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Consider the $1+1$ wave equation $$ u_{tt} -c^2 u_{xx} = 0 $$

The following page from my lecture notes explains d'Alembert's solution to this

notes

As you can see, they have used the substitution $$ \xi = x + ct \hspace{5 mm} \eta = x - ct $$

I understand how they have deduced that $u_x = u_\xi + u_\eta$. However, I'm unsure of how they obtained their result for $u_{xx}$.

I would have worked this out using the chain rule, as follows $$ u_{xx} = \frac{\partial}{\partial \xi} \left( u_\xi + u_\eta \right) \cdot \frac{\partial \xi}{\partial x} + \frac{\partial}{\partial \eta} \left( u_\xi + u_\eta \right) \cdot \frac{\partial \eta}{\partial x} = u_{\xi \xi} + u_{\eta \xi} + u_{\xi \eta} + u_{\eta \eta} $$ This is only equal to their derived result for $u_{xx}$ if we assume that $u_{\eta \xi} = u_{\xi \eta}$. This leads me on to my first question: (why) can we say that $u_{\eta \xi} = u_{\xi \eta}$?

Secnd question: further down in these notes, it says that the boundary condition $u_t(x,t) = \psi (x)$ tells us that $$ f'(x) - g'(x) = \frac{1}{c} \psi(x) $$

How was this result obtained? I cannot see how they have deduced this?

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1 Answer 1

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$u_{\xi\eta}$=$u_{\eta\xi}$ because the order of differentiation does not matter if the function is $C^2$, more information here.

And if we have

$$u=f(x+ct)+g(x-ct)=f(\xi)+g(\eta)$$

then

$$u_t=\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial t}=cf'(\xi)-cg'(\eta)$$ because $\frac{\partial \xi}{\partial t}=c$ and $\frac{\partial \eta}{\partial t}=-c.$

Thus

$$u_t(x,0)=f'(x)-g'(x)=\frac{1}{c}\psi(x)$$

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