MTH 333/533: Partial Di¤erential EquationsChapter 0 INTRODUCTION

Section 0.1. Brief Review of Multivariable Calculus

² Partial Derivative

Consider two variable function

z = f (x, y) .

For …xed y = y0, one can view the function as an one variable function g (x)de…ned by

g (x) = f (x, y0) ,

and then di¤erentiate the function g (x)

dg

dx(x) = g0 (x)

at any point x = x0. This derivative g0 (x0) is called the partial derivativeof f (x, y) with respect to x at (x0, y0) ,and is denoted as

∂f

∂x(x0, y0) = fx (x0, y0) = g0 (x0) .

Analogously, for any …xed y = y0, one can de…ne an function h (y) of onesingle variable y as

h (y) = f (x0, y)

and de…ned the partial derivative of f (x, y) with respect to y at (x0, y0)as

∂f

∂y(x0, y0) = fy (x0, y0) = h0 (y0) .

1

Example 0.1.1. (a) f = x2 + y3 + 3xy. Then

fx (x, y) = 2x+ 3y

fy (x, y) = 3y2 + 3x

fxx (x, y) = 2

fyy (x, y) = 6y

fxy (x, y) = (2x+ 3y)y = 3

fyx (x, y) =¡3y2 + 3x

¢x= 3

(b) g = ln (xy2) + ey + y sinx.Then

gx =1

xy2¡y2

¢+ y cosx =

1

x+ y cos x

gy =1

xy2(2xy) + ey + sinx =

2

y+ ey + sinx

² Partial Integral

The same principle of "freezing all but one variable" is used to de…nepartial inde…nite integrals

F (x, y0) =

Zf (x, y0) dx =

Zg (x) dx

where g (x) = f (x, y0) is a function of x only

Analogously,

G (x0, y) =

Zf (x0, y) dy =

Zh (y) dy

where h (y) = f (x0, y) is a function of y only

and Z b

a

f (x, y0) dx =

Z b

a

g (x) dxZ d

c

f (x0, y) dy =

Z d

c

h (y) dy.

2

Note that in the case of one variable functions, general antiderivatives comewith arbitrary constants. In the case of two or more variables, partial in-de…nite integrals with respect to x, for instance, often come with arbitraryfunctions of the other variables. For instance, f (x, y) = 3x2 + 2y. ThenZ

f (x, y) dx =

Z ¡3x2 + 2y

¢dx = x3 + 2xy + C (y)Z

f (x, y) dy =

Z ¡3x2 + 2y

¢dy = 3x2y + y2 +D (x)

where C (y) is an arbitrary function of y and D (x) is an arbitrary functionof x.

For functions with continuous partial derivatives, we have the followingproperties

d

dxF (x, y) =

d

dx

Zf (x, y) dx = f (x, y) ,

d

dyG (x, y) =

d

dy

Zf (x, y) dy = f (x, y)

d

dyF (x, y) =

d

dy

Zf (x, y) dx =

Zfy (x, y) dx

d

dxG (x, y) =

d

dx

Zf (x, y) dy =

Zfx (x, y) dy.

² Gradient and Directional Derivative

Gradient of two variable function f (x, y) is de…ned as the vector-valuedfunction

rf (x, y) = hfx (x, y) , fy (x, y)i .

For three or more variable functions, gradient is de…ned analogously. Forinstance, for f (x, y, z)

fx =∂f

∂x=

d

dxf (x, y, z) when y and z are considered as constant parameters,

fy =∂f

∂y=

d

dyf (x, y, z) when x and z are considered as constant parameters,

fz =∂f

∂z=

d

dzf (x, y, z) when x and y are considered as constant parameters,

3

andrf (x, y, z) = hfx (x, y, z) , fy (x, y, z) , fz (x, y, z)i .

Given a unit vector ~n = ha, b, ci , the directional derivative of f (x, y, z) at(x0, y0, z0) along direction ~n is the rate of change of f along ~n, and is denotedand formulated as

∂f

∂~n(x0, y0, z0) =

d

dtf (x0 + at, y0 + bt, z0 + ct) jt=0

= a∂f

∂x(x0, y0, z0) + b

∂f

∂y(x0, y0, z0) + c

∂f

∂z(x0, y0, z0)

= ~n ¢ rf (x0, y0, z0) .

The directional derivative ∂f/∂~n describes the rate of change of the functionf along that particular direction ~n. In general, for any vector ~v,

∂f

∂~v(x, y, z) = directional derivative of f in the direction

~v

j~vj=

~v

j~vj ¢ rf (x, y, z)

Example 0.1.2. Let f = xy2 + x3ez + sin (xyz) . Find (a) rf (x, y, z) ,

(b)∂f

∂~n(x, y, z) where ~n = h1, 2, ¡1i , (c)

∂f

∂~n(1, 1, 0) .

