MSO 203b Assignment-02 October 08 – 15, 2012.

1. Classify the following second order PDE as hyperbolic, parabolic or elliptic.

(a) uxx + (5 + 2y2)uxy + (1 + y2)(4 + y2)uyy = 0.

Solution: Hyperbolic. A = 1, B = (5 + 2y2)/2 and C = (1 + y2)(4 + y2), thus

B2 − AC = 25/4 + y4 + 5y2 − 4− 5y2 − y4 = 9/4 > 0.

(b) (Tricomi equation) yuxx + uyy = 0.

Solution: Mixed type. A = y, B = 0 and C = 1, thus B2 −AC = −y. Hence,it hyperbolic in the lower half-plane y < 0 and elliptic in upper half-plane y > 0.On y = 0, x-axis, it is degenerately parabolic.

(c) yuxx = xuyy.

Solution: Mixed type. A = y,B = 0, C = −x and B2 − AC = xy. Thus, itis hyperbolic when xy > 0 (I and III quadrant), degenerately parabolic whenxy = 0 (both x and y axes) and elliptic when xy < 0 (II and IV quadrant).

(d) uyy − xuxy + yux + xuy = 0.

Solution: Mixed type. A = 0, B = −x/2, C = 1 and B2 − AC = x2/4. Thus,it is hyperbolic for x 6= 0 (R2 \ x = 0) and degeneratly parabolic on they-axis(x = 0).

(e) y2uxx + 2xyuxy + x2uyy = 0.

Solution: Parabolic. Since A = y2, B = xy, C = x2 and B2 − AC = 0.

(f) uxx + 2xuxy + (1− y2)uyy = 0.

Solution: Mixed type. A = 1, B = x, C = 1− y2 and B2 −AC = x2 + y2 − 1.Hence it is parabolic in x2 + y2 = 1 (circle of radius one), elliptic in x2 + y2 < 1(disk of radius one) and hyperbolic in x2 + y2 > 1 (R2 with unit ball removed).

2. Rewrite the PDE’s in their canonical forms and solve them.

(a) uxx + 2√

3uxy + uyy = 0

MSO 203b Page 2 of 5 October 08 – 15, 2012.

Solution: A = 1, B =√

3, C = 1 and B2 − AC = 2 > 0. The equation ishyperbolic and the characteristic equation is

dy

dx=√

3±√

2.

Solving we get y − (√

3 ±√

2)x = c. We set w = y − (√

3 +√

2)x and z =y − (

√3−√

2)x. Then

ux = −(√

3 +√

2)uw − (√

3−√

2)uz

uy = uw + uz

uxx = (√

3 +√

2)2uww + 2uwz + (√

3−√

2)2uzz

uyy = uww + 2uwz + uzz

uxy = −(√

3 +√

2)uww − 2√

3uwz − (√

3−√

2)uwz)

Substituting into the governing equations, we get

uwz = 0

uw = F (w)

u =

∫F (w) dw + g(z)

u = f(w) + g(z).

Thus, the general solution is

u = f(y − (√

3 +√

2)x) + g(y − (√

3−√

2)x).

(b) x2uxx − 2xyuxy + y2uyy + xux + yuy = 0

Solution: A = x2, B = −xy, C = y2 and B2 − AC = 0. The equation isparabolic and the characteristic equation is

dy

dx=−yx

,

with solution as xy = c. Let w = xy and we choose z = x such that Jacobianis non-zero. Jacobian is non-zero for x 6= 0. Now if x = 0, the PDE reduces to

MSO 203b Page 3 of 5 October 08 – 15, 2012.

yuyy + uy = 0, which is already in canonical form. Thus, for x 6= 0,

ux = yuw + uz

uy = xuw

uxx = y2uww + 2yuwz + uzz

uyy = x2uww

uxy = uw + xyuww + xuwz

Substituting them into the PDE, we get

x2uzz + xuz = 0 =⇒ zuzz + uz = 0

Solving for u, we get

d

dz(zuz) = 0

zuz = f(w)

uz =f(w)

zu = f(w) ln(|z|) + g(w)

u = (ln |x|)f(xy) + g(xy).

(c) uxx − (2 sinx)uxy − (cos2 x)uyy − (cosx)uy = 0

Solution: A = 1, B = − sinx, C = − cos2 x and B2 − AC = 1 > 0. Theequation is hyperbolic and the characteristic equation is

dy

dx= − sinx± 1,

with solution as y∓ x− cosx = c. Let w = y + x− cosx and z = y− x− cosx.Then

ux = (1 + sinx)uw + (−1 + sin x)uz

uy = uw + uz

uxx = cosx(uw + uz) + (1 + sinx)2uww − 2 cos2 xuwz + (1− sinx)2uzz

uyy = uww + 2uwz + uzz

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Substituting into the PDE, we get

uwz = 0

uw = F (w)

u =

∫F (w) dw + g(z)

u = f(w) + g(z)

u = f(y + x− cosx) + g(y − x− cosx).

(d) uxx + 4uxy + 4uyy = 0

Solution: A = 1, B = 2, C = 4 and B2 − AC = 0. The equation is parabolicand the characteristic equation is

dy

dx= 2

with the solution 2x − y = c. Let w = 2x − y and we choose z = x such thatthe Jacobian is non-zero. Hence

ux = 2uw + uz

uy = −uw

uxx = 4uww + 4uwz + uzz

uyy = uww

uxy = −2uww − uwz

Substituting into the PDE, we get

uzz = 0

uz = f(w)

u = f(w)z + g(w)

u = xf(2x− y) + g(2x− y).

3. Show that the following two variable functions are harmonic.

(i) ax + by + c for given constants a and b (ii) xy(iii) x3 − 3xy2 (iv) x4 − 6x2y2 + y4

(v) x2 − y2 (vi) ex sin y(vii) ex cos y

MSO 203b Page 5 of 5 October 08 – 15, 2012.

Solution: Trivial differentiation.

4. Show that u(x, y) = ln r is harmonic in R2 \ 0, i.e., u is a solution of the Laplaceequation

∆u = 0 in R2 \ 0,

where r =√

x2 + y2. (Hint: Use the fact that r = |(x, y)| and work in polar coordinates).

Solution: Verification is easy.

5. Show that u(x, y) = 1r

is harmonic in R3\0, i.e., u is a solution of the Laplace equation

∆u = 0 in R3 \ 0,

where r =√

x2 + y2 + z2. (Hint: Use the fact that r = |(x, y, z)| and work in sphericalcoordinates).

Solution: Verification is easy.

6. Let Ω = (x, y) ∈ R2 : 0 < x < a, and 0 < y < b be a rectangle with boundary ∂Ω.Solve the Dirichlet problem:

(a) ∆u(x, y) = 0 (x, y) ∈ Ωu(x, y) = x + y (x, y) ∈ ∂Ω

Solution: u(x, y) = x + y satisfies the Laplace equation and the boundaryconditions. Hence, by maximum principle, u(x, y) = x + y is the solution

(b) ∆u(x, y) = 0 (x, y) ∈ Ωu(x, y) = xy (x, y) ∈ ∂Ω

Solution: Similar to (a) and u(x, y) = xy is the solution.