In many of our derivations or in differential equations we come across the terms Jacobian or Wronskian. For example, to check the linear independence of solutions of differential equations, we ensure that the Wronskian is non zero. What role do these play? What do they actually account for?
[Math] What exacty is the role played by Jacobian or Wronskian
determinantmatricesordinary differential equations
Related Solutions
Equations of Motion
What you are describing is Hamiltonian's view of the evolution of a dynamical system. $\mathbf q$ is the vector describing system's configuration with generalized coordinates (or degrees of freedom) in some abstract configuration space isomorphic to $\mathbb R^{s}$. For instance $s=3n$ particle coordinates for a system of $n$ free particles in Euclidean space. $\mathbf p$ is the vector of generalized momentum attached to instanteanous time rate of the configuration vector. For example for particles of mass $m$ in non-relavistic mechanics $\mathbf p= m \frac{d}{dt}\mathbf q$. Classical mechanics postulates that the simulteanous knowledge of both $\mathbf q(t)$ and $\mathbf p(t)$ at time $t$ is required to predict the system's temporal evolution for any time $t'>t$ (causality). So the complete dynamical state of the system is in fact described by the phase $\mathbf y =(\mathbf p, \mathbf q)$ which evolves in an abstract space called the phase space isomorphic to $\mathbb R^{2s}$. The fact that the knowledge of this phase is sufficient to predict the evolution demands on a mathematical point of view configuration $\mathbf q$ to evolve according to a system of $s$ ODEs of at most second order in time (known as Lagrange equations) or equivalently that phase evolves according to a system of $2s$ ODEs of the first order in time knwown as Hamilton's equations of motion,
$\frac{d\mathbf p}{dt}=-\frac{\partial H}{\partial \mathbf q}(\mathbf y,t)$
$\frac{d\mathbf q}{dt}=\frac{\partial H}{\partial \mathbf p}(\mathbf y,t)$
given some Hamilton's function $H(\mathbf y,t)$ describing the system. Physically it represents mechanical energy. For non-dissipative systems it does not depend over time explicitely. as a consequence of Liouville's theorem it is a conserved quantity along phase-space curves.
More generally, equations of motions appear to be formulated as a system of Euler-Cauchy's ODEs
$\frac{d\mathbf y}{dt} = \mathbf f(\mathbf y,t)$
\with $\mathbf f$ some vector-function over $\mathbb R^{2s}\times\mathbb R$ In accordance with Hamiltonian formalism, $\mathbf f$ appears to be $\mathbf f= (-\frac{\partial H}{\partial \mathbf q}, \frac{\partial H}{\partial \mathbf p}) $
Liouville's theorem
Now consider the phase flow $\mathbf y_t$, that is consider the one-parameter ($t\in \mathbb R$) group of transformations $\mathbf y_0\mapsto \mathbf y_t(\mathbf y_0)$ mapping initial phase at time $t=0$ to current one in phase space. This is another parametrization of the phase using $y_0$ as curvilinear coordinates. Suppose you have now hypothetical system replica with initial phase points distributed to fill some volume $\mathcal V_0$ in phase space. Liouville's theorem says that the cloud of points will evolve such as preserving its density along their curves in phase space, like an incompressible fluid flow, keeping the filled volume unchanged. Since
$\mathcal V(t)=\int_{\mathcal V_0} \det \frac{\partial \mathbf y}{\partial \mathbf y_0}(\mathbf y_0,t)\ \underline{dy_0} = \int_{\mathcal V_0} J(\mathbf y_0,t)\ \underline{dy_0}$
Now compute the volume-change time rate at any instant $t$ introducing Euler's form (1),
$\frac{d\mathcal V}{dt}=\int_{\mathcal V} \frac{\partial \mathbf f}{\partial t}+ \nabla_{\mathbf y_0} .\mathbf f(\mathbf y_0,t)\ \underline{dy_0}$. Setting this time rate to zero gives Liouville's theorem in local form:
$\frac{\partial \mathbf f}{\partial t}+ \nabla_{\mathbf y_0} .\mathbf f(\mathbf y_0,t)=0$.
Applying them to Hamilton's form,
$\frac{\partial }{\partial t}[\frac{\partial H}{\partial \mathbf p}-\frac{\partial H}{\partial \mathbf q}] + \frac{\partial}{\partial \mathbf q}\frac{\partial H}{\partial \mathbf p} - \frac{\partial}{\partial \mathbf p}\frac{\partial H}{\partial \mathbf q}=0$ (1)
For conservative systems in energy, $H(\mathbf y,t)$ does not depend upon time explicitely and the left-most term in (1) vanishes. Liouville's theorem can be generalized to any physical observable depending upon the phase of the system $A(\mathbf y,t)$ which is a conserved along the curves of the phase space,
$\frac{dA}{dt}= \frac{\partial A}{\partial t} + \frac{\partial A}{\partial \mathbf q}\frac{\partial H}{\partial \mathbf p} - \frac{\partial A}{\partial \mathbf p}\frac{\partial H}{\partial \mathbf q}=0$
The Wronskian
Now consider the situation of Euler's ODE can be linearized around phase $\mathbf y_0$. The vectorial function $\mathbf f$ now expresses as a matrix vector product with the phase. You have now a $2s\times 2s$ square linear system of ODEs.
$\frac{d\mathbf y}{dt} = {\mathbf f}(\mathbf y_0,0)+M_{\mathbf y_0}(t)(\mathbf y-\mathbf y_0)+\ldots$
with $M=\frac{\partial f}{\partial \mathbf y_0}$.
One can solve a new system of the form, $\mathbf y'=M_{\mathbf y_0}(t) \mathbf y$ for any translated phase around $\mathbf y_0$.
Consider $2s$ phase solutions of this system $(\mathbf y^1, \mathbf y^2,\ldots, \mathbf y^{2s})$. Then the wronskian
$W=\det(\mathbf y^1, \mathbf y^2,\ldots, \mathbf y^{2s})$ satisfies the first-order ODE,
$\frac{d}{dt}W= \mathrm{tr}(M_{y_0}) W$
and can be integrated as
$W(t) = W_0\exp \int _0^t\mathrm{tr}(M_{\mathbf y_0}(s)) ds$
Hope this helps.
A fundamental set of solutions to a differential equation is the basis of the solution space of the differential equation. Put in another way, every solution to a differential equation can be written as a linear combination of these fundamental solutions.
Secondly, the Wronskian being non-zero at a point $t$, tells us that the two solutions are linearly independent at that point. For two solutions to be the part of the basis for a solution space, we require them to be linearly independent.
Lastly, since the differential equation you are working with is of second order, the fundamental solution set consists of two linearly independent solutions. These two linearly independent solutions span the solution space (and hence form the basis for the solution space).
Best Answer
Here is the difference between the two concepts:
The Jacobian is an $m\times n$ matrix and it consists of first-order derivatives of all the variables of a given function $f$. The Jacobian matrix is an $m\times n$ matrix that gives the best linear approximation of $f$ near the point $x\in \mathbb{R}^n$. If we have a square matrix, then $f:\mathbb{R}^n\to\mathbb{R}^n$ and the Jacobian tells us that $f$ is invertible if the Jacobian at a point is non-zero.
On the other hand, the Wronskian is used to show that a set of solutions are linearly independent, as long as the Wronskian does not vanish. For functions $f_1,...,f_n$ the Wronskian is the determinant of an $n\times n$ matrix defined on an interval $x\in I$. Unlike the Jacobian, it includes higher derivatives than the first derivatives (here, we must have $(n-1)^{\text{th}}$ derivatives).
*See Wikipedia for more details.