higher order partial derivatives

magnifik

Consider the partial dierential equation, (y⁴-x²)u_xx - 2xyu_xy - y²u_yy = 1. We will make the substitution x = s² - t² 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

tiny-tim

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 + y²)/y, 2t = (x - y²)/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 ).

magnifik

errr.. i meant the solution for part a, which was putting x and y in terms of s and t. thanks!

magnifik

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?

HallsofIvy

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].

magnifik

THANK YOU!! makes so much sense now :)

tiny-tim

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.