using Laplace transform to solve heat equation

Along the whole positive $x$-axis, we have an heat-conducting rod, the surface of which is .  The initial temperature of the rod is 0 .  Determine the temperature function$u(x,\,t)$  when at the time  $t=0$

(a) the head  $x=0$  of the rod is set permanently to the constant temperature;

(b) through the head  $x=0$  one directs a constant heat flux.

The heat equation in one dimension reads

 $\displaystyle u_{xx}^{\prime\prime}(x,\,t)\;=\;\frac{1}{c^{2}}\,u_{t}^{\prime}% (x,\,t).$ (1)

In this we have

 $\displaystyle\mbox{(a) }\begin{cases}\mbox{boundary conditions}\;\,u(\infty,\,% t)=0,\qquad u(0,\,t)=u_{0},\\ \mbox{initial conditions}\qquad u(x,\,0)=0,\qquad\underbrace{u_{t}^{\prime}(x,% \,0)=0}_{\mbox{for\;}x\,>\,0}\end{cases}$

and

 $\displaystyle\mbox{(b) }\begin{cases}\mbox{boundary conditions}\;\,u(\infty,\,% t)=0,\quad u_{x}^{\prime}(0,\,t)=-k,\\ \mbox{initial conditions}\qquad u(x,\,0)=0,\quad\underbrace{u_{t}^{\prime}(x,% \,0)=0}_{\mbox{for\;}x\,>\,0}.\end{cases}$

For solving (1), we first form its Laplace transform (see the table of Laplace transforms)

 $U_{xx}^{\prime\prime}(x,\,s)\;=\;\frac{1}{c^{2}}[s\,U(x,\,s)-u(x,\,0)],$

which is a ordinary linear differential equation

 $U_{xx}^{\prime\prime}(x,\,s)\;=\;\left(\frac{\sqrt{s}}{c}\right)^{2}U(x,\,s)$

of order (http://planetmath.org/ODE) two.  Here, $s$ is only a parametre, and the general solution of the equation is

 $U(x,\,s)\;=\;C_{1}e^{\frac{\sqrt{s}}{c}x}+C_{2}e^{-\frac{\sqrt{s}}{c}x}$

(see this entry (http://planetmath.org/SecondOrderLinearODEWithConstantCoefficients)).  Since

 $U(\infty,\,s)\;=\;\int_{0}^{\infty}\!e^{-st}u(\infty,\,t)\,dt\;=\;\int_{0}^{% \infty}\!0\,dt\;\equiv\;0,$

we must have  $C_{1}=0$.  Thus the Laplace transform of the solution of (1) is in both cases (a) and (b)

 $\displaystyle U(x,\,s)\;=\;C_{2}e^{-\frac{\sqrt{s}}{c}x}.$ (2)

For (a), the second boundary condition implies   $\displaystyle U(0,\,s)=\frac{u_{0}}{s}$.  But by (2) we must have  $U(0,\,s)=C_{2}\!\cdot\!1$, whence we infer that  $\displaystyle C_{2}=\frac{u_{0}}{s}$.  Accordingly,

 $U(x,\,s)\;=\;u_{0}\cdot\frac{1}{s}e^{-\frac{x}{c}\sqrt{s}},$

which corresponds to the solution function

 $u(x,\,t)\;:=\;u_{0}\mbox{ erfc}\frac{x}{2c\sqrt{t}}$

of the heat equation (1).

For (b), the second boundary condition says that  $\displaystyle U_{x}^{\prime}(0,\,s)=-\frac{k}{s}$,  and since (2) implies that  $U_{x}^{\prime}(x,\,s)=-\frac{\sqrt{s}}{c}C_{2}e^{-\frac{\sqrt{s}}{c}x}$,  we can infer that now

 $C_{2}\;=\;\frac{ck}{s\sqrt{s}}.$

Thus

 $U(x,\,s)\;=\;\frac{ck}{s\sqrt{s}}e^{-\frac{x}{c}\sqrt{s}},$

which corresponds to

 $u(x,\,t)\;:=\;k\left[2c\sqrt{\frac{t}{\pi}}e^{-\frac{x^{2}}{4c^{2}t}}-x\mbox{ % erfc}\frac{x}{2c\sqrt{t}}\right]\!.$

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