HW1

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(預2024-09-02)
目前不知道是啥毀的：

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$\text{Problem 1}$

Consider the shower equation: $\dot{T}(t)=-k[T(t-\tau )-T_d]$ with $\tau >0$ and $k>0$.

1. Find $\alpha ,\beta \in \mathbb{R}$ in the rescaling $\tilde{T}(t):=T(\alpha t)-\beta$ that results in the simpler equation

   $$\dot{\tilde{T}}(t)=-\tilde{k}\tilde{T}(t-1).$$

   What is $\tilde{k}$?

   $\mathit{\text{Proof}}.$ By differentiating both sides of the definition $\tilde{T}(t):=T(\alpha t)-\beta$, we immediately get

   $$\dot{\tilde{T}}(t)=\alpha \dot{T}(\alpha t)=\alpha [-k(T(\alpha t-\tau )-T_d)],$$

   where the last equation is given by the original shower equation at the very beginning, with $t$ replaced by $\alpha t$. On the other hand,

   $$-\tilde{k}\tilde{T}(t-1)=-\tilde{k}[T(\alpha t-\alpha )-\beta ]=-\tilde{k}T(\alpha t-\alpha )+\tilde{k}\beta .$$

   Suppose that $\dot{\tilde{T}}(t)=-\tilde{k}\tilde{T}(t-1)$. Then

   $$\alpha [-k(T(\alpha t-\tau )-T_d)]=-\tilde{k}T(\alpha t-\alpha )+\tilde{k}\beta .$$

   By comparison of the parameters, we have

   $$\alpha =\tau ,\alpha k=\tilde{k},\alpha k{T}_d=\tilde{k}\beta .$$

   Therefore, $\alpha =\tau ,\beta =T_d$, and $k=\tilde{k}/\alpha$. $◼$

2. Consider the rescaled shower equation

   $$\dot{T}(t)=-T(t-1).$$

   Use the method of steps. to find $T(1)$ and $T(2)$ with $\varphi (t)=1$ on $[-1,0]$.

   $\mathit{\text{Proof.}}$

   1. $\dot{T}(t)=-\varphi (t-1)=-1$ when $t\in [0,1]$. Therefore, by the method of steps,
   $$T(t)=T(0)+{\int }_0^t\dot{T}(s)ds=-t+1$$

   on $[0,1]$. Hence $T(1)=0$. $◼$
   Denote $T_1(t)=-t+1$ on $[0,1]$.
   2. $\dot{T}(t)=-T_1(t-1)=-t+1$ when $t\in [1,2]$. Therefore, by the method of steps again,

   $$\begin{array}{rl}T(t)=T(1)+{\int }_1^t\dot{T}(s)ds& =T(1)-{\int }_1^t{T}_1(s-1)ds\\ \\ & =T(1)-{\int }_1^t[-(s-1)+1]ds\\ \\ & =T(1)+{\int }_1^t(s-2)ds& \\ \\ & =0+\frac{t^2}{2}-2t+\frac{3}{2}=\frac{t^2}{2}-2t+\frac{3}{2}.\end{array}$$

   Hence, $T(2)=-1/2$. $◼$

$\text{Problem 2}$

(stepwise regularity) Let $u(t)$ be the solution to the DDE

with a continuous prehistory $u(t)=\phi (t)$ for $t\in [-1,0]$. Let $f\in C^{\infty }({\mathbb{R}}^N\times {\mathbb{R}}^N,{\mathbb{R}}^N)$. Prove $u\in C^j([j-1,T_{max}(\phi )),{\mathbb{R}}^N)$ for $j\in \mathbb{N}$, where $T_{max}(\phi )$ is the maximal time interval of the existence of the solution.

$\mathit{\text{Proof.}}$ Note that we can write

Therefore, $u(t)$ is at least $C^0$ on $[-1,T_{max}(\phi ))$. Now, we prove that $u(t)$ is $C^j$ on $[j-1,T_{max}(\phi ))$ by induction. Suppose that $u(t)$ is $C^j$ on $[j-1,T_{max}(\phi ))$ when $j=k$, i.e., $u(t)$ is $C^k$ on $[k-1,T_{max}(\phi ))$. Then, when $j=k+1$,

• $u(t)$ is $C^k$ on $[k,T_{max}(\phi ))\subset [k-1,T_{max}(\phi ))$ by assumption and
• $u(t-1)$ is $C^k$ on $[k-1,T_{max}(\phi ))$ since $t-1\in [k-1,T_{max}(\phi ))$ for $t\in [k,T_{max}(\phi ))$.

