Simplify the follwoing Equation

[andrey21]

Sounds waves in a pipe of varying cross-section are described by the equation

V² d/dx (1/A dAu/dx) = d²u/dt²

Where A = 0.2+0.3x

So first I substituted A into the equation:

V² d/dx (1/(0.2+0.3x) d(0.2+0.3x)u/dx) = d²u/dt²

V² d/dx (0.3u/(0.2+0.3x) du/dx) = d²u/dt²This is as far as I can get, any help would be fantastic.

 

[grey_earl]

I suppose this is V² d/dx (1/A d(Au)/dx) = d²u/dt², true? Then expand first before substituting, which gives
d²u/dx² + d(ln A)/dx du/dx + d²(ln A)/dx² u = 1/V² d²u/dt².
Now substitute to get
d²u/dx² + 1/(x + 2/3) du/dx - 1/(x + 2/3)² u = 1/V² d²u/dt², which I think is about as simple as it gets.

(PS: This has a solution u(x,t) = [ C1 x² (1 + x)/(2 + 3 x) + C2 (1 + (2 + 3 x)²)/(2 + 3 x) + C3 (
(1 - (2 + 3 x)²))/(2 + 3 t) ] [ 1/2 C1 t² V² + C4 + t C5 ], where C1 to C5 are arbitrary constants. But that may not be the solution you are looking for.)

 

[hunt_mat]

So the equation you have is:
$$v^{2}\frac{\partial}{\partial x}\frac{1}{A(x)}\frac{\partial}{\partial x}(A(x)u)=\frac{\partial^{2}u}{\partial t^{2}}$$
Use:
$$\frac{\partial}{\partial x}(Au)=A\frac{\partial u}{\partial x}+0.3u$$
Likewise for the 1/A term too.

 

[andrey21]

Thank you for the replies, hunt_mat could u please explain your post a little more please I am confused

 

[hunt_mat]

I will explain the second point:
$$\frac{\partial}{\partial x}(A(x)u)=A\frac{\partial u}{\partial x}+u\frac{\partial A}{\partial x}=A\frac{\partial u}{\partial x}+0.3u$$

 

[andrey21]

Ok using what you have said I have obtained:

V² d/dx (0.3u/0.2+0.3x + du/dx)

V² d/dx (0.15u+ u/x + du/dx)

Is this correct??

 

[hunt_mat]

You have to use the quotient rule which is:
$$\frac{d}{dx}\left(\frac{X}{A}\right) =\frac{A\frac{dX}{dx}-X\frac{dA}{dx}}{A^{2}}$$
Where:
$$X=\frac{\partial }{\partial x}(A(x)u))$$
So, no. Your answer isn't correct.

 

[andrey21]

Ok so I should use the quotient rule on:

0.3u/0.2+0.3x

(0.2+0.3x) (0.3) - 0.3u (0.3) / (0.2+0.3x)²

0.6 +0.9x -0.9u / (0.2+0.3x)²

am I on the right track??

 

[hunt_mat]

Not quite, you should have:
(0.3du/dx-(0.3)^2u)/(0.2+0.3x)2

 

[andrey21]

So the answer is:

V² d/dx ((0.3du/dx-(0.3)²u)/(0.2+0.3x)² + du/dx)