lecture 2 28 short

7/31/2019 Lecture 2 28 Short

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Neumann Boundary Conditions Robin Boundary Conditions

The one dimensional heat equation:Neumann and Robin boundary conditions

Ryan C. Daileda

Trinity University

Partial Differential EquationsFebruary 28, 2012

Daileda The heat equation


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The heat equation with Neumann boundary conditions

Our goal is to solve:

ut=

c2uxx, 0 0.


N B d C di i R bi B d C di i


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Case 2: k = 0

The ODE (4) is simply X = 0 so that

X = Ax + B.

The boundary conditions (6) yield

0 = X(0) = X(L) = A.

Taking B = 1 we get the solution

X0 = 1.

The corresponding equation (5) for T is T = 0, which yields

T = C. We setT0 = 1.

These give the zeroth normal mode:

u0(x, t) = X0(x)T0(t) = 1.


N B d C diti R bi B d C diti


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Case 3: k = 2 < 0

The ODE (4) is now X + 2X = 0 with solutions

X = c1 cos x + c2 sin x.

The boundary conditions (6) yield

0 = X(0) = c1 sin 0 + c2 cos 0 = c2,

0 = X(L) = c1 sin L + c2 cos L.

The first of these gives c2 = 0. In order for X to be nontrivial, the

second shows that we also need

sin L = 0.




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Case 3: k = 2 < 0

This can occur if and only if L = n, that is

= n =n

L, n = 1, 2, 3, . . .

Choosing c1 = 1 yields the solutions

Xn = cos nx, n = 1, 2, 3, . . .

For each n the corresponding equation (5) for T becomes

T

= 2nT, with n = cn. Up to a constant multiple, the

solution isTn = e

2nt.




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Normal modes and superposition

Multiplying these together gives the nth normal mode

un(x, t) = Xn(x)Tn(t) = e2nt cos nx, n = 1, 2, 3, . . .

where n

=n

/L

and n

=c

n

.The principle of superposition now guarantees that for any choiceof constants a0, a1, a2, . . .

u(x, t) = a0u0 +

n=1

anun = a0 +

n=1

ane2nt cos nx (7)

is a solution of the heat equation (1) with the Neumann boundaryconditions (2).




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Initial conditions

If we now require that the solution (7) satisfy the initial condition(3) we find that we need

f(x) = u(x, 0) = a0 +

n=1

an cosnx

L, 0 < x < L.

This is simply the cosine series expansion of f (x). Using ourprevious results, we finally find that if f(x) is piecewise smooththen

a0 =1

L

L

0

f(x) dx, an =2

L

L

0

f(x)cosnx

Ldx, n 1.




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Conclusion

Theorem

If f(x) is piecewise smooth, the solution to the heat equation (1)with Neumann boundary conditions (2) and initial conditions (3) isgiven by

u(x, t) = a0 +

n=1

ane2nt cos nx,

where

n =n

L, n = cn,

and the coefficients a0, a1, a2, . . . are those occurring in the cosineseries expansion of f (x). They are given explicitly by

a0 =1

L

L

0

f(x) dx, an =2

L

L

0

f(x)cosnx

Ldx, n 1.




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Example 1

Example

Solve the following heat conduction problem:

ut =1

4uxx, 0 < x < 1 , 0 < t,

ux(0,

t) =

ux(1,

t) = 0, 0 0. This states that the bar radiates heat toits surroundings at a rate proportional to its currenttemperature.

Recall that conditions such as (9) are called Robinconditions.




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Separation of variables

As before, the assumption that u(x, t) = X(x)T(t) leads to theODEs

X kX = 0,

T

c2

kT = 0,

and the boundary conditions (8) and (9) imply

X(0) = 0,

X

(L) = X(L).

Also as before, the possibilities for X depend on the sign of theseparation constant k.




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Case 1: k = 0

We have X = 0 and so X = Ax + B with

0 = X(0) = B,

A = X

(L) = X(L) = (AL + B).

Together these give A(1 + L) = 0. Since , L > 0, we have A = 0and hence X = 0. Thus,

there are only trivial solutions in this case.




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Case 2: k = 2 > 0

Once again we have X 2X = 0 and

X = c1ex + c2e

x.

The boundary conditions become

0 = c1 + c2,

(c1eL c2e

L) = (c1eL + c2e

L).

The first gives c2 = c1, which when substituted in the secondyields2c1 cosh L = 2c1 sinh L.




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Case 2: k = 2 > 0

We may rewrite this as

c1 ( cosh L + sinh L) = 0.

The quantity in parentheses is positive (since , and L are), sothis means we must have c1 = c2 = 0. Hence X = 0 and

there are only trivial solutions in this case.




