Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 4 ppt

Chapter 37Separation of Variables Condi-tions The method of separation of variables is a useful technique for finding special solutions of partial differential equations.. The left end i

We calculate the derivatives of σ and τ

σx= 0 σy = 2y

τx= 2x τy = 0Then we calculate the derivatives of u

ux = 2xuτ

uy = 2yuσ

uxx = 4x2uτ τ + 2uτ

uyy = 4y2uσσ+ 2uσFinally we transform the equation to canonical form

y2uxx+ x2uyy = 0σ(4τ uτ τ + 2uτ) + τ (4σuσσ + 2uσ) = 0

d2u

dx2 = 0This equation has the solution,

u = ax + b

Applying the boundary conditions we see that

u = b

To determine the constant, we note that the heat energy in the rod is constant in time

Z 1 0

u(x, t) dx =

Z 1 0

u(x, 0) dx

Z 1 0

b dx =

Z 1 0

x dxThus the equilibrium solution is

u(x) = 1

2.

36.4 Hints

Hint 36.1

Hint 36.2

Hint 36.3

36.5 Solutions

Solution 36.1

For y = −1, the equation is parabolic For this case it is already in the canonical form, uxx = 0

For y 6= −1, the equation is elliptic We find new variables that will put the equation in the form uξψ =G(ξ, ψ, u, uξ, uψ)

dy

dx = ı

p(1 + y)2 = ı(1 + y)dy

1 + y = ıdxlog(1 + y) = ıx + c

1 + y = c eıx(1 + y) e−ıx = c

ξ = (1 + y) e−ıx

ψ = ξ = (1 + y) eıxThe variables that will put the equation in canonical form are

σ = ξ + ψ

2 = (1 + y) cos x, τ =

ξ − ψı2 = (1 + y) sin x.

We calculate the derivatives of σ and τ

σx = −(1 + y) sin x σy = cos x

τx = (1 + y) cos x τy = sin xThen we calculate the derivatives of u

ux = −(1 + y) sin(x)uσ+ (1 + y) cos(x)uτ

uy = cos(x)uσ+ sin(x)uτ

uxx = (1 + y)2sin2(x)uσσ+ (1 + y)2cos2(x)uτ τ − (1 + y) cos(x)uσ − (1 + y) sin(x)uτ

uyy = cos2(x)uσσ+ sin2(x)uτ τ

We substitute these results into the differential equation to obtain the canonical form.

uxx+ (1 + y)2uyy = 0(1 + y)2(uσσ+ uτ τ) − (1 + y) cos(x)uσ − (1 + y) sin(x)uτ = 0

2

utt = c2uxx− γu

utt+ 0utx− c2uxx+ γu = 0Since 02− (1)(−c2) > 0, the equation is hyperbolic

3

(qux)x+ (qut)t= 0

quxx+ 0uxt+ qutt+ qxux+ qtut = 0Since 02− qq < 0, the equation is elliptic

ξψ = y2, y =pξψ

ψ = expξψ, ex=pψ/ξ, x = 1

2log

 ψξ

∂

∂ξ +

sψξ

∂

∂ψ

φx = −ξφξ+ ψφψ, φy =

sξ

ψφξ+

sψ

Finally we transform the equation to canonical form.

Then we calculate the derivatives of φ.

φxx+ xφyy = 0(6(ξ + ψ))1/3φξψ+ (6(ξ + ψ))1/3φξψ+ (6(ξ + ψ))−2/3(φξ+ φψ) = 0

φξψ = − φξ+ φψ

12(ξ + ψ)For x > 0, the equation is elliptic The variables we defined before are complex-valued

We transform the equation to canonical form.

Chapter 37

Separation of Variables

Condi-tions

The method of separation of variables is a useful technique for finding special solutions of partial differential equations

We can combine these special solutions to solve certain problems Consider the temperature of a one-dimensional rod

of length h1 The left end is held at zero temperature, the right end is insulated and the initial temperature distribution

is known at time t = 0 To find the temperature we solve the problem:

∂u

∂t = κ

∂2u

∂x2, 0 < x < h, t > 0u(0, t) = ux(h, t) = 0

u(x, 0) = f (x)

1 Why h? Because l looks like 1 and we use L to denote linear operators

We look for special solutions of the form, u(x, t) = X(x)T (t) Substituting this into the partial differential equationyields

X(x)T0(t) = κX00(x)T (t)

T0(t)

