Well-Posedness

🔍 Finite-Time Blow-Up

Consider the IVP

$$\dot{x}=x^2,x(0)=1.$$

🐍 💻 📈

def fun(t, y):
    return y * y
t0, tf, y0 = 0.0, 3.0, array([1.0])
result = solve_ivp(fun, t_span=[t0, tf], y0=y0)
figure()
plot(result["t"], result["y"][0], "k")
xlim(t0, tf); xlabel("$t$"); ylabel("$x(t)$")

Local vs Global

🤕 Ouch.

There is actually no global solution.

However there is a local solution $x(t)$,

• defined for $t\in [t_0,\tau [$

• for some $\tau >t_0$.

Indeed, the function ${\displaystyle x(t):=\frac{1}{1-t}}$ satisfies

$$\dot{x}(t)=\frac{d}{dt}x(t)=-\frac{-1}{(1-t{)}^2}=(x(t){)}^2$$ and $x(0)=1.$

⚠️ But it’s defined (continuously) only for $t<1.$

🐍 💻 📈

tf = 1.0
r = solve_ivp(fun, [t0, tf], y0,
              dense_output=True)
figure()
t = linspace(t0, tf, 1000)
plot(t, r["sol"](t)[0], "k")
ylim(0.0, 10.0); grid();
xlabel("$t$"); ylabel("$x(t)$")

This local solution is also maximal:

You cannot extend this solution beyond $\tau =1.0$.

🧩 Maximal Solutions

Consider the IVP

$$\dot{x}=x^2,x(0)=x_0\ne 0.$$

1. 🧮

Find a closed-formed local solution $x(t)$ of the IVP.

🗝️ Hint: assume that $x(t)\ne 0$ then compute

$$\frac{d}{dt}\frac{1}{x(t)}.$$

2. 🧠

Make sure that your solutions are maximal.

🔓 Maximal Solutions

1. 🔓

As long as $x(t)\ne 0$,

$$\frac{d}{dt}\frac{1}{x(t)}=-\frac{\dot{x}(t)}{x(t{)}^2}=1.$$

By integration, this leads to

$$\frac{1}{x(t)}-\frac{1}{x_0}=-t$$

and thus provides

$$x(t)=\frac{1}{\frac{1}{x_0}-t}=\frac{x_0}{1-x_{0}t}.$$

which is indeed a solution as long as the denominator is not zero.

2. 🔓

• If $x_0<0$, this solution is valid for all $t\ge 0$ and thus maximal.

• If $x_0>0$, the solution is defined until $t=1/x(0)$ where it blows up. Thus, this solution is also maximal.

🔍 No Local Solution

Consider the scalar IVP with initial value $x(0)=(0,0)$ and right-hand side

$$f(x_1,x_2)=|\begin{array}{rl}(+1,0)& \text{if }{x}_1<0\\ (-1,0)& \text{if }{x}_1\ge 0.\end{array}$$

🐍 📈 No Local Solution

def f(x1x2):
    x1, x2 = x1x2
    dx1 = 1.0 if x1 < 0.0 else -1.0
    return array([dx1, 0.0])
figure()
x1 = x2 = linspace(-1.0, 1.0, 20)
gca().set_aspect(1.0)
quiver(*Q(f, x1, x2), color="k")

💎 No Local Solution

This system has no solution, not even a local one, when $x(0)=(0,0)$.

🧠 Proof

• Assume that $x:[0,\tau [\to \mathbb{R}$ is a local solution.

• Since $\dot{x}(0)=-1<0$, for some small enough $0<ϵ<\tau$ and any $t\in ]0,ϵ]$, we have $x(t)<0$.

• Consequently, $\dot{x}(t)=+1$ and thus by integration

  $$x(ϵ)=x(0)+{\int }_0^{ϵ}\dot{x}(t)dt=ϵ>0,$$

which is a contradiction.

🧩 Sigmoid

Consider the dynamical system

$$\dot{x}=\sigma (x):=\frac{1}{1+e^{-x}}.$$

📈

def sigma(x):
  return 1 / (1 + exp(-x))
figure()
x = linspace(-7.0, 7.0, 1000)
plot(x, sigma(x), label="$y=\sigma(x)$")
grid(True)

1. 🧮 Existence

Show that there is a (at least one) maximal solution to each initial condition.

2. 🧮 Global

Show that any such solution is global.

🔓 Sigmoid

1. 🔓 Existence

The sigmoid function $\sigma$ is continuous.

Consequently, 💎 Existence proves the existence of a (at least one) maximal solution.

2. 🔓 Global

Let $x:[0,\tau [\to \mathbb{R}$ be a maximal solution to the IVP. We have

$$0\le \dot{x}(t)=\sigma (x(t))\le 1,0\le t<\tau$$

and by integration,

$$|x(t)|\le |x(0)|+t$$

Thus, it cannot blow-up in finite time; by 💎 Maximal Solutions, it is global.

