Separation of Variables: More Examplesksuweb.kennesaw.edu/~plaval/math4310/sepvar_ex_slides.pdf ·...

Separation of Variables: More Examples

Philippe B. Laval

KSU

Today

Philippe B. Laval (KSU) Separation of Variables Today 1 / 33

Introduction

In the previous section, we explained the separation of variabletechnique and looked at some examples.

All the examples we looked at had the same PDE and boundaryconditions. Only the initial condition changed.

We now looked at more examples. The PDE will be the same as inthe previous section, that is the one-dimensional heat equation whereonly diffusion occurs, that is ut = kuxx . However, we will change theboundary conditions. In the previous section, we only consideredwhen the temperature of the rod at the ends was maintained at 0◦.Here will consider mixed boundary conditions.

We proceed by examples.


Heat Equation for a Rod with Insulated Ends

Recall, if the ends are insulated, there is no heat loss or gain at theboundary, so the flux at the boundary is 0.Let’s solve

PDE ut = kuxx 0 < x < L 0 < t <∞

BCux (0, t) = 0ux (L, t) = 0

0 < t <∞

IC u (x , 0) = φ (x) 0 ≤ x ≤ L

(1)

We follow the steps outlined in the previous section.


Step 1: Transforming the PDE in Two ODEs

As in the previous section, we seek a nontrivial solution of the formu (x , t) = X (x)T (t). Proceeding as in the previous section, we transformthe PDE in two ODEs. We get:

T ′ + λkT = 0

X ′′ + λX = 0

where λ is a real number.


Step 2: Solving the Time-Dependent ODE

T ′ + λkT = 0 is a well known ODE. Its solution is:

T (t) = C1e−λkt (2)




T (t) = C1e−λkt (2)


Step 2: Solving the Eigenvalue Problem

The next ODE to solve isX ′′ + λX = 0 (3)

Since we are looking for a nontrivial solution, we have to solve theboundary problem

X ′′ + λX = 0X ′ (0) = 0X ′ (L) = 0

(4)

DefinitionThe values of λ for which the boundary value problem in 4 has nontrivialsolutions are called eigenvalues. The corresponding solutions X (x) arecalled eigenfunctions.



There are three possibilities.

1 λ < 0. Results in a trivial solution (see homework), so we discard it.2 λ = 0. In this case, equation X ′′ + λX = 0 becomes X ′′ = 0 withsolutions X = Ax + B. The boundary conditions imply thatX (x) = B = B0. The resulting solution is u (x , t) = B0.

3 λ > 0. In this case, the general solution of X ′′ + λX = 0 is

X (x) = C2 sin√λx + C3 cos

√λx . The eigenvalues are λ =

(nπL

)2for n = 1, 2, 3, ... The corresponding eigenfunctions areX (x) = C3 cos

nπLx for n = 1, 2, 3, ...The resulting solutions are

u (x , t) = Bne−(nπL )

2kt cos

nπLx for n = 1, 2, 3, ...


Step 3: Finishing the Problem - Principle of Superposition

To satisfy the IC, we use the principle of superposition. Since we have ahomogeneous equation, a linear combination of all the solutions we haveso far is also a solution. We remember that we have solutions when bothλ = 0 and λ > 0. Hence,

u (x , t) = B0 +∞∑n=1

Bne−(nπL )

2kt cos

nπLx

which we can also write as

u (x , t) =∞∑n=0

Bne−(nπL )

2kt cos

nπLx


Step 3: Finishing the Problem

Using the IC, we see that φ (x) = u (x , 0) = B0 +∞∑n=1

Bn cosnπLx .

This is similar to the previous section, but the infinite series involves cosineinstead of sine. If you recall, what made the problem easy to solve was thenice property integrals involving sine functions have. Fortunately for us,the cosine function has similar property, which we give as a proposition.

Proposition

∫ L0 cos

mπxL

cosnπxLdx =

0 if m 6= nL2

if m = n 6= 0L m = n = 0



To find the coeffi cients An, we proceed as we did in the previous section.We start with

φ (x) = B0 + B1 cosπ

Lx + B2 cos

2πLx + B3 cos

3πLx + ...

We multiply each side by cosmπxL

for m = 0, 1, 2, 3, ...

