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How to solve delu/delx + delu/dely = 0?
Note: del means partial derivative
Use separation of variables to solve the boundary value problem,
delu/delx + delu/dely = 0, u(0,y) = 3 e^(-y) + e^(-2y)
I put u(x,y) = F(x) G(y)
Then I got u(x,y) = F(0) G(0) e^k(x-y), where k is a constant.
From this I get u(0,y) = F(0) G(0) e^(-ky)
But don't know how to proceed after this.
2 Answers
- kbLv 77 years agoFavorite Answer
Substituting u(x,y) = F(x) G(y) into the PDE yields
F'(x) G(y) = F(x) G'(y)
==> F'(x)/F(x) = G'(y)/G(y).
Since the left side is a function of x, while the right side is a function of y. we conclude that F'(x)/F(x) = G'(y)/G(y) = c foe some constant c.
Hence, we have
F' - cF = 0 and G' - cG = 0
==> F(x) = Ae^(cx) and G(y) = Be^(cy) for some A, B.
==> u_c(x, y) = D_c e^(cx) e^(cy), where D_c = AB.
Since the PDE is linear, sums of solutions (over c) form other solutions.
Since u(0, y) = 3 e^(-y) + e^(-2y), we need two terms, one for c = -1 and one for c = -2.
So, u(x, y) = D_(-1) e^(-x) e^(-y) + D_(-2) e^(-2x) e^(-2y)
Letting x = 0:
3e^(-y) + e^(-2y) = D_(-1) e^(-y) + D_(-2) e^(-2y)
==> D_(-1) = 3 and D_(-2) = 1, by equating like coefficients.
Hence,
u(x, y) = 3e^(-x) e^(-y) + 1e^(-2x) e^(-2y)
...........= 3e^(-(x+y)) + e^(-2(x+y)).
I hope this helps!
- Anonymous7 years ago
yes
