MHB Solving Transport Eq. for Level Curves: x=X(t)

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The discussion focuses on solving the transport equation ux(x,t) + c(x)ux(x,t) = 0 for level curves, specifically when the sound speed c(x) is not constant. The approach involves finding curves where the solution u remains constant by introducing a new variable r, leading to a system of ordinary differential equations (ODEs). The participant derived the solution x(t) = sinh-1(sinh(x0)e^(c(0)t)), but another participant suggests a simpler form, x = x0e^(c(0)t). The conversation emphasizes the importance of understanding how the initial input u0(x) deforms over time under varying sound speeds. Overall, the thread provides insights into the mathematical techniques for analyzing transport equations in non-constant media.
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I have this equation
ux(x,t) + c(x)ux(x,t) = 0 x>0

I want to obtain information on how the initial input uo(x)=u(x,o) would deform when the sound speed is not constant. c(x) is the sound speedi wanted to start this by finding a DE for the level curves x=X(t) so that i can solve in terms the initial point x0=X(0)(greater than or equal to) 0

but how can i solve for these level curves
 
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simo said:
I have this equation...

$\displaystyle u_{t}\ (x,t) + c(x)\ u_{x}\ (x,t) = 0,\ x>0$

I want to obtain information on how the initial input uo(x)=u(x,o) would deform when the sound speed is not constant. c(x) is the sound speedi wanted to start this by finding a DE for the level curves x=X(t) so that i can solve in terms the initial point x0=X(0)(greater than or equal to) 0

but how can i solve for these level curves

The standard approach for a PDE like...

$\displaystyle u_{t} + c(x,t)\ u_{x} = 0\ (1)$

... is to find curves along which u is constant. If we introduce a new variable r for which is $t=t(r)$ and $x=x(r)$ , then for chaining rule is...

$\displaystyle \frac{d u}{d r} = u_{t}\ \frac {d t}{d r} + u_{x}\ \frac{d x}{d r}\ (2)$

... and combining (1) and (2) we arrive to the ODE pair...

$\displaystyle \frac{d t}{d r} = 1,\ \frac{d x}{d r}= c(x,r)\ (3)$

In Your case c(*) is function of the x alone so that is...

$\displaystyle \int \frac{d x}{c(x)} = r + \gamma\ (4)$

... where $\gamma$ is an arbitrary constant...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
The standard approach for a PDE like...

$\displaystyle u_{t} + c(x,t)\ u_{x} = 0\ (1)$

... is to find curves along which u is constant. If we introduce a new variable r for which is $t=t(r)$ and $x=x(r)$ , then for chaining rule is...

$\displaystyle \frac{d u}{d r} = u_{t}\ \frac {d t}{d r} + u_{x}\ \frac{d x}{d r}\ (2)$

... and combining (1) and (2) we arrive to the ODE pair...

$\displaystyle \frac{d t}{d r} = 1,\ \frac{d x}{d r}= c(x,r)\ (3)$

In Your case c(*) is function of the x alone so that is...

$\displaystyle \int \frac{d x}{c(x)} = r + \gamma\ (4)$

... where $\gamma$ is an arbitrary constant...Kind regards

$\chi$ $\sigma$
I solved the equation and I had x(t) = sinh-1(sinhx0ec(0)t) is it possible to draw the characteristic curve of this function
 
simo said:
I solved the equation and I had x(t) = sinh-1(sinhx0ec(0)t)...

Ehm!... what's the reason why You don't write simply $\displaystyle x= x_{0}\ e^{\ c(0)\ t}$?...

Kind regards

$\chi$ $\sigma$
 

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