Heat equation in one dimension with constant heat supply

  • #1
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Homework Statement



A bar of length ##L## has an initial temperature of ##0^{\circ}C## and while one end (##x=0##) is kept at ##0^{\circ}C## the other end (##x=L##) is heated with a constant rate per unit area ##H##. Find the distribution of temperature on the bar after a time ##t##.

Homework Equations



Heat equation in one dimension:
[tex] \alpha^2u_{xx}=u_t [/tex]
Initial conditions:
[tex] u(0,t)=0 [/tex]
[tex] u(L,t)=\frac{k}{H}t [/tex]
[tex] u(x,0)=0 [/tex]

The Attempt at a Solution



If ##u(x,t)## can be written as ##u(x,t)=X(x)T(t)## we can separate variables so that the heat equation becomes ##\alpha^2X''(x)T(t)=T'(t)X(t)## which can be written as
[tex] \alpha^2\frac{X''(x)}{X(x)}=\frac{T'(t)}{T(t)} [/tex]
and since each side only depends on one variable the two ratios above are constant (because if, for example, I differentiate ##T'/T## with respect to ##x## I get ##0## so ##X''(x)/X(x)## is constant). So I get
[tex] \alpha^2\frac{X''(x)}{X(x)}=\gamma\Rightarrow{}X(x)=c_1cos\left(\frac{\sqrt{\gamma}}{\alpha}x\right)+c_2sin\left(\frac{\sqrt{\gamma}}{\alpha}x\right) [/tex]
and
[tex] \frac{T'(t)}{T(t)}=\gamma\Rightarrow{}T(t)=ce^{\gamma{}t} [/tex]
Now from the boundary conditions
[tex] u(0,t)=X(0)T(t)=ce^{\gamma{t}}\left[c_1cos\left(\frac{\sqrt{\gamma}}{\alpha}0\right)+c_2sin\left(\frac{\sqrt{\gamma}}{\alpha}0\right)\right]=0\Rightarrow{}c_1=0 [/tex]
and now here's where I got into trouble
[tex] u(L,t)=X(L)T(t)=ce^{\gamma{t}}\left[c_1\cos\left(\frac{\sqrt{\gamma}}{\alpha}L\right)+c_2\sin\left(\frac{\sqrt{\gamma}}{\alpha}L\right)\right]=\frac{k}{H}t [/tex]

But what am I supposed to do with
[tex] c_2\sin{\left(\frac{\sqrt{\gamma}}{\alpha}L\right)}=\frac {k}{Hce^{\gamma{t}}}t [/tex]
to keep going to find ##u##?
 
Last edited:

Answers and Replies

  • #2
pasmith
Homework Helper
2,056
683
To use separation of variables here, you need constant boundary conditions and a non-zero initial condition, which is not what you have. But you can obtain a problem with constant (indeed zero) boundary conditions and a non-zero initial condition.

Consider [tex]f(x,t) = \frac{ktx}{HL} + g(x).[/tex] This satisfies [itex]f(0,t) = 0[/itex] and [itex]f(L,t) = \frac{kt}{H}[/itex] provided [itex]g(0) = g(L) = 0[/itex]. If [itex]f[/itex] is to satisfy [itex]f_t = \alpha^2 f_{xx}[/itex] then we need [tex]
\frac{kx}{HL} = \alpha^2 g''.[/tex] That's a second-order ODE for [itex]g[/itex] and we have two boundary conditions which [itex]g[/itex] must satisfy, so that's OK.

It won't be the case that [itex]f(x,0) = g(x) = 0[/itex], but we can always set [itex]u(x,t) = f(x,t) + v(x,t)[/itex] where [itex]v[/itex] satisfies [tex]
v_t = \alpha^2 v_{xx}
[/tex] subject to [itex]v(0,t) = v(L,t) = 0[/itex] and [itex]v(x,0) = -g(x)[/itex].
 

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