Therefore, $\dot{u}(t)=f(u(t),u(t-1))$ is of $C^k$, and hence $u(t)$ is $C^{k+1}$ on $[k,T_{max}(\phi ))$.
By induction, the desired result holds for all $j\in \mathbb{N}$. $◼$

$\text{Problem 3}$

Consider the shower equation

with a discrete delay $\tau >0$ and a parameter $k>0$.

1. Since the equation is linear, we can substitute the "exponential Ansatz,"

   $$T(t):=e^{t\lambda },$$

   into the equation. Derive the characteristic equation for $\lambda$.

   $\mathit{\text{Proof.}}$ Plug $T(t)=e^{t\lambda }$ into the DDE yields $\lambda e^{t\lambda }=-k{e}^{\lambda (t-\tau )}$. Since $e^{t\lambda }>0$ for all $t$, we can divide $\lambda e^{t\lambda }=-k{e}^{\lambda (t-\tau )}$ both sides with $e^{\lambda }t$ (or set $t=0$) to get

   $$\lambda +k{e}^{-\lambda \tau }=0,$$

   i.e., the characteristic equation for $\lambda$. $◼$

2. Prove that for each delay $\tau >0$, there exist countably many $k_j>0,j\in \mathbb{N}$, such that the shower equation with $k=k_j$ has a periodic solution in the form

   $$T(t)=\mathrm{Re}(e^{it{\omega }_j})=cos({\omega }_{j}t),{\omega }_j\in \mathbb{R}.$$

   Here $\mathrm{Re}(z)$ denotes the real part of $z\in \mathbb{C}$. What is ${\omega }_j$? Since $T(t)\equiv 0$ is a (trivial) solution, conclude that two different solutions can intersect in $\mathbb{R}$.

   $\mathit{\text{Proof.}}$ Write $\lambda ={\lambda }_1+i{\lambda }_2$, where ${\lambda }_1,{\lambda }_2\in \mathbb{R}$ and $i=\sqrt{-1}$. With the fresh characteristic equation of $\lambda$ we derived the above yields

   $$({\lambda }_1+i{\lambda }_2)+k^{-({\lambda }_1+i{\lambda }_2)\tau }=0,$$

   which implies that $({\lambda }_1+i{\lambda }_2)+k{e}^{-{\lambda }_1\tau }[\cos ({\lambda }_2\tau )-i\sin ({\lambda }_2\tau )]=0$, and hence

   $$\begin{array}{rl}{\lambda }_1+k{e}^{-{\lambda }_1\tau }\cos ({\lambda }_2\tau )& =0\\ {\lambda }_2-k{e}^{-{\lambda }_1\tau }\sin ({\lambda }_2\tau )& =0\end{array}$$

   because $({\lambda }_1+k{e}^{-{\lambda }_1\tau }\cos ({\lambda }_2\tau ))+i({\lambda }_2-k{e}^{-{\lambda }_1\tau }\sin ({\lambda }_2\tau ))=0$. Since we wish that our solution has the form $cos({\omega }_{j}t),{\omega }_j\in \mathbb{R}$, we need ${\lambda }_1$ to be $0$. Hence we get

   $$\begin{array}{rl}k\cos ({\lambda }_2\tau )& =0\\ {\lambda }_2-k\sin ({\lambda }_2\tau )& =0\end{array}$$

   which implies that ${\lambda }_2\tau =(2n+1)(\pi /2)$, i.e., ${\lambda }_2=(2n+1)(\pi /2\tau )$ for $n\in \mathbb{Z}$, and

   $$k={\lambda }_2/\sin ({\lambda }_2\tau )=\frac{\frac{(2n+1)\pi }{2\tau }}{\sin ((2n+1)(\pi /2))}=\{\begin{array}{ll}(2n+1)(\pi /2\tau ),& \text{if }n\text{ even}\\ -(2n+1)(\pi /2\tau ),& \text{if }n\text{ odd}.\end{array}$$

