Example of special fluid-Method of characteristics

[mathmari]

Hey!

I have to solve the following system using the method of the characteristics.
$$u_t+uu_x+p_x=0$$
$$p_t+up_x+pu_x=0$$

I have done the following:

$$A=\begin{bmatrix}
u & 1\\
p & u
\end{bmatrix}$$

The eigenvalues are: $\lambda=u \pm \sqrt{p}$

- $\lambda=u+\sqrt{p}:$

$$\begin{bmatrix}
\gamma_1 & \gamma_2
\end{bmatrix}\begin{bmatrix}
-\sqrt{p} & 1\\
p & -\sqrt{p}
\end{bmatrix}=\begin{bmatrix}
0 & 0
\end{bmatrix}$$

We set $\gamma_1=1$, so $\gamma_2=\frac{1}{\sqrt{p}}$.

$$\gamma_1(u_t+uu_x+p_x) + \gamma_2(p_t+up_x+pu_x)=0$$

$$(u_t+uu_x+p_x) + \frac{1}{\sqrt{p}}(p_t+up_x+pu_x)=0$$

$$(u_t+\frac{1}{\sqrt{p}}p_t)+(uu_x+p_x+\frac{1}{\sqrt{p}}up_x+\sqrt{p}u_x)=0$$

$$(u_t+\frac{1}{\sqrt{p}}p_t)+(u+\sqrt{p})(u_x+\frac{1}{\sqrt{p}}p_x)=0$$

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

We set $\displaystyle{r_+=u+2 \sqrt{p}}$ and $\displaystyle{v_+=u+\sqrt{p}}$, so we get:

$$\frac{\partial{r_+}}{\partial{t}}+v_+ \frac{\partial{r_+}}{\partial{x}}=0$$

How can I continue?? (Wondering)

 

[jvicens]

Hello! It looks like you are on the right track with your calculations. The next step would be to find the characteristic curves for the system, which can be done by setting $r_+$ and $v_+$ as constants and solving for $x$ and $t$. Once you have the characteristic curves, you can use the method of characteristics to solve for the solution of the system. I would also suggest checking your calculations to make sure they are correct. Good luck!