We get

B0 =1L

∫ L

0φ (x) dx

and

Bn =2L

∫ L

0φ (x) cos

nπxLdx for n > 0



So, we have the following result

PropositionAssuming that such a representation exists, the Fourier cosine seriesrepresentation of the function φ (x) between 0 and L is

φ (x) = B0 +∞∑n=1

Bn cosnπLx (5)

where

B0 =1L

∫ L

0φ (x) dx (6)

and

Bn =2L

∫ L

0φ (x) cos

nπxLdx for n > 0 (7)



In conclusion, we have the following proposition:

PropositionThe solution to the IBVP

PDE ut = kuxx 0 < x < L 0 < t <∞

BCux (0, t) = 0ux (L, t) = 0

0 < t <∞

IC u (x , 0) = φ (x) 0 ≤ x ≤ L

is

u (x , t) = B0 +∞∑n=1

Bne−(nπL )

2kt cos

nπLx

where Bn for n = 0, 1, 2, 3, ... are the coeffi cients in the Fourier cosine

series representation of φ (x) that is φ (x) = B0 +∞∑n=1

Bn cosnπLx



Before looking at an example with a specific function for φ (x), we make afinal remark.

Remark

We can see from the solution u (x , t) = B0 +∞∑n=1

Bne−(nπL )

2kt cos

nπLx

that limt→∞

e−(nπL )

2kt = 0 and since cos

nπLx and Bn are bounded, it follows

that limt→∞

Bne−(nπL )

2kt cos

nπLx = 0 and therefore lim

t→∞u (x , t) = B0. Note

that B0 =1L

∫ L0 φ (x) dx represents the average value of φ (x), the initial

temperature distribution. So, the steady state solution is the average ofthe initial temperature distribution.


Example

ExampleSolve the IBVP

PDE ut = kuxx 0 < x < 2 0 < t <∞

BCux (0, t) = 0ux (2, t) = 0

0 < t <∞

IC u (x , 0) = x2 0 ≤ x ≤ 2


Example

We should have found the following:

x2 =43+∞∑n=1

(−1)n(4nπ

)2cos

nπ2x (a picture is shown in the next slide)

Therefore, the solution is

u (x , t) =43+∞∑n=1

(−1)n(4nπ

)2e−(

nπ2 )

2kt cos

nπ2x


Example

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.00

1

2

3

x

y

Fourier Cosine Series for x2black: x2

red: first two terms of the seriesgreen: first three terms of the seriespink: first 6 terms of the series


Heat Equation in a Thin Circular Ring

Here, we consider the heat equation in a circular ring. We make thefollowing assumptions:

1 We assume the ring has length 2L. It will be obvious in a while why

2L and not L. Note that the radius r of the ring will then be r =Lπ.

2 Assume the wire making up the ring is thin enough so that thetemperature is constant along cross sections. In other words, heatonly flows along the wire.

3 The lateral sides of the wire are perfectly insulated, there is no heatloss or gain.

Under these conditions, the ring will satisfy the one-dimensional heatequation where the distance x is the arc length along the wire. As far asheat flow is concerned, the ring will behave like a thin rod. Only theboundary conditions will be different. so, before we solve it the heatequation for the ring, we have to formulate the correct boundaryconditions.



To facilitate formulating the boundary conditions, we do the following:

Position our ring so that its center is at the origin, see the picture onthe next slide.

We measure arc length (x) counterclockwise from the point at whichthe ring intersects with the positive horizontal axis, P in the picture.Hence, the arc length, x , ranges from −L to L.We can think of the ring as a thin rod which has been bent into theshape of the ring.The ring is positioned so that the connectionhappens where the ring meets the negative horizontal axis, Q in thepicture. Hence, the connection happens at x = L and x = −L. Thesevalues correspond to our boundaries.

Where the two ends of the thin wire meet to form the ring, we willhave perfect thermal contact. Our boundaries are not reallyboundaries since they correspond to the same point.





Therefore, the IBVP which describes this problem, and the one we willsolve is

PDE ut = kuxx −L < x < L 0 < t <∞

BCu (−L, t) = u (L, t)ux (−L, t) = ux (L, t)

0 < t <∞

IC u (x , 0) = φ (x) −L ≤ x ≤ L

(8)

We follow the steps outlined in the previous previous examples.


Step 1: Transforming the PDE in Two ODEs

As in the previous section, we seek a nontrivial solution of the formu (x , t) = X (x)T (t). Proceeding as in the previous section, we transformthe PDE in two ODEs. We get:

T ′ + λkT = 0

X ′′ + λX = 0

where λ is a real number.




T (t) = C1e−λkt




T (t) = C1e−λkt



The next ODE to solve isX ′′ + λX = 0

Since we are looking for a nontrivial solution, we have to solve theboundary problem

X ′′ + λX = 0X (−L) = X (L)X ′ (−L) = X ′ (L)

(9)



There are three possibilities.

1 λ < 0. Results in a trivial solution (see homework), so we discard it.2 λ = 0. In this case, equation X ′′ + λX = 0 becomes X ′′ = 0 withsolutions X (x) = Ax + B. The boundary conditions imply thatX (x) = B = B0. The resulting solution is u (x , t) = B0.

