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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 3

Choose the <stream function> convention
$$
u=-\psi_y,
\qquad
v=\psi_x,
$$
which satisfies continuity identically. Introduce
$$
D=\left(\frac{\mu^2}{\rho B}\right)^{1/3},
\qquad
U_0=\left(\frac{B^2}{\rho\mu}\right)^{1/3},
$$
so that
$$
\Delta=Dx^{2/5},
\qquad
U=U_0x^{1/5},
\qquad
\eta=\frac{y}{Dx^{2/5}}.
$$
Take the <similarity solution>
$$
\psi=U\Delta f(\eta)=U_0D x^{3/5}f(\eta).
$$
Then
$$
u=-U_0x^{1/5}f'
$$
and
$$
v=U_0D x^{-2/5}
\left(\frac35f-\frac25\eta f'\right).
$$
A direct differentiation shows that the terms containing $\eta f'f''$ cancel:
$$
u u_x+v u_y
=U_0^2x^{-3/5}
\left[\frac15(f')^2-\frac35ff''\right].
$$
Also,
$$
u_{yy}=-\frac{U_0}{D^2}x^{-3/5}f''',
$$
while the delta scaling gives
$$
b(x)\delta(y)
=\frac{B}{D}x^{-3/5}\delta(\eta).
$$
The definitions of $D$ and $U_0$ imply
$$
\frac{\rho U_0D^2}{\mu}=1,
\qquad
\frac{BD}{\mu U_0}=1.
$$
After cancelling the common dimensional factor, the <similarity equation for a two-dimensional laminar plume> is
$$
\boxed{f'''+\frac15(f')^2-\frac35ff''=\delta(\eta)}.
$$
Thus
$$
\boxed{c_1=\frac15,\qquad c_2=-\frac35}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 38D / c / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 38D / c / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 38D / c / Solution

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