Lecture6. Partial Differential Equationsolgar/L6.pde.pdf · Lecture6. Partial Differential Equations ... The simplest form of partial differential equation is such that a solution

EP219 Lecture notes - prepared by- Assoc. Prof. Dr. Eser OLĞAR 2012-Spring

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Lecture6. Partial Differential Equations

6.1 Review of Differential Equation

We have studied the theoretical aspects of the solution of linear homogeneous differential equations with

constant coefficient equation 𝑎𝑦’’+ 𝑏𝑦’ + 𝑐𝑦 = 0 in practice.

We have the sum of a function and its derivatives equal to zero, so the derivatives must have the same

form as the function. Therefore we expect the function to be ex. If we insert this in the equation, we

obtain:

𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0

This is called the auxiliary equation of the homogeneous differential equation 𝑎𝑦’’ + 𝑏𝑦’ + 𝑐𝑦 = 0. If we solve the characteristic equation, we will see three different possibilities: Two real roots, double real

root and complex conjugate roots.

6.2 Partial Differential Equations

A differential equation in which partial derivatives occur is called a partial differential equation.

Mathematical models of physical phenomena involving more than one independent variable often include

partial differential equations. They also arise in such diverse areas as epidemiology (for example,

multivariable predator/prey models of AIDS), traffic flow studies and the analysis of economies.


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We will be primarily concerned in this part with three broadly defined kinds of phenomena: wave

motion, radiation or conduction of energy, and potential theory. Models of these phenomena involve

partial differential equations called, respectively, the wave equation, the heat equation, and the potential

equation, or Laplace's equation.

Partial differential equations are much more complicated compared with the ordinary differential

equations. There is no universal solution technique for nonlinear equations, even the numerical

simulations are usually not straightforward. Thus, we will mainly focus on the linear partial differential

equations and the equations of special interests in engineering and computational sciences. A partial

differential equation (PDE) is a relationship containing one or more partial derivatives. Similar to the

ordinary differential equation, the highest nth partial derivative is referred to as the order n of the partial

differential equation. The general form of a partial differential equation can be written as

𝜑(𝑥,𝑦,𝜕𝑢

𝜕𝑥,𝜕𝑢

𝜕𝑦,𝜕2𝑢

𝜕𝑥2,𝜕2𝑢

𝜕𝑦2,𝜕2𝑢

𝜕𝑥𝜕𝑦,……… ) = 0

where u is the dependent variable and x, y, ... are the independent variables. A simple example of partial

differential equations is the linear first order partial differential equation, which can be written as

𝑎 𝑥,𝑦 𝜕𝑢

𝜕𝑥+ 𝑏 𝑥,𝑦

𝜕𝑢

𝜕𝑦= 𝑓(𝑥,𝑦)

for two independent variables and one dependent variable u. If the right hand side is zero or simply

f(x, y) = 0, then the equation is said to be homogeneous. The equation is said to be linear if a, b and f are

unctions of x, y only, not u itself.

For simplicity in notations in the studies of partial differential equations, compact subscript forms are

often used in the literature. They are

𝑢𝑥 ≡𝜕𝑢

𝜕𝑥,𝑢𝑦 ≡

𝜕𝑢

𝜕𝑦,𝑢𝑥𝑥 ≡

𝜕2𝑢

𝜕𝑥2,𝑢𝑦𝑦 ≡

𝜕2𝑢

𝜕𝑦2,𝑢𝑥𝑦 ≡

𝜕2𝑢

𝜕𝑥𝜕𝑦

and thus we can write the general of one dimensional PDE as

a ux + b uy = f.

In this chapter generally we deal with second order differential equation.

6.3 Classification of PDE

A linear second-order partial differential equation can be written in the generic form in terms of two

independent variables x andy,

a uxx + b uyx + c uyy + g ux + h uy +k u = f.

where a, b, c, g, h, k and f are functions of x and y only. If f ( x, y, u) is also a function of u, then we say

that this equation is quasi-linear.

If, ∆= 𝑏2 − 4𝑎𝑐 < 0 the equation is elliptic. One famous example is the Laplace equation uxx + uyy = 0

If Δ > 0, it is hyperbolic. One example is the wave equation utt =c2 uxx.

If Δ = 0, it is parabolic. Diffusion and heat conduction equations are of the parabolic type ut =k uxx.

