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Essential Partial Differential Equations Examples 1

The exercises marked with ? are the most important.

(1.1) Check that the examples of explicit solutions to Poisson, wave heat and Laplacegiven in Lecture 1 are indeed solutions to the problems.

(1.2) ? Plot or sketch the following functions f : (0, 1) R and list the ones that arecontinuous on [0, 1].

(i) f1(x) = x2

(ii) f2(x) = x3/2

(iii) f3(x) = x1/2

(iv) f4(x) = x1/4

(v) f5(x) = x1/2

(vi) f6(x) = x1

Determine which of these functions are square integrable on (0,1). (That is,test for membership of the space L2(0, 1).)

For the functions that are square integrable on (0,1), identify those functionswhose derivative is also square integrable on (0,1). (This tests for membershipof the Sobolev space H1(0, 1).)

(1.3) ? Consider the norm p defined for p 1 by

xp =(|x1|p + + |xn|p

)1/p, x Rn.

(i) Let x = (1, 1)T . Compute xp for p = 1, 5, 10, 20. What do you thinklimp xp is?

(ii) Prove that xp converges to x := maxi |xi| as p.(iii) Show that a valid inner product is given by

x,y := x1y1 + . . .+ xnyn = xTy, x,y Rn.

(1.4) ? Suppose that V = L2(0, 1). Show that a valid inner product is given by

u, v := 1

0uv, u, v V.

(1.5) (i) Prove that for any x and y R and t [0, 1]

etx+(1t)y tex + (1 t)ey.

Hint: This is the statement that the exponential function is convex.(ii) Use the previous part to prove Youngs inequality which says that for any

a and b > 0 and 1 < p, q

2

Here the inner product and norms were defined in exercise 1.3. Hint: Con-sider proving

x

xp,

y

yq

1

which is equivalent.Note that essentially the same proof applies in the case of integrals and theL2 inner product to show that when 1 < p, q

3

Essential Partial Differential Equations Solutions 1

(1.1) Just follows directly from differentiation.(1.2) The six functions f1, f2, . . . , f6 are plotted below. The functions with positive

exponents are continuous over [0, 1]. The functions with negative exponent areunbounded in the limit x 0 so are not continuous.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

3

x

f(x)

x2

x3/2

x1/2

x1/4

x1/2

x1

To determine square integrability we need to check each function in turn to

see if 1

0 f2i < . The results are tabulated below. It is easy to see that all

functions x with > 1/2 are in L2(0, 1).

fi fi L2(0, 1) f i L2(0, 1)x2 X X

x32 X X

x12 X

x14 X

x12

x1

We can similarly check the square integrability of f i . In this case we see thatx has a derivative L2(0, 1) in the case > 1/2. Thus continuous functions neednot have square integrable first derivatives!

(1.3) (i) xp = 21/p. Thus, limp xp = 1.

4

(ii) To prove that xp x as p , We let x have maximum value inits first component so that x = |x1|. Then

xp = |x1|( n

i=1

(|xi|/|x1|)p)1/p

.

Notice all the terms in the sum are less than 1 and hence xp |x1|n1/p |x1| as p. Finally, since it is clear that xp |x1|, the result follows.

(iii) To show that , is a valid inner product we need to check the axioms: symmetry

x,y = x1y1 + . . .+ xnyn = y1x1 + . . .+ ynxn = y,x positivity

x,x = x210

+ . . .+ x2n0

0

uniquenessx,x = 0 x21

0

+ . . .+ x2n0

= 0 xi = 0, i.

linearityx + y,w = (x1 + y1)w1 + . . .+ (xn + yn)wn

= x1w1 + y1w1 + . . .+ xnwn + ynwn

= (x1w1 + . . .+ xnwn) + (y1w1 + . . .+ ynwn)

= x,w+ y,w (1.4) Checking the inner product axioms:

symmetry

(u, v) =

10uv =

10vu = (v, u)

positivity

(u, u) =

10

u20

0

uniqueness

(u, u) = 0 1

0u20

= 0 u = 0 almost everywere.

linearity

(u+ v,w) =

10{u+ v}w =

10uw +

10vw = (u,w) + (v, w)

(1.5) (i) Set f(t) = etx+(1t)y tex (1 t)ey. The problem then is to prove thatf(t) 0 for t (0, 1). Note that

f (t) = (x y)2etx+(1t)y 0,

5

and so by the mean value theorem f (t) is an increasing function. Nowsuppose that f(t) > 0 for some t (0, 1). Then

f(t) f(0) = f(t) > 0, f(1) f(t) = f(t) < 0.

Again by the mean value theorem there must be t1 (0, t) and t2 (t, 1)such that

f (t1) = (f(t) f(0))/t > 0, f (t2) = (f(1) f(t))/(1 t) < 0.

Therefore f (t1) > f(t2) which contradicts the fact that f

(t) is increasing.This completes the proof. In fact this statement holds if the exponentialfunction is replaced by any function with nonnegative second derivative andthe proof is the same.

