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

[simo1]

I have this equation
u_x(x,t) + c(x)u_x(x,t) = 0 x>0

I want to obtain information on how the initial input u_o(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

 

[chisigma]

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 u_o(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

$\chi$ $\sigma$

 

[simo1]

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(sinhx₀e^c(0)t) is it possible to draw the characteristic curve of this function

 

[chisigma]

I solved the equation and I had x(t) = sinh^-1(sinhx₀e^c(0)t)...

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

Kind regards

$\chi$ $\sigma$

 

[blue_raver22]

and how can i use the information to understand the deformation of the initial input uo(x) when the sound speed is not constant?

As a scientist, you are attempting to solve the transport equation for level curves using the given equation. The first step would be to find a differential equation for the level curves x=X(t). This can be done by taking the derivative of x=X(t) with respect to t, giving us dx/dt = dX/dt.

Next, we can substitute this into the given equation to obtain:

dX/dt * ux(X(t),t) + c(X(t)) * ux(X(t),t) = 0

We can then rearrange this equation to solve for ux(X(t),t):

ux(X(t),t) = -c(X(t)) * dX/dt

This equation gives us the value of ux at any point (X(t), t) on the level curve x=X(t). We can use this information to understand how the initial input uo(x) would deform when the sound speed is not constant.

To do this, we can consider a specific point x0=X(0) on the level curve. At this point, the initial input uo(x0) would be the value of uo at x0. As we move along the level curve, the sound speed c(x) and the derivative dX/dt will change, causing the value of ux(X(t),t) to change. This change in ux(X(t),t) will then affect the value of uo(x) at x0, resulting in a deformation of the initial input.

By solving the transport equation for level curves, we can obtain information on how the initial input uo(x) would deform when the sound speed is not constant. This can help us understand the behavior of the system and make predictions for future scenarios.