Examples of PDEs

Many physical processes in engineering are governed by three classic partial differential equations so they

are widely used in a vast range of applications. Among the most frequently encountered PDEs are the

following:

1. Laplace's equation ∇2𝜑 = 0

This very common and very important equation occurs in studies of

a. electromagnetic phenomena including electrostatics, dielectrics, steady


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currents, and magnetostatics,

b. hydrodynamics (irrotational flow of perfect fluid and surface waves),

с. heat flow, and

d. gravitation.

2. Poisson's equation ∇2𝜑 = −𝜌/𝜀0.

In contrast to the homogeneous Laplace equation, Poisson's equation is non-homogeneous with a source

term -𝜌/𝜀0.

3. The wave (Helmholtz) and time-independent diffusion equations

∇2𝜑 ∓ 𝑘2𝜑 = 0

These equations appear in such diverse phenomena as

a. elastic waves in solids including vibrating strings, bars, membranes,

b. sound or acoustics,

с electromagnetic waves, and

d. nuclear reactors.

4. The time-dependent diffusion equation

∇2𝜑 =1

𝑎2

𝜕𝜑

𝜕𝑡

and the corresponding four-dimensional forms involving the d'Alembertian, a four-dimensional analog of

the Laplacian in Minkowski space

5. The time-dependent wave equation, 𝜕2𝜑 = 0

6. The scalar potential equation ∂2𝜑 = −𝜌

𝜀0.

Like Poisson's equation this equation is nonhomogeneous with a source term 𝜌

𝜀0.

7. The Klein-Gordon equation 𝜕2𝜑 = −𝜇2𝜑 and the corresponding vector equations in which the

scalar function 𝜑 is replaced by a vector function. Other more complicated forms are common.

8. The Schrodinger wave equation

and

for the time-independent case.

9. The equations for elastic waves and viscous fluids and the telegraphy equation.

10. Maxwell's coupled partial differential equations for the electric and magnetic fields and those of

Dirac for relativistic electron wavefunctions.

6.4 Techniques for Solving PDEs

Different types of equations usually require different solution techniques. However, there are some

methods that work for most of the linearly partial differential equations with appropriate boundary

conditions on a regular domain. These methods include separation of variables, series expansions,

similarity solutions, hybrid methods, and integral transform methods.


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i) Solution by direct integration

The simplest form of partial differential equation is such that a solution can be determined by direct

partial integration.

Example 1:

Solve the differential equation uxx = 12 x2 (t+1) given that x = 0, u = cos(2t) and ux = sint.

Notice that the boundary conditions are functions of t and not just constants. Integrating the equation

uxx = 12 x2 (t+1) partially with respect to x, we have ux = 4 x

3 (t+1) + f(t) where the arbitrary function f(t)

takes the place of the normal arbitrary constant in ordinary integration. Integrating partially again with

respect to x gives

u = x4 (t+1) + x f(t)+ g(t)

where g(t) is a second arbitrary function.

To find the two arbitrary functions, we apply the given initial conditions that x = 0, u = cos(2t) and

ux = sint. Subtituting in the relevant equations gives

f(t) = sint and g(t) = cos(2t)

then the solution becomes

u = x4 (t+1) + x sint + cos(2t)

Example 2:

Solve the differential equation uxy = sin(x+y) given that at y = 0, ux =1 and at x = 0, u = (y-1)2.

Initial conditions and boundary conditions:

As with any differential equation, the arbitrary constants or arbitrary functions in any particular case are

determines from the additional information given concerning the variables of the equation. These extra

facts are called the initial conditions or, more generally the boundary conditions since they do not lways

refer to zero values of the independent variables.

Example 3:

Solve the differential equation uxy = sin(x)cos(y), subject to the boundary conditions that at y = π/2,

ux =2x and at x = π, u = 2sin(y).


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ii) Separation of Variables

The separation of variables attempts a solution of the form

u = X(x)Y(y)Z(z)T(t)

where X(x), Y(y), Z(z) and T(t) are functions of x, y, z, t, respectively. In order to determine these

functions, they have to satisfy the partial differential equation and the required boundary conditions. As a

result, the partial differential equation is usually transformed into two or three ordinary differential

equations (ODEs), and these ordinary differential equations often appear as eigenvalue problems. The

final solution is then obtained by solving these ODEs.

A solution that has this form is said to be separable in x, y, z and t, and seeking solutions of this form is

called the method of separation of variables.

As simple examples we may observe that, of the functions

(i) xyz2 sin bt, (ii) xy + zt, (iii) (x

2 + y

2)z cos ωt,

(i) is completely separable, (ii) is inseparable in that no single variable can be separated out from it and

written as a multiplicative factor, whilst (iii) is separable in z and t but not in x and y.