Solution. (a)

fx = y2 + 3x2ez + yz cos (xyz)

fy = 2xy + xz cos (xyz)

fz = x3ez + xy cos (xyz)

So

rf (x, y, z) = hfx (x, y, z) , fy (x, y, z) , fz (x, y, z)i­y2 + 3x2ez + yz cos (xyz) , 2xy + xz cos (xyz) , x3ez + xy cos (xyz)

®.

(b) Note that the unit vector in the direction of ~n is

1p1 + 4 + 1

h1, 2, ¡1i = 1p6

h1, 2, ¡1i .

4

So

∂f

∂~n(x, y, z) =

1p6

h1, 2,¡1i ¢ rf (x, y, z)

=1p6

©y2 + 3x2ez + yz cos (xyz) + 2 [2xy + xz cos (xyz)]¡ £

x3ez + xy cos (xyz)¤ª

=1p6

¡y2 + 3x2ez + 4xy ¡ x3ez

¢+1p6(yz + 2xz ¡ xy) cos (xyz)

(c)∂f

∂~n(1, 1, 0) =

1p6(1 + 3 + 4¡ 1) + 1p

6(¡1) = 6p

6=

p6

Note that this means that if we are moving along the straight line passingthrough (1, 1, 0) with direction ~n,then the rate of change for f at (1, 1, 0) isp6.

² Tangent plane and Linear Approximation

For a three variable function F (x, y, z), the equation

F (x, y, z) = 0

represents a surface. At any point (x0, y0, z0) on the surface, i.e.,

F (x0, y0, z0) = 0,

the plane passing through (x0, y0, z0) with the normal vector

rF (x0, y0, z0) = hFx (x0, y0, z0) , Fy (x0, y0, z0) , Fz (x0, y0, z0)iis called the tangent plane. Every line on the tangent passing through(x0, y0, z0) is a tangent line. The equation of the tangent plane is

rF (x0, y0, z0) ¢ hx ¡ x0, y ¡ y0, z ¡ z0i = 0or

(x ¡ x0)Fx (x0, y0, z0) + (y ¡ y0)Fy (x0, y0, z0) + (z ¡ z0)Fz (x0, y0, z0) = 0

Locally, the tangent plane just touches the surface only at (x0, y0, z0) ,and isthe linear approximation to the surface. In particular, the graph of a two-variable function z = f (x, y) can be represented by the following equation

F (x, y, z) = f (x, y)¡ z = 0.

5

So at any point (x0, y0) , the point on the graph is (x0, y0, z0) = (x0, y0, f (x0, y0))and the normal to the tangent is

rF (x0, y0, z0) = hfx (x0, y0) , fy (x0, y0) , ¡1i .

So the equation of tangent plane is

(x ¡ x0) fx (x0, y0) + (y ¡ y0) fy (x0, y0)¡ (z ¡ z0) = 0

or(z ¡ z0) = (x ¡ x0) fx (x0, y0) + (y ¡ y0) fy (x0, y0)

Since z0 = f (x0, y0) , it leads to

z = f (x0, y0) + (x ¡ x0) fx (x0, y0) + (y ¡ y0) fy (x0, y0) .

Example 0.2.3. Find the tangent to the ellipsoid

x2

2+ y2 + z2 = 1

atµ1,1

2,1

2

¶.

Solution: Set

F =x2

2+ y2 + z2 ¡ 1.

Then

rF (x, y, z) = hx, 2y, 2zi , rF

µ1,1

2,1

2

¶= h1, 1, 1i ,

and the tangent is

(x ¡ 1) +µ

y ¡ 1

2

¶+

µz ¡ 1

2

¶= 0

orx+ y + z ¡ 2 = 0.

6

² Line Integral and Green’s Formula

Let C be a parametric curve (For simplicity, we consider 2D cases):

C : ~r0 (t) = (f (t) , g (t)) , a < t < b.

We always assume that the curved is naturally oriented, i.e., the curvemoves along with its parameter t : starting at (f (a) , g (a)) and endingat (f (b) , g (b)) . The line integral (of the …rst kind) of a function P (x, y)along curve C is de…ned asZ

C

P (x, y) ds =

Z b

a

P (f (t) , g (t)) j~r0 (t)j dt

=

Z b

a

P (f (t) , g (t))

q(f 0 (t))2 + (g0 (t))2dt.

Suppose now that C is a simple closed curve with the positive ori-entation, i.e., C is the boundary of a connected domain D and itsorientation follows the right-hand rule. The line integral of a vector…eld hP (x, y) , Q (x, y)i along curve C is de…ned asZ

C

P (x, y) dx+Q (x, y) dy =

Z ZD

µ∂Q

∂x¡ ∂F

∂y

¶.

