Hamilton-Jacobi Equation and Hamiltonian Equation
📂Partial Differential Equations Hamilton-Jacobi Equation and Hamiltonian Equation There are two ways to derive the Hamilton equations. One is from the Euler-Lagrangian equations , and the other, which will be introduced in this article, is from the characteristic equations of the Hamilton-Jacobi equation.
Definition The following partial differential equation is called the general Hamilton-Jacobi equation .
G ( D u , u t , u , x , t ) = u t + H ( D u , x ) = 0
G(Du, u_{t}, u, x, t)=u_{t}+H(Du, x)=0
G ( D u , u t , u , x , t ) = u t + H ( D u , x ) = 0
t > 0 ∈ R t >0 \in \mathbb{R} t > 0 ∈ R x ∈ R n x \in \mathbb{R}^{n} x ∈ R n u : R n → R u : \mathbb{R}^{n} \to \mathbb{R} u : R n → R Here, the differential operator D D D follows the multi-index notation , and let us always consider differentiation with respect to the spatial variable x x x , that is, D = D x D=D_{x} D = D x , and D u = D x u = ( u x 1 , ⋯ , u x n ) Du=D_{x}u=(u_{x_{1}}, \cdots, u_{x_{n}}) D u = D x u = ( u x 1 , ⋯ , u x n ) . And H : R n × R n → R H : \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R} H : R n × R n → R is called the Hamiltonian .
Characteristic Equation For convenience, assume that H ∈ C ∞ ( R n × ( 0 , ∞ ) ) H \in C^{\infty} \big(\mathbb{R}^{n} \times (0,\infty) \big) H ∈ C ∞ ( R n × ( 0 , ∞ ) ) . Given such a Hamilton-Jacobi equation, let’s simplify the expression by combining space-time variables into one as y y y .
y = ( x , t ) = ( x 1 , ⋯ , x n , t )
y=(x,t)=(x_{1}, \cdots, x_{n}, t)
y = ( x , t ) = ( x 1 , ⋯ , x n , t )
Also, the time derivative and spatial derivative of u u u are together represented as q q q .
q = q ( D u , u t ) = q ( u x 1 , u x 2 , … , u x n , u t ) = ( p , p n + 1 ) = ( p 1 , p 2 , … , p n , p n + 1 )
\begin{align*}
q &=q(Du, u_{t}) =q(u_{x_{1}}, u_{x_{2}},\dots, u_{x_{n}}, u_{t})
\\ &= (p, p_{n+1}) =(p_{1}, p_{2}, \dots, p_{n}, p_{n+1})
\end{align*}
q = q ( D u , u t ) = q ( u x 1 , u x 2 , … , u x n , u t ) = ( p , p n + 1 ) = ( p 1 , p 2 , … , p n , p n + 1 )
Lastly, if we say z = u z=u z = u , the Hamilton-Jacobi equation can be represented as below.
G ( q , z , y ) = p n + 1 + H ( p , x ) = 0 ∀ ( q , z , y ) ∈ R n + 1 × R × ( R n × ( 0 , ∞ ) )
\begin{equation}
G(q, z, y)=p_{n+1}+H(p, x)=0 \quad \forall (q, z, y)\in\mathbb{R}^{n+1}\times \mathbb{R} \times \big( \mathbb{R}^n\times (0, \infty) \big)
\label{eq1}
\end{equation}
G ( q , z , y ) = p n + 1 + H ( p , x ) = 0 ∀ ( q , z , y ) ∈ R n + 1 × R × ( R n × ( 0 , ∞ ) )
Solving for the derivative of G G G , we get the following.
