How does the expression for p affect the PDE?

[andrey21]

A model for single land of traffic is given below:

p.dv/dx + v. dp/dx + dp/dt = 0

Where v = kx/p

Show that with the expression for p, the PDE becomes:

dp/dt = -k



Here is my attempt

v = kx/p

dv/dx = k/p

Sub into pde:

p (k/p) + (kx/p) .dp/dx + dp/dt = 0

k + (kx/p). dp/dx + dp/dt = 0

This is how far I can get, any help would be great.

 

[Mark44]

A model for single land of traffic is given below:

p.dv/dx + v. dp/dx + dp/dt = 0

Where v = kx/p

Show that with the expression for p, the PDE becomes:

dp/dt = -k



Here is my attempt

v = kx/p

dv/dx = k/p

Sub into pde:

p (k/p) + (kx/p) .dp/dx + dp/dt = 0

k + (kx/p). dp/dx + dp/dt = 0

This is how far I can get, any help would be great.

The equation v = kx/p gives v as a function of x and p. If x and p are independent of each other, then x is not a function of p, and p is not a function of x.

If we make the assumption that p and x are independent, then
$$\frac{\partial p}{\partial x} = 0$$

so your last equation reduces to
$$k + \frac{\partial p}{\partial t} = 0$$

or
$$\frac{\partial p}{\partial t} = -k$$

 

[andrey21]

Thank you Mark 44, as a follow up I am asked to establish the characteristics and what happens to traffic density along a characteristic?

I'm assuming I have to use the following:

dx/a = dt/b = du/c

Am I on the right track?

 

[Mark44]

Thank you Mark 44, as a follow up I am asked to establish the characteristics and what happens to traffic density along a characteristic?

I'm assuming I have to use the following:

dx/a = dt/b = du/c

Am I on the right track?

I don't know. How are u, a, b, and c related to the original problem? Also, refresh my memory as to what a characteristic is.

 

[andrey21]

Well given that I have established:

dp/dt = -k

the original pde can be written as:

K + 0 - k = 0

Therefore:

a = k b = 0 c = -k

 

[Mark44]

Well given that I have established:

dp/dt = -k

the original pde can be written as:

K + 0 - k = 0

Therefore:

a = k b = 0 c = -k

Your derivatives are throwing me off.

What we found was $$\frac{\partial p}{\partial t} = -k$$
Click the equation to see how I wrote it in LaTeX.

It looks to me like this:
K + 0 - k = 0

should be this:
k + 0 - k = 0

I still have no idea how a, b, and c (and u) tie into things.