7

Section 0.2. Brief Review on Ordinary Di¤erentialEquations

A …rst-order ordinary di¤erential equation (ODE) is an equation involvingtime t,unknown function y (t) of t, and its derivative, in the form

dy

dt= F (t, y) .

An solution is a function y (t) that satis…es the equation. In general, thereare in…nite many solutions. General solution means the set of all possible so-lutions. Under appropriate assumptions, for instance, F (t, y) has continuouspartial derivatives, the initial-value problem (IVP)

dy

dt= F (t, y) , y (t0) = y0

admits a unique solution. ODE is often used to model dynamic behavior:how a quantity or parameter changes with respect to time.

Geometrically, function F (t, y) creates a slope …eld:

Slope …eld of F (t, y) = ty2

ODEy0 = F (t, y)

8

is called integral of the slope …eld in the sense that any solution y = f (t) ofODE is a curve such that its tangent …eld (by drawing a tangent line segmentat every point of the curve) matches the vector …eld.

There are very limited types of ODE can be solved analytically.

1. Separable equations:dy

dt= f (y) g (t)

This ODE can be solved by integratingZdy

f (y)=

Zg (t) dt.

Example 0.2.1. Solve

y0 =t

y2

Solution:y2dy = tdtZ

y2dy =

Ztdt

1

3y3 =

1

2t2 + C

So general solution is

y =

µ3t2

2+ C

¶1/3

.

2. First-order linear equation

dy

dt+ P (t) y = Q (t) .

We use the method of integrating factor to …nd the general solution.

9

One can check that·y exp

µZ t

0

P (s) ds

¶¸0= y0 exp

µZ t

0

P (s) ds

¶+ yP (t) exp

µZ t

0

P (s) ds

¶= exp

µZ t

0

P (s) ds

¶(y0 + P (t) y)

= Q (t) exp

µZ t

0

P (s) ds

¶.

Therefore, by integrating both sides, we …nd

y exp

µZ t

0

P (s) ds

¶=

Z ·Q (t) exp

µZ t

0

P (s) ds

¶¸dt

and consequently

y = exp

µ¡

Z t

0

P (s) ds

¶Z ·Q (t) exp

µZ t

0

P (s) ds

¶¸dt.

(Sol Lin ODE)

In particular, when P (t) = P and Q (t) = Q are constants. then theabove solution is

y = exp

µ¡

Z t

0

Pds

¶Z ·Q exp

µZ t

0

Pds

¶¸dt

= exp (¡Pt)

ZQ exp (Pt) dt

= Q exp (¡Pt)

·1

Pexp (P t) + C

¸=

Q

P+ Ce¡P t.

3. Free Harmonic Oscillator (2nd order ODE):

y" + py0 + qy = 0

To solve this ODE, we try function

y = Ceλt.

10

Left-Hand Side = y" + py0 + qy

= Ceλtλ2 + pCeλtλ+ qCeλt

= Ceλt¡λ2 + pλ+ q

¢Right-Hand Side = 0

So λmust be a solution of the characteristic equation

λ2 + pλ+ q = 0.

4. Forced Harmonic Oscillator:

y" + py0 + qy = f (t)

General solution has the form

y = yh + yp

where yh is general solution for its homogeneous equation

y" + py0 + qy = 0

and yp is a particular solution. When

f (t) = P (t) eλt, P (t) is a polynomial function

then

yp = Q (t) eλt, Q (t) is an undetermined polynomial function.

Whenf (t) = At sinωt+B cosωt

where A (t) and B (t) are polynomials, then

yp = C sinωt+D cosωt

for some constants C and D.

5. First-order Linear Systems with Constant Coe¢cients:

~x0 (t) = A~x (t)

11

where A is a m £ n matrix and

~x (t) =

26664x1 (t)x2 (t)

...xn (t)

37775 .

We look for fundamental solutions in the form

~x (t) = eλt~u, ~u is a constant vector.

Then

LHS = ~x0 (t) =¡eλt~u

¢0= λeλt~u

RHS = A~x (t) = A¡eλt~u

¢= eλtA (~u) .

So ~x (t) = eλt~u is a solution i¤

eλtA (~u) = λeλt~u =) A (~u) = λ~u.

This implies that

λ is an eigenvalue and ~u is an eigenvector associated with λ.

6. Higher-order ODE can be converted to an equivalent …rst-order system.

For instance, the harmonic oscillator

y" + py0 + qy = 0

is equivalent to

y0 = u

u0 = ¡pu ¡ qy,

or ·yu

¸0=

·0 1

¡q ¡p

¸ ·yu

¸.

12

Section 0.3. Basic on Partial Di¤erential EquationsODE can only model dynamics of quantities or parameters independent

of location. When a dynamic system depends also upon on spatial location,we need partial di¤erential equations as mathematical models. A partialdi¤erential equation (PDE) is an equation involves an unknown functionu (x, y) and its partial derivatives:

F (x, y, u, ux, uy, uxx, uxy, uyy, uxxx, ....) = 0.