D q G ( q , z , y ) = ( G p 1 , ⋯ , G p n + 1 ) = ( H p 1 ( p , x ) , … , H p n ( p , x ) , 1 ) = ( D p H ( p , x ) , 1 ) D z G ( q , z , y ) = G z = 0 D y G ( q , z , y ) = ( G y 1 , ⋯ , G y n + 1 ) = ( H x 1 ( p , x ) , ⋯ , H x n ( p , x ) , H t ( p , x ) ) = ( D x H ( p , x ) , 0 )
\begin{align}
D_{q} G(q, z, y) &= (G_{p_{1}}, \cdots, G_{p_{n+1}})=\big(H_{p_{1}}(p,x), \dots , H_{p_{n}}(p,x), 1\big)=\big( D_{p} H(p,x), 1\big) \label{eq2}
\\ D_{z} G(q, z, y) &= G_{z}=0 \label{eq3}
\\ D_{y} G(q, z, y) &= \big( G_{y_{1}}, \cdots, G_{y_{n+1}} \big)=\big( H_{x_{1}}(p,x), \cdots, H_{x_{n}}(p,x), H_{t}(p,x) \big) =\big( D_{x}H (p,x), 0\big) \label{eq4}
\end{align}
D q G ( q , z , y ) D z G ( q , z , y ) D y G ( q , z , y ) = ( G p 1 , ⋯ , G p n + 1 ) = ( H p 1 ( p , x ) , … , H p n ( p , x ) , 1 ) = ( D p H ( p , x ) , 1 ) = G z = 0 = ( G y 1 , ⋯ , G y n + 1 ) = ( H x 1 ( p , x ) , ⋯ , H x n ( p , x ) , H t ( p , x ) ) = ( D x H ( p , x ) , 0 )
Moreover, the characteristic equation of G ( q , z , y ) G(q,z,y) G ( q , z , y ) is as follows.
{ q ˙ ( s ) = − D y G ( q ( s ) , z ( s ) , y ( s ) ) − D z G ( q ( s ) , z ( s ) , y ( s ) ) q ( s ) z ˙ ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) ) ⋅ q ( s ) y ˙ ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) )
\left\{
\begin{align*}
\dot{q}(s) &= -D_{y} G\big(q(s), z(s), y(s) \big)-D_{z} G\big(q(s), z(s), y(s) \big)q(s)
\\ \dot{z}(s) &= D_{q} G\big(q(s), z(s), y(s) \big) \cdot q(s)
\\ \dot{y}(s) &= D_{q} G\big(q(s), z(s), y(s) \big)
\end{align*}
\right.
⎩ ⎨ ⎧ q ˙ ( s ) z ˙ ( s ) y ˙ ( s ) = − D y G ( q ( s ) , z ( s ) , y ( s ) ) − D z G ( q ( s ) , z ( s ) , y ( s ) ) q ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) ) ⋅ q ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) )
Then q ˙ ( s ) \dot{q}(s) q ˙ ( s ) is as follows.
q ˙ ( s ) = − D y G ( q ( s ) , z ( s ) , y ( s ) ) − D z G ( q ( s ) , z ( s ) , y ( s ) ) q ( s ) = − D y G ( q ( s ) , z ( s ) , y ( s ) ) = − ( D x H ( p , x ) , 0 )
\begin{align*}
\dot{ q}(s) &= -D_{y} G\big(q(s), z(s), y(s) \big)-D_{z} G\big(q(s), z(s), y(s) \big)q(s)
\\ &= -D_{y} G\big(q(s), z(s), y(s) \big)
\\ &=- (D_{x} H(p,x), 0)
\end{align*}
q ˙ ( s ) = − D y G ( q ( s ) , z ( s ) , y ( s ) ) − D z G ( q ( s ) , z ( s ) , y ( s ) ) q ( s ) = − D y G ( q ( s ) , z ( s ) , y ( s ) ) = − ( D x H ( p , x ) , 0 )
The second equality is by ( eq2 ) \eqref{eq2} ( eq2 ) , and the third equality is by ( eq4 ) \eqref{eq4} ( eq4 ) . Since q = ( p , p n + 1 ) q=(p, p_{n+1}) q = ( p , p n + 1 ) , each component of q ˙ \dot{q} q ˙ is as follows.
p ˙ i ( s ) = − H x i ( p ( s ) , x ( s ) ) ( i = 1 , … , n ) p ˙ n + 1 ( s ) = 0
\begin{align*}
\dot{p}^{i}(s) &= -H_{x_{i}} \big( p(s), x(s) \big) &( i=1,\dots,n)
\\ \dot{p}^{n+1}(s) &= 0
\end{align*}
p ˙ i ( s ) p ˙ n + 1 ( s ) = − H x i ( p ( s ) , x ( s ) ) = 0 ( i = 1 , … , n )
z ˙ ( s ) \dot{z}(s) z ˙ ( s ) is as follows.
