chapter12.partialdifferentialequations(pdes)phtseng/aem/ch12.pdf · 2020-06-05 ·...

Chapter 12. Partial Differential Equations (PDEs)曾柏軒 (Po-Hsuan Tseng)Department of Electronic Engineering

National Taipei University of Technology

[email protected] 18, 2020 1 / 25

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Outline

1 Sec. 12.1 Basic Concepts of PDEs

2 Sec. 12.2 Modeling: Vibrating String, Wave Equation

3 Sec. 12.3 Solution by Separating Variables. Use of Fourier Series

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Outline

1 Sec. 12.1 Basic Concepts of PDEs2 Sec. 12.2 Modeling: Vibrating String, Wave Equation3 Sec. 12.3 Solution by Separating Variables. Use of Fourier Series

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12.1 Basic Concepts of PDEs

• A partial differential equation (PDE) is an equation involving one or more partialderivatives of an (unknown) function, call it u, that depends on two or more variables,often time t and one or several variables in space.

• The order of the highest derivative is called the order of the PDE.- Just as was the case for ODEs, second-order PDEs will be the most important ones inapplications.

• A PDE is linear if it is of the first degree in the unknown function u and its partialderivatives.

• We call a linear PDE homogeneous if each of its terms contains either u or one of itspartial derivatives.

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Example 1 (Sec. 12.1 Example 1, Important Second-Order PDEs)

One-dimensional wave equation:∂2u∂t2

= c2∂2u∂x2

One-dimensional heat equation:∂u∂t

= c2∂2u∂x2

Two-dimensional Laplace equation:∂2u∂x2

+∂2u∂y2

= 0

Two-dimensional Poisson equation:∂2u∂x2

+∂2u∂y2

= f(x, y)

Two-dimensional wave equation:∂2u∂t2

= c2(∂2u∂x2

+∂2u∂y2

)Three-dimensional Laplace equation:

∂2u∂x2

+∂2u∂y2

+∂2u∂z2

= 05 / 25

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• In general, the totality of solutions of a PDE is very large.

• ∂2u∂x2

+∂2u∂y2

= 0

- Solutions include u = x2 + y2, u = ex cos y, u = sin x cosh y, u = ln(x2 + y2), ... etc.• Additional conditions arising from the problem:

- Boundary conditions: the solution u assume given values on the boundary of the region R- Initial conditions: when time t is one of the variables, u (or ut = ∂u/∂t or both) may beprescribed at t = 0

• The superposition theorem holds for a homogeneous linear ODE. For PDEs the situationis quite similar:

Theorem 2 (Sec. 12.1, Theorem 1, Fundamental Theorem on Superposition)If u1 and u2 are solutions of a homogeneous linear PDE in some region R, thenu = c1u1 + c2u2 with any constants c1 and c2 is also a solution of that PDE in theregion R

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Example 3 (Sec. 12.1 Example 2, Solving uxx − u = 0 Like an ODE)Find solutions u of the PDE uxx − u = 0 depending on x and y.

• Since no y-derivatives occur, we can solve this PDE like u′′ − u = 0.• In Sec. 2.2 we would have obtained u = Aex + Be−x with constant A and B• Here A and B may be functions of y, so that the answer is

u(x, y) = A(y)ex + B(y)e−x

with arbitrary functions A and B.• We thus have a great variety of solutions. Check the result by differentiation.

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Example 4 (Sec. 12.1 Example 3, Solving uxy = −ux Like an ODE)Find solutions u = u(x, y) of the PDE uxy = −ux.

