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If I change coordinates from $(x,t)$ to $(u,v)$ via the formulas $u=x-ct$ and $v=x+ct$ then I want to show that $$\frac{\partial}{\partial x} - \frac{1}{c}\frac{\partial}{\partial t}=2\frac{\partial}{\partial u}$$ $$\frac{\partial}{\partial x}+\frac{1}{c}\frac{\partial}{\partial x}=2\frac{\partial}{\partial v}$$

but I'm not sure how to do this.

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  • $\begingroup$ Use a function and check how the operators act on it. $\endgroup$ Jan 30, 2017 at 20:55

2 Answers 2

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Somebody can offer a more comprehensive explanation, but I try to do. The operators, like these, are defined by its action over some suitable set of objects. Thus its properties (or relations) have to be proved taking instances of those objects and making the operators act on them. The partial derivative acts over functions following the known rules for derivation. Equalities for partial derivatives are true if are true these equalities when acting over any function.

$$\frac{\partial \psi (x(u,v),y(u,v))}{\partial x}=\frac{\partial \psi}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial \psi}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial \psi}{\partial u}\ + \frac{\partial \psi}{\partial v}$$ $$\frac{\partial \psi (x(u,v),y(u,v))}{\partial t}=\frac{\partial \psi}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial \psi}{\partial v}\frac{\partial v}{\partial t}=-c\frac{\partial \psi}{\partial u}\ + c\frac{\partial \psi}{\partial v}$$

$$\frac{1}{c}\frac{\partial \psi}{\partial t}=-\frac{\partial \psi}{\partial u}\ + \frac{\partial \psi}{\partial v}$$

$$\frac{\partial \psi}{\partial x}+\frac{1}{c}\frac{\partial \psi}{\partial t}=2\frac{\partial \psi}{\partial v}$$

$$\frac{\partial \psi}{\partial x}-\frac{1}{c}\frac{\partial \psi}{\partial t}=2\frac{\partial \psi}{\partial u}$$

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  • $\begingroup$ Somehow, it seems more natural to take the derivative of $\psi(x(u,v),y(u,v))$ with respect to $u$ or $v$... $\endgroup$
    – Fabian
    Jan 30, 2017 at 21:29
  • $\begingroup$ Not sure. To prove the equalities I saw two ways: invert the coordinate transformation solving a system of equations (as you can see in martini's answer) then derive or derive then solve a system of equations. $\endgroup$ Jan 30, 2017 at 21:35
  • $\begingroup$ All that I am saying is that it might be more natural to start from $F(u(x,y), v(x,y))$. $\endgroup$
    – Fabian
    Jan 31, 2017 at 6:26
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From $u = x-ct$ and $v = x+ct$ we get $$ x = \frac 12 (u+v), \quad t = \frac 1{2c} (v-u). $$ Hence, by the chain rule $$ \def\pd#1#2{\frac{\partial #1}{\partial #2}}\def\p#1{\pd{}#1}\p u = \pd xu \p x + \pd tu \p t = \frac 12 \p x - \frac 1{2c} \p t $$ and $$ \p v = \pd xv \p x + \pd tv \p t = \frac 12 \p x + \frac 1{2c} \p t. $$

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