κT (t) =

X00(x)X(x)Since the left side is only dependent on t, the right side in only dependent on x, and the relation is valid for all t and

x, both sides of the equation must be constant

T0

κT =

X00

X = −λHere −λ is an arbitrary constant (You’ll see later that this form is convenient.) u(x, t) = X(x)T (t) will satisfy thepartial differential equation if X(x) and T (t) satisfy the ordinary differential equations,

T0 = −κλT and X00= −λX

Now we see how lucky we are that this problem happens to have homogeneous boundary conditions 2 If the leftboundary condition had been u(0, t) = 1, this would imply X(0)T (t) = 1 which tells us nothing very useful abouteither X or T However the boundary condition u(0, t) = X(0)T (t) = 0, tells us that either X(0) = 0 or T (t) = 0.Since the latter case would give us the trivial solution, we must have X(0) = 0 Likewise by looking at the rightboundary condition we obtain X0(h) = 0

We have a regular Sturm-Liouville problem for X(x)

X00+ λX = 0, X(0) = X0(h) = 0The eigenvalues and orthonormal eigenfunctions are

hsin

 (2n − 1)π

, n ∈ Z+

2 Actually luck has nothing to do with it I planned it that way.

Now we solve the equation for T (t).

T0 = −κλnT

T = c e−κλntThe eigen-solutions of the partial differential equation that satisfy the homogeneous boundary conditions are

un(x, t) =

r2

Z h 0

sinpλnx



f (x) dx

Consider the temperature in a one-dimensional rod of length h The ends are held at temperatures α and β,respectively, and the initial temperature is known at time t = 0 Additionally, there is a heat source, s(x), that isindependent of time We find the temperature by solving the problem,

ut= κuxx+ s(x), u(0, t) = α, u(h, t) = β, u(x, 0) = f (x) (37.1)Because of the source term, the equation is not separable, so we cannot directly apply separation of variables Fur-thermore, we have the added complication of inhomogeneous boundary conditions Instead of attacking this problemdirectly, we seek a transformation that will yield a homogeneous equation and homogeneous boundary conditions

Consider the equilibrium temperature, µ(x) It satisfies the problem,

Z h 0

x<(x>− h)s(ξ) dξ,

µ(x) = α + (β − α)x

h− 1κh

(x − h)

Z x 0

ξs(ξ) dξ + x

Z h x

(ξ − h)s(ξ) dξ



Now we substitute u(x, t) = v(x, t) + µ(x) into Equation 37.1

We seek a solution for v(x, t) that is a linear combination of eigen-solutions of the heat equation We substitute theseparation of variables, v(x, t) = X(x)T (t) into Equation 37.2

T0

κT =

X00

X = −λThis gives us two ordinary differential equations

2

, Xn =

r2

hsin

nπxh

, n ∈ Z+

We solve for T (t)

Tn = c e−κ(nπ/h)2t.The eigen-solutions of the partial differential equation are

vn(x, t) =

r2

hsin

nπxh



e−κ(nπ/h)2t.The solution for v(x, t) is a linear combination of these

hsin

nπxh

hsin

nπxh



= f (x) − µ(x)

an=

r2h

Z h 0

sinnπxh

(f (x) − µ(x)) dx

The temperature of the rod is

hsin

nπxh



e−κ(nπ/h)2t

Con-ditions

Now consider the heat equation with a time dependent source, s(x, t)

ut= κuxx+ s(x, t), u(0, t) = u(h, t) = 0, u(x, 0) = f (x) (37.3)

In general we cannot transform the problem to one with a homogeneous differential equation Thus we cannot representthe solution in a series of the eigen-solutions of the partial differential equation Instead, we will do the next best thingand expand the solution in a series of eigenfunctions in Xn(x) where the coefficients depend on time

hsin

nπxh

, n ∈ Z+

We expand the heat source in the eigenfunctions

hsin

nπxh



sn(t) =

r2h

Z h 0

sinnπxh

s(x, t) dx,

We substitute the series solution into Equation 37.3.

hsin

nπxh

+

hsin

nπxh

hsin

nπxh



= f (x)

un(0) =

r2h

Z h 0

sinnπxh



f (x) dx ≡ fnThe temperature is given by

hsin

nπxh

,

un(t) = fne−κ(nπ/h)2t+

Z t 0

e−κ(nπ/h)2(t−τ )sn(τ ) dτ

Consider the temperature of a one-dimensional rod of length h The left end is held at the temperature α(t), theheat flow at right end is specified, there is a time-dependent source and the initial temperature distribution is known

at time t = 0 To find the temperature we solve the problem:

ut = κuxx+ s(x, t), 0 < x < h, t > 0 (37.4)u(0, t) = α(t), ux(h, t) = β(t) u(x, 0) = f (x)

