Additional problems, week 5
Example 1. Classify uxx+uxy +uyy =ux+uy+u as elliptic, parabolic or hyperbolic.
Example 2. Classify 8uxx + 6uyy + 4uzz +uxy + 2uxz +uyz = 0 as elliptic, parabolic or hyperbolic.
Example 3. Classify 2uxy−2uxz+ 2uyz+ 3ux−u= 0 as elliptic, parabolic or hyperbolic.
Example 4. Compute the principal part and solve the characterictic equation for (a) uxx−4uxy+ 4uyy+ 2uy +u= 0,
(b) uxx+ 2uxy−3uyy+ 3ux−u= 0, (c) e2yuxx−e2xuyy= 0
Example 5. Transform the pde
uxx+ 2uxz+uyy+ 2uyz+ 2uzz = 0 into the canonical form.
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Solutions for the additional problems Solution 1. We have
A= 1 1/2 1/2 1
!
and this results in
0 = (1−λ)2− 1 4 (1−λ)2 = 1
4 1−λ =±1
2 λ1 = 1
2, λ2 = 3 2. So the pde is elliptic.
Solution 2. For the pde 8uxx+ 6uyy+ 4uzz+uxy+ 2uxz+uyz = 0 we get
A=
8 0.5 1 0.5 6 0.5
1 0.5 4
.
A is positive definite if and only if all principal minors are positive. Here the principal minors are
det(A1) = 8 >0 det(A2) = 24− 1
4 >0 det(A3) = 196 + 1
4+ 1
4−2−1−6>0.
So A is positive definite and the pde is elliptic.
Solution 3. For2uxy −2uxz+ 2uyz+ 3ux−u= 0 we get
A=
0 1 −1 1 0 1
−1 1 0
2
with
det(A−λI) =−λ3−1−1 +λ+λ+λ
=−λ3+ 3λ−2
and the eigenvalues areλ1 = 1, λ2 = 1, λ3 =−2. The pde is hyperbolic.
Solution 4. The general approach enfold three steps:
(i) the characterictic equation
L0(x, ξ) = X
aij(x)ξiξj = 0,
(ii) computation of the characteristic directions, a characterictic direction is a vector ξ 6= 0 with L0(x, ξ) = 0,
(iii) computation of the characterictic curves/surfaces, these are the curves/surfaces S, for which the normal vector −→n(x) has a characterictic direction in every point of x∈S.
Solution for (a):
(i) L0 =ξ12−4ξ1ξ2+ 4ξ22= 0
(ii) L0 = (ξ1−2ξ2)2 = 0 has the solution ξ= (2,1)T
(iii) This results in S as the line 2x+y=cas the final solution.
Solution for (b):
(i) L0 =ξ12+ 2ξ1ξ2−3ξ22= 0 (ii) ξ1 = (1,1)T and ξ2 = (−3,1)T
(iii) characterictic curves are the lines orthogonal to these vectors, i.e. x+ y = c1 and
−3x+y=c2. Solution for (c):
(i) The characterictic equation L0 =e2yξ12−e2xξ22 = 0 can be rewritten in the form 0 = (eyξ1+exξ2)(eyξ1−exξ2)
0 =eyξ1±exξ2 0 =e−xξ1±e−yξ2
(ii) ξ1 = (ex,ey)T and ξ2 = (ex,−ey)T
(iii) the characterictic curves read exx+eyy=c1 and exx−eyy=c2.
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Solution 5. The matrix reads
A=
1 0 1 0 1 1 1 1 2
and the eigenvalues and -vector are
det(A−λI) = (1−λ)2(2−λ)−(1−λ)−(1−λ)
= (1−λ)2(2−λ)−2(1−λ)
= (1−λ)((1−λ)(2−λ)−2)
= (1−λ)(λ2−3λ+ 2−2)
= (1−λ)λ(λ−3) = 0
⇒λ1 = 0, λ2 = 1, λ3 = 3
v1 = 1
√3
1 1
−1
, v2 = 1
√2
1
−1 0
, v3 = 1
√6
1 1 2
.
The new coordinates read
ξ= 1
√3(z−x−y), η = 1
√2(y−x), ζ = 1
2(x+y+ 2z) and the transformed pde is
2uξξ+uηη + 5uζζ = 0.
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