Lecture 2 28 Short
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Transcript of 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
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Neumann Boundary Conditions Robin Boundary Conditions
The heat equation with Neumann boundary conditions
Our goal is to solve:
ut=
c2uxx, 0 0.
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N B d C di i R bi B d C di i
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Neumann Boundary Conditions Robin Boundary Conditions
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.
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Neumann Boundary Conditions Robin Boundary Conditions
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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Neumann Boundary Conditions Robin Boundary Conditions
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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Neumann Boundary Conditions Robin Boundary Conditions
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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Neumann Boundary Conditions Robin Boundary Conditions
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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Neumann Boundary Conditions Robin Boundary Conditions
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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Neumann Boundary Conditions Robin Boundary Conditions
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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Generalized Fourier coefficients
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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Generalized Fourier coefficients
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) +
Daileda The heat equation