cauchy problem of 1st order PDE from Partial Differential Equations
pure math
加一个一个Episodes
pde进可攻退可守
pure math
f:R→R,y=f(x),dy=f′(x)dxf:\mathbb{R}\rightarrow\mathbb{R},y=f(x),dy=f'(x)dxf:R→R,y=f(x),dy=f′(x)dx
f′(x)≠0⇒x=f−1(y)f'(x)\neq0 \Rightarrow x=f^{-1}(y)f′(x)=0⇒x=f−1(y)
A:Rn→Rn,y=Ax,dy=Adx.A: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}, y=Ax,dy=Adx. A:Rn→Rn,y=Ax,dy=Adx.
det(A)≠0⇒x=A−1ydet(A)\neq 0 \Rightarrow x=A^{-1}ydet(A)=0⇒x=A−1y
In general, you can not explicitly solve for the inverse!
f:Rn→Rn,y=f(x),dy=Dfdx.f:\mathbb{R}^n \rightarrow \mathbb{R}^n, y=f(x),dy=Dfdx.f:Rn→Rn,y=f(x),dy=Dfdx.
Df=[δ(y1,y2,y3,...yn)∂(x1,x2,x3,...,xn)]Df = [\frac{\delta(y_1,y_2,y_3,...y_n)}{\partial (x_1,x_2,x_3, ..., x_n)}]Df=[∂(x1,x2,x3,...,xn)δ(y1,y2,y3,...yn)]
Jacobian matrix
det(Df)≠0⇒x=f−1(y)det(Df) \neq 0 \Rightarrow x=f^{-1}(y)det(Df)=0⇒x=f−1(y)
在连续的函数中
一个点大于0
implies
一段大于0
theorem cometheorem gobut example last forever 这个代码的符号是在键盘的左上角的符号 三行五行的证明一定要会但是两页的证明不要看了一定要会用用久了一定会证明theoremcome,theoremgo,butexamplelastforevertheorem come, theorem go, but example last forevertheoremcome,theoremgo,butexamplelastforever
jacobian matrix
dx=xudu+xvdv,dy=yudu+yvdvdx=x_u du+x_v dv, dy = y_u du+y_v dvdx=xudu+xvdv,dy=yudu+yvdv
顺便吹爆这个math formula的插件，真香
ux+3y23uy=2,u(x,1)=1+xu_x+3y^\frac{2}{3}u_y=2, u(x,1)=1+xux+3y32uy=2,u(x,1)=1+x
dxdt=1,dydt=3y23,dudt=2\frac{dx}{dt}=1, \frac{dy}{dt}=3y^\frac{2}{3},\frac{du}{dt}=2dtdx=1,dtdy=3y32,dtdu=2
(x,y,u)∣t=0=(x0,y0,u0)=(s,1,1+s)(x,y,u)|_{t=0}=(x_0,y_0,u_0)=(s,1,1+s)(x,y,u)∣t=0=(x0,y0,u0)=(s,1,1+s)
x=t+s,y=(t+1)3,u=2t+1+sx=t+s,y=(t+1)^3, u = 2t+1+sx=t+s,y=(t+1)3,u=2t+1+s
∂(x,y)∂(t,s)\frac{\partial (x,y)}{\partial (t,s)}∂(t,s)∂(x,y)
D(q)=[p1+p2+2p3cosq2p2+p3cosq2p2+p3cosq2p2]D(q) = \begin{bmatrix} p_1+p_2+2p_3cosq_{2} & p_2+p_3cosq_2\\ p_2+p_3cosq_2 & p_2 \end{bmatrix} D(q)=[p1+p2+2p3cosq2p2+p3cosq2p2+p3cosq2p2]
C(q,q˙)=[−p3q˙2sinq2−p3(q˙1+q˙2)sinq2p3q˙1sinq20]C(q,\dot q) = \begin{bmatrix} -p_3\dot q_2sinq_2 &-p_3(\dot q_1+\dot q_2)sinq_2 \\ p_3\dot q_1sinq_2 & 0 \end{bmatrix} C(q,q˙)=[−p3q˙2sinq2p3q˙1sinq2−p3(q˙1+q˙2)sinq20]
=∣xtxsytys∣=∣113(t+1)20∣=−3(t+1)2≠0\frac{}{}=\begin{vmatrix} x_t & x_s \\ y_t & y_s\end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 3(t+1)^2 & 0\end{vmatrix} = -3(t+1)^2 \neq 0=∣∣∣∣xtytxsys∣∣∣∣=∣∣∣∣13(t+1)210∣∣∣∣=−3(t+1)2=0
t=y13−1,s=x+1−y13t=y^{\frac{1}{3}}-1, s=x+1-y^{\frac{1}{3}}t=y31−1,s=x+1−y31
eliminate s and t
u(x,y)=2t+1+s=x+y13u(x,y)=2t+1+s=x+y^{\frac{1}{3}}u(x,y)=2t+1+s=x+y31
t≠−1t\neq -1t=−1
t=−1⇒y=0isasingualarpointofD.E.t = -1 \Rightarrow y = 0 is a singualar point of D.E.t=−1⇒y=0isasingualarpointofD.E.
Dimensional Analysis: (from equation!)
[u][x]=[y]23[u][y]⇒[x]=[y]13\frac{[u]}{[x]}=[y]^{\frac{2}{3}}\frac{[u]}{[y]} \Rightarrow [x]=[y]^\frac{1}{3}[x][u]=[y]32[y][u]⇒[x]=[y]31
x→λ2x,y→λyx \rightarrow \lambda^2 x, y \rightarrow \lambda yx→λ2x,y→λy
u(u(u(
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