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

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

Title	using Laplace transform to solve heat equation
Canonical name	UsingLaplaceTransformToSolveHeatEquation
Date of creation	2015-05-30 6:55:05
Last modified on	2015-05-30 6:55:05
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Example
Classification	msc 35K20
Classification	msc 35Q99
Classification	msc 35K05
Synonym	using Laplace transform to solve partial differential equation
Related topic	LaplaceTransform