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Show that the wave equation $\partial_x^2 u - \partial_y^2 u = 0$ can be written

  1. in factored form: $(\partial_x - \partial_y)(\partial_x + \partial_y) = 0$
  2. as a first-order system $u_x = v_y, u_y = v_x$
  3. as $u_{st} = 0$ under the change of variables $s = x + y$ and $t = x - y$. Use the form (c) to determine the general solution.

Attempt:

  1. $\partial_x^2 u - \partial_y^2 u = \partial_x u_x - \partial_y u_y + \partial_x \dot \partial_y - \partial_y \partial_x = (\partial_x - \partial_y)(\partial_x + \partial_y)$. Can I assume smoothness to change the order of differentiation?
  2. I am not certain how to use the provided hint: $u_x = v_y, u_y = v_x$.
  3. Same as number 2, I am completely lost.
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    $\partial^2_x u$ is not the same as $\partial_x u \partial_xu$. – 2017-01-30
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    @JohnMa I apologize. I should have written $\partial_x u_x$. Is the argument logical if the symbols are edited? – 2017-01-30

1 Answers 1

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For the second part, they have used a confusing notation, because the $u$ in the desired form is not the same as the $u$ in the original equation. I will show how to get the coupled equations $w_x = v_y, w_y = v_x$.

Let $v(x,y) = \frac{\partial u(x,y)}{\partial x}$ and let $w(x,y) = \frac{\partial u(x,y)}{\partial y}$. Then

$$w_x = \frac{\partial^2 u(x,y)}{\partial x\partial y}\\ v_y = \frac{\partial^2 u(x,y)}{\partial y\partial x} $$ and making a smoothness assumption, you can change the order of partial differentiation, so these two expressions are equal: $w_x = v_y$. And

$$w_y = \frac{\partial^2 u(x,y)}{\partial y^2}\\ v_x = \frac{\partial^2 u(x,y)}{\partial x^2} \\ v_x - w_y = \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = 0 $$ For the third part, $x=\frac{s+t}{2}$ and $y=\frac{s-t}{2}$. Then $$ \frac{\partial u}{\partial x} = \frac12 \left( \frac{\partial u}{\partial s} + \frac{\partial u}{\partial t } \right) \\ \frac{\partial u}{\partial y} = \frac12 \left( \frac{\partial u}{\partial s} - \frac{\partial u}{\partial t } \right) $$ whence

$$ \frac{\partial^2 u}{\partial x^2} = \frac14 \left( \frac{\partial^2 u}{\partial s^2} + 2\frac{\partial^ u}{\partial s \partial t } + \frac{\partial^2 u}{\partial t^2} \right)\\ \frac{\partial^2 u}{\partial y^2} = \frac14 \left( \frac{\partial^2 u}{\partial s^2} - 2\frac{\partial^ u}{\partial t \partial s } + \frac{\partial^2 u}{\partial t^2} \right)\\ u_{xx}-u_{yy} = \frac14 \left(4\frac{\partial^ u}{\partial s \partial t } \right) = u_{st} $$

The solution to $u_{st}=0$ is $u(s,t)=f(s)+g(t)$, for arbitrary differentiable functions $f$ and $g$. This translates to a solution to the original equation $$ u(x,y) = f(x+y) + g(x-y)$$

Try it out on some particular cases, like $u(x,y) = (x+y)^3 - 2(x-y)^2$ and you will see ilt alwyas satisfies the original equation.

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    Thank you very much, @MarkFischler. Can you verify that my first part actually makes sense. – 2017-01-30
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    Yes it does. Although some smoothness conditions are necessary to justify reversing the order of differentiation, whenever the mixed second derivative exists and is finite and continuous, I believe you can do it inn either order. – 2017-01-30