🧩 Pendulum

Consider the pendulum, subject to a torque $c$

$$m{l}^2\ddot{\theta }+b\dot{\theta }+mg\ell \sin \theta =c(\theta ,\dot{\theta })$$

We assume that the torque provides a bounded power:

$$P:=c(\theta ,\dot{\theta })\dot{\theta }\le P_M<+\mathrm{\infty }.$$

1. 🧮

Show that for any initial state, there is a global solution $(\theta ,\dot{\theta })$.

🗝️ Hint. Compute the derivative with respect to $t$ of

$$E=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell \cos \theta .$$

🔓 Pendulum

1. 🔓

Since the system vector field

$$(\theta ,\dot{\theta })\to (\dot{\theta },(-b/m{\ell }^2)\dot{\theta }-(g/\ell )\sin \theta +c(\theta ,\dot{\theta })/m{\ell }^2)$$

is continuous, 💎 Existence yields the existence of a (at least one) maximal solution.

Additionally,

$$\begin{array}{rl}\dot{E}& =\frac{d}{dt}(\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell \cos \theta )\\ & =-b{\dot{\theta }}^2+c(\theta ,\dot{\theta })\dot{\theta }\\ & \le P_M<+\mathrm{\infty }.\end{array}$$

By integration

$$E(t)=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2(t)-mg\ell \cos \theta (t)\le E(0)+P_{M}t$$

Hence, since $|\cos \theta (t)|\le 1$,

$$|\dot{\theta }(t)|\le \sqrt{\frac{2E(0)}{m{\ell }^2}+\frac{2g}{\ell }+\frac{2P_M}{m{\ell }^2}t}$$

Thus, $\dot{\theta }(t)$ cannot blow-up in finite time. Since

$$|\theta (t)|\le |\theta (0)|+{\int }_0^t|\dot{\theta }(s)|ds,$$

$\theta (t)$ cannot blow-up in finite time either.

By 💎 Maximal Solutions, any maximal solution is global.

🧩 Linear Systems

Let $A\in {\mathbb{R}}^{n\times n}$.

Consider the dynamical system

$$\dot{x}=Ax,x\in {\mathbb{R}}^n.$$

1. 🧮

Show that

$$y(t):=\parallel x(t){\parallel }^2$$

is differentiable and satisfies

$$\dot{y}(t)\le 2\alpha y(t)$$

for some $\alpha \ge 0$. 🔓

2. 🧮

Let

$$z(t):=y(t)e^{-2\alpha t}.$$

Compute $\dot{z}(t)$ and deduce that

$$0\le y(t)\le y(0)e^{2\alpha t}.$$

3. 🧮

Prove that for any initial state $x(0)\in {\mathbb{R}}^n$ there is a corresponding global solution $x(t)$. 🔓

🔓 Linear Systems

1. 🔓

By definition of $y(t)$ and since $\dot{x}(t)=Ax(t)$,

$$\begin{array}{rl}\dot{y}(t)& =\frac{d}{dt}\parallel x(t){\parallel }^2\\ & =\frac{d}{dt}x(t{)}^{t}x(t)\\ & =\dot{x}(t{)}^{t}x(t)+x(t{)}^t\dot{x}(t)\\ & =x(t{)}^t{A}^{t}x(t)+x(t{)}^{t}Ax(t).\end{array}$$

Let $\alpha$ denote the largest singular value of $A$ (i.e. the operator norm $\parallel A\parallel$).

$$\alpha :={\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}(A)=\parallel A\parallel .$$

For any vector $u\in {\mathbb{R}}^n$, we have $$\parallel Au\parallel \le \parallel A\parallel \parallel u\parallel .$$

By the triangle inequality and the Cauchy-Schwarz inequality, we obtain

$$\begin{array}{rl}\dot{y}(t)& =\parallel x(t{)}^t{A}^{t}x(t)+x(t{)}^{t}Ax(t)\parallel \\ & \le \parallel (Ax(t){)}^{t}x(t)\parallel +\parallel x(t{)}^t(Ax(t))\parallel \\ & \le \parallel Ax(t)\parallel \parallel x(t)\parallel +\parallel x(t)\parallel \parallel Ax(t)\parallel \\ & \le \parallel A\parallel \parallel x(t)\parallel \parallel x(t)\parallel +\parallel x(t)\parallel \parallel A\parallel \parallel x(t)\parallel \\ & =2\parallel A\parallel y(t)\end{array}$$

and thus $\dot{y}(t)\le 2\alpha y(t)$ with $\alpha :=\parallel A\parallel .$

2. 🔓

Since $y(t)=\parallel x(t){\parallel }^2$, the inequality $0\le y(t)$ is clear.