Let ux = P,uxy = −ux ⇒ Py = −P ⇒ Py

P = −1 ⇒ dPdy

1P = −1

⇒ 1PdP = −dy

⇒∫

1PdP = −

∫dy

⇒ lnP = −y+ c̃(x)⇒ P = e−y+c̃(x) = c(x)e−y

⇒ u(x, y) =∫c(x)e−ydx = f(x)e−y + g(y) where f(x) =

∫c(x)dx

Here, f(x) and g(y) are arbitrary

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Outline


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12.2 Modeling: Vibrating String, Wave Equation

• We want to derive the PDE modeling small transverse vibrations of an elastic string, suchas a violin string

• We place the string along the x-axis, stretch it to length L, and fasten it at the ends x = 0and x = L

• We then distort the string, and at some instant, call it t = 0, we release it and allow it tovibrate

Problem 5 (Vibrating String Problem)Determine the vibrations of the string, to find its deflection u(x, t) at any point x andat any time t > 0

• u(x, t) will be the solution of a PDE that is the model of our physical system to be derived.

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Figure 1: Deflected string at fixed time t

• Physical Assumptions:- Mass density is constant ρ- Perfectly elastic: does not offer any resistance to bending- Gravitational force on the string can be neglected- Every particle of the string moves strictly vertically- Deflection and slope at every point of the string always remain small

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• Derivation of the PDE of the Model (“Wave Equation”) from Forces:1 No horizontal motion: net horizontal force is zero

T1 cosα = T2 cosβ = T

2 Newton’s 2nd Law (F = ma)

T2 sinβ − T1 sinα = ρ△x∂2u∂t2

⇒ T2 sinβT2 cosβ

− T1 sinαT1 cosα

= tanβ − tanα =ρ△xT

∂2u∂t2

Notice that tanβ = ∂u∂x

∣∣x+△x and tanα = ∂u

∂x

∣∣x are slope at x+△x and x

1

△x

(∂u∂x

∣∣∣∣x+△x

−∂u∂x

∣∣∣∣x

)=

ρ

T∂2u∂t2

When△x → 0,∂2u∂x2

=1

c2∂2u∂t2

, where c2 = T/ρ12 / 25

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Outline


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12.3 Solution by Separating Variables. Use of Fourier Series

Problem 6 (One-dimensional wave equation)• One-dimensional wave equation for the unknown deflection u(x, t) of the string:

∂2u∂t2

= c2∂2u∂x2

c2 = T/ρ (1)

which satisfies two boundary conditions:

(a) u(0, t) = 0 (b) u(L, t) = 0 ∀t ≥ 0 (2)

and two initial conditions:

(a) u(x, 0) = f(x) (b)∂

∂tu(x, 0) = g(x) 0 ≤ x ≤ L (3)

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Step 1. Method of separating variables: setting u(x, t) = F(x)G(t) to obtain two ODEsfrom the wave equation

u(x, t) = F(x)G(t)

⇒ ∂2u∂t2

= FG̈ and∂2u∂t2

= F′′G

FG̈ = c2F′′GG̈c2G

=F′′

FHence both sides must be constant

G̈c2G

=F′′

F= k

Two ordinary DEs can be obtained as

F′′ − kF = 0 (4)

G̈− c2kG = 0 (5)15 / 25

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Step 2. Satisfying the Boundary Conditions (2)We determine solutions F and G of (16) and (5) so that u = FG satisfies the boundaryconditions (2)

u(0, t) = F(0)G(t) = 0 and u(L, t) = F(L)G(t) = 0 ∀tIf G(t) = 0, then u = FG = 0, which is of no interest. Hence G(t) ̸= 0 Then the problem ofcombining with (2) becomes a sub-problem:

Problem 7F′′ − kF = 0, F(0) = F(L) = 0

• Assuming that k = 0, F(x) = ax+ b

⇒

{F(0) = b = 0

F(L) = aL+ b = 0

b = 0 ⇒ a = 0 ⇒ F = 0, which is of no interest16 / 25

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• For positive k = µ2,⇒ s2 − µ2 = 0 ⇒ s = ±µTherefore, the general solution,

F(x) = Aeµx + Be−µx

⇒

{F(0) = A+ B = 0

F(L) = AeµL + Be−µL = 0

⇒ A = B = 0

⇒ F = 0, which is of no interest

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12.3 Solution by Separating Variables. Use of Fourier Series• For negative v = −p2,

⇒ s2 + p2 = 0 ⇒ s = ±ipTherefore, the general solution,

F(x) = A cos px+ B sin px

⇒

{F(0) = A+ B · 0 = 0

y(L) = A cos pL+ B sin pL = 0

A = 0 ⇒ B sin pL = 0

Thus, pL = nπ so that p = nπL . Setting B = 1, we thus obtain infinitely many solutions

F(x) = Fn(x), where

Fn(x) = sinnπLx n = 1, 2, ...

