Consider the partial dierential equation, (y^{4}-x^{2})u_{xx} - 2xyu_{xy} - y^{2}u_{yy} = 1. We will make the substitution x = s^{2} - t^{2} and y = s - t, to simplify (a) Solve for s and t as functions of x and y the farthest point i got to was x = s^2 - t^2 = (s+t)(s-t) = y(s+t) y = s - t s+t = x/y i don't know what to do after that.. i have the solution, but i have no idea how to get to it. the solution is s = x + y^2 / 2y t = x - y^2 / 2x
Hi magnifik! No, you're mssing the point … that isn't the solution, it's just writing s and t in terms of x and y … s+t = x/y, s-t = y, so 2s = x/y + y = (x + y^{2})/y, 2t = (x - y^{2})/y. To get the solution, you need to find u_{ss} u_{st} and u_{tt} (and I expect u_{st} will be zero, which will make the solution fairly easy ).
but i don't understand how you got from s+t = x/y, s-t = y to 2s and 2t...how did you combine the previous two equations to get that?
Excuse me, tiny-tim, but the only question asked was "(a) Solve for s and t as functions of x and y". Presumably, there is a "b" but so far all we want to do is solve for s and t. magnifik, you have [itex]x= s^2- t^2[/itex] and [itex]y= s- t[/itex]. From [itex]y= s- t[/itex] you can say that [itex]s= y+ t[/itex] and, putting that into [itex]x=s^2- t^2[/itex], [itex]x= y^2+2yt+ t^2- t^2= y^2+ 2yt[/itex]. Solve that equation for t, then replace t by that expression in [itex]s= y+ t[/itex].
erm … I added, and subtracted! Add s+t to s-t, you get 2s; subtract, you get 2t … this is a common transformation, and you should be familiar with it.