + \xi(t) \frac{d}{dt}(e^{\int_{0}^{t} \phi(s) ds})&\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \\ (\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds} + \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t)&\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \tag{*}\\ (\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds}&\leqslant 0 \\ \frac{d}{dt}\xi(t) &\leqslant 0 \\ \end{align} $$ , where we used in the last step that \(\forall x: e^{x}>0 \) and (*) \(\frac{d}{dt}\int_0^t\phi(s)ds) = \phi(t) \) follows from the Fundamental Theorem of Calculus.

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

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

Problem 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 \( u \in C^2(S^1 \times [0,\infty)) \) exists.

Solution i)

We want to show uniqueness of the solution to 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 \(S^1 \times [0,T] \) (thus the function is limited on the intervall) and dominated convergence theorem.

(Sketch) Since the function is continous and limited, it is Riemann integrable. Thus consider a series of (Riemann) step functions (each of them is Riemann integrable) which will converge monotonic to our considered function. So we can even apply the monotone convergence theorem.

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 in \(x\), 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 \(u \equiv v\).

Solution ii)

A solution to the (IVP) is: $$ u(x,t)=w_0/(1-w_0t)$$ since: $$ \begin{align} \exists T > 0  : u &\in C^2(S^1 \times [0,T]) \\ \frac{\partial}{\partial t} u(x,t)&=\frac{\partial}{\partial t} w_0/(1-w_0t)=\frac{w_0^2}{(1-w_0t)^2}=(u(x,t))^2 \\ \frac{\partial^2}{\partial x^2} u(x,t)&=0 \\ u(x,0)&=w_0 \\ \end{align} $$ This must be the unique solution as we have seen in 4b ii).

Solution iii)

If \(w_0 \in (-\infty,0] \) there is no problem. But for initial data \( w_0 \in (0,\infty) \) we see that \( u(x,t) \) is not defined at \( t=1/w_0 \in (0,\infty)\) and therefore not at every time \( t \in [0,\infty)\). So there is no global solution \( u \in C^2(S^1 \times [0,\infty)) \) for \(w_0 \in (0,\infty) \).