Clarification on the given PDE problem

[chwala]

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.

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.