Solving the wave equation with change of variables approach

[chwala]

Homework Statement

Kindly see the attachment below

Relevant Equations

D'Alembert approach (change of variables )

I am refreshing on the pde's, and i am trying to understand how the textbook was addressing change of variables, i find it a bit confusing. I will share the textbook approach, then later share my own understanding on change of variables approach. Here is the textbook approach;

My approach on this change of variables would look like this,
Let the pde be of the form;
$au_{xx}+bu_{xt}+cu_{tt}+du_x+eu_t+fu=0$
Consider the pde; $u_{tt}= c^2u_{xx}$ then we may use the change of variables as indicated by literature into working out the solution...

Now letting, $ξ = x +ct$ and $η= x-ct$, then it follows that,
$u_x$=$u_ξ⋅ξ_x + u_η⋅η_x$ =$u_ξ + η_x$
$u_{xx}$ = $u_x⋅u_ξ⋅ξ_x + u_x⋅u_η⋅η_x$ = $u_x⋅u_ξ+u_x⋅u_η$ = $[u_ξ⋅ξ_x + u_η⋅η_x]⋅u_ξ +[u_ξ⋅ξ_x + u_η⋅η_x]⋅u_η$ = $u_{ξξ} + 2u_{ξη} + u_{ηη}$
Similarly, i can show that,
$u_t$ =$cu_ξ - cu_η$ ... Note that $[u_t$ =$u_ξ ⋅ξ_t + u_η ⋅η_t]$
$u_{tt}$= $u_t⋅u_ξ⋅ξ_t +u_t ⋅u_η⋅ η_t$=... $c^2u_{ξξ}-2c^2u_ξ⋅u_η +c^2u_{ηη}$
also,
$u_{xt}$= $u_x⋅u_ξ⋅ξ_t + u_x⋅u_η⋅η_t$

My point is that, since i understand this quite well then i should not bother with the textbook approach as both ways would work towards the same solution...right?

 

[BvU]

HI,

You left out a chunk of the text in the book, and now compare the section intended to show that the classification of PDEs is independent of the coordinate system $-$ for which they consider a change of variable $-$ with one specific change of variable that works for your wave equation.

So in your case both ways are equivalent, yes.

$\$

 

[chwala]

This is a continuation of the same 'subject' content...now looking at the highlighted part...

I want to be certain that i am getting it right,...on substituting for $u_{xx}$ and $u_{tt}$we shall end up with;
$-4c^2u_{ξη}=0$
which simplifies to $u_{ξη}=0$...

the other parts seem to be straightforward; ie

we can get the highlighted solution by integrating,
$cφ^{'}(x)-cΦ^{'}(x)$=$g(x)$
On integration, we shall get;
$cφ(x)-cΦ(x)$=$\int_{x_0}^x g(s)ds$...from this we can establish that;
$φ(x) - Φ(x)$=$\frac {1}{c}$ $\int_{x_0}^x g(s)ds$...(1)
$φ(x) + Φ(x)$=$f(x)$...(2)

solving (1) and (2) simultaneously yields the highlighted solution $Φ(x)$ ... the second solution can also be found in a similar approach.

In the second approach to solving this, i.e by use of... separation of variables... using $u(x,t) = X(x) T(t)$ seems to be straightforward...phew