[數撚圍爐區] 祝手巴日日無工開, 想出紙永遠失敗 (54)
kai龍(苗撚)
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Assume左先，唔得先算。
Sor, t is a parameter，自膠。
t(x,y)
t(x,y)
有variable coefficient嘅PDE⋯⋯”a” will not be a constant in this case……
unique就job done。唔unique可以點？
我會用computer格硬嚟
。
我會用computer格硬嚟
。唔unique engineer physicist咪啪個physical condition落去逼到佢unique
e.g. heat equation oh unboinded domain
e.g. heat equation oh unboinded domain
同一個方法solve
只係d path唔再係直線
只係d path唔再係直線
得閒搵個律師撚下佢地就化
佢地最驚律師再唔係搬去BW
我既經驗BW 個D人無咁西
佢地最驚律師再唔係搬去BW我既經驗BW 個D人無咁西
Sorry, Now it becomes an ODE:
In this case, the characteristic line is dx/dt=a.
In this case, the characteristic line is dx/dt=a.
同一個方法，不過個solution detail難搞D囉。
suppose u is a C^1 solution to the PDE, then
x' = x
y' = -y
u' - 3u = 2x^2
which gives
x = x(0) e^t
y = y(0) e^{-t}
u = -2x(0)^2 e^{2t} + Ce^{3t} = -2x^2 + Cx^3 (This is the only C^1 solution on x =/= 0.)
now if x(0) = 1 and y(0) = y_0, then
y_0 = u(1, y_0) = -2[x(0)]^2 + C [x(0)]^3
which implies C = 2 + y_0 = 2 + xy (not that xy = x(0) y(0) = y_0 from what we have solved above)
So the unique C^1 solution is
u(x, y) = -2x^2 + (2 + xy) x^3 = -2x^2 + 2x^3 + x^4 y
x' = x
y' = -y
u' - 3u = 2x^2
which gives
x = x(0) e^t
y = y(0) e^{-t}
u = -2x(0)^2 e^{2t} + Ce^{3t} = -2x^2 + Cx^3 (This is the only C^1 solution on x =/= 0.)
now if x(0) = 1 and y(0) = y_0, then
y_0 = u(1, y_0) = -2[x(0)]^2 + C [x(0)]^3
which implies C = 2 + y_0 = 2 + xy (not that xy = x(0) y(0) = y_0 from what we have solved above)
So the unique C^1 solution is
u(x, y) = -2x^2 + (2 + xy) x^3 = -2x^2 + 2x^3 + x^4 y
"And from the definition of total derivative, the equation is now:
u_t+dx/dt*u_x=Du/Dt=0 along dx/dt=a."
this is just chain rule
total derivative係第二樣野黎
u_t+dx/dt*u_x=Du/Dt=0 along dx/dt=a."
this is just chain rule
total derivative係第二樣野黎
btw the characteristics covers the whole right half plane for the initial data given and the paths are mutually disjoint
so we dont need to adjust the solution.
so we dont need to adjust the solution.
