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Find a solution of the PDE $u_{tt} - c^2 u_{xx} = 0$, (where $c$ is a constant) in the half plane $t > 0$ with initial conditions $u(x, 0) = g_0(x)$ and $u_t(x, 0) = g_1(x)$.

$$\therefore \mathcal{L} = \partial^2_t - c^2 \partial^2_x = (\partial_t + c \partial_x)(\partial_t - c \partial_x)$$ (Linear/Differential Operator.)

Apparently, this factorisation suggests the change of variables

$$y = x - ct, s = x + ct$$

How/Why does factoring the linear/differential operator suggest a specific change of variables? I see the obvious similarity between the change of variables above and the linear operator of the wave equation, but, speaking in terms of the general concept/theory, this was not explained before this example.

Thank you.

Postscript: Please note that this is within the context of a section on characteristics of second-order PDEs (method of characteristics for Second-Order PDEs).

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    $\begingroup$ Could it be that it should $y = ct - x$ ? $\endgroup$
    – denklo
    Aug 10, 2018 at 12:06
  • $\begingroup$ @denklo Yes! Thank you for spotting that. :) $\endgroup$
    – user575678
    Aug 10, 2018 at 12:07

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From your comments it seems like you're more interested in some intuition about why to make that change of variables aside from the similarity of the two? Forgive me if I have misinterpreted. Also the answer of denklo is the more complete one, as I'm really just saying in words what that last eigensystem is saying.

The most intuitive way I grasp it is from a physics perspective: that factorization suggests a type of symmetry, that what happens going leftwards is the same as that going rightwards.

Changing to $x \pm c t$ is to follow the information packets traveling left and rightwards with speed $c$. In physical terms, it's a change of frame of reference to that of the waveform, which because the system has no diffusion or dispersion, must retain its structure.

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    $\begingroup$ Well, I do want intuition, but that intuition can also be gained from a mathematical explanation. The problem is that whatever mathematical explanation is provided needs to have sufficient explanation to make it understandable to a person who doesn't already understand the concept. $\endgroup$
    – user575678
    Aug 10, 2018 at 14:40
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Just give it a try: $$ (\partial_t+c \partial_x)y = 0 = (\partial_t -c \partial_x)s $$

$$ (\partial_t - c \partial_x)y = -2c = -(\partial_t + c \partial_x)s $$ which means $$ (\partial_t - c \partial_x) = -2c \partial_y \text{ and } (\partial_t + c \partial_x) = 2c \partial_s. $$ This will strongly simplify your differential equation: $$ \mathcal{L} = -4 c^2 \partial_y\partial_s $$ which means g only depends on s or y, if $\mathcal{L}g = 0$.

In general, having $$ \mathcal{L} = (a_{11}\partial_x + a_{12}\partial_t)(a_{21}\partial_x+a_{22}\partial_t) $$ by the ansatz $$ y = b_{11}x + b_{21}t \text{ and } s = b_{12}x + b_{22}t $$ you set $$ (a_{11}\partial_x + a_{12}\partial_t)y = 1 $$ $$ (a_{21}\partial_x + a_{22}\partial_t)y = 0 $$ $$ (a_{11}\partial_x + a_{12}\partial_t)s = 0 $$ $$ (a_{21}\partial_x + a_{22}\partial_t)s = 1 $$ This is equivalent to $$ \left(\matrix{a_{11} && a_{12}\\a_{21}&&a_{22}}\right) \left(\matrix{b_{11}\\b_{21}} \right)= \left(\matrix{1\\0} \right) $$ and $$ \left(\matrix{a_{11} && a_{12}\\a_{21}&&a_{22}}\right) \left(\matrix{b_{12}\\b_{22}} \right)= \left(\matrix{0\\1} \right) $$ In other words $(b_{ij})_{ij} = (a_{ij})^{-1}_{ij}$.

Then the transform $y = b_{11}x + b_{21}t$ and $s = b_{12}x + b_{22}t$ will simplify your eqation as above.

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  • $\begingroup$ Give what a try? I don’t understand what your answer is saying. $\endgroup$
    – user575678
    Aug 10, 2018 at 12:43
  • $\begingroup$ I'm just suggesting to you to calculate $(\partial_t+ c \partial_x)y$ and the others, to convince youself of $ \partial_t + c\partial_x \propto \partial_y$... $\endgroup$
    – denklo
    Aug 10, 2018 at 12:53
  • $\begingroup$ Hmm, but does that address my question? $\endgroup$
    – user575678
    Aug 10, 2018 at 12:57
  • $\begingroup$ "How/Why does factoring the linear/differential operator suggest a specific change of variables?" : because now you have a new, orthogonal coordinate system wich trivially solves your equation, as i tried to explain in my response. $\endgroup$
    – denklo
    Aug 10, 2018 at 13:25
  • $\begingroup$ I extended my response, does it fit your question better now? $\endgroup$
    – denklo
    Aug 10, 2018 at 13:46

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