10.3 The Laplace transform
For a function on \(t \geq 0\) of at most exponential growth, the Laplace transform is \[ \bar f(s) = \int _0^\infty f(t)e^{-st}\,dt \quad (\operatorname {Re}s \text { large}) , \] with \(\overline {f'}(s) = s\bar f(s) - f(0)\) and \(\overline {f * g} = \bar f\,\bar g\) for the one-sided convolution \(\int _0^tf(t - \tau )g(\tau )\,d\tau \). It is the natural tool in time because it absorbs the initial value, and it handles boundary data that switch on at \(t = 0\).
Example 10.3.1. Let the end \(x = 0\) of a half-infinite rod, initially at temperature \(0\), be held at temperature \(h(t)\) from \(t = 0\): \(u_t = \kappa u_{xx}\) for \(x > 0\), \(u(x, 0) = 0\), \(u(0, t) = h(t)\). Transforming in \(t\), \(s\bar u = \kappa \bar u_{xx}\), so \(\bar u = A(s)e^{-x\sqrt {s/\kappa }}\) for a solution bounded in \(x\), and the boundary condition gives \[ \bar u(x, s) = \bar h(s)\,e^{-x\sqrt {s/\kappa }} . \] For \(h = 1\), \(\bar h = 1/s\), and the inverse transform of \(e^{-x\sqrt {s/\kappa }}/s\) is \(\operatorname {erfc}\bigl (x/\sqrt {4\kappa t}\bigr )\) (Software 10.4.3); this is \(1\) minus the image solution \(\operatorname {erf}(x/\sqrt {4\kappa t})\) of a problem in Section 9, as it must be, since \(1 - u\) starts at \(1\) and is held at \(0\) at the end. For general \(h\), the convolution rule gives Duhamel’s formula \[ u(x, t) = \int _0^th(\tau )\,\frac {\partial }{\partial t}\operatorname {erfc}\frac {x}{\sqrt {4\kappa (t - \tau )}}\,d\tau , \] a superposition of responses to the switching on of the boundary temperature at each earlier time.
The inverse Laplace transform is the contour integral \(f(t) = \frac 1{2\pi i}\int _{\gamma - i\infty }^{\gamma + i\infty }\bar f(s)e^{st}\,ds\), the Fourier inversion formula along a vertical line. In practice it is found from tables, from the residues of \(\bar f\), or numerically by deforming the contour into the left half-plane so that \(e^{st}\) decays rapidly (Talbot’s method).
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