3
$\begingroup$

I'm reading about solving the wave equation and I found the method a little dodgy.

$$L(u)=u_{tt} -c^2 u_{xx} = 0$$

They then 'factorised' this as you would the difference of two squares, to obtain

$$L(u) = \left ( \frac{\partial}{\partial t} - c \frac{\partial}{\partial x} \right )\left ( \frac{\partial}{\partial t} + c \frac{\partial}{\partial x} \right )u$$

Then the individual brackets are equated to zero to solve, as two separate transport equations.

But to me the step of 'factoring' an operator and 'letting it equal zero' seems totally unjustified and arbitrary. Is there a more careful way of explaining this, or was this done intentionally because the justification is too difficult for this level (learning how to solve various PDE's)?

$\endgroup$

1 Answer 1

4
$\begingroup$

it is not factoring like a multiply but like a composition of derivatives. Just develop it and see ! :-)

$\endgroup$
1
  • 2
    $\begingroup$ (note that in Fourier domain it would really be a multiply, for Fourier transforms differential equations into polynomials). $\endgroup$ Nov 1, 2015 at 12:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .