- #1
fahraynk
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<< Mentor Note -- thread moved from the technical math forums >>
I am getting stuck on this partial differential equation.
Ut = Uxx - U + x ; 0<x<1
U(0,t) = 0
U(1,t) = 1
U(x,0) = 0
Here is my work so far :
U = e-tw + x
gives the new eq wt=wxx
to get rid of boundary conditions :
w=x+W
Wt=Wxx
W(0,t) = 0
W(1,t)=0
W(x,0)=-x
W=X(x)T(t)
Plug that in, and I get these :
T'=μT
X''=μx
w = e-(nπ)2t[ansin(nπx)]
an = -2∫xsin(nπx) = 2cos(nπ)/nπ
w = x + W
w = x +(2/π)Σ(1/n)cos(nπ)sin(nπx)e-(nπ)2t
u = e-tw + x
u = x + e-t(x +(2/π)Σ(1/n)cos(nπ)sin(nπx)e-(nπ)2t)
But the books answer is :
u(x,t) = x - (2/π)e-t* [ e-π2tsin(πx) - (1/2) e-2π2tsin(2πx)+...]
What am I doing wrong?
I am getting stuck on this partial differential equation.
Ut = Uxx - U + x ; 0<x<1
U(0,t) = 0
U(1,t) = 1
U(x,0) = 0
Here is my work so far :
U = e-tw + x
gives the new eq wt=wxx
to get rid of boundary conditions :
w=x+W
Wt=Wxx
W(0,t) = 0
W(1,t)=0
W(x,0)=-x
W=X(x)T(t)
Plug that in, and I get these :
T'=μT
X''=μx
w = e-(nπ)2t[ansin(nπx)]
an = -2∫xsin(nπx) = 2cos(nπ)/nπ
w = x + W
w = x +(2/π)Σ(1/n)cos(nπ)sin(nπx)e-(nπ)2t
u = e-tw + x
u = x + e-t(x +(2/π)Σ(1/n)cos(nπ)sin(nπx)e-(nπ)2t)
But the books answer is :
u(x,t) = x - (2/π)e-t* [ e-π2tsin(πx) - (1/2) e-2π2tsin(2πx)+...]
What am I doing wrong?
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