5 7.36, §5 Back

We need to find the original to

$$\hat u = - \frac{a}{s+ap} \hat f e^{-sy/a}$$

Looking in the tables:

$$\frac{1}{s+ap} \quad \stackrel{\hbox{Table 6.4, \char93 3... ...lbar\joinrel \Relbar\joinrel\Longrightarrow} \quad e^{-apx}$$

The other factor is a shifted function $f$, restricted to the interval that its argument is positive:

$$e^{-sy/a} \hat f \quad \stackrel{\hbox{Table 6.3, \char93 ... ...rel\Longrightarrow} \quad \bar f\left(x - \frac{y}{a}\right)$$

With the bar, I indicate that I only want the part of the function for which the argument is positive. This could be written instead as

$$f\left(x - \frac{y}{a}\right) H \left(x - \frac{y}{a}\right)$$

where the Heaviside step function $H(x)=0$ if $x$ is negative and 1 if it is positive.

Figure 34: Function $\bar f$.

Use convolution, Table 6.3, # 7. again to get the product.

$$u(x,y) = - \int_0^x a \bar f\left(\xi - \frac{y}{a}\right) e^{-ap(x-\xi)} \d\xi$$

This must be cleaned up. I do not want bars or step functions in my answer.

I can do that by restricting the range of integration to only those values for which $\bar f$ is nonzero. (Or $H$ is nonzero, if you prefer)

Figure 35: Function $\bar f$.

Two cases now exist:

$$u(x,y) = - \int_{y/a}^x a f\left(\xi - \frac{y}{a}\right) e^{-ap(x-\xi)} \d\xi \qquad (x > \frac ya)$$

$$u(x,y) = 0 \qquad (x < \frac ya)$$

It is neater if the integration variable is the argument of $f$. So, define $\phi=\xi - y/a$ and convert:

$$u(x,y) = - \int_{0}^{x-y/a} a f\left(\phi\right) e^{-apx+py+ap\phi} \d\phi \qquad (x > \frac ya)$$

$$u(x,y) = 0 \qquad (x < \frac ya)$$

This allows me to see which physical $f$ values I actually integrate over when finding the flow at an arbitrary point:

Figure 5: Supersonic flow over a membrane.