z ˙ ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) ) ⋅ q ( s ) = ( D p H ( p ( s ) , x ( s ) ) , 1 ) ⋅ ( p ( s ) , p n + 1 ( s ) ) = D p H ( p ( s ) , x ( s ) ) ⋅ p ( s ) + p n + 1 ( s ) = D p H ( p ( s ) , x ( s ) ) ⋅ p ( s ) − H ( p ( s ) , x ( s ) )
\begin{align*}
\dot{z}(s) &= D_{q} G\big(q(s), z(s), y(s) \big) \cdot q(s)
\\ &= \Big( D_{p}H\big(p(s), x(s) \big), 1 \Big)\cdot\big( p(s), p_{n+1}(s) \big)
\\ &= D_{p} H\big( p(s), x(s)\big)\cdot p(s) +p_{n+1}(s)
\\ &= D_{p} H\big( p(s), x(s)\big)\cdot p(s) -H\big( p(s), x(s)\big)
\end{align*}
z ˙ ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) ) ⋅ q ( s ) = ( D p H ( p ( s ) , x ( s ) ) , 1 ) ⋅ ( p ( s ) , p n + 1 ( s ) ) = D p H ( p ( s ) , x ( s ) ) ⋅ p ( s ) + p n + 1 ( s ) = D p H ( p ( s ) , x ( s ) ) ⋅ p ( s ) − H ( p ( s ) , x ( s ) )
The second equality is by ( eq2 ) \eqref{eq2} ( eq2 ) , and the fourth equality is by ( eq1 ) \eqref{eq1} ( eq1 ) . y ˙ ( s ) \dot{y}(s) y ˙ ( s ) is as follows.
y ˙ ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) ) = ( D p H ( p , x ) , 1 )
\begin{align*}
\dot{y}(s) &= D_{q}G\big(q(s), z(s), y(s) \big)
\\ &= \big(D_{p}H(p,x), 1 \big)
\end{align*}
y ˙ ( s ) = D q G ( q ( s ) , z ( s ) , y ( s ) ) = ( D p H ( p , x ) , 1 )
Since y = ( x , t ) y=(x,t) y = ( x , t ) , each component of y ˙ = ( x ˙ , t ˙ ) \dot{y}=(\dot{x}, \dot{t}) y ˙ = ( x ˙ , t ˙ ) is as follows.
{ x ˙ ( s ) = D p H ( p ( s ) , x ( s ) ) t ˙ ( s ) = 1
\begin{cases} \dot{x}(s) = D_{p}H\big( p(s), x(s) \big)
\\ \dot{t}(s)=1 \end{cases}
{ x ˙ ( s ) = D p H ( p ( s ) , x ( s ) ) t ˙ ( s ) = 1
From the results above, we can consider s s s to be equivalent to t t t . By synthesizing everything we calculated so far, we obtain the characteristic equations of the Hamilton-Jacobi equation as follows.
p ˙ ( s ) = − D x H ( p ( s ) , x ( s ) ) z ˙ ( s ) = D p H ( p ( s ) , x ( s ) ) ⋅ p ( s ) − H ( p ( s ) , x ( s ) ) x ˙ ( s ) = D p H ( p ( s ) , x ( s ) )
\begin{align*}
\dot{p}(s) &= -D_{x}H \big( p(s), x(s) \big)
\\ \dot{z}(s) &= D_{p} H\big( p(s), x(s)\big)\cdot p(s) -H\big( p(s), x(s)\big)
\\ \dot{x}(s) &= D_{p}H\big( p(s), x(s) \big)
\end{align*}
p ˙ ( s ) z ˙ ( s ) x ˙ ( s ) = − D x H ( p ( s ) , x ( s ) ) = D p H ( p ( s ) , x ( s ) ) ⋅ p ( s ) − H ( p ( s ) , x ( s ) ) = D p H ( p ( s ) , x ( s ) )
Here, specifically groups the first and third equations as Hamilton’s equations .
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