Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 11

Table of contents
- a 11
  - Solution a
- b 11
  - Solution b
- c 11
  - Solution c
- d 11
  - Solution d

a

Solution

Applying the multivariable chain rule at fixed

y

gives

(\frac{\partial f}{\partial x})_{y} = (\frac{\partial f}{\partial u})_{v} (\frac{\partial u}{\partial x})_{y} + (\frac{\partial f}{\partial v})_{u} (\frac{\partial v}{\partial x})_{y} .

The held-fixed variables distinguish the two coordinate descriptions; the same scalar function is being differentiated.

b

Solution

Differentiate the first-order chain rule again, including derivatives of the coordinate coefficients. The second derivatives under a two-variable coordinate change are

f_{xx} = f_{uu} u_{x}^{2} + 2 f_{uv} u_{x} v_{x} + f_{vv} v_{x}^{2} + f_{u} u_{xx} + f_{v} v_{xx},

and

f_{yy} = f_{uu} u_{y}^{2} + 2 f_{uv} u_{y} v_{y} + f_{vv} v_{y}^{2} + f_{u} u_{yy} + f_{v} v_{yy} .

The last two terms account for derivatives of the coordinate functions; they vanish for an affine map. Smoothness justifies equality of the two mixed partial derivatives.

c

Solution

For polar coordinates with

u > 0

,

\partial_{x} = cos v \partial_{u} - (sin v / u) \partial_{v}

and

\partial_{y} = sin v \partial_{u} + (cos v / u) \partial_{v}

. Write

c = cos v

,

s = sin v

. Applying these operators twice, including their variable coefficients, gives the Cartesian second derivatives in polar coordinates:

f_{xx} = c^{2} f_{uu} - \frac{2 sc}{u} f_{uv} + \frac{s ^{2}}{u ^{2}} f_{vv} + \frac{s ^{2}}{u} f_{u} + \frac{2 sc}{u ^{2}} f_{v},

f_{yy} = s^{2} f_{uu} + \frac{2 sc}{u} f_{uv} + \frac{c ^{2}}{u ^{2}} f_{vv} + \frac{c ^{2}}{u} f_{u} - \frac{2 sc}{u ^{2}} f_{v} .

In particular their sum is the Laplacian in polar coordinates,

f_{uu} + f_{u} / u + f_{vv} / u^{2}

.

d

Solution

The substitution gives

x^{2} - y^{2} = u^{2} cos 2 v

. For this function

f_{uu} = 2 cos 2 v

,

f_{u} / u = 2 cos 2 v

, and

f_{vv} / u^{2} = - 4 cos 2 v

, whose sum is zero. Thus it satisfies the polar form of Laplace's equation for

u > 0

. The apparent coordinate singularity at

u = 0

does not make the original Cartesian polynomial singular; it is a smooth harmonic function through the origin.

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