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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.

Definition1

The following partial differential equation is called the general Hamilton-Jacobi equation.

G(Du,ut,u,x,t)=ut+H(Du,x)=0 G(Du, u_{t}, u, x, t)=u_{t}+H(Du, x)=0

  • t>0Rt >0 \in \mathbb{R}
  • xRnx \in \mathbb{R}^{n}
  • u:RnRu : \mathbb{R}^{n} \to \mathbb{R}

Here, the differential operator DD follows the multi-index notation, and let us always consider differentiation with respect to the spatial variable xx, that is, D=DxD=D_{x}, and Du=Dxu=(ux1,,uxn)Du=D_{x}u=(u_{x_{1}}, \cdots, u_{x_{n}}). And H:Rn×RnRH : \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R} is called the Hamiltonian.

Characteristic Equation

For convenience, assume that HC(Rn×(0,))H \in C^{\infty} \big(\mathbb{R}^{n} \times (0,\infty) \big). Given such a Hamilton-Jacobi equation, let’s simplify the expression by combining space-time variables into one as yy.

y=(x,t)=(x1,,xn,t) y=(x,t)=(x_{1}, \cdots, x_{n}, t)

Also, the time derivative and spatial derivative of uu are together represented as qq.

q=q(Du,ut)=q(ux1,ux2,,uxn,ut)=(p,pn+1)=(p1,p2,,pn,pn+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*}

Lastly, if we say z=uz=u, the Hamilton-Jacobi equation can be represented as below.

G(q,z,y)=pn+1+H(p,x)=0(q,z,y)Rn+1×R×(Rn×(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}

Solving for the derivative of GG, we get the following.

DqG(q,z,y)=(Gp1,,Gpn+1)=(Hp1(p,x),,Hpn(p,x),1)=(DpH(p,x),1)DzG(q,z,y)=Gz=0DyG(q,z,y)=(Gy1,,Gyn+1)=(Hx1(p,x),,Hxn(p,x),Ht(p,x))=(DxH(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}

Moreover, the characteristic equation of G(q,z,y)G(q,z,y) is as follows.

{q˙(s)=DyG(q(s),z(s),y(s))DzG(q(s),z(s),y(s))q(s)z˙(s)=DqG(q(s),z(s),y(s))q(s)y˙(s)=DqG(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.

Then q˙(s)\dot{q}(s) is as follows.

q˙(s)=DyG(q(s),z(s),y(s))DzG(q(s),z(s),y(s))q(s)=DyG(q(s),z(s),y(s))=(DxH(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*}

The second equality is by (eq2)\eqref{eq2}, and the third equality is by (eq4)\eqref{eq4}. Since q=(p,pn+1)q=(p, p_{n+1}), each component of q˙\dot{q} is as follows.

p˙i(s)=Hxi(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*}

z˙(s)\dot{z}(s) is as follows.

z˙(s)=DqG(q(s),z(s),y(s))q(s)=(DpH(p(s),x(s)),1)(p(s),pn+1(s))=DpH(p(s),x(s))p(s)+pn+1(s)=DpH(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*}

The second equality is by (eq2)\eqref{eq2}, and the fourth equality is by (eq1)\eqref{eq1}. y˙(s)\dot{y}(s) is as follows.

y˙(s)=DqG(q(s),z(s),y(s))=(DpH(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*}

Since y=(x,t)y=(x,t), each component of y˙=(x˙,t˙)\dot{y}=(\dot{x}, \dot{t}) is as follows.

{x˙(s)=DpH(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}

From the results above, we can consider ss to be equivalent to tt. By synthesizing everything we calculated so far, we obtain the characteristic equations of the Hamilton-Jacobi equation as follows.

p˙(s)=DxH(p(s),x(s))z˙(s)=DpH(p(s),x(s))p(s)H(p(s),x(s))x˙(s)=DpH(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*}

Here, specifically groups the first and third equations as Hamilton’s equations.


  1. Lawrence C. Evans, Partial Differential Equations (2nd Edition, 2010), p113-114 ↩︎