The following examples are to illustrate the method of solution by studying the wave equation, the heat

equation and the Laplace equation.

a) The wave equation

The wave equation is

𝑐2∇2𝑢 =𝜕2𝜑

𝜕2𝑡= 𝑢𝑡𝑡

governs a wide variety a wave phenomena such as electromagnetic waves, water waves, supersonic flow,

pulsatile blood flow, acoustics, elastic waves in solids, and vibrating strings and membranes. In this

introductory section we derive the wave equations governing the vibrating string and vibrating membrane

and outline, in the exercises, several other such cases leading to wave equations.

Vibrating String with Zero Initial Velocity

Consider an elastic string of length L, fastened at its ends on the x axis at x = 0 and x = L. The string is

displaced, then released from rest to vibrate in the x, y plane. We want to find the displacement function

y(x, t), whose graph is a curve in the x, y plane showing the shape of the string at time t. If we took a

snapshot of the string at time t, we would see this curve. The boundary value problem for the

displacement function is


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The Fourier method, or separation of variables, consists of attempting a solution of the form y(x, t) =

X(x)T(t) . Substitute this into the wave equation to obtain

XT" = c2X" T,

where T' = dT/dt and X' = dX/dx. Then

The left side of this equation depends only on x, and the right only on t . Because x and t are independent,

we can choose any to we like and fix the right side of this equation at the constant value T"(to)/c2 T(to),

while varying x on the left side. Therefore X"/X must be constant for all x in (0, L). But then T"/c2 T must

equal the same constant for all t > 0. Denote this constant - . (The negative sign is customary and

convenient, but we would arrive at the same final solution if we used just ) . A is called the separation

constant, and we now have

Then

X" + X = 0 and T" + c2T = 0.

The wave equation has separated into two ordinary differential equations.

Now consider the boundary conditions. First,

y(0, t) = X(0)T(t) = 0

for t ≥ 0. If T(t) = 0 for all t ≥ 0, then y(x, t) = 0 for 0 ≤ x ≤ L and t ≥ 0. This is indeed the solution if

f(x) = 0, since in the absence of initial velocity or a driving force, and with zero displacement, the string

remains stationary for all time. However, if T(t) ≠ 0 for any time, then this boundary condition can be

satisfied only if

X(0)=0

Similarly,

y(L, t) = X(L)T(t) = 0

for t ≥ 0 requires that

X(L)=0.

We now have a boundary-value problem for X:

X" + X = 0 ; X (0) = X (L) = 0.

The values of for which this problem has nontrivial solutions are the eigenvalues of this problem, and

the corresponding nontrivial solutions for X are the eigenfunctions. By solving we get

The eigenfunctions are nonzero constant multiples of

𝑋𝑛 𝑥 = 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿

for n = 1, 2, . . . . At this point we therefore have infinitely many possibilities for the separatio n


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constant and for X(x) .

Now turn to T(t). Since the string is released from rest,

This requires that T' (0) = 0. The problem to be solved for T is therefore

T" + c2 T = 0; T' (0) = 0.

However, we now know that A can take on only values of the form n2π

2/L

2, so this problem is really

T" +( n2π

2 c

2/L

2)T = 0; T’(0) = 0.

The differential equation for T has general solution

𝑇 𝑡 = 𝑎 𝑐𝑜𝑠 𝑛𝜋𝑐𝑡

𝐿 + 𝑏 𝑠𝑖𝑛

𝑛𝜋𝑐𝑡

𝐿

Now

𝑇 ′ 0 =𝑛𝜋𝑐

𝐿𝑏 = 0

so b = 0. We therefore have solutions for T(t) of the form

𝑇𝑛 𝑡 = 𝑐𝑛 𝑐𝑜𝑠 𝑛𝜋𝑐𝑡

𝐿

for each positive integer n, with the constants c, as yet undetermined.

We now have, f o r n = 1, 2, . . .., functions

𝑦𝑛 𝑥, 𝑡 = 𝑐𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 𝑐𝑜𝑠

𝑛𝜋𝑐𝑡

𝐿

Each of these functions satisfies both boundary conditions and the initial condition yt(x, 0) = 0. We need

to satisfy the condition y(x, 0) = f(x).

