# The general solution to the heat equation

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?

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

The definition of $erfc$ is
$$erfc(x) = \frac{2}{\sqrt \pi} \int_{0}^{x} e^{-\mu^2} d\mu.$$
So substituting the value of $\varphi$
$$u = \frac{2}{\sqrt \pi} \int_{x^2 / \sqrt (\kappa t)}^{\infty} \exp \Bigg (\lambda (t-\frac{x^2}{4 \kappa \mu^2} ) -\mu^2\Bigg ) d\mu$$ $$= \frac{e^{\lambda t}}{2} \Bigg (e^{-x/\sqrt(\lambda/\kappa)} erfc \Big (\frac{x}{2 \sqrt(\kappa t)}- \sqrt (\lambda t) \Big) + e^{-x/\sqrt(\lambda/\kappa)} erfc \Big (\frac{x}{2 \sqrt(\kappa t)}+ \sqrt (\lambda t) \Big) \Bigg )$$