Firstly, we have some weather stations
Spatial Coordinates : s ∈ Ω ⊆ R D s \begin{aligned}
\text{Spatial Coordinates}: && &&
\mathbf{s}&\in\Omega\subseteq\mathbb{R}^{D_s}
\end{aligned} Spatial Coordinates : s ∈ Ω ⊆ R D s In our case, the vector, s \mathbf{s} s
Temporal Coordinates : t ∈ T ⊆ R + \begin{aligned}
\text{Temporal Coordinates}: && &&
t &\in \mathcal{T} \subseteq \mathbb{R}^+
\end{aligned} Temporal Coordinates : t ∈ T ⊆ R + There are two time domains that we have access to.
For the global mean surface temperature (GMST ), we have a time series that is available from 1800 till present day (2020).
For the daily maximum temperature measurements, we have a time series that is available from 1960 till present day (2020).
First, we have our measurements which are maximum temperature values.
Measurements : y = y ( t , s ) y : R + → R t ∈ T y ⊆ R + s ∈ Ω ⊆ R D s \begin{aligned}
\text{Measurements}: && &&
y &= y(t,\mathbb{s}) && &&
y:\mathbb{R}^+\rightarrow\mathbb{R}
&& &&
t \in \mathcal{T}_y\subseteq \mathbb{R}^+ &&
\mathbf{s}\in\Omega\subseteq\mathbb{R}^{D_s}
\end{aligned} Measurements : y = y ( t , s ) y : R + → R t ∈ T y ⊆ R + s ∈ Ω ⊆ R D s Next, we have our GMST measurements which is the temperature anomaly (in degrees Celsius) over our time domain.
This is a single time series so there is no dependency on spatial coordinates.
Covariate : x = x ( t ) x : R + → R x ∈ T x ⊆ R + \begin{aligned}
\text{Covariate}: && &&
x &= x(t) && &&
x:\mathbb{R}^+\rightarrow\mathbb{R}
&& &&
x \in \mathcal{T}_x\subseteq \mathbb{R}^+
\end{aligned} Covariate : x = x ( t ) x : R + → R x ∈ T x ⊆ R + The DMT is formulated as an ordinary differential equation (ODE).
First, we will define it as a system of ODEs whereby we have a state variable
State : z = [ x y ] , z ∈ R 2 \begin{aligned}
\text{State}: && &&
\mathbf{z} &=
\begin{bmatrix}
x \\ y
\end{bmatrix}, && &&
\mathbf{z}\in\mathbb{R}^2
\end{aligned} State : z = [ x y ] , z ∈ R 2 Now, we can define an equation of motion which describes the temporal dynamics of the system.
Equation of Motion : d z d t = f ( z , t , θ ) , f : R 2 × R + × Θ → R \begin{aligned}
\text{Equation of Motion}: && &&
\frac{d\mathbf{z}}{dt} &= \boldsymbol{f}(\mathbf{z},t,\theta),
&& &&
\boldsymbol{f}:\mathbb{R}^2 \times \mathbb{R}^+ \times \Theta \rightarrow \mathbb{R}
\end{aligned} Equation of Motion : d t d z = f ( z , t , θ ) , f : R 2 × R + × Θ → R We also have initial measurements of the system
Initial Values : z ( 0 ) = [ x ( 0 ) y ( 0 ) ] : = z 0 \begin{aligned}
\text{Initial Values}: && &&
\mathbf{z}(0) &=
\begin{bmatrix}
x(0) \\ y(0)
\end{bmatrix}
:=
\mathbf{z}_0
\end{aligned} Initial Values : z ( 0 ) = [ x ( 0 ) y ( 0 ) ] := z 0 From the fundamental theory of calculus, we know that the solution of said ODE is a temporal integration wrt time
z t = z 0 + ∫ 0 t f ( z 0 , t , θ ) d t \begin{aligned}
\mathbf{z}_t = \mathbf{z}_0 + \int_0^t \boldsymbol{f}(\mathbf{z}_0, t, \theta)dt
\end{aligned} z t = z 0 + ∫ 0 t f ( z 0 , t , θ ) d t Conventionally, we use ODE solvers like Euler, Heun, or Runge-Kutta.
z t = ODESolve ( f , z 0 , t , θ ) \mathbf{z}_t = \text{ODESolve}(\boldsymbol{f}, \mathbf{z}_0, t, \theta) z t = ODESolve ( f , z 0 , t , θ ) Non-Dimensionalization ¶ We will reparameterize this ODE to remove some dependencies on time.
The above equation is divided by
d y d t × d t d x = d y d x = f ( y , t , θ ) g ( x , t , θ ) : = h ( y , x , θ ) \frac{dy}{dt}\times \frac{dt}{dx}
= \frac{dy}{dx}
= \frac{f(y,t,\theta)}{g(x,t,\theta)}
:= h(y,x,\theta) d t d y × d x d t = d x d y = g ( x , t , θ ) f ( y , t , θ ) := h ( y , x , θ ) Parameterization ¶ There are many special forms of ODEs which are known from the literature.
1st Order ODE : f ( y , x , θ ) = f 1 ( x ) − f 2 ( x ) ⋅ y \begin{aligned}
\text{1st Order ODE}: && &&
\boldsymbol{f}(y,x,\theta) &=
\boldsymbol{f}_1(x) - \boldsymbol{f}_2(x)\cdot y
\end{aligned} 1st Order ODE : f ( y , x , θ ) = f 1 ( x ) − f 2 ( x ) ⋅ y An example form would the following:
f ( y , x , θ ) = a 0 + a 1 x + a 2 y \boldsymbol{f}(y,x,\theta) =
a_0 + a_1 x + a_2 y f ( y , x , θ ) = a 0 + a 1 x + a 2 y Constant Form .
The first form assumes that we have a constant change in DMT wrt the GMST
Constant EOM : f ( y , x , θ ) = a 0 Linear Solution : y ( x ) = y 0 + a 0 t \begin{aligned}
\text{Constant EOM}: && &&
\boldsymbol{f}(y,x,\theta)
&=
a_0 \\
\text{Linear Solution}: && &&
y(x) &=
y_0 + a_0 t
\end{aligned} Constant EOM : Linear Solution : f ( y , x , θ ) y ( x ) = a 0 = y 0 + a 0 t Linear Form .
The first form assumes that we have a constant change in DMT wrt the GMST
Linear EOM : f ( y , x , θ ) = a 0 + a 1 t Quadratic Solution : y ( x ) = y 0 + a 0 t + 1 2 a 1 t 2 \begin{aligned}
\text{Linear EOM}: && &&
\boldsymbol{f}(y,x,\theta)
&=
a_0 + a_1 t\\
\text{Quadratic Solution}: && &&
y(x) &=
y_0 + a_0 t + \frac{1}{2}a_1t^2
\end{aligned} Linear EOM : Quadratic Solution : f ( y , x , θ ) y ( x ) = a 0 + a 1 t = y 0 + a 0 t + 2 1 a 1 t 2 Multiplicative Form .
The first form assumes that we have a constant change in DMT wrt the GMST
Linear EOM : f ( y , x , θ ) = a 2 y Exponential Solution : y ( x ) = y 0 exp ( a 2 t ) \begin{aligned}
\text{Linear EOM}: && &&
\boldsymbol{f}(y,x,\theta)
&=
a_2 y\\
\text{Exponential Solution}: && &&
y(x) &=
y_0 \exp \left( a_2t \right)
\end{aligned} Linear EOM : Exponential Solution : f ( y , x , θ ) y ( x ) = a 2 y = y 0 exp ( a 2 t )