It may be possible to choose some n so that yn(x, t) is the solution for some choice of cn. For example,

suppose the initial displacement is

𝑓 𝑥 = 14𝑠𝑖𝑛 3𝜋𝑥

𝐿

Now choose n = 3 and c3 = 14 to obtain the solution

𝑦 𝑥, 𝑡 = 14 𝑠𝑖𝑛 3𝜋𝑥

𝐿 𝑐𝑜𝑠

3𝜋𝑐𝑡

𝐿

This function satisfies the wave equation, the conditions y(0) = y(L) = 0, the initial condition

y(x, 0) = 14sin(3 x/L), and the zero initial velocity condition yt (x, 0) = 0

However, depending on the initial displacement function, we may not be able to get by simply by

picking a particular n and cn. For example, if we initially pick the string up in the middle and have initial

displacement function

(as in Figure ), then we can never satisfy y(x, 0) = f(x) with one of the y’n s. Even if we try a finite linear

combination


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we cannot choose c1, . . . , cN to satisfy y(x, 0) = f(x) for this function, since f(x) cannot be written as a

finite sum of sine functions. We are therefore led to attempt an infinite superposition

𝑦 𝑥, 𝑡 = 𝑐𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 𝑐𝑜𝑠

𝑛𝜋𝑐𝑡

𝐿

∞

𝑛=1

We must choose the cN to satisfy

𝑦 𝑥, 0 = 𝑐𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿

∞

𝑛=1

We can do this! The series on the right is the Fourier sine expansion of f(x) on [0, L]. Thus choose the

Fourier sine coefficients

𝑐𝑛 =2

𝐿 𝑓(𝑥)𝑠𝑖𝑛

𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

With this choice, we obtain the solution

𝑦 𝑥, 𝑡 = 2


𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 𝑐𝑜𝑠

𝑛𝜋𝑐𝑡

𝐿

∞

𝑛=1

This strategy will work for any initial displacement function f which is continuous with a piecewise

continuous derivative on [0, L], and satisfies f(0) = f(L) = 0. These conditions ensure that the Fourier sine

series of f(x) on [0, L] converges to f(x) for 0 ≤ x ≤ L.

In specific instances, where f(x) is given, we can of course explicitly compute the coefficients in this

solution . For example, if L = π and the initial position function is f(x) = x cos(5x/π) on [0, π], then the nth

coefficient in the solution is

𝑐𝑛 =2

𝜋 cos

5𝑥

2 sin

𝑛𝜋𝑥

𝐿 𝑑𝑥 =

𝜋

0

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𝜋

𝑛(−1)𝑛+1

𝜋(5 + 2𝑛)2(5 − 2𝑛)2

The solution for this initial displacement function, and zero initial velocity, is

𝑦 𝑥, 𝑡 = 8

𝜋

𝑛(−1)𝑛+1

𝜋(5 + 2𝑛)2(5 − 2𝑛)2 sin(𝑛𝑥)cos(𝑛𝑐𝑡)

∞

𝑛=1

Vibrating String with Given Initial Velocity and Zero Initial Displacement

Now consider the case that the string is released from its horizontal position (zero initial displacement),

but with an initial velocity given at x by g(x) . The boundary value problem for the displacement function

is


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We begin as before with separation of variables. Put y(x, t) = X(x)T(t). Since the partial differential

equation and boundary conditions are the same as before, we again obtain

X" + X = 0; X (0) = X(L) = 0,

with eigenvalues

and eigenfunctions constant multiples of

𝑋𝑛 𝑥 = sin 𝑛𝜋𝑥

𝐿

Now, however, the problem for T is different and we have

y(x, 0) = 0 = X(x)T(0),

so T(0) = 0. The problem for T is

T" +( n2π

2 c

2/L

2)T = 0; T(0) = 0.

(In the case of zero initial velocity we had T'(0) = 0) . The general solution of the differential

equation for T is

𝑇 𝑡 = 𝑎 cos 𝑛𝜋𝑐𝑡

𝐿 + 𝑏 sin

𝑛𝜋𝑐𝑡

𝐿

Since T(0) = a = 0, solutions for T(t) are constant multiples of sin(nπct/L) . Thus, for n = 1 , 2, . . . , we

have functions

𝑦𝑛 𝑥, 𝑡 = 𝑐𝑛 sin 𝑛𝜋𝑥

𝐿 cos

𝑛𝜋𝑐𝑡

𝐿

Each of these functions satisfies the wave equation, the boundary conditions and the zero initial

displacement condition. To satisfy the initial velocity condition yt (x, 0) = g(x), we generally must attempt

a superposition

𝑦 𝑥, 𝑡 = 𝑐𝑛 sin 𝑛𝜋𝑥

𝐿 sin

𝑛𝜋𝑐𝑡

𝐿

∞

𝑛=1

Assuming that we can differentiate this series term by term, then

𝜕𝑦

𝜕𝑥 𝑥, 0 = 𝑐𝑛 sin

𝑛𝜋𝑥

𝐿 𝑛𝜋𝑐

𝐿= 𝑔(𝑥)

