How do we conclude to the last relation?

[mathmari]

Hey!

I am looking at an example of the characteristic system of hyperbolic equations.

One part of the example is the following:

$\displaystyle{v=\text{ constant }, v=u_1+\sqrt{\frac{a}{b}}u_2}$, when $\displaystyle{\frac{dx}{dt}=\sqrt{ab}}$

$\displaystyle{x=\sqrt{ab}t+c \Rightarrow c=x- \sqrt{ab}t}$

$\displaystyle{v}$ is constant when $\displaystyle{x-\sqrt{ab}t}$ is constant.

That means that $$u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}$$Could you explain me how we conclude to the last relation?? (Wondering)

 

[mathmari]

One part of the example is the following:

$\displaystyle{v=\text{ constant }, v=u_1+\sqrt{\frac{a}{b}}u_2}$, when $\displaystyle{\frac{dx}{dt}=\sqrt{ab}}$

$\displaystyle{x=\sqrt{ab}t+c \Rightarrow c=x- \sqrt{ab}t}$

$\displaystyle{v}$ is constant when $\displaystyle{x-\sqrt{ab}t}$ is constant.

That means that $$u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}$$

At the beginning we had set $\displaystyle{v=u_1+\sqrt{\frac{a}{b}}u_2}$ , then we got (by replacing in an other equation):
$$v_t +\sqrt{ab}v_x =0$$

Then we know that $v$ is constant when $\displaystyle{\frac{dx}{dt} =\sqrt{ab}}$ , that means when $\displaystyle{c=x−\sqrt{ab} t }$, so when $\displaystyle{x-\sqrt{ab}t}$ is constant.

$$v: \text{ constant } \Rightarrow v=C ′ \Rightarrow u_1+\sqrt{\frac{a}{b}}u_2=C ′$$

So do we consider that this $C ′$ is equal to $\displaystyle{\frac{f(x-\sqrt{ab}t)}{2}}$ ??

Do we suppose that there is a function $f$ such that at the point $(x-\sqrt{ab}t)$ it's equal to $2C ′ $ ??

Or is there an other way how we conclude that $\displaystyle{u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}}$?? (Wondering)

 

[I like Serena]

Since no one has answered yet... can I ask for clarification? (Blush)

What is the actual problem statement?

And how are all these symbols defined? (Wasntme)

 

[mathmari]

Since no one has answered yet... can I ask for clarification? (Blush)

What is the actual problem statement?

And how are all these symbols defined? (Wasntme)

I have posted the whole example at the thread http://mathhelpboards.com/differential-equations-17/need-explanation-hyperbolic-system-equations-10732.html

(post #5)

(Cool)

 

[blue_raver22]

To conclude to the last relation, we can follow the steps given in the example. First, we have the equation $\displaystyle{v=u_1+\sqrt{\frac{a}{b}}u_2}$ and the condition $\displaystyle{\frac{dx}{dt}=\sqrt{ab}}$. This means that $\displaystyle{v}$ must be constant when $\displaystyle{x-\sqrt{ab}t}$ is constant.

Next, we can solve for $\displaystyle{x}$ in terms of $\displaystyle{t}$ by taking the integral of $\displaystyle{\frac{dx}{dt}=\sqrt{ab}}$ with respect to $\displaystyle{t}$. This gives us the equation $\displaystyle{x=\sqrt{ab}t+c}$, where $\displaystyle{c}$ is a constant of integration.

Substituting this expression for $\displaystyle{x}$ into the original equation for $\displaystyle{v}$, we get $\displaystyle{v=u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}}$, where $\displaystyle{f}$ is a function of $\displaystyle{x-\sqrt{ab}t}$.

Finally, we can rearrange this equation to get the last relation, $\displaystyle{u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}}$, which shows the relationship between $\displaystyle{u_1}$ and $\displaystyle{u_2}$ when $\displaystyle{x-\sqrt{ab}t}$ is constant. This conclusion is based on the given conditions and the mathematical operations used to solve for $\displaystyle{x}$ and substitute it into the original equation.