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Case 3: k = 2 < 0

From X + 2X = 0 we find

X = c1 cos x + c2 sin x

and from the boundary conditions we have

0 = c1,

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

Together these imply that

c2 ( cos L + sin L) = 0.




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Case 3: k = 2 < 0

Since we want nontrivial solutions (i.e. c2 = 0), we must have

cos L + sin L = 0

which can be rewritten as

tan L =

.

This equation has an infinite sequence of positive solutions

0 < 1 < 2 < 3 <




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Case 3: k = 2 < 0

The figure below shows the curves y = tan L (in red) andy = / (in blue).

The -coordinates of their intersections (in pink) are the values 1,2, 3, . . .




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Remarks

From the diagram we see that:

For each n(2n 1)/2L < n < n/L.

As n n (2n 1)/2L.

Smaller values of and L tend to accelerate this convergence.




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Normal modes

As in the earlier situations, for each n 1 we have the solution

Xn = sin nx

and the corresponding

Tn = e2nt, n = cn

which give the normal modes of the heat equation with boundaryconditions (8) and (9)

un(x, t) = Xn(x)Tn(t) = e2nt sin nx.




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Superposition

Superposition of normal modes gives thegeneral solution

u(x, t) =n=1

cnun(x, t) =n=1

cne2nt sin nx.

Imposing the initial condition u(x, 0) = f(x) gives us

f(x) =n=1

cn sin nx.

This is a generalized Fourier series for f(x). It is different fromthe ordinary sine series for f(x) since

n is not a multiple of a common value.




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Generalized Fourier coefficients

To compute the generalized Fourier coefficients cn we will use

the following fact.

Proposition

The functions

X1(x) = sin 1x, X2(x) = sin 2x, X3(x) = sin 3x, . . .

form a complete orthogonal set on [0, L].

Complete means that all sufficiently nice functions can be

represented via generalized Fourier series. This is aconsequence of Sturm-Liouville theory, which we will studylater.

We can verify orthogonality directly, and will use this toexpress the coefficients cn as ratios of inner products.




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Assuming orthogonality for the moment, for any n 1 we have thefamiliar computation

f, Xn =

m=1

cm sin mx, sin nx=

m=1

cmsin mx, sin nx

= cnsin nx, sin nx

= cnXn, Xn

since the inner products with m = n all equal zero.




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It follows immediately that the generalized Fourier coefficients aregiven by

cn =f, Xn

Xn, Xn

=

L

0

f(x)sin nx dx

L0

sin2 nx dx

.

For any given f(x) these integrals can typically be computedexplicitly in terms of n.

The values of n, however, must typically be found vianumerical methods.




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Conclusion

Theorem

The solution to the heat equation (1) with Robin boundaryconditions (8) and (9) and initial condition (3) is given by

u(x, t) =

n=1

cne2nt sin nx,

where n is the nth positive solution to

tan L =

,

n = cn, and the coefficients cn are given by

cn =

L

0f(x)sin nx dx

L0 sin2 nx dx .Daileda The heat equation



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Example 2

Example

Solve the following heat conduction problem:

ut =1

25uxx, 0 < x < 3 , 0 < t,

u(0, t) = 0, 0 < t,

ux(3, t) = 1

2u(3, t), 0 < t,

u(x, 0) = 1001 x3 , 0 < x < 3.

We have c = 1/5, L = 3, = 1/2 and f(x) = 100(1 x/3).




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Example 2

The integrals defining the Fourier coefficients are

100

30

1 x

3sin nx dx =

100(3n sin3n)

32n

and 30

sin2 nx dx =3

2+ cos2 3n.

Hence

cn = 200(3n sin3n)32n (3 + 2 cos

2 3n).




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Example 2

We can therefore write out the full solution as

u(x, t) =n=1

200(3n sin3n)32n (3 + 2 cos

2 3n)e

2nt/25 sin nx,

where n is the nth positive solution to tan 3 = 2.




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Example 2

Remarks:

In order to use this solution for numerical approximation orvisualization, we must compute the values n.

This can be done numerically in Maple, using the fsolve

command. Specifically, n can be computed via the commandfsolve(tan(mL)=-m/k,m=(2n-1)Pi/(2L)..nPi/L);

where L and k have been assigned the values of L and ,respectively.

These values can be computed and stored in an Arraystructure, or one can define n as a function using the ->operator.




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Example 2

Here are approximate values for the first 5 values of n and cn.

n n cn1 0.7249 47.04492 1.6679 45.14133 2.6795 21.3586

4 3.7098 19.34035 4.7474 12.9674

Therefore

u(x, t) = 47.0449e0.0210t sin(0.7249x) + 45.1413e0.1113t sin(1.6679x)

+ 21.3586e0.2872t sin(2.6795x) + 19.3403e0.5505t sin(3.7098x)

+ 12.9674e0.9015t sin(4.7474x) +


lecture 2 28 short

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