∞

𝑛=1

This is the Fourier sine expansion of g(x) on [0, L]. Choose the entire coefficient of sin(nπx/L) to be the

Fourier sine coefficient of g(x) on [0, L] :

𝑐𝑛𝑛𝜋𝑐

𝐿=

2

𝐿 𝑔 𝑥 sin

𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0


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Or

𝑐𝑛 =2

𝑛𝜋𝑐 𝑔 𝑥 sin

𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

The solution is

𝑦 𝑥, 𝑡 = 2

𝑛𝜋𝑐 𝑔 𝑥 sin

𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

sin 𝑛𝜋𝑥

𝐿 sin

𝑛𝜋𝑐𝑡

𝐿

∞

𝑛=1

For example, suppose the string is released from its horizontal position with an initial

velocity given by g(x) = x (1 + cos(a π x/L)). Compute

𝑔 𝑥 sin 𝑛𝜋𝑥

𝐿 𝑑𝑥 = 𝑥(1 + cos(𝜋𝑥/𝐿)) sin

𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

𝐿

0

The solution corresponding to this initial velocity function is

If we let c = 1 and L = π, the solution becomes

Example 4:


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b) The Heat equation

The conduction of heat in a uniform bar depends on the initial distribution of temperature and on the

physical properties of bar, i.e. the thermal conductivity and specific heat of material, and the mass per unit

length of the bar.

Ends of the Bar Kept at Temperature Zero

Suppose we want the temperature distribution u(x, t) in a thin, homogeneous (constant density) bar of

length L, given that the initial temperature in the bar at time zero in the cross section at x perpendicular to

the x axis is f(x). The ends of the bar are maintained at temperature zero for all time.

The one dimensional heat equation is written as

where k = K/ is a positive constant depending on the material of the bar. The number k is called the

diffusivity of the bar.

The boundary value problem modeling this temperature distribution is

We will use separation of variables. Substitute u(x, t) = X(x)T(t) into the heat equation to get

X T' = k X" T

or

The left side depends only on time, and the right side only on position, and these variables are

independent. Therefore for some constant ,

Now

u(0, t) = X(0)T(t) = 0.

If T(t) = 0 for all t, then the temperature function has the constant value zero, which occurs if the initial

temperature f(x) = 0 for 0 ≤ x ≤ L. Otherwise T(t) cannot be identically zero, so we must have X(0) = 0.

Similarly, u(L, t) = X(L)T(t) = 0 implies that X(L) = 0. The problem for X is therefore

X" +X = 0; X(0) = X(L) = 0.

We seek values of A (the eigenvalues) for which this problem for X has nontrivial solutions (the

eigenfunctions).

This problem for X is exactly the same one encountered for the space-dependent function in separating

variables in the wave equation. There we found that the eigenvalues are

for n = 1, 2, . . ., and corresponding eigenfunctions are nonzero constant multiples of


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𝑋𝑛 𝑥 = 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿

The problem for T becomes

T’ +( n2π

2 k /L

2)T = 0

which has general solution

𝑇𝑛 𝑡 = 𝑐𝑛 𝑒−𝑛2𝜋2𝑘𝑡/𝐿2

For n = 1, 2, • • • , we now have functions

𝑢𝑛 𝑥, 𝑡 = 𝑐𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 𝑒−𝑛

2𝜋2𝑘𝑡/𝐿2

which satisfy the heat equation on [0, L] and the boundary conditions u(0, t) = u(L, t) = 0. There remains

to find a solution satisfying the initial condition. We can choose n and cn, so that

𝑢𝑛 𝑥, 0 = 𝑐𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 = 𝑓(𝑥)

only if the given initial temperature function is a multiple of this sine function. This need not be the case.

In general, we must attempt to construct a solution using the superposition

𝑢𝑛 𝑥, 𝑡 = 𝑐𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 𝑒−𝑛

2𝜋2𝑘𝑡/𝐿2∞𝑛=1 .