   Since our $k$ are greater than $0$, we must narrow down our candidates of $n$ to be positive even integers, $0$, and negative odd integers. Finally, we can state our result:
   Set

   $$\begin{array}{rl}V& =\{v:v=0,\text{or }v\text{ is a positive even integer or a negative odd integer}\}\\ & =\{v_j{\}}_{j=1}^{\mathrm{\infty }}.\end{array}$$

   Let $\text{sign}(x)=1$ if $x$ is nonnegative and $\text{sign}(x)=-1$ if $x$ is negative. Then, for each delay $\tau >0$, there exist countably many $k_j=\text{sign}(v_j)(2v_j+1)(\pi /2\tau )>0,j\in \mathbb{N}$, such that the DDE with $k=k_j$ has a periodic solution in the form

   $$T(t)=\cos ({\omega }_{j}t)=\cos ((2v_j+1)(\pi /2\tau )t).({\omega }_j=(2v_j+1)(\pi /2\tau ))$$

   With these periodic solutions and the trivial solution, we see that the solutions of the same equation can intersect with each other. (?!!!!) $◼$

3. Why the scalar shower equation has periodic solutions no matter how small the delay $\tau >0$ is?
   多了一個大於$0$的$\tau$之後，$\lambda$的特徵方程($\lambda +k{e}^{-\lambda \tau }=0$)因為那個$\tau$的關係導致多出一個$\exp$而具有複數解(在沒有delay的時候，特徵方程為$\lambda +k=0$)。所以當我們使用exponential ansatz來解DDE的時候，$\lambda$如果有imaginary part，就會跑出$\cos$和$\sin$這兩個週期函數。

$\text{Problem 4}$

We know that $x(t)=x_0/(1-x_{0}t)$ solves the ODE $\dot{x}(t)=x^2(t)$ with the initial condition $x(0)=x_0>0$, and blows up at a finite time $t=1/x_0$ , that is, $li{m}_{t\nearrow 1/x_0}x(t)=\infty$.

1. Prove that the DDE $u(t)=u^2(t-\tau )$ does not have finite-time blow-up solutions for any delay $\tau >0$ and prehistory $\phi \in C^0([-\tau ,0],\mathbb{R}$). Hint: Use the method of steps.

   $\mathit{\text{Proof.}}$ $\dot{u}(t)=u^2(t-\tau )={\phi }^2(t-\tau )$ is continuous on $[0,\tau ]$, hence

   $$u(t)={\int }_0^t{\phi }^2(s-\tau )ds+\phi (0)$$

   is continuous and finite on $[0,\tau ]$. Now we prove that $u(t)$ is continuous and finite on $[(j-1)\tau ,j\tau ]$ for all $j\in \mathbb{N}$. Suppose that the result holds for $j=k$. Then when $j=k+1$,
   - $u^2(s-\tau )$ is continuous for all $s\in [(j-1)\tau ,t]$ since $(s-\tau )\in [(k-1)\tau ,k\tau ]$, and
   - $u((j-1)\tau )=u(k\tau )$ is finite.

   Therefore, $u(t)={\int }_{(j-1)\tau }^t{u}^2(s-\tau )ds+u((j-1)\tau )$ is continuous and finite on $[(j-1)\tau ,j\tau ]=[k\tau ,(k+1)\tau ]$.
   By induction, the result holds for all $j\in \mathbb{N}$. Hence $u(t)$ is finite all the time. $◼$

2. Explain the behavior of solutions of the DDE with $\phi (t)\equiv 1$ as $\tau \searrow 0$.

   $\mathit{\text{Proof.}}$ The "slope" in the nth step is given by the (n-1)th step, where each step has length $\tau$. When $\tau$ becomes smaller and smaller, each step "shrinks," and the "slope" at each $t$ becomes larger and hence grows faster than before. Therefore, the solution will explode to $\mathrm{\infty }$ as $\tau \searrow 0$.
   (For example, assume that the original $\tau$ is $1$, then the "slope" at $3/4$ is $1$, which is given by the 0th step. When $\tau$ becomes $1/2$, the "slope" at $3/4$ becomes larger than $1$, which is given by the 1st step.) $◼$

(2024-09-11)

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