Separation of Variables Technique

Separation of Variables Technique

Philippe B. Laval

KSU

Current Semester

Philippe B. Laval (KSU) Separation of Variables Current Semester 1 / 20

Introduction

Separation of variables is a very powerful method used to solve certainPDEs. It transforms a PDE in two ODEs.We will describe the problems to which it applies. Then, we will explainthe method, and explain how to use it to solve certain PDEs.


When can we Use Separation of Variables?

The technique of separation of variables we are about to describe is one ofthe oldest known techniques to solve PDEs (IBVPs). It dates from theperiod of the French mathematician Joseph Fourier (1768-1830). He madegreat contributions toward solving the heat equation. This techniqueapplies to problems with the following characteristics:

1 The PDE must be linear and homogeneous.2 The boundary conditions are mixed and homogeneous. More precisely,they are of the form

αux (0, t) + βu (0, t) = 0

γux (L, t) + δu (L, t) = 0

where α, β, γ, δ are constants and we are assuming our spatialdomain is a thin rod of length L.


Separation of Variables: General Idea

When using this technique, we look for simple-type real-valued solutionsu (x , t) of the form

u (x , t) = X (x)T (t) (1)

where X is a function of x only and T is a function of t only, hence thename separation of variables. We will see that in fact, we can find aninfinite number of these solutions which satisfy both the PDE and theBCs. These simple functions un (x , t) = Xn (x)Tn (t) are calledfundamental solutions. The solution which will satisfy the PDE, the BCsand the IC will be of the form∑

AnXn (x)Tn (t)

where An are constants to determine.


Separation of Variables: Some remarks

Before we continue, let us also make a few remarks for this section.

1 When using the notation f ′ to denote the derivative of f it isunderstood that f is a function of only one variable and f ′ meansderivative with respect to that variable. For example, if T (t) is afunction of t only and X (x) is a function of x only then T ′ (t) meansderivative of T with respect to t but X ′ (x) means derivative of Xwith respect to x .

2 Given a function f (x), f (x) ≡ 0 means that f (x) = 0 for every x .


Separation of Variables

We will explain how the technique is applied by looking at a specificexample and outlining the steps to follow. We will solve the followingproblem:

PDE ut = kuxx 0 < x < L 0 < t <∞

BCu (0, t) = 0u (L, t) = 0

0 < t <∞

IC u (x , 0) = φ (x) 0 ≤ x ≤ L

(2)

The solution we seek must satisfy three conditions: the PDE, the BCs,and the IC. The technique we outline has three steps. First, assuming thesolution is of the form u (x , t) = X (x)T (t), we replace the PDE by twoODEs. Next, we find solutions of these ODEs which satisfy the BCs.Finally, in step three, we combine all the solutions found in step 2 usingthe principle of superposition and finish solving by using IC.


Step 1: Transforming the PDE in Two ODEs

With u (x , t) = X (x)T (t), the PDE ut = kuxx becomes

X (x)T ′ (t) = kX ′′ (x)T (t). Which can be written asT ′ (t)kT (t)

=X ′′ (x)X (x)

. This in turn implies

T ′ (t)kT (t)

=X ′′ (x)X (x)

= −λ. Hence the two ODEs:

T ′ + λkT = 0 (3)

X ′′ + λX = 0




X (x)T ′ (t) = kX ′′ (x)T (t). Which can be written as

T ′ (t)kT (t)

=X ′′ (x)X (x)


T ′ (t)kT (t)

=X ′′ (x)X (x)


T ′ + λkT = 0 (3)

X ′′ + λX = 0





=X ′′ (x)X (x)


T ′ (t)kT (t)

=X ′′ (x)X (x)


T ′ + λkT = 0 (3)

X ′′ + λX = 0





=X ′′ (x)X (x)


T ′ (t)kT (t)

=X ′′ (x)X (x)


T ′ + λkT = 0 (3)

X ′′ + λX = 0





=X ′′ (x)X (x)


T ′ (t)kT (t)

=X ′′ (x)X (x)


T ′ + λkT = 0 (3)

X ′′ + λX = 0


Step 2: Solving the Time-Dependent ODE

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

T (t) = C1e−λkt (4)


Step 2: Solving the Time-Dependent ODE

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

T (t) = C1e−λkt (4)


Step 2: Solving the Eigenvalue Problem

The next ODE to solve isX ′′ + λX = 0 (5)

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

X ′′ + λX = 0X (0) = 0X (L) = 0

(6)

DefinitionThe values of λ for which the boundary value problem in 6 has nontrivialsolutions are called eigenvalues. The corresponding solutions X (x) arecalled eigenfunctions.



There are three possibilities.

