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On p. 10 of Stein and Shakarchi's Fourier Analysis book, they consider a standing wave $u(x, t)$ and make the change of variables $\xi = x + t, \eta = x - t$, and define $v(\xi, \eta) = u(x, t)$. The change of variables formula shows that

$\displaystyle\frac{\partial^2 v}{\partial \xi \partial \eta} = 0$

They then claim that "integrating this relation twice gives $v(\xi, \eta) = F(\xi) + G(\eta)$". I don't see why this last claim is true: couldn't we for example have $v(\xi, \eta) = \xi \eta$ which satisfies the same relation?

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It doesn't. It satisfies $\frac{\partial^2 v}{\partial \xi \partial \eta} = 1$. Do you understand what is meant by "integrating this relation twice"? You integrate with respect to one variable (which gives a constant of integration depending on the other variable), then the other. –  Qiaochu Yuan Feb 28 '12 at 4:15
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It does not satisfy the same relation, because

$$\frac{\partial^2 (\xi\eta)}{\partial \xi \partial \eta}=1.$$

Integrating the relation wrt $\xi$ gives

$$\frac{\partial v}{\partial \eta}=g$$

where $g$ is constant wrt $\xi$ but not necessarily $\eta$, so we write $g=g(\eta)$. Now integrate wrt $\eta$ for

$$v = F(\xi)+G(\eta) $$

where $G$ is the antiderivative of $g$ and $F=F(\xi)$ is a constant wrt $\eta$ but not necessarily $\xi$.

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