cauchy problem of 1st order PDE from Partial Differential Equations
作者：互联网
pure math
加一个一个Episodes
pde进可攻退可守
pure math
f : R → R , y = f ( x ) , d y = f ′ ( x ) d x f:\mathbb{R}\rightarrow\mathbb{R},y=f(x),dy=f'(x)dx f: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 : R n → R n , y = A x , d y = A d x . A: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}, y=Ax,dy=Adx. A:Rn→Rn,y=Ax,dy=Adx.
d e t ( A ) ≠ 0 ⇒ x = A − 1 y det(A)\neq 0 \Rightarrow x=A^{-1}y det(A)=0⇒x=A−1y
In general, you can not explicitly solve for the inverse!
f : R n → R n , y = f ( x ) , d y = D f d x . f:\mathbb{R}^n \rightarrow \mathbb{R}^n, y=f(x),dy=Dfdx. f:Rn→Rn,y=f(x),dy=Dfdx.
D f = [ δ ( y 1 , y 2 , y 3 , . . . y n ) ∂ ( x 1 , x 2 , x 3 , . . . , x n ) ] 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
d e t ( D f ) ≠ 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 come
theorem go
but 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
t h e o r e m c o m e , t h e o r e m g o , b u t e x a m p l e l a s t f o r e v e r theorem come, theorem go, but example last forever theoremcome,theoremgo,butexamplelastforever
jacobian matrix
d x = x u d u + x v d v , d y = y u d u + y v d v dx=x_u du+x_v dv, dy = y_u du+y_v dv dx=xudu+xvdv,dy=yudu+yvdv
顺便吹爆这个math formula的插件，真香
u x + 3 y 2 3 u y = 2 , u ( x , 1 ) = 1 + x u_x+3y^\frac{2}{3}u_y=2, u(x,1)=1+x ux+3y32uy=2,u(x,1)=1+x
d x d t = 1 , d y d t = 3 y 2 3 , d u d t = 2 \frac{dx}{dt}=1, \frac{dy}{dt}=3y^\frac{2}{3},\frac{du}{dt}=2 dtdx=1,dtdy=3y32,dtdu=2
( x , y , u ) ∣ t = 0 = ( x 0 , y 0 , u 0 ) = ( 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 = 2 t + 1 + s x=t+s,y=(t+1)^3, u = 2t+1+s x=t+s,y=(t+1)3,u=2t+1+s
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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 ˙ ) = [ − p 3 q ˙ 2 s i n q 2 − p 3 ( q ˙ 1 + q ˙ 2 ) s i n q 2 p 3 q ˙ 1 s i n q 2 0 ] 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]
= ∣ x t x s y t y s ∣ = ∣ 1 1 3 ( t + 1 ) 2 0 ∣ = − 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 = y 1 3 − 1 , s = x + 1 − y 1 3 t=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 ) = 2 t + 1 + s = x + y 1 3 u(x,y)=2t+1+s=x+y^{\frac{1}{3}} u(x,y)=2t+1+s=x+y31
t ≠ − 1 t\neq -1 t=−1
t = − 1 ⇒ y = 0 i s a s i n g u a l a r p o i n t o f D . 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 ] 2 3 [ u ] [ y ] ⇒ [ x ] = [ y ] 1 3 \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 → λ 2 x , y → λ y x \rightarrow \lambda^2 x, y \rightarrow \lambda y x→λ2x,y→λy
u ( u( u(
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标签：mathbb,p3,frac,Differential,dx,dy,Partial,du,order 来源： https://blog.csdn.net/Dequn_Teng_CSDN/article/details/114526075