We can combine jump conditions for mass flux
- \rho (v_x - \lambda_f) = \rho^*(v_x^* - \lambda_f)
- \rho^* (v_x^* - \lambda_s) = \rho^{**}(v_x^{**} - \lambda_s)
along with the symmetry requirements (v_x^{**} = v_y^{**} = 0) to get
- \rho(v_x - \lambda_f) = \rho^*(\lambda_s - \lambda_f)-\rho^{**} \lambda_s
as well as jump conditions for y-momentum
- \rho(v_x - \lambda_f) v_y - B_xB_y = \rho^*(v_x^* - \lambda_f) v_y^* - B_x B_y^*
- \rho^*(v_x^* - \lambda_s) v_y^* - B_xB_y^* = -B_x B_y^{**}
to get
- \rho(v_x - \lambda_f) v_y - B_xB_y = \rho^*v_y^*(\lambda_s - \lambda_f)-B_x B_y^{**}
and the jump conditions for magnetic flux
- B_y (v_x - \lambda_f) - B_x v_y = B_y^* (v_x^* - \lambda_f) - B_x v_y*
- B_y^*(v_x^* - \lambda_s) - B_x v_y* = -B_y^{**} \lambda_s
to get
- B_y (v_x - \lambda_f) - B_x v_y = B_y^*(\lambda_s - \lambda_f) - B_y^{**} \lambda_s
This gives us
- \rho(v_x - \lambda_f) = \rho^*(\lambda_s - \lambda_f)-\rho^{**} \lambda_s
- \rho(v_x - \lambda_f) v_y - B_xB_y = \rho^*v_y^*(\lambda_s - \lambda_f)-B_x B_y^{**}
- B_y (v_x - \lambda_f) - B_x v_y = B_y^*(\lambda_s - \lambda_f) - B_y^{**} \lambda_s
Now in our setup, the flow and the field are parallel, so we have B_y v_x = B_x v_y. Combining this with the third equation gives us
- B_y^{**} \lambda_s + B_y^* ( \lambda_f - \lambda_s ) = B_y \lambda_f
Which is evident from the geometry of the problem.
So we have
- \rho(v_x - \lambda_f) = \rho^*(\lambda_s - \lambda_f)-\rho^{**} \lambda_s
- \rho(v_x - \lambda_f) v_y - B_xB_y = \rho^*v_y^*(\lambda_s - \lambda_f)-B_x B_y^{**}
- B_y^{**} \lambda_s + B_y^* ( \lambda_f - \lambda_s ) = B_y \lambda_f
- -B_y \lambda_f = B_y^* (v_x^* - \lambda_f) - B_x v_y*
Now we can combine the 1st and 3rd equations and the 2nd and 4th equations to get
- \rho^*(v_x^* - \lambda_f) (v_y - v_y^*) = B_x(B_y - B_y^*)
- - \rho^{**}\lambda_s v_y^* = B_x ( B_y^* -B_y^{**})
So we have the following six equations for 8 unknowns (including the 2 wave speeds)
- \rho(v_x - \lambda_f) = \rho^*(\lambda_s - \lambda_f)-\rho^{**} \lambda_s
- \rho(v_x - \lambda_f) v_y - B_xB_y = \rho^*v_y^*(\lambda_s - \lambda_f)-B_x B_y^{**}
- B_y^{**} \lambda_s + B_y^* ( \lambda_f - \lambda_s ) = B_y \lambda_f
- -B_y \lambda_f = B_y^* (v_x^* - \lambda_f) - B_x v_y*
- \rho^*(v_x^* - \lambda_f) (v_y - v_y^*) = B_x(B_y - B_y^*)
- - \rho^{**}\lambda_s v_y^* = B_x ( B_y^* -B_y^{**})
Because of symmetry, the streamline coming in from the right, must remained pinned to the point along the contact. And given that Bx is a constant, we have
- B_y^{**} = B_x \tan \theta^{**}
- B_y^{*} = B_x \tan \theta^{*}
- B_y = B_x \tan \theta
and that
\tan \theta^{**} \lambda_s + \tan \theta^{*} \left ( \lambda_f - \lambda_s \right) = \lambda_f \tan \theta
We also have
- v_y^{**} = v_x^{**}=0
- v_y^* = v_x^* \tan \theta_v^* = v_x^* \tan \theta^* + \lambda_s \tan \theta^{**} + \lambda_s \tan \theta^*
v_y^* = v_x^*\tan \theta^* + \lambda_f \left ( \tan \theta - \tan \theta^* \right ) + 2 \lambda_s \tan \theta^*
- v_y = v_x \tan \theta
- v_x = c_s M \cos \theta
- B_x = \sqrt{\rho} v_A \tan \theta
Ideally, our givens would be
- M
- \mu = \frac{1}{\beta}
- \theta
And we can set (without loss of generality)
and we would need to solve for
- \rho^*
- \rho^{**}
- P^{*}
- P^{**}
- v_x^{*}
- \lambda_f
- \lambda_s
- \theta^*
Conservation of mass gives
\rho (v_x - \lambda_f) = \rho^* \left ( v_x^* - \lambda_f \right )
and conservation of momentum gives
- \rho (v_x - \lambda_f)^2 + P+B_x^2 \left ( \tan^2 \theta - 1 \right ) = \rho^* (v_x^* - \lambda_f)^2 + P^* + B_x^2 \left ( \tan^2 \theta^* - 1 \right )
- \rho (v_x - \lambda_f) v_x \tan \theta - B_x^2 \tan \theta= \rho^*(v_x^* - \lambda_f) \left (v_x^* \tan \theta + \lambda_f \left ( \tan \theta - \tan \theta^* \right ) + 2 \lambda_s \tan \theta^* \right )
and conservation of magnetic flux gives
- B_x \tan \theta \left (v_x - \lambda_f \right ) - B_x v_x \tan \theta = B_x \tan \theta^* \left (v_x^* - \lambda_f \right ) - B_x \left ( v_x^*\tan \theta^* + \lambda_f \left ( \tan \theta - \tan \theta^* \right ) + 2 \lambda_s \tan \theta^* \right )