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Solved Problems

We substitute these results into the differential equation to obtain the canonical
form.

u xx + 2u xy − 3u + 2u + 6u = 0
yy
x
y
16u ξψ + 8u = 0.
ξ
Exercise 65. Classify and transform the following equation into canonical form.

2
u xx + (1 + y) u yy = 0

Solution: For y = −1, the equation is parabolic. For this case it is already in the
canonical form, u xx = 0.
For y 6= −1, the equation is elliptic. We find new variables that will put the
equation in the form u ξψ = G(ξ, ψ, u, u , u ).
ξ
ψ
dy p
2
= i (1 + y) = i(1 + y)
dx
dy
= idx
1 + y
log(1 + y) = ix + c

1 + y = c e ix

(1 + y) e −ix = c
ξ = (1 + y) e −ix

ψ = ξ = (1 + y) e ix

The variables that will put the equation in canonical form are

ξ + ψ ξ − ψ
σ = = (1 + y) cos x, τ = = (1 + y) sin x.
2 i2

We calculate the derivatives of σ and τ.

σ = −(1 + y) sin x σ = cos x
y
x
τ = (1 + y) cos x τ = sin x
x
y
Then we calculate the derivatives of u.

u = −(1 + y) sin(x)u + (1 + y) cos(x)u τ
σ
x
u = cos(x)u + sin(x)u τ
σ
y
2
2
2
2
u =(1+y) sin (x)u +(1+y) cos (x)u −(1 + y) cos(x)u − (1 + y) sin(x)u τ
xx
σ
σσ
ττ
2
2
u yy = cos (x)u σσ + sin (x)u ττ
We substitute these results into the differential equation to obtain the canonical
form.
2
u xx + (1 + y) u yy = 0
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