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[parent] using Laplace transform to solve heat equation (Example)

Along the whole positive $x$ -axis, we have an ideal heat-conducting rod, the surface of which is isolated. The initial temperature of the rod is 0 degrees. 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

(1)

In this we have
(a)    boundary conditions   initial conditions    

and
(b)    boundary conditions   initial conditions    

For solving (1), we first form its Laplace transform (see the table of Laplace transforms) $$U_{xx}''(x,\,s) \;=\; \frac{1}{c^2}[s\,U(x,\,s)-u(x,\,0)],$$ which is a simple ordinary linear differential equation $$U_{xx}''(x,\,s) \;=\; \left(\frac{\sqrt{s}}{c}\right)^2U(x,\,s)$$ of order two. Here, $s$ is only a parametre, and the general solution of the equation is $$U(x,\,s) \;=\; C_1e^{\frac{\sqrt{s}}{c}x}+C_2e^{-\frac{\sqrt{s}}{c}x}$$ (see this entry). 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)
(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 $\displaystyle U_x'(0,\,s) = -\frac{k}{s}$ , and since (2) implies that $U_x'(x,\,s) =-\frac{\sqrt{s}}{c}C_2e^{-\frac{\sqrt{s}}{c}x}$ , 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 $$u(x,\,t) \;:=\; k\left[2c\sqrt{\frac{t}{\pi}}e^{-\frac{x^2}{4c^2t}}-x\,\mbox{erfc}\frac{x}{2c\sqrt{t}}\right]\!.$$

[Not ready . . .]




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See Also: Laplace transform

Other names:  using Laplace transform to solve partial differential equation
Keywords:  PDE

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Cross-references: implies, boundary condition, solution, equation, general solution, parametre, linear differential equation, table of Laplace transforms, Laplace transform, dimension, heat equation, flux, function, surface, positive
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This is version 6 of using Laplace transform to solve heat equation, born on 2009-07-11, modified 2009-07-19.
Object id is 11836, canonical name is UsingLaplaceTransformToSolveHeatEquation.
Accessed 828 times total.

Classification:
AMS MSC35K05 (Partial differential equations :: Parabolic equations and systems :: Heat equation)
 35K20 (Partial differential equations :: Parabolic equations and systems :: Boundary value problems for second-order, parabolic equations)
 35Q99 (Partial differential equations :: Equations of mathematical physics and other areas of application :: Miscellaneous)

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