1 λ < 0. In this case, the general solution of equation 5 isX (x) = C2e

√−λx + C3e−

√−λx . The boundary conditions imply that

only the trivial solution is possible. So, we discard this case.2 λ = 0. In this case, equation 5 becomes X ′′ = 0 with solutionsX = C2X + C3. The boundary conditions imply that only the trivialsolution is possible. So, we discard this case.

3 λ > 0. In this case, the general solution of equation 5 isX (x) = C2 sin

√λx + C3 cos

√λx . We will see below that there are

solutions for certain values of λ in this case.



To summarize, we have found that

T (t) = C1e−λkt

X (x) = C2 sin√λx + C3 cos

√λx

where λ, C1, C2, and C3 are constants with λ > 0. It follows that

u (x , t) = T (t)X (x)

u (x , t) = e−λkt(A sin

√λx + B cos

√λx)

(7)

We still have to find the constants A, B, and λ.



u (x , t) = e−λkt(A sin

√λx + B cos

√λx)solves not only the PDE, it

must also satisfy the BCs.

u (0, t) = 0 =⇒ B = 0

u (L, t) = 0 =⇒ sin√λL = 0⇐⇒

√λL = ±π, ±2π, ±2π... in order

to have a nontrivial solution.

Hence the values of λ for which the eigenvalue problem has a solutionare √

λ = ±nπLor λ =

(nπL

)2Thus, we have found an infinite number of solutions

un (x , t) = Ane−(nπL )

2kt sin

(nπxL

)for n = 1, 2, 3, ... (8)


Step 3: Finishing the Problem - Principle of Superposition

Since all the un (x , t) are solutions, and our PDE is a second order linearhomogeneous PDE, we know from before that any linear combination ofthese functions is also a solution. So, we look at the solution

u (x , t) =∞∑n=1

un (x , t)

=∞∑n=1

Ane−(nπL )

2kt sin

(nπxL

)(9)

We find the coeffi cients An so the IC is also satisfied that is

u (x , 0) = φ (x)


Step 3: Finishing the Problem

The IC implies

φ (x) =∞∑n=1

An sin(nπxL

)(10)

The question is can an arbitrary function φ (x) be represented as aninfinite series as shown in equation 10. We will give a complete answer tothis question in the next chapter. For now, let us assume that if φ iscontinuous, then it can be represented as shown in equation 10. The nextquestion is how do we find the coeffi cients An. This is actually not toohard. The following proposition helps us in doing that.

Proposition

∫ L0 sin

(mπxL

)sin(nπxL

)dx =

{0 if m 6= nL2

if m = n. This property is known

as orthogonality.



We can now find An. From

φ (x) =∞∑n=1

An sin(nπxL

)= A1 sin

πxL+ A2 sin

2πxL+ A3 sin

3πxL+ ...

We multiply each side by sinmπxL

where m is an arbitrary integer, we get

φ (x) sinmπxL

= A1 sinπxLsinmπxL+A2 sin

2πxLsinmπxL+A3 sin

3πxLsinmπxL+...

We then integrate each side with respect to x , from 0 to L and get

Am =2L

∫ L0 φ (x) sin

mπxLdx



DefinitionThe expansion

φ (x) =∞∑n=1

An sin(nπxL

)= A1 sin

πxL+ A2 sin

2πxL+ A3 sin

3πxL+ ...

where An =2L

∫ L0 φ (x) sin

nπxLdx is called the Fourier sine expansion of

φ (x). It is named after Joseph Fourier, whom we mentioned above.



In conclusion, we have the following proposition:

PropositionThe solution to the IBVP 2 that is

PDE ut = kuxx 0 < x < L 0 < t <∞

BCu (0, t) = 0u (L, t) = 0

0 < t <∞

IC u (x , 0) = φ (x) 0 ≤ x ≤ L

is

u (x , t) =∞∑n=1

Ane−(nπL )

2kt sin

(nπxL

)where

An =2L

∫ L

0φ (x) sin

nπxLdx



RemarkWe finish with a few remarks and an example.

1 As t gets larger, e−(nπL )

2kt gets closer and closer to 0. Thus for large

t, the terms in the series∞∑n=1

Ane−(nπL )

2kt sin

(nπxL

)get small very

fast.2 Step 1 depends on the PDE in the problem being solved, step 2depends on the BCs and step 3 on the IC. So, when solving twoproblems which have the same PDE, the result at the end of step 1will be the same. So, there is no need to repeat step 1 twice.Similarly, if the two problems have the same PDE and the same BCs,then their steps 1 and 2 will be the same. Again, no need to redothem twice.



ExampleSolve

PDE ut = uxx 0 < x < 1 0 < t <∞

BCu (0, t) = 0u (1, t) = 0

0 < t <∞

IC u (x , 0) = sin 3πx 0 ≤ x ≤ 1


Exercises

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


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

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Separation of Variables Technique