(ii) We follow the hint and use some properties of the natural log:

ab = eln(ab) = eln(a)+ln(b) = ep1ln(ap)+q1ln(bq).

Now apply the result of the previous part with x = ln(ap), y = ln(bq), andt = p1 so that 1 t = q1. We obtain from this

ab p1eln(ap) + q1eln(bq) = ap

p+bq

q

which is the desired result.(iii) Following the hint consider

x

xp,

y

yq

=

nj=1

xjxp

yjyq

nj=1

|xj |xp

|yj |yq

nj=1

(1

p

|xj |p

xpp+

1

q

|yj |q

yqq

)=

1

p+

1

q= 1.

This completes the proof.(1.6) (i) To show that A is a valid norm we simply check the axioms:

(positive) For any x X, x 6= 0

Axx

0,

and so

A = supx 6=0

Axx

0

(definite)() If A = 0 then

A = supx 6=0

0xx

= 0.

6

() If A = 0 then

0 =Axx

for all x X, x 6= 0. Thus Ax = 0 for all x, which implies Ax = 0for all x. This means A = 0. (scaling)

A = supx 6=0

Axx

= || supx6=0

Axx

= ||A

(triangle inequality)

A+B = supx6=0

(A+B)xx

= supx 6=0

Ax+Bxx

supx 6=0

(Axx

+Bxx

) A + B.

(ii) For the second part, we denote

f1(A) = inf{M : Ax Mx, x X},

f2(A) = supx 6=0

Axx

.

To show that f1(A) f2(A) note that by definition

f2(A) Axx

xf2(A) Ax

for all x X. Thus f1(A) f2(A). To show that f1(A) f2(A) note that for any > 0 and all x X

Ax (f1(A) + )x.

This implies that

Axx

(f1(A) + )

for all x X with x 6= 0. Thus f2(A) (f1(A) + ) for any > 0,and taking to zero we find f2(A) f1(A).

7

(iii) For the last part we have

AB = supx 6=0

ABxx

supx 6=0

ABxx

(Part (ii))

AB (Definition of B).

8

Essential Differential Equations Examples 2

The exercises marked with ? are the most important. On this example sheet we arecareful to distinguish between functions and the distributions they represent. If u is afunction, then Fu is the distribution represented by u. You should note carefully that thisis not general practice, and it is quite common to simply write u for both the functionand the corresponding distribution, and refer to the distributional derivative Fu as thederivative u.

(2.1) ? Construct a weak formulation of the following boundary value problem:

u(x) + u(x) = f(x), x (0, 1); u(0) = 0, u(1) = 0.Show that the weak solution is uniquely defined. Hint: construct a proof bycontradiction.

(2.2) What is the distributional derivative of the Dirac delta?(2.3) Show that the distributional derivative is linear. That is, if F and G D(a, b)

and , R, show that(F + G) = F + G.

(2.4) Suppose that F D(a, b) and take any C(a, b). Then F defines adistribution by

F () = F ().

Prove that the product rule holds:

(F ) = F + F.

(2.5) Let H be the Heaviside function:

H(x) =

{1 if x > 0,0 if x < 0

defined on any interval (a, b) containing 0, and for any c (a, b) define Hc(x) =H(x c). Find the distributional derivative of the distribution FHc D(a, b)represented by Hc. Does Hc have a weak derivative?

(2.6) ? Suppose a < 0 < b, C(a, b), and c (a, b). Letu(x) = (x)Hc(x)

where Hc is the Heaviside function from exercise 2.5. Use exercises 2.4 and 2.5to find the distributional derivative of Fu. Under what condition does u have aweak derivative?

(2.7) Suppose that u is a piecewise smooth function on the interval [a, b]. Then it canbe written as a sum

u = 0(x) +nj=1

j(x)Hcj (x)

for some set of j C[a, b] and cj (a, b) (assume that cj 6= ck for j 6= k).Show that the distributional derivative Fu is given by

Fu = F0 +

nj=1

FjHcj + j(cj)cj .

9

Under what condition will u have a weak derivative?(2.8) ? Consider the following piecewise continuous functions h : (1, 1) R:

h1(x) =

{1, 1 < x < 0;1, 0 < x < 1.

h2(x) =

{x, 1 < x < 0;x, 0 < x < 1.

Show that both functions are in L2(1, 1). What are the distributional deriva-tives Fh1 and Fh2? Is either function in the Sobolev space H

1(1, 1)?

10

Essential Partial Differential Equations Solutions 2

(2.1) We consider the problem

u + u = f, u(0) = u(1) = 0.

Multiplying by a test function v H10 (0, 1) and integrating by parts gives

1

0uv dx+

10uv dx =

10fv dx

[uv]1

0+

10u(x)v(x) dx+

10u(x)v(x) dx =

10f(x)v(x) dx.

The first term is zero because v H10 (0, 1) means that v(0) = v(1) = 0. Denot-ing the standard L2(0, 1) inner product and associated norm by (, ) and ,respectively, the weak formulation is:

Given F H10 (0, 1), find u H10 (0, 1)