exam 2 solved

7/30/2019 Exam 2 Solved

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Math 412-501 October 20, 2006

Exam 2: Solutions

Problem 1 (50 pts.) Solve the heat equation in a rectangle 0 < x < , 0 < y < ,u

t=

2u

x2+

2u

y2

subject to the initial condition

u(x,y, 0) = (sin2x + sin 3x)sin y

and the boundary conditions

u(0, y , t) = u(, y, t) = 0, u(x, 0, t) = u(x,,t) = 0.

Solution: u(x,y,t) = e5t sin2x sin y + e10t sin3x sin y.

We search for the solution of the initial-boundary value problem as a superposition of solutionsu(x,y ,t) = (x)h(y)G(t) with separated variables of the heat equation that satisfy the boundaryconditions. Substituting u(x,y ,t) = (x)h(y)G(t) into the heat equation, we obtain

(x)h(y)G (t) = (x)h(y)G(t) + (x)h(y)G(t),

G (t)

G(t)=

(x)

(x)+

h(y)

h(y).

Since any of the expressionsG (t)

G(t),

(x)

(x), and

h(y)

h(y)depend on one of the variables x,y ,t and does

not depend on the other two, it follows that each of these expressions is constant. Hence

(x)

(x)= , h

(y)

h(y)= , G

(t)

G(t)= ( + ),

where and are constants. Then

= , h = h, G = ( + )G.

Conversely, if functions , h, and G are solutions of the above ODEs for the same values of and ,then u(x,y ,t) = (x)h(y)G(t) is a solution of the heat equation.Substituting u(x,y ,t) = (x)h(y)G(t) into the boundary conditions, we get

(0)h(y)G(t) = ()h(y)G(t) = 0, (x)h(0)G(t) = (x)h()G(t) = 0.

It is no loss to assume that neither nor h nor G is identically zero. Then the boundary conditionsare satisfied if and only if (0) = () = 0, h(0) = h() = 0.

To determine , we have an eigenvalue problem

= , (0) = () = 0.

This problem has eigenvalues n = n2

, n = 1, 2, . . . . The corresponding eigenfunctions are n(x) =sin nx.

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To determine h, we have the same eigenvalue problem

h = h, h(0) = h() = 0.

Hence the eigenvalues are m = m2, m = 1, 2, . . . . The corresponding eigenfunctions are hm(y) =

sin my.The function G is to be determined from the equation G = ( + )G. The general solution of

this equation is G(t) = c0e(+)t, where c0 is a constant.

Thus we obtain the following solutions of the heat equation satisfying the boundary conditions:

unm(x,y ,t) = e(n+m)tn(x)hm(y) = e

(n2+m2)t sin nx sin my, n, m = 1, 2, 3, . . .

A superposition of these solutions is a double series

u(x,y ,t) =n=1

m=1

cnme(n2+m2)t sin nx sin my,

where cnm are constants. To determine the coefficients cnm, we substitute the series into the initialcondition u(x,y, 0) = (sin 2x + sin 3x)sin y:

(sin 2x + sin 3x)sin y =n=1

m=1

cnm sin nx sin my.

It is easy to observe that c2,1 = c3,1 = 1 while the other coefficients are equal to 0. Therefore

u(x,y ,t) = e5t sin2x sin y + e10t sin3x sin y.

Problem 2 (50 pts.) Solve Laplaces equation inside a quarter-circle 0 < r < 1,0 < < /2 (in polar coordinates r, ) subject to the boundary conditions

u(r, 0) = 0, u(r,/2) = 0, |u(0, )| < , u(1, ) = f().

Solution: u(r, ) =

n=1cnr

2n sin2n, where

cn =4

/20

f()sin2n d, n = 1, 2, . . .

Laplaces equation in polar coordinates (r, ):

2u

r2+

1

r

u

r+

1

r22u

2= 0.

We search for the solution of the boundary value problem as a superposition of solutions u(r, ) =h(r)() with separated variables of Laplaces equation that satisfy the three homogeneous boundaryconditions. Substituting u(r, ) = h(r)() into Laplaces equation, we obtain

h(r)() +1

rh(r)() +

1

r2h(r)() = 0,

r2h(r) + rh(r)h(r)

= ()

().

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Since the left-hand side does not depend on while the right-hand side does not depend on r, it followsthat

r2h(r) + rh(r)

h(r)=

()

()= ,

where is a constant. Then

r2h(r) + rh(r) = h(r), = .

Conversely, if functions h and are solutions of the above ODEs for the same value of , thenu(r, ) = h(r)() is a solution of Laplaces equation in polar coordinates.

Substituting u(r, ) = h(r)() into the homogeneous boundary conditions, we get

h(r)(0) = 0, h(r)(/2) = 0, |h(0)()| < .

It is no loss to assume that neither h nor is identically zero. Then the boundary conditions aresatisfied if and only if (0) = (/2) = 0, |h(0)| < .

To determine , we have an eigenvalue problem

= , (0) = (/2) = 0.

