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

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

In this we have

$\text{(a) }\{\begin{array}{cc}\text{boundary conditions}u(\mathrm{\infty },t)=0,u(0,t)=u_0,\hfill & \\ \text{initial conditions}\mathit{\hspace{1em}\hspace{1em}}u(x,\mathrm{\hspace{0.17em}0})=0,\underset{\text{for }x>\mathrm{\hspace{0.17em}0}}{\underbrace{u_t^{\prime }(x,\mathrm{\hspace{0.17em}0})=0}}\hfill & \end{array}$

and

$\text{(b) }\{\begin{array}{cc}\text{boundary conditions}u(\mathrm{\infty },t)=0,u_x^{\prime }(0,t)=-k,\hfill & \\ \text{initial conditions}\hspace{1em}\hspace{1em}u(x,\mathrm{\hspace{0.17em}0})=0,\underset{\text{for }x>\mathrm{\hspace{0.17em}0}}{\underbrace{u_t^{\prime }(x,\mathrm{\hspace{0.17em}0})=0}}.\hfill & \end{array}$

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

$$U_{xx}^{\prime \prime }(x,s)=\frac{1}{c^2}[sU(x,s)-u(x,\mathrm{\hspace{0.17em}0})],$$

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

$$U(\mathrm{\infty },s)={\int }_0^{\mathrm{\infty }}{e}^{-st}u(\mathrm{\infty },t)𝑑t={\int }_0^{\mathrm{\infty }}0𝑑t\equiv \mathrm{\hspace{0.33em}0},$$

we must have  $C_1=0$.  Thus the Laplace transform of the solution of (1) is in both cases (a) and (b)

$U(x,s)=C_2{e}^{-\frac{\sqrt{s}}{c}x}.$(2)

For (a), the second boundary condition implies   $U(0,s)={\displaystyle \frac{u_0}{s}}$.  But by (2) we must have  $U(0,s)=C_2\cdot 1$, whence we infer that  $C_2={\displaystyle \frac{u_0}{s}}$.  Accordingly,

$$U(x,s)=u_0\cdot \frac{1}{s}{e}^{-\frac{x}{c}\sqrt{s}},$$

which corresponds to the solution function

$$u(x,t):=u_0\text{ erfc}\frac{x}{2c\sqrt{t}}$$

of the heat equation (1).

For (b), the second boundary condition says that  $U_x^{\prime }(0,s)=-{\displaystyle \frac{k}{s}}$,  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

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