Now we need

𝑢 𝑥, 0 = 𝑐𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 = 𝑓(𝑥)

∞

𝑛=1

which we recognize as the Fourier sine expansion of f(x) on [0, L]. Thus choose

𝑐𝑛 =2


𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

With this choice of the coefficients, we have the solution for the temperature distribution function:

𝑢 𝑥, 𝑡 = 2


𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

𝑠𝑖𝑛 𝑛𝜋𝑥

𝐿 𝑒−𝑛


∞

𝑛=1

Example 5:

Suppose the initial temperature function is constant A for 0 < x < L, while the temperature at the ends is

maintained at zero. To write the solution for the temperature distribution function, we need to compute

𝑐𝑛 =2

𝐿 𝐴𝑠𝑖𝑛

𝑛𝜋𝑥

𝐿 𝑑𝑥 =

2𝐴

𝑛𝜋 1 − cos 𝑛𝜋 =

2𝐴

𝑛𝜋 1 − (−1)n

𝐿

0

Then, the solution is

𝑢 𝑥, 𝑡 = 2𝐴

𝑛𝜋 1 − (−1)n 𝑠𝑖𝑛

𝑛𝜋𝑥

𝐿 𝑒−𝑛


∞

𝑛=1

Since 1 - (-1)n is zero if n is even, and equals 2 if n is odd, we need only sum over the odd integers and

can write


15

𝑢 𝑥, 𝑡 =4𝐴

𝜋

1

2𝑛 − 1 𝑠𝑖𝑛

(2𝑛 − 1)𝜋𝑥

𝐿 𝑒−(2𝑛−1)2𝜋2𝑘𝑡/𝐿2

∞

𝑛=1

Temperature in a Bar with Insulated Ends

Consider heat conduction in a bar with insulated ends, hence no energy loss across the ends. If the initial

temperature is f(x), the temperature function is modeled by the boundary value problem

Attempt a separation of variables by putting u(x, t) = X(x)T(t). We obtain, as in the preceding subsection,

X" + X = 0,T' +k T=0.

Now

implies (except in the trivial case of zero temperature) that X'(0) = 0. Similarly,

implies that X'(L) = 0. The problem for X(x) is therefore

X" + X = 0 ; X' (0) = X' (L) = 0 .

The eigenvalues are

for n = 0, 1, 2, . . ., with eigenfunctions nonzero constant multiples of

𝑋𝑛 𝑥 = 𝑐𝑜𝑠 𝑛𝜋𝑥

𝐿

The equation for T is now

T’ +( n2π

2 k /L

2)T = 0

When n = 0 we get

To(t) = constant.

For n = 1, 2, . . . ,

𝑇𝑛 𝑡 = 𝑐𝑛 𝑒−𝑛2𝜋2𝑘𝑡/𝐿2

We now have functions

𝑢𝑛 𝑥, 𝑡 = 𝑐𝑛 𝑐𝑜𝑠 𝑛𝜋𝑥

𝐿 𝑒−𝑛



16

for n = 0, 1, 2, . . ., each of which satisfies the heat equation and the insulation boundary conditions. To

satisfy the initial condition we must generally use a superposition

𝑢 𝑥, 𝑡 =1

2𝑐0 + 𝑐𝑛 𝑐𝑜𝑠

𝑛𝜋𝑥

𝐿 𝑒−𝑛


∞

𝑛=1

Here we wrote the constant term (n=0) as co/2 in anticipation of a Fourier cosine expansion. Indeed, we

need

𝑢 𝑥, 0 = 𝑓(𝑥) =1

2𝑐0 + 𝑐𝑛 𝑐𝑜𝑠

𝑛𝜋𝑥

𝐿

∞

𝑛=1

the Fourier cosine expansion of f(x) on [0, L] . (This is also the expansion of the initial temperature

function in the eigenfunctions of this problem.) We therefore choose

𝑐𝑛 =2

𝐿 𝑓(𝑥)𝑐𝑜𝑠

𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

With this choice of coefficients, equation of differential equation gives the solution of this boundary value

problem as

𝑢 𝑥, 𝑡 =1

2𝑐0 +

2

𝐿 𝑓(𝑥)𝑐𝑜𝑠

𝑛𝜋𝑥

𝐿 𝑑𝑥

𝐿

0

𝑐𝑜𝑠 𝑛𝜋𝑥

𝐿 𝑒−𝑛


∞

𝑛=1

Example 6:

Suppose the left half of the bar is initially at temperature A, and the right half is kept at temperature zero.

