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© 2011 Zachary S Tseng 1

MATH 251 Work sheet / Things to know

WORK SHEET #5 PARTIAL DIFFERENTIAL EQUATIONS

Chapter 10

1. Second order linear partial differential equations

(Recall from chapter 1) What is a PDE?

In this class we shall only study the family of second order linear PDEs using the method

of separation of variables to obtain Fourier series solutions.

What are the 3 types of second order linear PDEs?

A partial differential equation doesn’t have a single general solution like an ordinary

differential equation does. To find general solution we need both the equation and a set

of boundary conditions. Different boundary conditions give different general solutions.

What are boundary conditions?

What is an initial-boundary value problem?


Checklist: the steps of the method of separation of variables

Solving a PDE is a rather lengthy process containing several steps. Below is a list of

topics that we will cover in this chapter. You need to know both the subject matter of

each topic, as well as how they are interrelated and to put the steps together to solve a

PDE initial-boundary value problem.

Steady-state solution (if the boundary conditions are nonhomogeneous)

Separation of variables

Two-point boundary value problem -- eigenvalues and eigenfunctions.

Constructing the general solution

Fourier series

- Finding Fourier coefficients

- Fourier convergence theorem

- Cosine and sine series extensions

- Use it to solve for the particular solution


2. One-dimensional heat conduction equation

What is the standard form of 1-dimensional homogeneous heat conduction equation?

What is the meaning of the constant coefficient in the equation?

Know the physical meaning of the given boundary conditions (more on this later). In

general, given the same PDE, different boundary conditions will result in different

general solutions.

The first set of boundary conditions is for a bar that has both ends kept at constant zero

temperature: u(0, t) = and u(L, t) =

Putting together the equation, the boundary conditions, and an (arbitrary) initial

condition, we have our first initial-boundary value problem:


3. Separation of variables

Assume the solution is in the form: u(x, t) = X(x) T(t)

The various partial derivatives can be easily rewritten in terms of X and T. (It is not

necessary to memorize the new expressions as long as you know how partial

differentiation works.)

u = ux = ut =

uxx = utt = uxt = utx =

Use the substitutions above to rewrite the equation in terms of X and T. Algebraically re-

arrange the equation so that each side contains only one variable. Then the expressions

on both sides of the equation are equated to a constant of separation. Once this is done

the equalities can be rewritten into 2 separate equations of one variable each.

Let us do this for the heat conduction equation:

Ex.10.3.1 Separate u − 3 utx + 2 utt = 0, ux(0, t) = 0, u(L, t) = 0.

Not every PDE is separable. An example of an equation that is not separable is

uxx − 2 utx − utt = 0.

(Why not?)


The homogeneous boundary conditions (i.e., they are = 0) need to be separated similarly.

How to separate the boundary conditions?

Substitutions u(a, t) = ux(a, t) =

Ex.10.3.2 Separate each set of boundary conditions.

(a) ux(0, t) = 0, u(L, t) = 0.

(b) ux(0, t) + 5u(0, t) = 0, ux(8, t) = 0.

Combine the separated ODEs and the set of separated boundary conditions, we obtain a

system of two simultaneous linear differential equations. There is also a set of boundary

conditions that is associated with one (and only one) of the ODEs.

What is the system of equations (of one independent variable each)?


4. Two-point boundary value problem: finding eigenvalues and eigenfunctions

Once both the partial differential equation and boundary conditions are separated, we

have a system of 2 linear equations. Except we don’t just have 1 system, we actually

have a family of infinitely many of them – the systems differ from each other by the

value of the separation constant. In addition, one of the two ordinary differential

equations in each system also has boundary conditions attached to it. Our immediate

goal is to find the nonzero solutions (a.k.a., the non-trivial solutions), if any, of such a

system that also satisfy the boundary conditions. Not every value of the separation

constant will result in a nonzero solution, only a certain subset of them will.

To do so, we will look for the nonzero solutions of the equation in the system having the

two boundary conditions (hence the name, two-point boundary value problem). We need

to find nonzero solutions of the boundary value problem (BPV). Not every value of λ

would give an equation that has a nonzero solution of the BPV. In fact, most don't. But

there are some. Those values that do yield nonzero solutions are called the eigenvalues

of the boundary value problem. The actual nonzero solutions are their corresponding

eigenfunctions. The process of finding the eigenvalues and eigenfunctions of a BPV is a

very important computation step that you should pay close attention to, both the result

and the process.

