5 7.24, §5 Back

We need to find the original function $u$ corresponding to the transformed

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

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

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

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 $f(t)g(t)$. The convolution theorem Table 6.3 # 7 is needed:

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