Since $z(t)=y(t)e^{-2\alpha t}$,

$$\begin{array}{rl}\dot{z}(t)& =\frac{d}{dt}y(t)e^{-2\alpha t}\\ & =\dot{y}(t)e^{-2\alpha t}+y(t)(-2\alpha e^{-\alpha t})\\ & =(\dot{y}(t)-2\alpha y(t))e^{-2\alpha t}\\ & \le 0.\end{array}$$

By integration

$$\begin{array}{rl}y(t)e^{-2\alpha t}=z(t)& =z(0)+{\int }_0^t\dot{z}(s)ds\\ & \le z(0)=y(0),\end{array}$$

hence

$$y(t)\le y(0)e^{2\alpha t}.$$

3. 🔓

The vector field $$x\in {\mathbb{R}}^n\to Ax$$ is continuous, thus by 💎 Existence there is a maximal solution $x:[0,t_{\mathrm{\infty }}[$ for any initial state $x(0).$

Moreover,

$$\parallel x(t)\parallel =\sqrt{\parallel y(t)\parallel }\le \sqrt{y(0)e^{2\alpha t}}=\parallel x(0)\parallel e^{\alpha t}.$$

Hence there is no finite-time blow-up and the maximal solution is global.

🔍 Non-Uniqueness

The IVP

$$\dot{x}=\sqrt{x},x(0)=0$$

has several maximal (global) solutions.

Proof

For any $\tau \ge 0$, $x_{\tau }$ is a solution:

$$x_{\tau }(t)=|\begin{array}{ll}0& \text{if}t\le \tau ,\\ 1/4\times (t-\tau {)}^2& \text{if}t>\tau .\end{array}$$

🔍 Prey-Predator

Let

$$\begin{array}{rcl}\dot{x}& =& \alpha x-\beta xy\\ \dot{y}& =& \delta xy-\gamma y\end{array}$$

with $\alpha =2/3$, $\beta =4/3$, $\delta =\gamma =1.0$.

🐍

alpha = 2 / 3; beta = 4 / 3; delta = gamma = 1.0

def fun(t, y):
    x, y = y
    u = alpha * x - beta * x * y
    v = delta * x * y - gamma * y
    return array([u, v])

💻

tf = 3.0
result = solve_ivp(
  fun, 
  t_span=(0.0, tf), 
  y0=[1.5, 1.5], 
  max_step=0.01)
x, y = result["y"][0], result["y"][1]

📈

def display_streamplot():
    ax = gca()
    xr = yr = linspace(0.0, 2.0, 1000)
    def f(y):
        return fun(0, y)
    streamplot(*Q(f, xr, yr), color="grey")

📈

def display_reference_solution():
    for xy in zip(x, y):
        x_, y_ = xy
        gca().add_artist(Circle((x_, y_), 
                         0.2, color="#d3d3d3"))
    gca().add_artist(Circle((x[0], y[0]), 0.1, 
                     color="#808080"))
    plot(x, y, "k")

📈

def display_alternate_solution():
    result = solve_ivp(fun, 
                       t_span=[0.0, tf],
                       y0=[1.5, 1.575], 
                       max_step=0.01)
    x, y = result["y"][0], result["y"][1]
    plot(x, y, "k--")

📈

figure()
display_streamplot()
display_reference_solution()
display_alternate_solution()
axis([0,2,0,2]); axis("square")

🧩 Continuity

Let $h\ge 0$ and $x^h(t)$ be the solution of the IVP

$$\dot{x}=x,x^h(0)=1+h.$$

1. 🧮

Let $ϵ>0$ and $\tau \ge 0$.

Find the largest $\delta >0$ such that $|h|<\delta$ ensures that $$\text{for any }t\in [t_0,\tau ],|x^h(t)-x^0(t)|\le ϵ$$

2. 🧮

What is the behavior of $\delta$ when $\tau$ goes to infinity?

🔓 Continuity

2. 🔓

The solution $x^h(t)$ to the IVP is $$x^h(t)=(1+h)e^t.$$ Hence, $$|x^h(t)-x^0(t)|=|(1+h)e^t-e^t|=|h|e^t$$ $$\underset{t\in [0,\tau ]}{max}|x^h(t)-x^0(t)|=|h|e^{\tau }.$$

Thus, the smallest $\delta$ such that $|h|\le \delta$ yields $$\underset{t\in [0,\tau ]}{max}|x^h(t)-x^0(t)|\le ϵ.$$ is $\delta =\epsilon e^{-\tau }.$

2. 🔓

For any $\epsilon >0$, $$\underset{\tau \to +\mathrm{\infty }}{lim}\delta =0.$$

🧩 Continuity Issues

Consider the IVP $$\dot{x}=\sqrt{|x|},x(0)=x_0\in \mathbb{R}.$$

1. 💻 📈

Solve numerically this IVP for $t\in [0,1]$ and $x_0=0$ and plot the result.