Thus

Then

𝑐0 =2

𝐿 𝐴𝑑𝑥 = 𝐴𝐿/2

0

and, for n = 1, 2, . . .,

𝑐𝑛 =2

𝐿 𝐴𝑐𝑜𝑠

𝑛𝜋𝑥

𝐿 𝑑𝑥 =

2𝐴

𝑛𝜋sin(𝑛𝜋/2)

𝐿/2

0

The solution for this temperature function is

𝑢 𝑥, 𝑡 =1

2𝐴 +

2𝐴

𝑛𝜋sin(𝑛𝜋/2) 𝑐𝑜𝑠

𝑛𝜋𝑥

𝐿 𝑒−𝑛


∞

𝑛=1

Now sin(nπ/2) is zero if n is even. Further, if n = 2k - 1 is odd, then sin(nπ/2) = (-1)k+ 1

. The solution may

therefore be written

𝑢 𝑥, 𝑡 =1

2𝐴 +

2𝐴

𝑛𝜋

(−1)𝑛+1

2𝑛 − 1 𝑐𝑜𝑠

(2𝑛 − 1)𝜋𝑥

𝐿 𝑒−(2𝑛−1)2𝜋2𝑘𝑡/𝐿2

∞

𝑛=1


17

a) The Potential (Laplace) equation

Harmonic Functions and the Dirichlet Problem

The partial differential equation

𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2= 0

is called Laplace's equation in two dimensions. In 3-dimensions this equation is

𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2+𝜕2𝑢

𝜕𝑧2= 0

The Laplacian 2 (read "del squared") is defined in 2-dimensions by

∇2=𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2

and in three dimensions by

∇2=𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2+𝜕2𝑢

𝜕𝑧2

In this notation, Laplace's equation ∇2𝑢 = 0

A function satisfying Laplace's equation in a certain region is said to be harmonic on that region. For

example,

x2 - y

2

and

2xy

are both harmonic over the entire plane.

Laplace's equation is encountered in problems involving potentials, such as potentials for force fields in

mechanics, or electromagnetic or gravitational fields. Laplace's equation is also known as the steady-state

heat equation. The heat equation in 2- or 3-space dimensions is

𝑘∇2𝑢 =𝜕𝑢

𝜕𝑡

In the steady-state case (the limit as 𝑡 → ∞) the solution becomes independent of t, so 𝜕𝑢

𝜕𝑡= 0 and the heat

equation becomes Laplace's equation.

In problems involving Laplace's equation there are no initial conditions. However, we often encounter the

problem of solving

∇2𝑢(𝑥,𝑦) = 0

for (x, y) in some region D of the plane, subject to the condition that

u(x, y) = f(x, y),

for (x, y) on the boundary of D. This boundary is denoted D. Here f is a function having given values on

D, which is often a curve, or made up of several curves (Figure). The problem of determining a harmonic

function having given boundary values is called a Dirichlet problem, and f is called the boundary data of

the problem. There are versions of this problem in higher dimensions, but we will be concerned primarily

with dimension 2.


18

Typical boundary D of a region D.

The difficulty of a Dirichlet problem is usually dependent on how complicated the region D is. In general,

we have a better chance of solving a Dirichlet problem for a region that possesses some type of symmetry,

such as a disk or rectangle. We will begin by solving the Dirichlet problem for some familiar regions in

the plane.

Dirichlet Problem for a Rectangle

Let R be a solid rectangle, consisting of points (x, y) with 0 ≤ x ≤ L, 0 ≤ y ≤ K. We want to find a function

that is harmonic at points interior to R, and takes on prescribed values on the four sides of R, which form

the boundary R of R.

This kind of problem can be solved by separation of variables if the boundary data is nonzero on only one

side of the rectangle. We will illustrate this kind of problem, and then outline a strategy to follow if the

boundary data is nonzero on more than one side.

Example 7:

Consider the Dirichlet problem

2u(x,y)=0 for 0 < x < L, 0 < y < K,

u(x,0)=0 for 0 ≤ x ≤ L,

u(0, y) = u(L, y) = 0 for 0 ≤ y ≤ K,

u(x, K) = (L - x) sin(x) for 0 ≤ x ≤ L.

where the boundary data are shown in figure.

Let u(x, y) = X(x)Y(y) and substitute into Laplace's equation to obtain

Then

X" + X = 0 and Y"- Y = 0.

From the boundary conditions,

u(x, 0) = X(x)Y(0) = 0

Boundary data given on boundary sides of the rectangle.