Transformation to a homogeneous equation Because of the inhomogeneous boundary conditions, we cannotdirectly apply the method of separation of variables However we can transform the problem to an inhomogeneousequation with homogeneous boundary conditions To do this, we first find a function, µ(x, t) which satisfies theboundary conditions We note that

µ(x, t) = α(t) + xβ(t)does the trick We make the change of variables

u(x, t) = v(x, t) + µ(x, t)

in Equation 37.4

vt+ µt= κ (vxx+ µxx) + s(x, t)

vt= κvxx+ s(x, t) − µtThe boundary and initial conditions become

hsin

 (2n − 1)πx2h



Direct eigenfunction expansion Alternatively we could seek a direct eigenfunction expansion of u(x, t)

hsin

 (2n − 1)πx2h



Note that the eigenfunctions satisfy the homogeneous boundary conditions while u(x, t) does not If we choose anyfixed time t = t0 and form the periodic extension of the function u(x, t0) to define it for x outside the range (0, h), then

this function will have jump discontinuities This means that our eigenfunction expansion will not converge uniformly.

We are not allowed to differentiate the series with respect to x We can’t just plug the series into the partial differentialequation to determine the coefficients Instead, we will multiply Equation37.4, by an eigenfunction and integrate from

x = 0 to x = h To avoid differentiating the series with respect to x, we will use integration by parts to move derivativesfrom u(x, t) to the eigenfunction (We will denote λn=(2n−1)π2h 

2

.)

r2h

Z h 0

sin(pλnx)(ut− κuxx) dx =

r2h

Z h 0

sin(pλnx)s(x, t) dx

u0n(t) −

r2

hκ

p

λn

Z h 0

uxcos(pλnx) dx = sn(t)

u0n(t) −

r2

hκ(−1)

nux(h, t) +

r2

hκ

p

λnhu cos(pλnx)i

h 0

+

r2

hκλn

Z h 0

u sin(pλnx) dx = sn(t)

u0n(t) −

r2

hκ(−1)

nβ(t) −

r2

hκ

p

λnα(t) + (−1)nβ(t)

+ sn(t)

Now we have an ordinary differential equation for each of the un(t) We obtain initial conditions for them using theinitial condition for u(x, t)

Z h 0

sin(pλnx)f (x) dx ≡ fn

Thus the temperature is given by

u(x, t) =

r2h

hκ

Z t 0

e−κλn(t−τ )pλnα(τ ) + (−1)nβ(τ )

dτ

Consider an elastic string with a free end at x = 0 and attached to a massless spring at x = 1 The partial differentialequation that models this problem is

To find the eigenvalues we consider the following three cases:

λ < 0 The general solution is

ψ = a cosh(√

−λx) + b sinh(√−λx)

ψ0(0) = 0 ⇒ b = 0.

ψ(1) + ψ0(1) = 0 ⇒ a cosh(√

−λ) + a√−λ sinh(√−λ) = 0

⇒ a = 0

Since there is only the trivial solution, there are no negative eigenvalues

λ = 0 The general solution is

ψ = ax + b

ψ0(0) = 0 ⇒ a = 0

ψ(1) + ψ0(1) = 0 ⇒ b + 0 = 0

Thus λ = 0 is not an eigenvalue

λ > 0 The general solution is

√

λ sin(

√λ) = 0

⇒ cos(

√λ) =

√

λ sin(

√λ)

2 4 6 8 10 -2

2 4 6 8

Figure 37.1: Plot of x and cot x

The solution for φ is

From the initial value we have

Here is an outline detailing the method of separation of variables for a linear partial differential equation for u(x, y, z, )

1 Substitute u(x, y, z, ) = X(x)Y (y)Z(z) · · · into the partial differential equation Separate the equation intoordinary differential equations

2 Translate the boundary conditions for u into boundary conditions for X, Y , Z, The continuity of u may giveadditional boundary conditions and boundedness conditions

3 Solve the differential equation(s) that determine the eigenvalues Make sure to consider all cases The functions will be determined up to a multiplicative constant

eigen-4 Solve the rest of the differential equations subject to the homogeneous boundary conditions The eigenvalues will

be a parameter in the solution The solutions will be determined up to a multiplicative constant

