# Partial differential equation-wave equation(2)

• Sep 4th 2009, 09:54 AM
Steve Lam
Partial differential equation-wave equation(2)
Can you help me to solve equation....
((∂²u)/(∂x²))(x,t)=(1/(c²))((∂²u)/(∂x²))(x,t)
Use the equation above to determine the dimension of the constant c
But why ((∂²u)/(∂x²))=LL⁻²and ((∂²u)/(∂x²))=LT⁻²........
How can it evolution?
• Sep 4th 2009, 11:02 AM
Jester
Quote:

Originally Posted by Steve Lam
Can you help me to solve equation....
((∂²u)/(∂x²))(x,t)=(1/(c²))((∂²u)/(∂x²))(x,t)
Use the equation above to determine the dimension of the constant c
But why ((∂²u)/(∂x²))=LL⁻²and ((∂²u)/(∂x²))=LT⁻²........
How can it evolution?

The wave in equation in general is

$\displaystyle \frac{\partial ^2 u}{\partial t^2} = c^2 \frac{\partial ^2 u}{\partial x^2}$

If the dimensions of t is T and x L, the due to the equality and the presence of u on both sides, the dimensions of c^2 is

$\displaystyle c^2 = \frac{L^2}{T^2}$, the dimension of c is like m/s and often c is called the wave speed.

As for the solution, if you introduce new coordinates

$\displaystyle r = x + c \,t$ and $\displaystyle s = x - c \,t$, then the PDE transforms to

$\displaystyle \frac{\partial ^2 u}{\partial r \partial s} = 0$

which has as it's olution

$\displaystyle u = f(r) + g(s)$
or

$\displaystyle u = f(x+c \,t) + g(x-c \,t)$
• Sep 5th 2009, 09:24 AM
Steve Lam
But why ((∂²u)/(∂x²))=LL⁻².....
Thanks!
At the same time, why (∂²u)=L and (∂x²)=L^2
Thanks!(Talking)
• Sep 5th 2009, 11:18 PM
Coomast
Quote:

Originally Posted by Steve Lam
Can you help me to solve equation....
((∂²u)/(∂x²))(x,t)=(1/(c²))((∂²u)/(∂x²))(x,t)
Use the equation above to determine the dimension of the constant c
But why ((∂²u)/(∂x²))=LL⁻²and ((∂²u)/(∂x²))=LT⁻²........
How can it evolution?

You have written an error in your PDE. Anyway the correct equation is:
$\displaystyle \frac{\partial^2u}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2u}{\partial t^2}$
The first term can be written as:
$\displaystyle \frac{\partial^2u}{\partial x^2}=\frac{1}{\partial x}\left( \frac{\partial u}{\partial x} \right)$
The dimensions of these are now:
$\displaystyle \frac{1}{L}\left( \frac{V}{L} \right)$
In which I used V as the dimension of the dependent variable. This can be anything the PDE stands for, a pressure, a voltage, a current, ....
So using this on the complete PDE you get:
$\displaystyle \frac{1}{L}\left(\frac{V}{L}\right) =\frac{1}{C^2} \frac{1}{T}\left( \frac{V}{T} \right)$
C is here the dimension of c. From which we can extract the dimension of c, i.e. C as:
$\displaystyle C=\frac{L}{T}$

coomast
• Sep 6th 2009, 08:54 AM
Steve Lam
Thank you! I am so happy!(Giggle)