Partial Differential Equation

1. x = (1/2)(u^2 − v^2), y = uv.
diff w.r.t. x

1=u(∂u/∂x)-v(∂v/∂x)
0=v(∂u/∂x)+u(∂v/∂x)

solving, get
(∂u/∂x)=u/(u^2+v^2)
(∂v/∂x)= -v/(u^2+v^2)

Similary, we have
0=u(∂u/∂y)-v(∂v/∂y)
1=v(∂u/∂y)+u(∂v/∂y)

so
(∂u/∂y)=v/(u^2+v^2)
(∂v/∂y)=u/(u^2+v^2)

Consider a function f,
fx=fu(∂u/∂x)+fv((∂v/∂x)+fz(∂z/∂x)
= u/(u^2+v^2) fu -v/(u^2+v^2) fv

fxx = u^2/(u^2+v^2)^2 fuu + v^2/(u^2+v^2)^2 fvv -2uv/(u^2+v^2)^2 fuv
+(u(u^2+v^2)-2u(u^2-v^2))/(u^2+v^2)^3 fu + (v(u^2+v^2)+2v(u^2-v^2))/(u^2+v^2)^3 fv

similarly, we get
fyy=v^2/(u^2+v^2)^2 fuu + u^2/(u^2+v^2)^2 fvv +2uv/(u^2+v^2)^2 fuv
+(u(u^2+v^2)-4uv^2)/(u^2+v^2)^3 fu + (v(u^2+v^2)-4vu^2 )/(u^2+v^2)^3 fv

Therefore,
Δf = fxx + fyy + fzz
=(1/(u^2+v^2))(fuu + fvv)+fzz
(note that all other terms cancel out)

in other words,
Δ = (1/(u^2+v^2))(∂^2 / ∂u^2 + ∂^2 / ∂v^2) + (∂^2 / ∂z^2)

2.Ut - U + XUx + ((X^2)/2)Uxx = 0,
Let U(x,t)=f(x)g(t). Then the equation becomes

f(x)g'(t)-f(x)g(t)+xf'(x)g(t)+(x^2 /2)f''(x)g(t)=0
f(x)g'(t)=f(x)g(t)-xf'(x)g(t)-(x^2 /2)f''(x)g(t)=(f(x)-xf'(x)-(x^2 /2)f''(x))g(t)
g'(t) / g(t) = (f(x)-xf'(x)-(x^2 /2)f''(x)) / f(x)

Since L.H.S depends only on t and R.H.S depends only on x, both sides must equal to some constant C.

g'(t) = Cg(t)
f(x)-xf'(x)-(x^2 /2)f''(x) = Cf(x)

2008-01-31 22:41:00 補充：
第2題答案的怪獸是derivative的符號。

2008-01-31 22:46:07 補充：
There is a simplier method for Q1.fu = ufx + vfy, fuu = u^2 fxx + uv fxy + fx + uv fxy + v^2 fyyfv = -vfx + ufy, fvv = v^2 fxx - uv fxy -fx - uv fyx + u^2 fyyso,fuu + fvv = (u^2 + v^2)(fxx+fyy)

2008-01-31 22:46:20 補充：
Δf = fxx + fyy + fzz=(1/(u^2+v^2))(fuu + fvv)+fzzBut this method is a little bit tricky, and I prefer the hard way given above.