Solving the wave equation with change of variables approach

chwala
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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;

1638706553231.png

1638706654514.png

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?
 
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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.

##\ ##
 
This is a continuation of the same 'subject' content...now looking at the highlighted part...
1638762517432.png

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
1638764275397.png


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
 
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