MATH 220: Problem Set 3

Solutions

Problem 1. Let C(R) be given by:

(x) =

0, x < 1,1 + x, 1 < x < 0,1 x, 0 < x < 1,0, x > 1,

(1)

so that it verifies 0, (x) = 0 if |x| 1 andR (x)dx = 1.

Consider (j)j1 constructed as j(x) = j(jx), so that j(x) = 0 if |x| 1/j,and

R j(x)dx = 1, for all j 1.

(1) Let us show that j 0 in D(R). By definition, it means that we

need to prove that, for all Cc (R),Rj(x)(x)dx 0() = (0). So lets

prove that for every > 0, there exists j0 such that for all j j0,Rj(x)(x)dx (0)

< . (2)Let > 0 and Cc (R). Since is continuous, there exists > 0 such

that for all |x| < , |(x) (0)| < . Then consider j0 such that 1/j0 < .Since we know that j(x) = 0 for all |x| 1/j,

R j(x)dx = 1, and j(x) 0

for all x, for all j j0 we have:Rj(x)(x)dx (0)

= Rj(x)(x)dx

Rj(x)(0)dx

1/j1/j

j(x) |(x) (0)| dx

1/j1/j

j(x)dx = .

(3)

(2) Let Cc (R) be such that (x) = 1 for all |x| < 1. Then, for j 1,

Rj(x)

2(x)dx =

01/j

j2(1 + jx)2dx+

1/j0

j2(1 jx)2dx = 2j3. (4)

1

Therefore, as j +,Rj(x)

2(x)dx +, and{2j ()

}+j=1

does not

converge.

Consequently,{2j

}+j=1

does not converge to any distribution since{2j ()

}+j=1

does not converge for the very we exhibited.

(3) We have just shown that j 0, but{2j ()

}+j=1

does not converge

to any distribution. Therefore there is no continuous extension of the mapQ : f 7 f2 on C(R) to D(R).

Problem 2. In t 0, we consider the conservation law

ut + f(u)x = 0, u(x, 0) = (x), (5)

and suppose that u is C1 except that it has a jump discontinuity along a C1-curve given by x = (t).

Let us show that u is a weak solution of the PDE (5) if and only if u+ andu solve the PDE in the classical sense.First, lets suppose that u is a weak solution. Then, for all Cc (R [0,+)) +

0

+

( ut + f(u)x) dx dt+

+

(x)(0, x)dx = 0. (6)

Now let choose such that (x, 0) = 0, and break up the first integral intothe regions + and . We then get

0 =

+0

+

( ut + f(u)x) dx dt

=

( ut + f(u)x) dx dt+

+

( u+t + f(u+)x) dx dt

(7)Now since u is C

1 on , we can use the classical divergence theorem andthe fact that has compact support and (x, 0) = 0 to write

( ut + f(u)x) dx dt =

(( u)t + (f(u))x)dx dt

x=(t)

( u2 + f(u)1) ds,

(8)where = (1, 2) is the outward unit normal to (so also the opposite

of the outward normal to +).

Since u is a weak solution, and u is smooth on either side of the curvex = (t), u is a strong solution of (5) in and

2

(( u)t + (f(u))x)dx dt = 0. (9)

Therefore, for all smooth function , we havex=(t)

( u2 + f(u)1) dsx=(t)

( u+2 + f(u+)1) ds = 0, (10)

which leads to

u((t), t)2 + f(u((t), t))1 = u+((t), t)2 + f(u+((t), t))1, (11)

and

f(u((t), t)) f(u+((t), t)) = 21

(u((t), t) u+((t), t))= (t)(u((t), t) u+((t), t)),

(12)

which is exactly the Rankine-Hugoniot jump condition.Hence u+ and u solve the PDE in the classical sense.

Conversely, lets suppose that u+ and u solve the PDE in the classical sense.Then (9) and (12) hold. By performing the previous operations in reverse orderand using similar arguments, we get that u is indeed a weak solution.

Now lets look at the Burgers equation with particular initial condition: ut + uux = 0, x R, t > 0,u(x, 0) = (x) = { 1, x < 00, x > 0

(13)

and lets find a weak solution using the previous analysis.From the method of characteristics for first-order quasi-linear PDEs, we knowthat we have an implicit solution where u is smooth, namely u = (x ut).Moreover u is constant along the projected characteristic curves given by xr(t) =(r)t+ r.

If r < 0, then (r) = 1, which implies that the characteristic curves are

xr(t) = t+ r, r < 0, (14)

and the solution u = u should equal 1 along those curves,

If r > 0, then (r) = 0, which implies that the characteristic curves are

xr(t) = r, r > 0, (15)

and the solution u = u+ should equal 0 along those curves.

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Furthermore, at the jump, we have the Rankine-Hugoniot jump condition:

(u)2

2 (u+)

2

2= (t)(u u+), (16)

which reduces to

(t) =1

2. (17)

Moreover, the curve x = (t) contains the point (x, t) = (0, 0) (we indeedhave a discontinuity for the initial function at x = 0). Therefore the curve of

discontinuity is given by x =t

2and the weak solution is:

u(x, t) =

1, x t

2

, (18)

which means that the discontinuity (shock) is moving at speed 1/2.

