- #1
mathmari
Gold Member
MHB
- 5,049
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I have to solve the following hyperbolic system of equations using the method of characteristics:
$$\left.\begin{matrix}
\frac{\partial{u_1}}{\partial{t}}+\frac{\partial{u_1}}{\partial{x}}+x^2 \frac{\partial{u_2}}{\partial{x}}=0\\
\frac{\partial{u_2}}{\partial{t}}+t^2 \frac{\partial{u_1}}{\partial{x}}+\frac{\partial{u_2}}{\partial{x}}=0
\end{matrix}\right\}$$
$ A=\begin{bmatrix}
1 & x^2\\
t^2 & 1
\end{bmatrix}$
$\displaystyle{u_t+Au_x=0}$
To check if it is hyperbolic, we have to find the eigenvalues:$\displaystyle{det(A-\lambda I)=(1-\lambda)^2-(xt)^2=0 \Rightarrow \lambda=1 \pm xt}$
$\gamma^T=\begin{bmatrix}
\gamma_1 & \gamma_2
\end{bmatrix}$: left eigenvectors
$$\gamma^T(A-\lambda I)=0 \Rightarrow \begin{bmatrix}
\gamma_1 & \gamma_2
\end{bmatrix} \begin{bmatrix}
xt & x^2\\
t^2 & xt
\end{bmatrix}=\begin{bmatrix}
0 & 0
\end{bmatrix}$$
We set $\gamma_1=1$, so $\gamma_2=-\frac{x}{t}$
We take the linear combination of the equations:
$$\gamma_1(\frac{\partial{u_1}}{\partial{t}}+\frac{\partial{u_1}}{\partial{x}}+x^2 \frac{\partial{u_2}}{\partial{x}})+\gamma_2 ( \frac{\partial{u_2}}{\partial{t}}+t^2 \frac{\partial{u_1}}{\partial{x}}+\frac{\partial{u_2}}{\partial{x}} )=0$$
$$(\frac{\partial{u_1}}{\partial{t}}-\frac{x}{t}\frac{\partial{u_2}}{\partial{t}})+((1-xt)\frac{\partial{u_1}}{\partial{x}}+(x^2-\frac{x}{t})\frac{\partial{u_2}}{\partial{x}})=0$$
I cannot write it as followed:
$$\frac{\partial}{\partial{t}}(u_1-\frac{x}{t}u_2)+(1-xt) \frac{\partial}{\partial{t}}(u_1-\frac{x}{t}u_2)=0$$
right??
So I cannot write it as a total derivative of something...
How can I continue then??
I have to solve the following hyperbolic system of equations using the method of characteristics:
$$\left.\begin{matrix}
\frac{\partial{u_1}}{\partial{t}}+\frac{\partial{u_1}}{\partial{x}}+x^2 \frac{\partial{u_2}}{\partial{x}}=0\\
\frac{\partial{u_2}}{\partial{t}}+t^2 \frac{\partial{u_1}}{\partial{x}}+\frac{\partial{u_2}}{\partial{x}}=0
\end{matrix}\right\}$$
$ A=\begin{bmatrix}
1 & x^2\\
t^2 & 1
\end{bmatrix}$
$\displaystyle{u_t+Au_x=0}$
To check if it is hyperbolic, we have to find the eigenvalues:$\displaystyle{det(A-\lambda I)=(1-\lambda)^2-(xt)^2=0 \Rightarrow \lambda=1 \pm xt}$
$\gamma^T=\begin{bmatrix}
\gamma_1 & \gamma_2
\end{bmatrix}$: left eigenvectors
- $\lambda=1-xt:$
$$\gamma^T(A-\lambda I)=0 \Rightarrow \begin{bmatrix}
\gamma_1 & \gamma_2
\end{bmatrix} \begin{bmatrix}
xt & x^2\\
t^2 & xt
\end{bmatrix}=\begin{bmatrix}
0 & 0
\end{bmatrix}$$
We set $\gamma_1=1$, so $\gamma_2=-\frac{x}{t}$
We take the linear combination of the equations:
$$\gamma_1(\frac{\partial{u_1}}{\partial{t}}+\frac{\partial{u_1}}{\partial{x}}+x^2 \frac{\partial{u_2}}{\partial{x}})+\gamma_2 ( \frac{\partial{u_2}}{\partial{t}}+t^2 \frac{\partial{u_1}}{\partial{x}}+\frac{\partial{u_2}}{\partial{x}} )=0$$
$$(\frac{\partial{u_1}}{\partial{t}}-\frac{x}{t}\frac{\partial{u_2}}{\partial{t}})+((1-xt)\frac{\partial{u_1}}{\partial{x}}+(x^2-\frac{x}{t})\frac{\partial{u_2}}{\partial{x}})=0$$
I cannot write it as followed:
$$\frac{\partial}{\partial{t}}(u_1-\frac{x}{t}u_2)+(1-xt) \frac{\partial}{\partial{t}}(u_1-\frac{x}{t}u_2)=0$$
right??
So I cannot write it as a total derivative of something...
How can I continue then??