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$u\in C^2$ is a solution of the one-dimensional wave equation $u_{tt}=u_{xx}$ with initial values $u(0,x)=f(x)$ and $u_t(0,x)=g(x)$

Now I a little bit confused with the following:

I define a function $v\in C^2(\Omega)$ where $\Omega=\{(a,b)\in\mathbb R^2 | a+b>0\}$ as $v(a,b)=u(a+b,a-b)$

Question 1: I do not see the relation that $u$ is a solution iff $v$ satifies $v_{ab}=0$

My thought was the following: I define $t=a+b, x=a-b$ then $v_{ab}=u_{tx}(t,x)=0$ and from this fact it should somehow be followed that $u_{tt}=u_{xx}$

Question 2: How can it be followed from above that every solution has the form $u(t,x)=F(t-x)+G(t+x)$. What exactly is $F$ and $G$ ?

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up vote 3 down vote accepted

This is based on D'Almbert's solution to the wave equation:

$$u(x,t) = F(x+t) + G(x -t)$$

It follows by substitution into the original equation. You may also consider the transformation $\alpha = x+t$, $\beta=x-t$, so that $x=(\alpha + \beta)/2$, $t=(\alpha-\beta)/2$. Then let

$$ u\left (\frac{\alpha+ \beta}{2}, \frac{\alpha- \beta}{2}\right) = v(\alpha, \beta) $$

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial \alpha} \frac{\partial \alpha}{\partial x} +\frac{\partial v}{\partial \beta} \frac{\partial \beta}{\partial x} = \frac{\partial v}{\partial \alpha} + \frac{\partial v}{\partial \beta}$$

$$\frac{\partial u}{\partial t} = \frac{\partial v}{\partial \alpha} \frac{\partial \alpha}{\partial t} +\frac{\partial v}{\partial \beta} \frac{\partial \beta}{\partial t} = \frac{\partial v}{\partial \alpha} - \frac{\partial v}{\partial \beta}$$

By differentiating again in this matter, the wave equation may then be expressed as

$$\frac{\partial^2 v}{\partial \alpha^2} + 2 \frac{\partial^2 v}{\partial \alpha \partial \beta} + \frac{\partial^2 v}{\partial \beta^2} = \frac{\partial^2 v}{\partial \alpha^2} - 2 \frac{\partial^2 v}{\partial \alpha \partial \beta} + \frac{\partial^2 v}{\partial \beta^2} $$

From which follows

$$ v_{\alpha \beta} = 0$$

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Thank you, but two questions: Using my transformation $u_x=v_a+v_b$ also ? u_tt=? – Voyage Jan 13 '13 at 15:03
If I am getting you right, I answered your questions in my answer. The transformation is really just an application of the chain rule. In terms of operators, $\partial/\partial x = \partial/\partial \alpha + \partial/\partial \beta$, so $(\partial/\partial x)^2 = (\partial/\partial \alpha + \partial/\partial \beta)^2$. Similar for $t$. – Ron Gordon Jan 13 '13 at 15:17
For completeness: Question no. 2 hasn't been answered. An argument of unicity must be performed in order to ensure that every solution has the desired form. – Pragabhava Jan 13 '13 at 15:36
@Pragabhava: Not sure how to go beyond $\partial G(\beta)/\partial \alpha = \partial F(\alpha)/\partial \beta = 0$. Perhaps you have something else to add? – Ron Gordon Jan 13 '13 at 16:02
@rlgordonma I'd say that the fact that $v_{\alpha \beta} = 0$ implies that $v(\alpha,\beta) = F(\beta) + G(\alpha)$ which, in turn, means that $u(x,t) = F(x-t) + G(x+t)$ (this can be ensured because the transformation $(x,t) \to (\alpha,\beta)$ is invertible), and given that the wave equation with initial conditions has a unique solution (provided $f$ and $g$ are in some space, details left to the OP), this has to be it. Finally, using the initial conditions, one can determine $F$ and $G$ in terms of $f$ and $g$, and derive d'Alambert solution. – Pragabhava Jan 13 '13 at 16:14

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