PDE 偏微 wave equation

(E) utt-uxx=0 , a homogeneous waveequation having two sets of characteristic curves: x-t=constant andx+t=constant so the general solution to this equation is u(x,t)=v(x-t)+w(x+t)for arbitrary functions v and w.Also because of this general solution form, one can easily provea parallelogram property that this u(x,t) will satisfy: If A, B, C, and D arevertices of a parallelogram using two x-t=const and two x+t=const edges, thenu(A)+u(C)=u(B)+u(D), where AC and BD are two diagonals. In particular wecan first choose A=(2/3,2), B=(0,4/3), C=(1/3,1), and D=(1,5/3). Since B falls on the line x=0, D falls on theline x=1, by boundary conditions, we have u(B)=0 and u(D)=0, thus u(A)=-u(C) oru(2/3,2)=-u(1/3,1). Next we choose another parallelogram: A’=(1/3,1), B’=(0,2/3),C’=(2/3,0), and D’=(1,1/3). [Note A’=C]. A similar argument shows that u(1/3,1)=-u(2/3,0)[ B’ is on x=0, D’ is on x=1]. Hence the question we like to solve becomesu(2/3,2)=u(2/3,0)=?Now we can use D’Alembert’s formula to solve u(2/3,0). Actually this point is right on t=0, whereInitial conditions are prescribed èu(2/3,0)= (2/3)^2(1-2/3)=4/27. So u(2/3,2)=4/27.Note: We cannot apply D’Alembert’s formula to u at (2/3,2)directly because the formula depends on data which falls outside of [0,1] onx-axis (t=0). Instead, using parallelogram property can bring down the points (likewe did above) so D’Alembert’s formula becomes applicable.

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U(2/3,2)=? if Utt=Uxx; 0<x<1U(x,0)=(1-x)x^2Ut(x,0)=(1-x)^2U(0,t)=U(1,t)=0
(1) Let U=U(V,W)V=x+t, W=x-t => Vx=Wx=1, Vt=-Wt=1
(2) Uxx=?Ux=Uv*Vx+Uw*Wx=Uv+UwUxx=(Uv+Uw)v*Vx+(Uv+Uw)w*Wx=Uvv+Uwv+Uvw+Uww=Uvv+2Uvw+Uww......Uvw=Uwv
(3) Utt=?Ut=Uv*Vt+Uw*Wt=Uv-UwUtt=(Uv-Uw)v*Vt+(Uv-Uw)w*Wt=(Uv-Uw)v-(Uv-Uw)w=Uvv-Uwv-Uvw+Uww=Uvv-2Uvw+Uww
(4) Uxx=UttUvv+2Uvw+Uww=Uvv-2Uvw+Uww=> Uvw=0 => Uv=1 => dU(v)/dv=h(v) => U(v)=∫h(v)dv+N(w)=M(v)+N(w)=> U(x,t)=M(x+t)+N(x-t)
(5) 邊界條件Let U=a(x+t)^3+b(x+t)^2+c(x+t)+d(x-t)^3+e(x-t)^2+f(x-t)+gU(x,0)=ax^3+bx^2+cx+dx^3+ex^2+fx+g=(a+d)x^3+(b+e)x^2+(c+f)x+g=(1-x)x^2=-x^3+x^2+0+0=> a+d=-1, b+e=1, c+f=0, g=0
(6) 初始條件Ut=3a(x+t)^2+2b(x+t)+c-3d(x-t)^2-2e(x-t)-fUt(x,0)=3ax^2+2bx+c-3dx^2-2ex-f=3(a-d)x^2+2(b-e)x+(c-f)=(1-x)^2=x^2-2x+1=> a-d=1/3, b-e=-1, c-f=1
(7) 解出係數 & 解答a+d=-1, a-d=1/3 => a=-1/3, d=-2/3b+e=1, b-e=-1 => b=0, e=1c+f=0, c-f=1 => c=1/2, f=-1/2=> U(x,t)=-(x+t)^3/3+(x+t)/2-2(x-t)^3/3+(x-t)^2-(x-t)/2
(8) U(2/3,2)=?=-(x+t)^3/3+(x+t)/2-2(x-t)^3/3+(x-t)^2-(x-t)/2=-(2/3+2)^3/3+(2/3+2)/2-2(2/3-2)^3/3+(2/3-2)^2-(2/3-2)/2=-(8/3)^3/3+(8/3)/2-2(-4/3)^3/3+(-4/3)^2-(-4/3)/2=-512/81+64/6-128/81+16/9+2/3=-640/81+121/9=(1089-640)/81=449/81......ans