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Solved Problems

x
Exercise 26. Solve using characteristics: uu + u = 1, u x=y = .
x
y
2
Solution:
x

uu + u = 1, u x=y = 2 (2.1)
y
x
We form du .
dy
du dx
= u x + u y
dy dy
We compare this with Equation 2.1 to obtain differential equations for x and u.

dx du
= u, = 1. (2.2)
dy dy

The initial data is
α
x(y = α) = α, u(y = α) = . (2.3)
2

We solve the differenial equation for u (2.2) subject to the initial condition (2.3).
α
u(x(y), y) = y −
2

The differential equation for x becomes

dx α
= y − .
dy 2

We solve this subject to the initial condition (2.3).

1
2
x(y) = (y + α(2 − y))
2
This defines the characteristic starting at the point (α, α). We solve for α.

2
y − 2x
α =
y − 2

We substitute this value for α into the solution for u.

y(y − 4) + 2x
u(x, y) =
2(y − 2)

This solution is defined for y 6= 2. This is because at (x, y) = (2, 2), the charac-

teristic is parallel to the line x = y.

Exercise 27. Solve using characteristics:

(y + u)u + yu = x − y, u y=1 = 1 + x.
y
x

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