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Heat equation in one dimension with constant heat supply
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[QUOTE="V0ODO0CH1LD, post: 4847644, member: 400805"] [h2]Homework Statement [/h2] 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##. [h2]Homework Equations[/h2] 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] [h2]The Attempt at a Solution[/h2] 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##? [/QUOTE]
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Heat equation in one dimension with constant heat supply
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