This problem has eigenvalues n = (2n)2, n = 1, 2, . . . . The corresponding eigenfunctions are n() =

sin2n.The function h is to be determined from the equation r2h+ rh = h and the boundary condition

|h(0)| < . We may assume that is one of the above eigenvalues so that > 0. Then the generalsolution of the equation is h(r) = c1r

+ c2r, where =

and c1, c2 are constants. The boundary

condition |h(0)| < holds if c2 = 0.Thus we obtain the following solutions of Laplaces equation satisfying the three homogeneous

boundary conditions:

un(r, ) = r2n sin2n, n = 1, 2, . . .

A superposition of these solutions is a series

u(r, ) =

n=1cnr

2n sin2n,

where c1, c2, . . . are constants. Substituting the series into the boundary condition u(1, ) = f(), weget

f() =

n=1cn sin2n.

The right-hand side is a Fourier sine series on the interval [0, /2]. Therefore the boundary conditionis satisfied if this is the Fourier sine series of the function f() on [0, /2]. Hence

cn =4

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f()sin2n d, n = 1, 2, . . .

Bonus Problem 3 (40 pts.) Consider a regular Sturm-Liouville eigenvalue problem

+ = 0, (0) = 0, (1) + h(1) = 0,

where h is a real constant.

(i) For what values of h is = 0 an eigenvalue?Solution: h = 0.

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In the case = 0, the general solution of the equation + = 0 is a linear function (x) =

c1x + c2, where c1, c2 are constants. Substituting it into the boundary conditions (0) = 0 and

(1) + h(1) = 0, we obtain equalities c1 = 0, c1 + h(c1 + c2) = 0. They imply that c1 = hc2 = 0. If

h = 0, it follows that c1 = c2 = 0, hence there are no eigenfunctions with eigenvalue = 0. Ifh = 0then (x) = 1 is indeed an eigenfunction.

(ii) For what values of h are all eigenvalues positive?

Solution: h > 0.

In the case < 0, the general solution of the equation + = 0 is

(x) = c1 cosh x + c2 sinh x,

where = > 0 and c1, c2 are constants. Note that

(x) = c1 sinh x + c2 cosh x.

The boundary condition (0) = 0 is satisfied if and only if c2 = 0. Substituting (x) = c1 cosh xinto the boundary condition (1) + h(1) = 0, we obtain

c1 sinh + hc1 cosh = 0,

c1( tanh + h) = 0.

If tanh = h, it follows that c1 = 0, hence there are no eigenfunctions with eigenvalue = 2.If tanh = h then (x) = cosh x is indeed an eigenfunction.

The function f() = tanh is continuous. It is easy to see that f(0) = 0 and f() > 0 for > 0.Since tanh 1 as +, we have that f() + as +. It follows that f takes allpositive values on (0,).By the above the eigenvalue problem has a negative eigenvalue if and only if h < 0. As shown inthe solution to the part (i), = 0 is an eigenvalue only for h = 0. Hence all eigenvalues are positiveif and only if h > 0.

The fact that for any h 0 all eigenvalues are nonnegative can also be obtained using the Rayleighquotient. If is an eigenfunction corresponding to an eigenvalue then

=

10

+

10|(x)|2 dx

10|(x)|2 dx

.

The boundary conditions imply that

10

= (0)(0) (1)(1) = h|(1)|2.

Hence 0 provided that h 0.

(iii) How many negative eigenvalues can this problem have?

Solution: One negative eigenvalue for h < 0.

Let f() = tanh . As shown in the solution to the part (ii), < 0 is an eigenvalue if and onlyif f() = h, where = 2, > 0. Observe that

f() = tanh + tanh = tanh +

cosh2 .

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In particular, f() > 0 for > 0. Since f is continuous, f(0) = 0, and f() + as +,it follows that f is a one-to-one map of the interval (0,) onto itself. Therefore for any h < 0 theeigenvalue problem has exactly one negative eigenvalue.

(iv) Find an equation for positive eigenvalues.

Solution: tan

=h

.

In the case > 0, the general solution of the equation + = 0 is

(x) = c1 cos x + c2 sin x,

where =

and c1, c2 are constants. Note that

(x) = c1 sin x + c2 cos x.

The boundary condition (0) = 0 is satisfied if and only if c2 = 0. Substituting (x) = c1 cos x intothe boundary condition (1) + h(1) = 0, we obtain

c1 sin + hc1 cos = 0,

c1(h cos sin ) = 0.If h cos = sin , it follows that c1 = 0, hence there are no eigenfunctions with eigenvalue = 2.If h cos = sin then (x) = cos x is indeed an eigenfunction.

Thus h cos

=

sin

is an equation for positive eigenvalues. Note that for any positivesolution of this equation we have cos

= 0. Indeed, if cos = 0 then sin = 1 and

sin

= 0. It follows that for > 0 this equation is equivalent to

tan

=h

.

(v) Find the asymptotics of n as n .Solution:

n (n 1) as n .

Positive eigenvalues are found from the equation tan

= h/

. The function f1() = tan iscontinuous, strictly increasing and assumes all real values on each of the intervals (m/2, m+/2),m = 0, 1, 2, . . . .

In the case h > 0, the function f2() = h/ is continuous and strictly decreasing on (0,). Itfollows that the equation f1() = f2() has exactly one solution in each of the intervals (0, /2) and(m /2, m + /2), m = 1, 2, . . . . In this case all eigenvalues are positive, hence (n 1) /2

exam 2 solved

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