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Here, s and α are the new independent variables. Solving these ODE’s yields

s
x = as y = bs + α u = e f(α).

We can solve for s and α and substitute these variables into the expression for u.

s = x/a α = y − bx/a u = e x/a f(y − bx/a)

Thus the solution to the partial differential equation is

u(x, y) = e x/a f(y − bx/a) .

Example 2.4 As another example of the method, consider the following problem
from “Handbook of Differential Equations” by Zwillinger.

2
u x + x u y = −yu, u(0, y) = f(y).
The corresponding system of ODE’s and initial conditions are

∂x ∂y 2 du
= 1 = x = −yu
∂s ∂s ds
x(s = 0) = 0 y(s = 0) = α u(s = 0) = f(α)

Solving the ODE’s and substituting in the values of the variables yields

∂y
x = s = s 2
∂s
s 3 du 3
s
x = s y = + α = − + α u
3 ds 3
4
3
x s
s = x α = y − u = f(α) exp − − sα
3 12

Thus the solution is
3 4
x x
u(x, y) = f y − exp − xy .
3 4

2.6 Similarity Methods

2.6.1 Introduction.

Consider the partial differential equation (not necessarily linear)

∂u ∂u
F , , u, t, x = 0.
∂t ∂x

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