Post by matthorr on Nov 17, 2015 13:21:50 GMT
I had some difficulty finding the right \( \hat{A} \) that was also messing up 8.3
Here is how I figured it out:
Finding \(S\) and \(S^{-1}\) isn't too bad. Reverse engineer \( \hat{A} = S^{-1}AS \) and write out the terms.
Expand the momentum equation terms by chain rule (cancel only the \(\partial{c^2}\) term) and separate them into terms you want in \( \hat{A} \) (from the step above) and the "rest" of the terms ( split \(2u\frac{\partial{u}}{\partial{x}}\) into two terms). Then, take the mass continuity equation and multiply it by \(\frac{u}{\rho}\). Substitute this into for the \( \frac{u\partial{\rho}}{\rho\partial{t}} \) term and all of the extra terms go away.
\( \hat{A} = u \quad\quad \rho\)
\(\frac{c^2}{\rho} \quad\quad u \)
Here is how I figured it out:
Finding \(S\) and \(S^{-1}\) isn't too bad. Reverse engineer \( \hat{A} = S^{-1}AS \) and write out the terms.
Expand the momentum equation terms by chain rule (cancel only the \(\partial{c^2}\) term) and separate them into terms you want in \( \hat{A} \) (from the step above) and the "rest" of the terms ( split \(2u\frac{\partial{u}}{\partial{x}}\) into two terms). Then, take the mass continuity equation and multiply it by \(\frac{u}{\rho}\). Substitute this into for the \( \frac{u\partial{\rho}}{\rho\partial{t}} \) term and all of the extra terms go away.
\( \hat{A} = u \quad\quad \rho\)
\(\frac{c^2}{\rho} \quad\quad u \)