# Substitution

Some second-order ODEs can be solved by making a substitution

\begin{equation}
u = \dd{}{y}{x}
\end{equation}

Then, you effectively "integrate twice": once to solve for *u*, then once more
to get *y* from *u*.

```{example} Kinematics
Newton's second law is $F = ma$, where *F* is the applied force, *m* is the
mass, and *a* is the acceleration of a body. We also know that acceleration
is the derivative of the velocity *v*, which is itself the first derivative
of position *x*. Hence, Newton's second-law is a second-order ODE:

\begin{equation}
m \dd{2}{x}{t} = F
\end{equation}

Solve for the position of a body *x* as a function of time *t* experience a
constant gravitational force $F = -mg$, where *g* is the acceleration due to
gravity.

---

Substitute the gravitational force and simplify

\begin{align}
m \dd{2}{x}{t} &= -mg \\
\dd{2}{x}{t} &= -g
\end{align}

Make the substitution $v = \d{x}/\d{t}$, then integrate because the ODE is
separable

\begin{align}
\dd{}{v}{t} &= -g \\
\int \d{v} &= \int -g \d{t} \\
v &= -g t + c_1
\end{align}

Replace *v* and integrate again:

\begin{align}
\dd{}{x}{t} &= -g t + c_1 \\
\int \d{x} &= \int\left(-g t + c_1 \right)\d{t} \\
x &= -\frac{1}{2} g t^2 + c_1 t + c_2
\end{align}

If the initial position is $x(0) = x_0$ and initial velocity is
$v(0)=x'(0)=v_0$, then

\begin{align}
x(0) &= c_2 = x_0 \\
v(0) &= c_1 = v_0
\end{align}

so

\begin{equation}
x(t) = -\frac{1}{2} g t^2 + v_0 t + x_0
\end{equation}

This is the classic equation of ballistic motion!
```

````{example} Incompressible flow in a cylinder
Steady, laminar pressure-driven flow in a cylindrical pipe:

```{image} ./_images/pipe_flow.jpg
:alt: Pipe flow
:width: 300px
:align: center
```

is governed by the simplified Navier-Stokes equation:

\begin{equation}
\mu \frac{1}{r} \dd{}{}{r} \left( r \dd{}{u_z}{r} \right) = -\frac{\Delta P}{L}
\end{equation}

where $u_z$ is the velocity along the pipe axis, $\Delta P$ is the difference
between the pressure at the inlet and at the outlet, *L* is the length of the
pipe, and $\mu$ is the dynamic viscosity.


The pipe walls have no-slip boundary conditions, meaning the velocity is zero
there. Derive an expression for $u_z(r)$.

---

Make the substitution $v = r \d{u_z}/\d{r}$, then rearrangeso

\begin{align}
\mu \frac{1}{r} \dd{}{v}{r} &= -\frac{\Delta P}{\mu L} \\
\int \d{v} &= \int -\frac{1}{\mu} \frac{\Delta P}{L} r \d{r} \\
v &=  -\frac{1}{2\mu} \frac{\Delta P}{L} r^2 + c_1
\end{align}

Then, substitute for *v*, separate, and integrate again:

\begin{align}
r \dd{}{u_z}{r} &= -\frac{1}{2\mu} \frac{\Delta P}{L} r^2 + c_1 \\
\int \d{u_z} &= \int \Biggl( -\frac{1}{2\mu} \frac{\Delta P}{L} r +
  \frac{c_1}{r} \Biggr) \d{r} \\
u_z &= -\frac{1}{4\mu} \frac{\Delta P}{L} r^2 + c_1 \ln r + c_2
\end{align}

The walls have no-slip boundary conditions so $u_z(R) = 0$. Additionally, the
pipe must have radial symmetry so $u_z'(0) = 0$. Applying these boundary
conditions requires:

\begin{align}
\lim_{r \to 0} u_z'(r) &= \lim_{r \to 0} \frac{c_1}{r} = 0 \\
u_z(R) &= -\frac{1}{4\mu} \frac{\Delta P}{L} R^2 + c_1 \ln R + c_2 = 0
\end{align}

The first equation requires $c_1 = 0$. The second equation then gives

\begin{equation}
c_2 = \frac{1}{4\mu} \frac{\Delta P}{L} R^2
\end{equation}

All together,

\begin{equation}
u_z(r)=\frac{1}{4 \mu}\frac{\Delta P}{L}(R^2-r^2)
\end{equation}

This is the classic Hagen-Poiseuille flow profile.
````