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1. Solve the one dimensional heat equation for a rod of length 1 with the following boundary and initial conditions:
BC: [tex]\partial[/tex]u(0,t)/[tex]\partial[/tex]t = 0
[tex]\partial[/tex]u(1,t)/[tex]\partial[/tex]t = 0 these are the wrong boundary conditions (see below)
Actual BC: [tex]\partial[/tex]u(0,t)/[tex]\partial[/tex]x = 0
[tex]\partial[/tex]u(1,t)/[tex]\partial[/tex]x = 0
IC:
u(x,0) = { 1 if 0[tex]\leq[/tex]x[tex]\leq[/tex].5
u(x,0) = { 0 if .5<x[tex]\leq[/tex]1
2. [tex]\partial[/tex]u/[tex]\partial[/tex]t = a2[tex]\partial[/tex][tex]^{2}[/tex]u/[tex]\partial[/tex]x[tex]^{2}[/tex]
3.
I used separation of variables and applied the boundary conditions to get the following:
u(x,t) = e(-(a2)(n*Pi)2*t) * (B*cos(n*Pi*x)) where
n= 1, 2, 3... and B is an unknown constant.
To find B I tried applying the initial conditions and that's where I got stuck because I got.
u(x,0) = 0 = B*cos(n*Pi*x) if 0[tex]\leq[/tex]x[tex]\leq[/tex].5
u(x,0) = 1 = B*cos(n*Pi*x) if .5<x[tex]\leq[/tex]1
Does this mean that B has two values depending on x? And if is so are those values:
B = 0 for 0[tex]\leq[/tex]x[tex]\leq[/tex].5
B = 1/cos(n*Pi*x) for .5<x[tex]\leq[/tex]1
As a follow-up, will the steady state of this heat equation problem be a piecewise function?
BC: [tex]\partial[/tex]u(0,t)/[tex]\partial[/tex]t = 0
[tex]\partial[/tex]u(1,t)/[tex]\partial[/tex]t = 0 these are the wrong boundary conditions (see below)
Actual BC: [tex]\partial[/tex]u(0,t)/[tex]\partial[/tex]x = 0
[tex]\partial[/tex]u(1,t)/[tex]\partial[/tex]x = 0
IC:
u(x,0) = { 1 if 0[tex]\leq[/tex]x[tex]\leq[/tex].5
u(x,0) = { 0 if .5<x[tex]\leq[/tex]1
2. [tex]\partial[/tex]u/[tex]\partial[/tex]t = a2[tex]\partial[/tex][tex]^{2}[/tex]u/[tex]\partial[/tex]x[tex]^{2}[/tex]
3.
I used separation of variables and applied the boundary conditions to get the following:
u(x,t) = e(-(a2)(n*Pi)2*t) * (B*cos(n*Pi*x)) where
n= 1, 2, 3... and B is an unknown constant.
To find B I tried applying the initial conditions and that's where I got stuck because I got.
u(x,0) = 0 = B*cos(n*Pi*x) if 0[tex]\leq[/tex]x[tex]\leq[/tex].5
u(x,0) = 1 = B*cos(n*Pi*x) if .5<x[tex]\leq[/tex]1
Does this mean that B has two values depending on x? And if is so are those values:
B = 0 for 0[tex]\leq[/tex]x[tex]\leq[/tex].5
B = 1/cos(n*Pi*x) for .5<x[tex]\leq[/tex]1
As a follow-up, will the steady state of this heat equation problem be a piecewise function?
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