Back to the current two-point BPV, in terms of x, we obtained earlier, what are its

eigenvalues and corresponding eigenfunctions?

X ″ + λX = 0, X(0) = 0, X(L) = 0


Ex.10.4.1 Find all eigenvalues and corresponding eigenfunctions of

X ″ + λX = 0, X ′(0) = 0, X(L) = 0


5. General solution

Each of the eigenvalues can be plugged into the other equation in the system (the one

without the 2-point boundary conditions, usually in terms of t) to give an instance of a

simple linear ordinary differential equation. Solve each of the equation thus obtained for

a solution T(t).

Solve for T(t):

How to combine X(x) and T(t) to form a fundamental solution?

The general solution of the PDE initial-boundary problem is just the linear combination

of all the fundamental solutions, an infinite sum of arbitrary multiples of the solution

found above.

What is the form of the general solution?

Now that the general solution is found, we can apply the initial condition:

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

We would then solve for the unique particular solution, but cannot do so just yet.

(Why not?)


6. Fourier series

A periodic function f (x), of period T = 2L, can be rewritten into an infinite series of

sinusoidal functions (a trigonometric series), called a Fourier series. The Fourier series

takes the form:

f (x) =

Where L = T/2, half of a period of f (x).

The coefficients a's (including the constant) are called:

The coefficients b's are called:

The coefficients can be computed directly using the Euler-Fourier formulas:

am =

bn =

Ex.10.6.1 Find a Fourier series representing f (x) such that

f (x) = x2, −1 < x < 1, f (x + 2) = f (x).


7. Fourier convergence theorem

Know what the theorem says. To what value will a Fourier series representing a

piecewise continuous periodic function converge at any point t (even if the function it

represents is undefined at that point)?

Ex.10.7.1 Take the Fourier series found in Ex.10.6.1. To what value does it converge

when, respectively, x = −1, 1, 2, 3.5?

8. Even and odd periodic functions

Review the properties (algebraic and calculus) of even functions and odd functions.

What is a Fourier cosine series? What does it look like?

What is a Fourier sine series? What does it look like?

9. Even and odd periodic extensions

Given a function f (x) defined on an interval of finite length, it is possible to construct

periodic functions whose domain contains all real numbers and whose behavior duplicate

that of f (x) on the latter's (finite length) domain. The two most useful (and the easiest to

construct) constructions of such nature are the even periodic (or Fourier cosine series)

extension and the odd periodic (or Fourier sine series) extension.


Even periodic extension

How is it done?

Odd periodic extension

How is it done?


10. Finding the particular solution

We are now finally able to find the particular solution of the heat conduction problem.

We can now see that, in order to use the initial condition, it has to be expanded into a

periodic function over (−∞, ∞). In the current case, it needs to be expanded into a Fourier

sine series (i.e., use the odd periodic extension technique). Then we can equate both

sides and extract the coefficients of the particular solution.

How do you that you need the odd periodic extension, instead of the even one?

How is it done? What are the coefficients Cn of the particular solution?

Ex.10.10.1 Solve the heat-conduction initial-boundary value problem.

25uxx = ut , 0 < x < 6, t > 0,

u(0, t) = 0, u(6, t) = 0,

u(x, 0) = 4sin(πx) − 7sin(5πx/3).


In general, the procedure works for any piecewise continuous initial condition.

Ex.10.10.2 Suppose the example above has, instead, an initial condition of

u(x, 0) = f (x) = 32. That is, a constant initial temperature distribution of 32 degrees.

How would you solve for the particular solution?


11. One-dimensional heat conduction equation − insulated ends

Now we shall apply what we have learned to solve a different heat-conduction initial-

boundary value problem. In this example, the bar has both ends sealed by a perfect

insulator so that no heat can move across either end. Note the new boundary conditions

that describe this new setting.

The new initial-boundary value problem is:

Outline of its solution:

i. The (same) equation separates the same way as before. How do the new boundary

conditions separate?

ii. What is the new two-point boundary value problem? What are its eigenvalues and

the corresponding eigenfunctions?


iii. The equation of T is solved the same way, except there is an additional case for

n = 0. What are the solutions T(t)?

iv. What is the general solution of the insulated-ends problem?

v. How to solve the coefficients and find the particular solution?

