Clarification on the given PDE problem

[chwala]

TL;DR

see attached.

My interest is on the highlighted part only...my understanding is that one should use simultaneous equation... unless there is another way hence my post query.

In my working i have;

$y=\dfrac{2ξ+η}{10}$ and $x=\dfrac{2η-ξ}{10}$ giving us;

$x+3y=\dfrac{2η-ξ+6ξ+3η}{10}=\dfrac{5ξ+5η}{10}=\dfrac{ξ+η}{2}$ cheers guys.

 

[pbuk]

Why do you always have to question things like this? You have two equations for $\xi$ and $\eta$ in $x$ and $y$ and you want to find two equations for $x$ and $y$ in $\xi$ and $\eta$: this is simultaneous equations by definition. You should do more thinking for yourself and not constantly seek assurance, this is not the way to develop confident problem solving skills.

 

[chwala]

@pbuk noted mate ...am slowly developing confidence...the pde's can be intimidating at times...not for the faint of hearts. Cheers...

 

[pasmith]

TL;DR Summary: see attached.

My interest is on the highlighted part only...my understanding is that one should use simultaneous equation... unless there is another way hence my post query.

View attachment 319956

I don't think this approach is the best approach to applying the method of characteristics.

Firstly I don't think it extends to non-constant coefficients, and secondly it leads to more complicated airthmetic in finding the integrating factor and solving the transformed PDE than does the approach of setting $$x_\xi = -2,\qquad y_\xi = 4$$ and then choosing $x_\eta$ and $y_\eta$ such that $\eta$ does not appear expressly in the transformed equation, leading to $$u_{\xi} + 5u = e^{10\xi + (x_\eta + 3y_\eta)\eta} = e^{10\xi}.$$ This approach also works for non-constant coefficient problems.

 

[chwala]

Thanks @pasmith ...i prefer the integrating factor method shown in the text for the time being... i need to try and study the approach that you are suggesting.