= Solution
For steady uniform-density inviscid flow with no body force, the <Euler equations> read $(\mathbf u\cdot\nabla)\mathbf u=-\nabla p/\rho$. Dotting with $\mathbf u$ gives
$$
\mathbf u\cdot\nabla\left(\frac12u^2+\frac p\rho\right)=0.
$$
Thus the <Bernoulli equation> states that $u^2/2+p/\rho$ is constant along each <streamline>; the constants need not be the same on unrelated <streamlines>.
In the slowly varying tube approximation, mass conservation gives $\pi(r-h)^2u=\pi R^2V$, so $r-h=R\sqrt{V/u}$. The upstream Bernoulli constant and the wall <pressure> law give
$$
p-p_0=\frac\rho2(V^2-u^2)=k(r-R),\qquad
\frac rR=1+\frac\lambda4\left[1-(u/V)^2\right],\quad\lambda=\frac{2\rho V^2}{kR}.
$$
Subtracting the internal radius obtains
$$
\boxed{\frac hR=H(s)=1-s^{-1/2}+\frac\lambda4(1-s^2),\qquad s=u/V>0}.
$$
This is <choking by wall thickness in an elastic tube>. The function tends to $-\infty$ as either $s\downarrow0$ or $s\to\infty$, and $H(1)=0$. Its derivative is
$$
H'(s)=\frac12s^{-3/2}-\frac\lambda2s,
\qquad H'(s)=0\iff s=s_c=\lambda^{-2/5}.
$$
It is positive before $s_c$ and negative afterwards, so the unique maximum is
$$
\boxed{\frac{h_c(\lambda)}R=H(s_c)=1+\frac\lambda4-\frac54\lambda^{1/5},\qquad
u_c=V\lambda^{-2/5}}.
$$
The maximum is nonnegative because the arithmetic-geometric mean inequality gives $4+\lambda\ge5\lambda^{1/5}$; it is zero only at $\lambda=1$. If any prescribed thickness exceeds $h_c$, no positive velocity can satisfy the mass/Bernoulli conditions there, so no flow of this assumed form is possible. Equality is the double-root critical case.
For each $h<h_c$, strict monotonicity on $(0,s_c)$ and $(s_c,\infty)$, together with the two negative-infinite limits, gives exactly two positive roots. The upstream condition is $s\to1$ as $h\to0$. Therefore
$$
\boxed{\begin{array}{c|c|c}
\lambda&\text{selected root}&\text{response to increasing }h\\\hline
0<\lambda<1&s<s_c&u>V\text{ increases}\\
\lambda>1&s>s_c&u<V\text{ decreases}
\end{array}}.
$$
For $\lambda<1$, the upstream value $1$ lies on the rising branch; for $\lambda>1$, it lies on the falling branch. Smooth continuation preserves this choice because the branches meet only at $h_c$, which is excluded by the strict bound. At $\lambda=1$ the upstream state is already critical: $h_c=0$, so the requirement $0\le h<h_c$ cannot hold, and any positive wall thickness is incompatible with the approximation. With $h=0$ the critical root is $u=V$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-3-elastic-tube-choking.png]
{title=Velocity branches and critical wall thickness for steady flow through an elastic tube}
The graph shows the unique maximum and the upstream point $(s,H)=(1,0)$ on the three parameter regimes. Only the portion with $H\ge0$ represents nonnegative wall thickness. The vertical maximum line separates the two roots when the prescribed thickness lies strictly below the maximum.
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