Richmarc5

Riemann Invariants

Riemann invariants are transformations made on a a system of quasi-linear first order partial differential equations(pdes) to make them more easily solvable. Riemann invariants are constant along the the characteristic curves of the partial differential equaitons where they obtain the name invariant. They were first obtained by Bernhard Riemann in his work on plane waves in gas dynamics(1858).

Mathematical Theory

Consider the set of Hyperbolic partial differential equations of the form

$l_{i}\left(A_{ij}{\frac {\partial u_{j}}{\partial x}}+a_{ij}{\frac {\partial u_{j}}{\partial x}}\right)+l_{j}b_{j}=0$ where $A_{ij}$ and $a_{ij}$ are the elements of the matrices A and a and where $l_{i}$ and $b_{i}$ are elements of vectors. It will be asked if it is possible to rewrite this equation to

$m_{j}\left(\beta {\frac {\partial u_{j}}{\partial x}}+\alpha {\frac {\partial u_{j}}{\partial x}}\right)+l_{j}b_{j}=0$

To do this curves wil be introduced in the (x,t) plane defined by the vector field $(\alpha ,\beta )$. The term in the brackets will be rewritten interms of a total derivative where x,t are parameterised as $x=X(\eta ),t=T(\eta )$

${\frac {\partial u_{j}}{\partial \eta }}=T'{\frac {\partial u_{j}}{\partial t}}+X'{\frac {\partial u_{j}}{\partial x}}$ comparing the last two equations we find $\alpha =X'(\eta ),\beta =T'(\eta )$

which can be now written in characteristic form $m_{j}{\frac {du_{j}}{\eta }}+l_{j}b_{j}$

where we must have the conditions $l_{j}A_{ij}=m_{j}T'$, $l_{i}a{ij}=m_{j}T'$ where $m_{j}$ ca be eliminated to give the necessary condition $l_{i}(A_{ij}X'-a_{ij}T')=0$ so for a nontrival solution is the determinate $|A_{ij}X'-a_{ij}T'|=0$

For Riemann invariants we are concerned with the case when the matrix A is an identity matrix to form

${\frac {\partial u_{i}}{\partial x}}+a_{ij}{\frac {\partial u_{j}}{\partial x}}=0$

notice this is homogeneous due to the vector b being zero. In cahracteristic form the system is

$l_{i}{\frac {du_{i}}{dt}}=0$ with ${\frac {dx}{dt}}=\lambda$

Where l the left eigenvector of the matrix A and the $\lambda 's$ the characteristic speeds are the eigenvalues of the matrix A which satisfy

$|A-\lambda \delta _{ij}|=0$

To simplify these characteristic equations we can make the transformations such that ${\frac {dr_{i}}{dt}}=l_{i}{\frac {du_{i}}{dt}}$

which form

$\mu l_{i}du_{i}=dr_{i}$

An integrating factor $\mu$ can be multiplied in to help integrate this. So the system now has the characteristic form

${\frac {dr_{i}}{dt}}$ on ${\frac {dx}{dt}}=\lambda _{i}$

which is equivalent to the diagonal system

$r_{t}^{k}+\lambda _{k}r_{x}^{k}=0,$ $k=1,...,N.$

The solution of this system can be given by the generalized hodograph method .,Tsarev 1985)

Example

Consider the shallow water equations

$\rho _{t}+\rho u_{x}+u\rho _{x}=0$

$u_{t}+uu_{x}+(c^{2}/\rho )\rho _{x}=0$

write this system in matrix form

$\left({\begin{matrix}\rho \\u\end{matrix}}\right)_{t}+\left({\begin{matrix}u&\rho \\{\frac {c^{2}}{\rho }}&u\end{matrix}}\right)\left({\begin{matrix}\rho \\u\end{matrix}}\right)_{x}=\left({\begin{matrix}0\\0\end{matrix}}\right)$

where the matrix a from the analysis above the eigenvalues and eigenvectors need to be found.The eigenvalues are found to satisfy

$\lambda ^{2}-2u\lambda +u^{2}-c^{2}=0$ to give

$\lambda =u\pm c$

and the eigenvectors are found to be $\left({\begin{matrix}1\\{\frac {c}{\rho }}\end{matrix}}\right),\left({\begin{matrix}1\\-{\frac {c}{\rho }}\end{matrix}}\right)$

where the riemann invariants are

$r_{1}=u+\int {\frac {c}{\rho }}d\rho ,$

$r_{2}=u-\int {\frac {c}{\rho }}d\rho ,$

In shallow water equations there is the relation $c={\sqrt {\rho }}$ to give the riemann invariants

$r_{1}=u-2{\sqrt {\rho }},$

$r_{1}=u+2{\sqrt {\rho }},$

to give the equations

${\frac {\partial r_{1}}{\partial t}}+(u+{\sqrt {\rho }}){\frac {\partial r_{1}}{\partial x}}=0$

${\frac {\partial r_{2}}{\partial t}}+(u-{\sqrt {\rho }}){\frac {\partial r_{2}}{\partial x}}=0$

Which can be solved by the hodograph transformation... If the matrix form of the system of pde's is in the form $A{\frac {\partial v}{\partial t}}+B{\frac {\partial v}{\partial x}}=0$

Then it may be possible to multiply across by the inverse matrix $A^{-1}$ so long as the matrix Determinant of A is not zero