3 λ > 0. In this case, the general solution of X ′′ + λX = 0 is

X (x) = C2 sin√λx + C3 cos

√λx . The eigenvalues are λ =

(nπL

)2for n = 1, 2, 3, ... The corresponding eigenfunctions areX (x) = C2 sin

nπLx + C3 cos

nπLx for n = 1, 2, 3, ... The resulting

solutions are u (x , t) = e−(nπL )

2kt(An sin

nπLx + Bn cos

nπLx)for

n = 1, 2, 3, ...


Step 3: Finishing the Problem - Principle of Superposition

To satisfy the IC, we use the principle of superposition. Since we have ahomogeneous equation, a linear combination of all the solutions we haveso far is also a solution. We remember that we have solutions when bothλ = 0 and λ > 0. Hence,

u (x , t) = B0 +∞∑n=1

e−(nπL )

2kt(An sin

nπLx + Bn cos

nπLx)

which we can also write as

u (x , t) =∞∑n=0

Bne−(nπL )

2kt cos

nπLx +

∞∑n=1

Ane−(nπL )

2kt sin

nπLx



Using the IC, we see that

φ (x) = u (x , 0) (10)

= B0 +∞∑n=1

(An sin

nπLx + Bn cos

nπLx)

This time, the series has both sine and cosine functions. Also, x nowranges from −L to L. Fortunately for us, the properties we used beforewhich made the problem simpler also hold in this case. We give them as aproposition.



Proposition

Let m and n be nonnegative integers. Then the following is true:

1∫ L−L cos

mπxL

cosnπxLdx =

0 if m 6= nL if m = n 6= 02L m = n = 0

2∫ L−L sin

(mπxL

)sin(nπxL

)dx =

{0 if m 6= nL if m = n 6= 0

3∫ L−L sin

(mπxL

)cos

nπxLdx = 0



To find the coeffi cients An, we proceed as we did in the previous examples.We start with

φ (x) = B0 + A1 sinπ

Lx + B1 cos

π

Lx + A2 sin

2πLx + B2 cos

2πLx + ... (11)

Then, for each m = 1, 2, 3, ..., we multiply both sides with sinmπLx and

integrate from −L to L, then for each m = 0, 1, 2, 3, ... we multiply bothsides with cos

mπLand integrate from −L to L.



PropositionAssuming that such a representation exists, the Fourier seriesrepresentation of the function φ (x) between −L and L is

φ (x) = B0 +∞∑n=1

(An sin

nπLx + Bn cos

nπLx)

where

An =1L

∫ L

−Lφ (x) sin

nπLxdx for m = 1, 2, 3, ...

B0 =12L

∫ L

−Lφ (x) dx

Bn =1L

∫ L

−Lφ (x) cos

nπxLdx



PropositionThe solution to the IBVP

PDE ut = kuxx −L < x < L 0 < t <∞

BCu (−L, t) = u (L, t)ux (−L, t) = ux (L, t)

0 < t <∞

IC u (x , 0) = φ (x) −L ≤ x ≤ L

is

u (x , t) = B0 +∞∑n=1

e−(nπL )

2kt(An sin

nπLx + Bn cos

nπLx)

where the coeffi cients A1,A2,A3, ... and the coeffi cients B0,B1,B2,B3, ...are given above.


Summary

In the examples and explanation of this section and the previous one, wewere solving the one-fimensional heat equation ut = kuxx . Using theseparation of variables method, we were seeking a solution of the formu (x , t) = X (x)T (t). We outline what these examples have in commonand also where they differ.

1 The separation of variable method resulted in having to solve twoODEs:

T ′ + λkT = 0

X ′′ + λX = 0

2 For the first ODE, we did not use BCs, we only found a generalsolution which was the same for all the examples:

T (t) = C1e−λkt

3 For the second ODE, we used BCs. We noticed that the BCs of theODE were similar to those of the PDE in the following sense. Theywere both homogeneous. If the BCs of the PDE involved derivatives,so did the BCs of the ODE. More specifically, if the BCs of the PDEwere u (0, t) = 0 and u (L, t) = 0 then the BCs of the ODE wereX (0) = 0 and X (L) = 0. if the BCs of the PDE were ux (0, t) = 0and ux (L, t) = 0 then the BCs of the ODE were X ′ (0) = 0 andX ′ (L) = 0. And so on.

4 The values of λ for which the second ODE has solutions are calledeigenvalues and the corresponding solutions are calledeigenfunctions. What these eigenvalues and eigenfunctions aredepends greatly on the boundary conditions. The table belowsummarizes the various possibilities.



Exercises

See the problems at the end of my notes on separation of variables: moreexamples.


http://science.kennesaw.edu/~plaval/math4310/sepvar_ex.pdf

http://science.kennesaw.edu/~plaval/math4310/sepvar_ex.pdf

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