5 7.24, §5 Back

We need to find the original function corresponding to the transformed

$\begin{displaymath} \hat u = - \sqrt{\frac{\kappa}{s}} e^{-\sqrt{s/\kappa} x} \hat g \end{displaymath}$

We do not really know what $\hat g$ is, just that it transforms back to . However, we can find the other part of $\hat u$ in the tables.

$\begin{displaymath} - \sqrt{\frac{\kappa}{s}} e^{-\sqrt{s/\kappa} x} \quad ... ...rrow} \quad - \sqrt{\frac{\kappa}{\pi t}} e^{-x^2/4\kappa t} \end{displaymath}$

How does $\hat g$ times this function transform back? The product of two functions, say $\hat f(s)\hat g(s)$ , does not transform back to . The convolution theorem Table 6.3 # 7 is needed:

$\begin{displaymath} u(x,t) = - \int_0^t \sqrt{\frac{\kappa}{\pi (t-\tau)}} e^{-x^2/4\kappa(t-\tau)} g(\tau) \d\tau \end{displaymath}$