(How do you know to use the even periodic extension of the initial condition?)


5uxx = ut , 0 < x < 8, t > 0,

ux(0, t) = 0, ux(8, t) = 0,

u(x, 0) = 20 + 6cos(2πx) − 7cos(5πx/2).


Ex.10.11.2 Suppose the example 10.11.1 has, instead, an initial condition of

u(x, 0) = f (x) = x2 + x. How would you solve for the particular solution?

12. One-dimensional heat conduction equation − two ends kept at arbitrary constant

temperatures: an example of nonhomogeneous boundary conditions

Let us now see what happens when the boundary conditions are nonzero (known as

nonhomogeneous boundary conditions). To keep things simple, our latest boundary

conditions are two arbitrary constants (either one could still be zero, but we do not

assume this). The new set of boundary conditions models a bar whose two ends are not

insulated, but rather kept at certain constant temperatures.

The new initial-boundary value problem is:


i. The equation will separate the same way. But the boundary conditions don't separate.

(Why not?) What do we do now?

Introducing the idea of decomposing the solution into a steady-state and a transient

parts:

u(x, t) = v(x) + w(x, t).

Where v(x) is

and w(x,t) is


ii. Find the steady-state solution. What is v(x) for this problem?

iii. Once the steady-state solution has been found, it can be set aside. Its effects should

be removed from the initial and boundary conditions. How to do this? What (new

problem) does the initial-boundary value problem become? We will need to solve it for

the transient part of the solution.

iv. Solve this new problem, which really is an old problem that we have already solved,

to find the transient solution w(x, t). What is the transient solution?

v. Combine the steady-state and transient parts to form the general solution:

u(x,t) =


vi. How to solve the coefficients and find the particular solution? (Remember to

subtract v(x) from the given initial condition first.) Do you expand the initial

temperature distribution into an even or an odd periodic function? Why?


10uxx = ut , 0 < x < 2, t > 0,

u(0, t) = 60, u(2, t) = 40,

u(x, 0) = 60 − 10x − 30sin(2πx) + 40sin(9πx/2).


13. Steady-state solution

A solution of a PDE may be a sum of two parts. One is dependent of time (the transient

solution). The other is independent of time. A steady-state solution is the time-

independent part.

That is, u(x, t) =

For the heat-conduction equation, as well as for the wave equation, the steady-state

solution introduced by nonhomogeneous boundary conditions is in the form of a

polynomial of degree 1.

v(x) =

u(x0, t) = v(x0), and ux(x0, t) = v′(x0)

Ex.10.13.1 Find the steady-state solution satisfying the given nonhomogeneous boundary

conditions.

(a) ux(0, t) = 5, u(10, t) = 30

(b) u(0, t) − 2ux(0, t) = 0, u(4, t) = 30

For heat conduction problems, the steady-state solution is always the limit reached by

u(x,t) as t → ∞.


14. One-dimensional wave equation: motion of a vibrating string

What is the standard form of 1-dimensional homogeneous wave equation?

What is the meaning of u(x, t)?

What is the meaning of the constant coefficient in the equation?

With both ends of the string firmly fixed in place, how does the initial-boundary value

problem appear? Note the problem has two initial conditions. Also note and understand

the meaning of the boundary conditions of the problem. They look familiar, but they

mean quite differently from the instance when they appeared with heat conduction

equation.)

We have learned all the tools necessary to solve this problem from scratch. Let us do so.

Outline of solution, undamped 1-dim homogeneous wave equation:

i. How does the wave equation separate? How do the boundary conditions separate?

ii. What is the two-point boundary value problem? It is a problem that we have seen

before. What are its eigenvalues and corresponding eigenfunctions?


iii. Now solve the equation of t. It is a simple second order linear equation. What are

the solutions T(t)?

iv. What is the general solution of the undamped wave equation with fixed ends? Note

that it has two distinct sets of (infinitely many) coefficients.

v. How to solve the coefficients and find the particular solution? There are two initial

conditions, which are used to solve the two sets of coefficients.

15. Wave equation: special cases of zero initial velocity or zero initial displacement

Because the way the coefficients are calculated (see above), the solution of the wave

equation takes on a quite simpler form if either of its two initial conditions is zero.

Special case 1: Zero initial velocity, ut(x, 0) = 0

The general solution becomes

u(x,t) =

The coefficients Bn = 0, for all n, while An are found as usual.