Problem 3. Consider the Burgers equationut + uux = 0, x R, t > 0,

u(x, 0) = (x) =

0, x < 0x

, 0 < x <

1, x >

(19)

Lets solve the equation. Once again, we know that u is constant along theprojected characteristic curves xr(t) = (t)t+ r.

If r < 0, then (r) = 0, which implies that the characteristic curves are

xr(t) = r, r < 0, (20)

and the solution u(x, t) = 0 along those curves,

If 0 < r < , then (r) = r

, which implies that the characteristic curvesare

xr(t) =r

+ r, 0 < r < , (21)

and the solution u(x, t) =r

=

x

t+ along those curves,

If r > , then (r) = 1, which implies that the characteristic curves are

xr(t) = t+ r, r > , (22)

and the solution u(x, t) = 1 along those curves.

4

For t 0, the characteristic curves do not intersect one another and thesolution is then defined as

u(x, t) =

0, x < 0x

t+ , 0 < x < t+

1, x > t+

(23)

Now as 0, it converges to

v(x, t) =

0, x < 0x

t, 0 < x < t

1, x > t

(24)

v is indeed a weak solution of the Burgers equation with (x) = H(x).To verify this, we use the characterization given in Problem 4., meaning thatwe consider here the domains 1 = x < 0, 2 = 0 < x < t and 3 = x > t,corresponding to the domains where v is C1. We easily verify that the PDEholds pointwise in 123, that the initial condition holds at t = 0 and thatwe have the Rankine-Hugoniot jump condition at each interface (v is continuoushere, so there is nothing to verify).

If we are looking for another weak solution which is piecewise constant, usingthe same procedure as in Problem 4., we get another solution given by

w(x, t) =

0, x t

2

(25)

However, w does not satisfy the entropy condition since

f (w) = w = 0,and

f (w+) = w+ = 1,but

f(w) f(w+)w w+

=1

2,

(26)

while the solution u of Problem 4. satisfies it since

f (u) = u = 1,and

f (u+) = u+ = 0,while

f(w) f(w+)w w+

=1

2.

(27)

5

Problem 4.

(1)uxx uxy 2uyy = 0. (28)

We have A =

(1 1/21/2 2

). Therefore det(A) = 21/4 = 9/4 < 0,

and Tr(A) = 1 + 2 = 3 < 0. Therefore the eigenvalues of A are non zero and ofopposite signs: Hyperbolic PDE.

(2)uxx 2uxy + uyy = 0. (29)

We have A =

(1 11 1

). Therefore det(A) = 0. Therefore at least one

of the eigenvalues of A is zero: Degenerate PDE.

(3)uxx + 2uxy + 2uyy = 0. (30)

We have A =

(1 11 2

). Therefore det(A) = 2 1 = 1 > 0 and Tr(A) =

3 > 0. Therefore the eigenvalues of A are non zero and of the same sign: EllipticPDE.

Problem 5.

(1) Lets find the general C2 solution of the PDE

uxx uxt 6utt = 0, (31)

by reducing it to a system of first order PDEs (by the way, this is an ellipticPDE).We are looking for a, b, c, d that formally verify:

xx xt 6tt = (ax + bt) (cx + dt)= acxx + (ad+ bc)xt + bdtt.

(32)

So we get the (under-determined) system: ac = 1ad+ bc = 1bd = 6.

(33)

From the first equation, let us simply take a = c = 1. Then the systemreduces to {

d+ b = 1bd = 6, (34)

6

which gives b = 2 and d = 3.

Therefore we can write uxx uxt 6utt = 0 as (x + 2t) (x 3t)u = 0.Now let v = (x 3t)u. Then v verifies vx + 2vt = 0 (first order linear PDE!).And we know that the solution writes v(x, t) = h(t 2x) for some h C1.Now for u we have the system:

ux 3ut = h(t 2x). (35)

Using the method of characteristics, we get the following equations: xr(s) = 1, xr(0) = 0,tr(s) = 3, tr(0) = r,vr(s) = h(tr(s) 2xr(s)), vr(0) = (r),

(36)

for some function C2.Therefore we have xr(s) = s, tr(s) = 3s+ r, and

vr(s) = h(3s+ r 2s) = h(5s+ r), (37)

and by integrating from s = 0, we get:

vr(s) =

s0

h(5s + r)ds + (r)

= 15

5s+rr

h(y)dy + (r),(38)

after a change of variables. Now, since we have s = x and r = t + 3x, wefinally get:

u(x, t) =1

5

t+3xt2x

h(y)dy + (t+ 3x)

= f(t+ 3x) + g(t 2x),(39)

for some f, g C2.

Reciprocally, we verify that u of the form u(x, t) = f(t+ 3x) + g(t 2x) forf, g C2 indeed solves the PDE.

(2) For an arbitrary Cc (R2) we have to show that

u(xx xt 6tt) = v(xx xt 6tt) + w(xx xt 6tt) = 0. (40)

But from Problem 2 of Pset 2, we have:

v(xx xt 6tt) = v((x 3t) (x + 2t)) = 0, (41)

and similarly,

w(xx xt 6tt) = w((x + 2t) (x 3t)) = 0. (42)

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