(adsbygoogle = window.adsbygoogle || []).push({}); 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 = a^{2}[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 thats 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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# Homework Help: 1D Heat Equation BC: Both ends insulated, IC: piecewise function

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