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Solved Problems

Solution:

(y + u)u + yu = x − y, u y=1 = 1 + x (2.4)
y
x
We differentiate u with respect to s.

du dx dy
= u x + u y
ds ds ds

We compare this with Equation 2.4 to obtain differential equations for x, y and u.

dx dy du
= y + u, = y, = x − y
ds ds ds

We parametrize the initial data in terms of s.

x(s = 0) = α, y(s = 0) = 1, u(s = 0) = 1 + α

We solve the equation for y subject to the inital condition.

y(s) = e s

This gives us a coupled set of differential equations for x and u.

dx s du s
= e +u, = x − e
ds ds

The solutions subject to the initial conditions are

s
−s
s
x(s) = (α + 1) e − e , u(s) = α e + e −s .
s
We substitute y(s) = e into these solutions.

1 1
x(s) = (α + 1)y − , u(s) = αy +
y y

We solve the first equation for α and substitute it into the second equation to

obtain the solution.
2 + xy − y 2
u(x, y) =
y
This solution is valid for y > 0. The characteristic passing through (α, 1) is

−s
s
s
x(s) = (α + 1) e − e , y(s) = e .
Hence we see that the characteristics satisfy y(s) ≥ 0 for all real s.

Exercise 28.
Consider the partial differential equation

∂u ∂u
2 3
ut + x − t x u = 0.
∂t ∂x

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