Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2008 / 8

Solution

e^{2 y} u_{xx} - u_{yy}

e^{2 y} (d y)^{2} - (d x)^{2} = 0,

d x / d y = \pm e^{y}

. Integration gives

x + e^{y} = constant

and

x - e^{y} = constant

. Thus take

ξ = x + e^{y}

η = x - e^{y}

- 2 e^{y} \neq = 0

, and their image is

ξ > η

For

u (x, y) = U (ξ, η)

, the chain rule gives

u_{xx} = U_{ξξ} + 2 U_{ξ η} + U_{ηη}, u_{y} = e^{y} (U_{ξ} - U_{η}),

u_{yy} = e^{2 y} (U_{ξξ} - 2 U_{ξ η} + U_{ηη}) + e^{y} (U_{ξ} - U_{η}) .

4 e^{2 y} U_{ξ η} = 0, U = F (ξ) + G (η) .

y = 0

F (x + 1) + G (x - 1) = x

. Differentiating this condition in

x

and using

u_{y} = 1

gives

F^{'} (x + 1) + G^{'} (x - 1) = 1, F^{'} (x + 1) - G^{'} (x - 1) = 1.

Hence

F^{'} = 1

and

G^{'} = 0

. Substitution into the value condition fixes their combined constant and yields

u (x, y) = x + e^{y} - 1.

Direct substitution verifies the PDE and both initial data. The cancellation illustrates a hyperbolic equation reduced by an exponential characteristic coordinate.

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