19

so Y(0) = 0. Similarly,

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

The problem for X(x) is a familiar one, with eigenvalues n = n2π

2/L

2 and eigenfunctions that are nonzero

constant multiples of sin(nπx/L).

The problem for Y is now

Y’’ -( n2π

2/L

2)Y = 0; Y(0)=0

Solutions of this problem are constant multiples of sinh(nπy/L).

For each positive integer n = 1, 2, . . . , we now have functions

𝑢𝑛 𝑥,𝑦 = 𝑏𝑛 sin 𝑛𝜋𝑥

𝐿 sinh

𝑛𝜋𝑦

𝐿

which are harmonic on the rectangle, and satisfy the zero boundary conditions on the top, bottom and left

sides of the rectangle. To satisfy the boundary condition on the side y = K, we must use a superposition

𝑢 𝑥,𝑦 = 𝑏𝑛 sin 𝑛𝜋𝑥

𝐿 sinh

𝑛𝜋𝑦

𝐿

∞

𝑛=1

Choose the coefficients so that

𝑢 𝑥,𝐾 = 𝑏𝑛 sin 𝑛𝜋𝑥

𝐿 sinh

𝑛𝜋𝐾

𝐿

∞

𝑛=1

= (𝐿 − 𝑥)sin(𝑥)

This is a Fourier sine expansion of (L - x) sin(x) on [0, L], so we must choose the entire coefficient to be

the sine coefficient:

𝑏𝑛 sinh(𝑛𝜋𝑥

𝐿) =

2

𝐿 (𝐿 − 𝑥)sin(𝑥)sin(

𝑛𝜋𝑥

𝐿)𝑑𝑥

𝐿

0

𝑏𝑛 sinh(𝑛𝜋𝐾

𝐿) = 4𝐿2

𝑛𝜋 1 − (−1)𝑛cos(𝐿)

𝐿4 − 2𝐿2𝑛2𝜋2 + 𝑛4𝜋4

Then,

𝑏𝑛 =4𝐿2

sinh(𝑛𝜋𝑥𝐿 )

𝑛𝜋 1 − (−1)𝑛cos(𝐿)

(𝐿2−𝑛2𝜋2)2

The solution is

𝑢 𝑥,𝑦 = 4𝐿2

sinh(𝑛𝜋𝑥𝐿 )

𝑛𝜋 1 − (−1)𝑛cos(𝐿)

(𝐿2−𝑛2𝜋2)2 𝑠𝑖𝑛

𝑛𝜋𝑥

𝐿 𝑠𝑖𝑛𝑕

𝑛𝜋𝑦

𝐿

∞

𝑛=1

Dirichlet Problem for a Disk

We will solve the Dirichlet problem for a disk of radius R centered about the origin. In polar coordinates,

the problem is

Laplace's equation in polar coordinates is

It is easy to check that the functions


20

1, rn cos(nθ), and r

n sin(nθ)

are all harmonic on the entire plane. Thus attempt a solution

To satisfy the boundary condition, we need to choose the coefficients so that

But this is just the Fourier expansion of f(θ) on [-π, π], leading us to choose

and

Then

and

The solution is

This can also be written

or

Example 8:

Solve the Dirichlet problem for

The solution is


21

Example 8:

Solve the Dirichlet problem for

Convert the problem to polar coordinates, using x = rcos(θ) and y = r sin(θ). Let

u(x, y) = u(rcos(θ), rsin(θ)) = U(r, θ).

The condition on the boundary, where r = 3, becomes

The solution is

Now

and

Therefore

To convert this solution back to rectangular coordinates, use the fact that

Then


22

Example 9:


23

6.5 Exercises

1.

Answer:

2.

Answer:


24

3.

Answer:

4.

Answer:

5.

Answer:

6.

Answer:


25

7.

Answer:

8.

Answer:

9.

Answer:

10.

Answer:

11.

Answer:

12.

Answer:


26

13.

Answer:

14.

Answer:

15.

Answer:

16.

Answer:

17.

Answer:

18.

Answer:

19.

Answer:

20.

Answer:

21.

Answer:

22.

Answer:

23.

Answer:

References:

1. Advanced Engineering Mathematics, International Student Edition, Peter V. O'Neil.

2. Advanced Engineering Mathematics, K.A. Stroud, 4th edition.

3. Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson and S. J. Bence, 3th

edition.

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Lecture6. Partial Differential Equationsolgar/L6.pde.pdf · Lecture6. Partial Differential Equations ... The simplest form of partial differential equation is such that a solution