# Thread: Partial differential equation-wave equation(2)

1. ## 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?

2. 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

$
\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

$
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

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

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

which has as it's olution

$
u = f(r) + g(s)
$

or

$
u = f(x+c \,t) + g(x-c \,t)
$

3. But why ((∂²u)/(∂x²))=LL⁻².....
Thanks!
At the same time, why (∂²u)=L and (∂x²)=L^2
Thanks!

4. 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:
$\frac{\partial^2u}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2u}{\partial t^2}$
The first term can be written as:
$\frac{\partial^2u}{\partial x^2}=\frac{1}{\partial x}\left( \frac{\partial u}{\partial x} \right)$
The dimensions of these are now:
$\frac{1}{L}\left( \frac{V}{L} \right)$
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:
$\frac{1}{L}\left(\frac{V}{L}\right) =\frac{1}{C^2} \frac{1}{T}\left( \frac{V}{T} \right)$
C is here the dimension of c. From which we can extract the dimension of c, i.e. C as:
$C=\frac{L}{T}$

coomast

5. Thank you! I am so happy!