1
$\begingroup$

In the book, Conduction of Heat in Solids by Carslaw and Jaeger (Section 2.5), the general solution for the heat equation

$$u_t = \kappa u_{xx}$$ with the boundary conditions of $$u(x,0)=0\\ u(0,t)=\phi (t)$$

is given

$$ u = \frac{2}{\sqrt \pi} \int_{x / 2\sqrt (\kappa t)}^{\infty} \phi \Bigg (t-\frac{x^2}{4 \kappa \mu^2} \Bigg )e^{-\mu^2} d\mu $$ where $$\mu = \frac{x}{2\sqrt(\kappa(t-\lambda))}$$

Then, a series of special cases is given. For example, if $\phi (t) = e^{\lambda t}$ the above solution transforms to

$$ u = \frac{e^{\lambda t}}{2} \Bigg (e^{-x/\sqrt(\lambda/\kappa)} erfc \Big (\frac{x}{2 \sqrt(\kappa t)}- \sqrt (\lambda t) \Big) + e^{-x/\sqrt(\lambda/\kappa)} erfc \Big (\frac{x}{2 \sqrt(\kappa t)}+ \sqrt (\lambda t) \Big) \Bigg )$$

My question is how is this equation derived from the above-mentioned general solution?

$\endgroup$
  • $\begingroup$ Are you working on $x \ge 0$? $\endgroup$ – DisintegratingByParts Apr 27 '18 at 19:26
  • $\begingroup$ @DisintegratingByParts yes $x$ is non-negative. $\endgroup$ – Googlebot Apr 27 '18 at 22:23
  • $\begingroup$ @FelixMarin the typo was fixed. It is $x/2\sqrt{\kappa t}$ $\endgroup$ – Googlebot Apr 27 '18 at 22:23
2
$\begingroup$

The definition of $erfc$ is

$$erfc(x) = \frac{2}{\sqrt \pi} \int_{0}^{x} e^{-\mu^2} d\mu.$$

So substituting the value of $\varphi$

$$u = \frac{2}{\sqrt \pi} \int_{x^2 / \sqrt (\kappa t)}^{\infty} \exp \Bigg (\lambda (t-\frac{x^2}{4 \kappa \mu^2} ) -\mu^2\Bigg ) d\mu$$ $$= \frac{e^{\lambda t}}{2} \Bigg (e^{-x/\sqrt(\lambda/\kappa)} erfc \Big (\frac{x}{2 \sqrt(\kappa t)}- \sqrt (\lambda t) \Big) + e^{-x/\sqrt(\lambda/\kappa)} erfc \Big (\frac{x}{2 \sqrt(\kappa t)}+ \sqrt (\lambda t) \Big) \Bigg ) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.