Special case 2: Zero initial displacement, u(x, 0) = 0

The general solution becomes

u(x,t) =

The coefficients An = 0, for all n, while Bn are found as usual.

Ex.10.15.1 Solve the one-dimensional wave equation initial-boundary problem

25uxx = utt , 0 < x < 2π, t > 0,

u(0, t) = 0, and u(2π, t) = 0,

u(x, 0) = 10sin(x) − 5sin(3x) + 20sin(7x/2),

ut(x, 0) = 0.

Ex.10.15.2 Suppose the example above has, instead, an initial displacement of 0,

u(x, 0) = f (x) = 0, and a constant initial velocity, ut(x, 0) = g(x) = 100. How would you

solve for the particular solution?


16. The D'Alembert solution of the wave equation

Besides the Fourier series solution of the wave equation, there is also an older, geometric-

oriented, solution to the wave equation. The so-called D'Alembert solution readily gives

us some interesting insight into the behavior of the solution of the wave equation.

For the special case of zero initial velocity, the D'Alembert solution is given by:

u(x,t) =

The coefficients An are found the same way as before, by expanding the initial

displacement into an odd periodic function of period L. then extract its Fourier sine

coefficients.

What does the D'Alembert solution say regarding the behavior of the solution of the wave

equation?

Ex.10.16.1 (Ex.10.15.1) Solve the one-dimensional wave equation initial-boundary

problem using the D'Alembert's formula. Then use the addition formula of sine to check

that the solution is equivalent to the Fourier series solution found in Example 10.15.1.

25uxx = utt , 0 < x < 2π, t > 0,

u(0, t) = 0, and u(2π, t) = 0,

u(x, 0) = 10sin(x) − 5sin(3x) + 20sin(7x/2),

ut(x, 0) = 0.


17. Two-dimensional Laplace equation [in rectangular coordinates]

What is the standard form of 2-dimensional Laplace equation (also known as the

potential equation)?

Note the lack of a time variable. The solution u(x, y) is some kind of steady-state

potential function. There is not an initial condition. But there are extra boundary

conditions in the problem. Two of the boundary conditions will be used to find the

general solution. While the others will serve the role usually played by the initial

conditions, to be used to solve for the particular solution.

For the purpose of this discussion, we will study the solution of the Laplace equation

over a rectangular region whose 4 sides are represented by the boundary conditions:

u(x, 0) = 0, u(x, b) = 0, u(0, y) = 0, and u(a, y) = f (y).

The boundary value problem can, therefore, be visualized by the sketch:


i. How does the equation separate? How do the boundary conditions separate? (There

is not a unique way to do separate the equation. The boundary conditions would tell us

what the most convenient way is to separate the equation.)

ii. What is the two-point boundary value problem? It is a familiar one that we've seen

twice before. Recall its eigenvalues and corresponding eigenfunctions.


iii. Solve the second equation, in this case it is the equation of x.

iv. What is the general solution of this boundary value problem?

Alternately, we can write it in terms of hyperbolic function:

u(x,y) =

v. How to solve the coefficients and find the particular solution?


Summary: Solving Second Order Linear Partial Differential Equations

The Method of Separation of Variables:

0. (If the boundary conditions are nonhomogeneous) Solve for the

steady-state solution, v(x), which is a function of x only that satisfies

both the PDE and the boundary conditions. Afterwards rewrite the

problem’s boundary and initial conditions to subtract out the

contribution from the steady-state solution. Therefore, the problem is

now transformed into one with homogeneous boundary conditions.

1. Separate the PDE into ODEs of one independent variable each.

Rewrite the boundary conditions so they associate with only one of

the variables.

2. One of the ODEs is a part of a two-point boundary value problem.

Solve this problem for its eigenvalues and eigenfunctions.

3. Solve the other ordinary differential equation.

4. Multiply the results from steps (2) and (3), and sum up all the

products to find the general solution respect to the given

homogeneous boundary conditions. Add to it the steady-state solution

(from step 0, if applicable) to find the overall general solution.

5. Expand the initial condition into a suitable (e.g. a sine or a cosine

series) Fourier series. Then compare it against u(0, t) to find the

coefficients for the particular solution.

math251-worksheet 5 … · title: microsoft word - math251-worksheet 5 author: zach created date:...

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