5 The eigen-solutions are the product of the solutions of the ordinary differential equations φn = XnYnZn· · · The solution of the partial differential equation is a linear combination of the eigen-solutions

u(x, y, z, ) =Xanφn

6 Solve for the coefficients, an using the inhomogeneous boundary conditions

37.8 Exercises

Exercise 37.1

Solve the following problem with separation of variables

ut− κ(uxx+ uyy) = q(x, y, t), 0 < x < a, 0 < y < bu(x, y, 0) = f (x, y), u(0, y, t) = u(a, y, t) = u(x, 0, t) = u(x, b, t) = 0

Hint, Solution

Exercise 37.2

Consider a thin half pipe of unit radius laying on the ground It is heated by radiation from above We take the initialtemperature of the pipe and the temperature of the ground to be zero We model this problem with a heat equationwith a source term

ut = κuxx+ A sin(x)u(0, t) = u(π, t) = 0, u(x, 0) = 0

Hint, Solution

Exercise 37.3

Consider Laplace’s Equation ∇2u = 0 inside the quarter circle of radius 1 (0 ≤ θ ≤ π2, 0 ≤ r ≤ 1) Write the problem

in polar coordinates u = u(r, θ) and use separation of variables to find the solution subject to the following boundaryconditions



= 0, ∂u

∂r(1, θ) = g(θ)Under what conditions does this solution exist?

u(x, 0, t) = 0 u(x, 1, t) = 0,and initial temperature

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

1 Reduce this to a set of 3 ordinary differential equations using separation of variables

2 Find the corresponding set of eigenfunctions and give the solution satisfying the given initial condition

Hint, Solution

Hint, Solution

Exercise 37.8

Solve the Laplace’s equation by separation of variables

∆u ≡ uxx+ uyy = 0, 0 < x < 1, 0 < y < 1,u(x, 0) = f (x), u(x, 1) = 0, u(0, y) = 0, u(1, y) = 0Here f (x) is an arbitrary function which is known

Hint, Solution

Exercise 37.9

Solve Laplace’s equation in the unit disk with separation of variables

∆u = 0, 0 < r < 1u(1, θ) = f (θ)

The Laplacian in cirular coordinates is

with separation of variables.

Here w(x, t) and f (x) are prescribed functions

φ(x, 0) = 0, φ(0, t) = t, φx(l, t) = −cφ(l, t)

Work out the series expansion for the given boundary and initial conditions.

with separation of variables.

Hint, Solution

Exercise 37.19

Consider the equilibrium temperature distribution in a two-dimensional block of width a and height b There is a heatsource given by the function f (x, y) The vertical sides of the block are held at zero temperature; the horizontal sidesare insulated To find this equilibrium temperature distribution, solve the potential equation,

uxx+ uyy = f (x, y), 0 < x < a, 0 < y < b,u(0, y) = u(a, y) = 0, uy(x, 0) = uy(x, b) = 0,with separation of variables

utt+ a2uxxxx= 0, 0 < x < L, t > 0,u(x, 0) = f (x), ut(x, 0) = g(x),u(0, t) = uxx(0, t) = 0, u(L, t) = uxx(L, t) = 0,with separation of variables

Hint, Solution

Exercise 37.21

The temperature along a magnet winding of length L carrying a current I satisfies, (for some α > 0):

ut= κuxx+ I2αu

The ends of the winding are kept at zero, i.e.,

Hint, Solution

Exercise 37.22

The ”e-folding” time of a decaying function of time is the time interval, ∆e, in which the magnitude of the function

is reduced by at least 1e Thus if u(x, t) = e−αtf (x) + e−βtg(x) with α > β > 0 then ∆e = β1 A body with heatconductivity κ has its exterior surface maintained at temperature zero Initially the interior of the body is at the uniformtemperature T > 0 Find the e-folding time of the body if it is:

a) An infinite slab of thickness a

b) An infinite cylinder of radius a

c) A sphere of radius a

Note that in (a) the temperature varies only in the z direction and in time; in (b) and (c) the temperature varies only

in the radial direction and in time

d) What are the e-folding times if the surfaces are perfectly insulated, (i.e., ∂u∂n = 0, where n is the exterior normal

at the surface)?

Hint, Solution

The diffusivity, κ(t), is a known, positive function.

Hint, Solution

Find the ensuing motion.

c) Compare the kinetic energies of each harmonic in the two solutions Where should the string be struck in order

to maximize the energy in the nth harmonic in each case?

sin πtδ
, for |x − ξ| < d, 0 < t < δ,