Then, solve it again for $x_0=0.1$, $x_0=0.01$, etc. and plot the results.

2. 🔬

Does the solution seem to be continuous with respect to the initial value?

3. 🧠

Explain this experimental result.

🔓 Continuity Issues

1. 🔓

def fun(t, y):
  x = y[0]
  dx = sqrt(abs(y))
  return [dx]
tspan = [0.0, 3.0]
t = linspace(tspan[0], tspan[1], 1000)

2. 🔓

The solution does not seem to be continuous with respect to the initial value since the graph of the solution seems to have a limit when $x_0\to 0^{+}$, but this limit is different from $x(t)=0$ which is the numerical solution when $x_0=0$.

3. 🔓

The jacobian matrix of the vector field is not defined when $x=0$, thus the continuity was not guaranted to begin with. Actually, uniqueness of the solution does not even hold here, see 🔍 Non-Uniqueness. The function $x(t)=0$ is valid when $x_0=0$, but so is $$x(t)=\frac{1}{4}{t}^2$$ and the numerical solution seems to converge to the second one when $x_0\to 0^{+}$.

🧩 Prey-Predator

Consider the system

$$\begin{array}{rcl}\dot{x}& =& \alpha x-\beta xy\\ \dot{y}& =& \delta xy-\gamma y\end{array}$$

where $\alpha$, $\beta$, $\delta$ and $\gamma$ are positive.

1. 🧮

Prove that the system is well-posed.

2. 🧮 🧠

Prove that all maximal solutions such that $x(0)>0$ and $y(0)>0$ are global and satify $x(t)>0$ and $y(t)>0$ for every $t\ge 0$.

Hint 🗝️. Compute the ODE satisfied by $u=\ln x$ and $v=\ln y$ and then the derivative w.r.t. time of $$V:=\delta e^u-\gamma u+\beta e^v-\alpha v.$$

🔓 Prey-Predator

🔓 1.

The jacobian matrix of the system vector field $$f(x,y)=(\alpha x-\beta xy,\delta xy-\gamma y)$$ is defined and continuous: $$\frac{\partial f}{\partial (x,y)}=\left[\begin{array}{rr}\alpha -\beta y& -\beta x\\ \delta y& \delta x-\gamma \end{array}\right]$$ thus the sytem is well-posed.

🔓 2.

The (continuously differentiable) change of variable $$F:(x,y)\mapsto (u,v):=(\ln x,\ln y)$$ is a bijection between ${]0,+\mathrm{\infty }[}^2$ and ${\mathbb{R}}^2$.

Since $$\frac{d}{dt}\ln x=\frac{\dot{x}}{x},\frac{d}{dt}\ln y=\frac{\dot{y}}{y}$$ the prey-predator ODE is equivalent to $$\begin{array}{rcl}\dot{u}& =& \alpha -\beta e^v\\ \dot{v}& =& \delta e^u-\gamma \end{array}$$

Accordingly, $$\begin{array}{rl}\frac{d}{dt}V& =\delta e^u\dot{u}-\gamma \dot{u}+\beta e^v\dot{v}-\alpha \dot{v}\\ & =(\delta e^u-\gamma )\dot{u}+(\beta e^v-\alpha \dot{v})\\ & =(\delta e^u-\gamma )(\alpha -\beta e^v)+(\beta e^v-\alpha )(\delta e^u-\gamma )\\ & =0\end{array}$$

Therefore $V(u(t),v(t))$ is constant.

Now, the function $$\varphi (u):=\delta e^u-\gamma u,\psi (v):=\beta e^v-\alpha v$$ are continuous and $$\underset{|u|\to +\mathrm{\infty }}{lim}\varphi (u)=+\mathrm{\infty },\underset{|v|\to +\mathrm{\infty }}{lim}\varphi (v)=+\mathrm{\infty }.$$ As $V(u,v)=\varphi (u)+\psi (v)$, $$\underset{\parallel (u,v)\parallel \to +\mathrm{\infty }}{lim}V(u,v)=+\mathrm{\infty }.$$

Consequently, since $V(x(t),y(t))$ is constant, the solution $(u(t),v(t))$ cannot blow up (either in finite or infinite time).

Therefore the solution $(u(t),v(t))$ is global as is the solution in the original variables $(x(t),y(t))$.

Since $(x,y)=F^{-1}(u,v)$ and the domain of $F$ is ${]0,+\mathrm{\infty }[}^2$, $x(t)>0$ and $y(t)>0$ for any $t\ge 0$.