7.2 A parabolic example

This example illustrates Laplace transform solution for a parabolic partial differential equation.

7.2.1 The physical problem

Find the flow velocity in a viscous fluid being dragged along by an accelerating plate.

Figure 7.1: Viscous flow next to a moving plate

7.2.2 The mathematical problem

Figure 7.2: Viscous flow next to a moving plate

• Semi-infinite domain $\bar \Omega$: $0\le x < \infty$
• Unknown vertical velocity $u=u(x,t)$
• Parabolic
• One homogeneous initial condition
• One Neumann boundary condition at $x=0$ and a regularity constraint at $x=\infty$
• Constant kinematic viscosity $\kappa$

Try a Laplace transform in $t$.

7.2.3 Transform the problem

Transform the partial differential equation:

$$\strut u_t = \kappa u_{xx} \quad \stackrel{\hbox{Table 6... ... - \overline{\smash{u(x,0)}\vphantom{.}} = \kappa \hat u_{xx}$$

Transform the boundary condition:

$$u_x = g(t) \quad \Relbar\joinrel\Relbar\joinrel\Relb... ...\Relbar\joinrel\Longrightarrow \quad \hat u_x = \hat g(s)$$

7.2.4 Solve the transformed problem

Solve the partial differential equation:

$$s \hat u = \kappa \hat u_{xx}$$

This is a constant coefficient ordinary differential equation in $x$, with $s$ simply a parameter. Solve from the characteristic equation:

$$s = \kappa k^2 \quad\quad\Rightarrow\quad\quad k = \pm \sqrt{s/\kappa}$$

$$\hat u = A e^{\sqrt{s/\kappa} x} + B e^{-\sqrt{s/\kappa} x}$$

Apply the boundary condition at $x=\infty$ that $u$ must be regular there:

$$A=0$$

Apply the given boundary condition at $x=0$:

$$\hat u_x = \hat g(s) \quad\quad\Rightarrow\quad\quad - B \sqrt{\frac{s}{\kappa}} = \hat g$$

Solving for $B$ and plugging it into the solution of the ordinary differential equation, $\hat u$ has been found:

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

7.2.5 Transform 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.

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

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:

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