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• Notice now k = −p2 = −(nπL )2. Thus, G̈− c2kG = 0 ⇒ G̈− c2(−p2)G = 0

⇒ G̈+ (cp)2G = 0

Let λn = cp = cnπL , the ODE of G(t) becomes a problem as:

⇒ G̈+ λ2nG = 0

Since s2 + λ2n = 0 ⇒ s = ±λni, a general solution is

Gn(t) = Bn cosλnt+ B∗n sinλnt

Here the solution of (1) combining with (2) can be written as

un(x, t) = Gn(t)Fn(x) = (Bn cosλnt+ B∗n sinλnt) sin

nπLx n = 1, 2, ...

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Step 3. Solution of the Entire Problem. Fourier Series:The deflection u(x, t) has a solution in the following form:

u(x, t) =∞∑n=1

(Bn cosλnt+ B∗n sinλnt) sin

nπLx

Since the deflection should satisfy the initial condition u(x, 0) = f(x), where 0 ≤ x ≤ L

u(x, 0) =∞∑n=1

Bn sinnπLx = f(x)

We must choose the Bn’s so that u(x, 0) becomes the Fourier sine series of f(x),

Bn =2

L

∫ L

0f(x) sin

nπxL

dx n = 1, 2, ...

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Since the deflection should satisfy the initial condition ut(x, 0) = g(x),

ut(x, 0) =

[ ∞∑n=1

(−Bnλn sinλnt+ B∗nλn cosλnt) sin

nπLx

] ∣∣t=0

(6)

=

∞∑n=1

B∗nλn sin

nπLx = g(x)

We must choose the Bn’s so that u(x, 0) becomes the Fourier sine series of f(x),

B∗nλn =

2

L

∫ L

0g(x) sin

nπxL

dx

Since λn = cnπ/L, we obtain

B∗n =

2

cnπ

∫ L

0g(x) sin

nπxL

dx n = 1, 2, ...

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When the initial velocity g(x) is identically zero, B∗n = 0, and the solution reduces to

un(x, t) = Gn(t)Fn(x) = Bn cosλnt sinnπLx λn =

cnπL

It is possible to sum this series, that is, to write the result in a closed or finite form.

coscnπL

t sinnπLx =

1

2

[sin{nπ

L(x− ct)}+ sin{nπ

L(x+ ct)}

]Consequently, we may write the solution in the form

u(x, t) =1

2

∞∑n=1

Bn sin{nπL(x− ct)}+ 1

2

∞∑n=1

Bn sin{nπL(x+ ct)}

These two series are those obtained by substituting x− ct and x+ ct, respectively, for thevariable x in the Fourier sine series for f(x)

u(x, t) =1

2[f∗(x− ct) + f∗(x+ ct)]

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Example 8 (Sec. 12.3 Example 1, Vibrating String if the Initial Deflection IsTriangular)Find the solution of the wave equation (1) satisfying (2) and corresponding to the triangularinitial deflection

f̂(x) =

2kLx 0 < x <

L2

2kL(L− x)

L2< x < L

and initial velocity zero.

Since g(x) = 0, we have B∗n = 0. We need to determine Bn so that u(x, 0) becomes the Fourier

sine series of f(x). Thus,

u(x, t) =8kπ2

[sin

π

Lx cos

πcLt− 1

32sin

3π

Lx cos

3πcL

t+ ...

](7)

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Figure 2: Solution u(x, t) in for various values of t 24 / 25...

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Question & Answer

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chapter12.partialdifferentialequations(pdes)phtseng/aem/ch12.pdf · 2020-06-05 ·...

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