PDE 偏微分方程 yu_x- 2xyu_y =2xu

偏微分方程PDE,說明解題過程,有看不懂的地方Ans:y*Ux-2xy*Uy=2x*U邊界條件: U(0,yo)=yo^3Let dx/ds=y.....(1) dy/ds=-2xy.....(2)(1),(2)兩者合併:dy/ds=-2x*y=-2x*(dx/ds) => dy+2xdx=0兩邊積分: y(s)+x(s)^2=C.....(3)移項: C-x(s)^2=y(s)=dx/ds.....Eq.(1)邊界條件: s=0, y=yo => dx/ds=yo-x^2Let U(x,y)=(y+x^2)^m/y^n.....Eq.(3)邊界條件: U(0,yo)=yo^m/yo^n=yo^3=> m=n+3n=1 => m=4=> U(x,y)=(y+x^2)^4/y


2014-03-04 16:27:23 補充：
修正:

y*Ux-2xy*Uy=2x*U

邊界條件: U(0,yo)=yo^3

全微分: dU/ds=Ux*(dx/ds)+Uy*(dy/ds)

=Ux*y+Uy*(-2xy).....dx/ds=y, dy/ds=-2xy

=原式左邊

=2x*U.....原式右邊

=-(dy/yds)*U.....由dy/ds=-2xy

0=dU/ds+(dy/yds)*U

=dU/U+dy/y

ln(C)=∫dU/U+dy/y=ln(U*y)

y*U=C => U(x,y)=C/y

2014-03-04 16:27:54 補充：
但是C=y+x^2.....由Eq.(3)

=> U(x,y)=C/y=(y+x^2)^m/y

邊界條件: U(0,yo)=yo^m/yo=yo^3 => m=4

=> U(x,y)=(y+x^2)^4/y.....ans

2014-03-04 18:34:35 補充：
繼續修正:

這兩個C不一樣

C=y+x^2.....由Eq.(3)

U(x,y)=C1/y

=C^4/y........C1=C^4

=(y+x^2)^4/y......ans

2014-03-05 19:24:40 補充：
重新整理:

全微分: dU/ds=Ux*(dx/ds)+Uy*(dy/ds)

=Ux*y+Uy*(-2xy).......Let dx/ds=y, dy/ds=-2xy

=原式左邊

=2x*U.......原式右邊

=-(dy/yds)*U.......由dy/ds=-2xy

2014-03-05 19:25:01 補充：
0=dU/ds+(dy/yds)*U

=dU/U+dy/y

ln(C)=∫dU/U+dy/y=ln(U*y)

y*U=C1 => U(x,y)=C1/y

2014-03-05 19:25:22 補充：
現在回頭來看dx/ds=y, dy/ds=-2xy, 兩者相除:

dx/dy=-y/2xy=-1/2x => 2xdx+ydy=0

兩邊積分: y+x^2=C2

2014-03-05 19:25:48 補充：
邊界條件: U(0,yo)=yo^3

Let C1=C2^n

U(x,y)=C1/y

=C2^n/y

=(y+x^2)^n/y

=(yo+0)^n/yo

=yo^n/yo

=yo^(n-1)

=yo^3 => n=4


=> U(x,y)=(y+x^2)^4/y

越來越多關切正解的觀眾。若肯稍候一二日，我貼全解。

2014-03-05 00:56:19 補充：
首先為你的問題正名(E) yu_x -2xyu_y=2xu ----au_x +b u_y=c is a quasi-linear 1st order pdefor u=u(x,y), where a, b, c are functions of (x,y,u); With u=y^3 when x=0 (and 1<=y<=2) à the solution is to satisfy the initialcondition(不叫邊界條件) (IC) u(0, y)=y^3.(E) +(IC) ,如同ode裡的initial value problem希望解可以是唯一存在雖然有時不完全如願, 在這裡它的正式名稱是歌西問題(Cauchy problem for (E)).其次，解歌西問題有一套標準流程叫做method of characteritics:A. A. 根據(E) +(IC) 寫下等價的odesystem with initial conditions: (1) x_s =a=y, with x(0,t)=0; (2) y_s=b=-2xy, with y(0,t)=t;(3) u_s=c=2xu, with u(0,t)=t^3其中 x, yand u暫設為二參數 s,t 之函數.B. B. 竭盡所能將系統A中x, yand u解出, 則u=u(s,t) 和想要達到的終點u=u(x,y)就不遠了.C. C. B. 的最後一步需仰賴 x(s,t) and y(s,t) 可反求出公式(s,t)-à(x,y) [implicit function theorem] 才算完成.閣下題供的所謂解答裡,開頭和以上systemA 類似但沒有使用(IC).如此每次積分不知如何處置產生的積分常數. 於是最後是找不到唯一的公式 (s,t)-à(x,y) . 我認為是後半段支支唔唔草草收場的原因.第一個why: (1)乘以(2x)+(2) à2xx_s+y_s=0 àd/ds [x(s,t)]^2+d/ds[y(s,t)]=0 à [x(s,t)]^2+y(s,t)= C=C(t), theintegral constant, but he does not know how to get unique C;第二個why: plug the y he got into (1) .System A 不好求解是因為 xequation 裡有y and y equation裡有x, u equation裡又有x. 因此建議這樣的處理 (E): divided by (2xy) à (E’) (1/(2x)u_x- u_y=(1/y)u 如此改變equationsin ode system while initial conditionsstate the same: (1’) x_s=(1/2x); (2’)y_s=-1; (3’) u_s=(1/y).Solving (1’)and (2’) with their IC’s è x(s,t)=s^(1/2); y(s,t)=t-s both are unique sol’n.;Then plug the unique y to (3’) which will then be a separable odewhile t is considered a parameter, together with ICè u(s,t)= t^4/|t-s| . Reading from x and y we find t=(y+x^2) , ands=x^2, so s-t=y. Thusu(x,y)=(y+x^2)^4/y. 大功告成. Your why #3 !