MHB How do we conclude to the last relation?

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mathmari
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Hey! :o

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)
 
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mathmari said:
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)
 
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 like Serena said:
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)
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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