MHB Method of characteristics-How can I continue?

[mathmari]

Hey!

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

• $\lambda=1-xt:$

To find the eigenvector, we do the following:

$$\gamma^T(A-\lambda I)=0 \Rightarrow \begin{bmatrix}
\gamma_1 & \gamma_2
\end{bmatrix} \begin{bmatrix}
xt & x^2\\
t^2 & xt
\end{bmatrix}=\begin{bmatrix}
0 & 0
\end{bmatrix}$$

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 like Serena]

I cannot write it as followed:

$$\frac{\partial}{\partial{t}}(u_1-\frac{x}{t}u_2)+(1-xt) \frac{\partial}{\partial{x}}(u_1-\frac{x}{t}u_2)=0$$

right??

So I cannot write it as a total derivative of something...

How can I continue then??

Hi! (Wasntme)

You can write it like that only if $\frac x t$ is constant.
So I think that is what you need:
$$x = C t$$

 

[mathmari]

Hi! (Wasntme)

You can write it like that only if $\frac x t$ is constant.
So I think that is what you need:
$$x = C t$$

Ahaa... (Thinking)

• $\lambda=1+xt$

$\displaystyle{\gamma_1=1, \gamma_2=\frac{x}{t}}$

$$\gamma_1(\frac{\partial{u_1}}{\partial{t}}+\frac{\partial{u_1}}{\partial{x}}+x^2 \frac{\partial{u_2}}{\partial{x}})+\gamma_2 ( \frac{\partial{u_2}}{\partial{t}}+t^2 \frac{\partial{u_1}}{\partial{x}}+\frac{\partial{u_2}}{\partial{x}} )=0$$$$(\frac{\partial{u_1}}{\partial{t}}+\frac{x}{t}\frac{\partial{u_2}}{\partial{t}})+(1+xt)(\frac{\partial{u_1}}{\partial{x}}+\frac{x}{t}\frac{\partial{u_2}}{\partial{x}})=0$$

$$(\frac{\partial{u_1}}{\partial{t}}+C\frac{\partial{u_2}}{\partial{t}})+(1+Ct)(\frac{\partial{u_1}}{\partial{x}}+C\frac{\partial{u_2}}{\partial{x}})=0$$

$$\frac{\partial}{\partial{t}}(u_1+Cu_2)+(1+Ct) \frac{\partial}{\partial{x}}(u_1+Cu_2)=0$$

We set $\displaystyle{v_+=u_1+Cu_2}$:

$$\frac{\partial}{\partial{t}}v_++(1+Ct) \frac{\partial}{\partial{x}}v_+=0$$

$$\frac{dv_+}{dt}=\frac{\partial{v_+}}{\partial{t}}+\frac{dx}{dt}\frac{\partial{v_+}}{\partial{t}}=0 \text{ when } \frac{dx}{dt}=1+Ct$$

To solve the equation $\displaystyle{\frac{dx}{dt}=1+Ct }$, do I consider $C$ as a constant?? Or do I have to replace it with $\displaystyle{\frac{x}{t}}$?? (Wondering)

 

[I like Serena]

$$(\frac{\partial{u_1}}{\partial{t}}+\frac{x}{t}\frac{\partial{u_2}}{\partial{t}})+(1+xt)(\frac{\partial{u_1}}{\partial{x}}+\frac{x}{t}\frac{\partial{u_2}}{\partial{x}})=0$$

$$(\frac{\partial{u_1}}{\partial{t}}+C\frac{\partial{u_2}}{\partial{t}})+(1+Ct)(\frac{\partial{u_1}}{\partial{x}}+C\frac{\partial{u_2}}{\partial{x}})=0$$

Wah?

To solve the equation $\displaystyle{\frac{dx}{dt}=1+Ct }$, do I consider $C$ as a constant?? Or do I have to replace it with $\displaystyle{\frac{x}{t}}$?? (Wondering)

Yep. Like a constant. (Mmm)

 

[mathmari]

Wah?

It should be as followed, shouldn't it?? (Blush) (Sweating)

$$(\frac{\partial{u_1}}{\partial{t}}+\frac{x}{t}\frac{\partial{u_2}}{\partial{t}})+(1+xt)(\frac{\partial{u_1}}{\partial{x}}+\frac{x}{t}\frac{\partial{u_2}}{\partial{x}})=0$$

$$(\frac{\partial{u_1}}{\partial{t}}+C\frac{\partial{u_2}}{\partial{t}})+(1+Ct^2)(\frac{\partial{u_1}}{\partial{x}}+C\frac{\partial{u_2}}{\partial{x}})=0$$

$$\frac{\partial}{\partial{t}}(u_1+Cu_2)+(1+Ct^2) \frac{\partial}{\partial{x}}(u_1+Cu_2)=0$$

We set $\displaystyle{v_+=u_1+Cu_2}$:

$$\frac{\partial}{\partial{t}}v_++(1+Ct^2) \frac{\partial}{\partial{x}}v_+=0$$

$$\frac{dv_+}{dt}=\frac{\partial{v_+}}{\partial{t}}+\frac{dx}{dt}\frac{\partial{v_+}}{\partial{t}}=0 \text{ when } \frac{dx}{dt}=1+Ct^2$$

$\displaystyle{\frac{dx}{dt}=1+Ct^2 \Rightarrow dx=(1+Ct^2)dt \Rightarrow x=t+\frac{C}{3}t^3+a \Rightarrow a=x-t-\frac{C}{3}t^3}$

So $v_+$ is constant when $x-t-\frac{C}{3}t^3$ is constant, so
$$v_+=f(x-t-\frac{C}{3}t^3)$$Or have I done something wrong?? (Wondering)