partial differential equations

OMG - that takes me back 20 years - will have to look up my Uni notes on this one - is that 2nd year Uni Pure Mathematics?

Didn't find them any help - That was hard!!!

Let U = f(x,y)

U = f(x,y) = -1/2x^2 + x - xy + [d + e y ^3 + f y ^ 2 +gy]

or let G(y) be any finite, continuous, closed function (polynomial, transindental etc) purely of y (no x's or infinities in the formulea!)

The full answer is U = f(x,y) = -1/2x^2 + x - xy + G(y)

where G(y) - or say d, e, f and g (or any higher polynomial) can have any value because they drop out on the first partial derivative of x.


Lets test it an see!

dU/dx = -x + 1 -y ( called f.x(x,y) )
d2U/dxdy = -1 ( called f.xy(x,y) )

So d2U/dxdy + dU/dx = - x - y

Or

d2U/dxdy + dU/dx +x + y = 0

* * *

How did I do this?

Well its simple to see

d2U/dxdy + dU/dx = - x - y So integrate both side by x giving, Call this equation Y

dU/dy + U = -1/2 x^2 - xy Call this equation Z

This means the difference between U and U.y is at most a second degree polynomial in terms of x and y. This implies U is at most a third degree polynominal of x and y, with alot of collapsing terms that must satisfy equation Z.

So I build f(x,y) in general terms to see which coefficents must be 0 to satisfy equation Z.

Let U = f(x,y) be

ax^3 + bx^2 + cx +d + ey^3 +fy^2 + gy + ix^3y^3 + jx^2y^3 + kxy^3 + lx^3y^2 + mx^3y + nxy

So f.x(x,y) = du/dx =

3ax^2 +2bx +c + 0 + 0 + 0 + 0 + 3ix^2y^3 + 3ky^3 + 3lx^2y^2 +3mx^2y + ny

and f.xy(x,y) = d2U/dxdy =

9ix^2y^2 + 6jxy^2 +3k +6lx^2y +3m + n

So fx(x,y) + fx.y(x,y) =

3ax^2 + 2bx +c + 3ix^2y^3 +ky^3 + 3lx^2y^2 + 3mx^2y +ny + 9ix^2y^2 + 6jxy^2 +3k + 6lx^2y 3mx^2 +n

Ugly but we're in the final straight - group terms by like coefficents to get

x^2(3a-3m) + x(2b) + (c +k +n) + x^2y^2(3i) + xy^3(2j) + x^2y^2(3l + 9i) + x^2y(3m + 6l) + xy^2(6j) + y(n) + y^3(3k)

Now by equation Y - this must equal -x - y, so solve for the coeffecients that must be zero, to give:

3a = -3m
2b = -1
c + 3k + n = 0
i = 0
j = 0
3l + 9i = 0 so l = 0
3m + 6l = 0, so m = 0 and hence a = 0
j = 0
k = 0
3k + n = -1, so as k = 0 then n = -1
c + 3k +n = 0, and therfore c + 0 + -1 = 0 hence c = 1

Note terms d, e, f, g are't defined at all so they can be any and all real values!

Leaving you with f(x,y) = -1/2x^2 + x - xy + d + e y ^3 + f y ^ 2 +g y

or f(x,y) = -1/2x^2 + x - xy +G(y)

* * *

Now I need a beer!

Thank you.
This is 2nd year Electrical engineerig

You're most welcome! ECT is a bugger of a hard course, it get easier - alot easier in 3rd year!