For instance,

xux + yuy = 3u (03.1)

yuxx + xuyy ¡ 3xy = 0 (03.2)

ut ¡ x2uxx = t2 + x (03.3)

ut = (ux)2 (03.4)

are all partial di¤erential equations. A partial di¤erential equation is said tobe linear if F is linear in u and its partial derivatives, i.e., with …xed x andy

F (x, y, u, ux, uy, uxx, uxy, uyy, uxxx, ....) is a linear function.

For instance, PDE (03.1), (03.2) and (03.3) are linear, but (03.4) is not. Inother words, to be linear, the highest degree (combined power of u and itspartial derivatives) is one. The highest-order of partial derivatives involvedis called the order of a PDE. For instance, (03.1) and 03.4) are …rst-orderPDEs. PDE (03.2) and (03.3), on the other hand, are second-order PDEs.A solution of a PDE is a function u (x, y) that, together with its partialderivatives, satis…es the equation. We shall focus on linear PDEs.

Example 0.3.1. Verify that the function u (x, y) = x3 + y3 is a solutionof (03.1).

Solution: We di¤erentiate the function and …nd

ux = 3x2, uy = 3y

2.

So

LHS = xux + yuy = 3x3 + 3y3

RHS = 3u = 3¡x3 + y3

¢.

13

Since LHS = RHS for all x, y, the function u (x, y) = x3 + y3 is a solution.A solution of a PDE that includes every other solution is called the gen-

eral solution. For ODEs, we recall that general solutions involves arbitraryconstants. The number of arbitrary constants in general equal to the orderof equation. For instance, general solutions of …rst-order ODEs involve oneconstant; for 2nd-order ODEs, general solution involve two constants. ForPDE, the situation becomes much more complicated.

Example 0.3.2. Find the general solution of

uxy = cosx+ 2y.

Solution: We integrate the equation with respect to y …rst (with …xed x) andthen w.r.t. x (with …xed y) :

ux = y cos x+ y2 + C (x)

u = y sinx+ xy2 +

ZC (x) +D (y)

= y sinx+ xy2 + C1 (x) +D (y) .

Thus, in this example the general solution involves two arbitrary functionsof one variable.

Example 0.3.3. Find the general solution of

x2uxy + xuy = y

Solution: Setv = uy.

The equation reduces tox2vx + xv = y

orvx +

1

xv =

y

x2.

Note that for each …xed y,this is actually an linear ODE with x = t

P =1

x, Q =

y

x2.

14

Using the formula discussed before, we obtain

v = exp

µ¡

Z x

1

P (s) ds

¶Z ·Q (x) exp

µZ x

1

P (s) ds

¶¸dx

= exp

µ¡

Z x

1

1

sds

¶Z ·y

x2exp

µZ x

1

1

sds

¶dx

¸= exp (¡ lnx)

Z h y

x2exp (lnx) dx

i=1

x

Z h y

x2xdx

i=

y

x

Z ·1

xdx

¸=

y

x(lnx+ C (y)) ,

oruy = v =

y

x(lnx+ C (y)) .

Integrating with respect to y, with …xed x, we obtain the general solution

u =

Zy

x(lnx+ C (y)) dy

=

Z ³y

xlnx+

y

xC (y)

´dy

=lnx

x

Zydy +

1

x

ZyC (y) dy

=lnx

x

µy2

2+D (x)

¶+

C1 (y) +D1 (x)

x

=y2 ln x

2x+

C1 (y)

x+

D1 (x)

x+lnx

xD (x)

=y2 ln x

2x+

C1 (y)

x+D2 (x) ,

where C1 (y) and D2 (x) are arbitrary functions.

² Generally speaking, for PDEs, general solutions involve arbitrary func-tions of one less variables.

² Transport Equations (three variables):

ut (x, y, t) + ha (x, y, t) , b (x, y, t)i ¢ ru (x, y, t) = 0

15

(where r = h∂x, ∂yi , with respect to the spatial variables x, y only), ormore explicitly

ut + aux + buy = 0. (Transport Equ)

The problem models transport motion in a velocity …led ¡ha, bi . Whenha, bi is a conservative vector …eld, i.e., it is the gradient of a functionF,

ha, bi = rF (x, y) ,

it is called equation of conservation laws:

ut +rF (x, y) ¢ ru = 0.

When a and b are constants, the general solution has the form,

u = h (x ¡ at, y ¡ bt)

for any function h (x, y). Indeed,

ut = hx (¡a) + hy (¡b)

ux = hx, uy = hy.

So

LHS = ut + aux + buy

= hx (¡a) + hy (¡b) + ahx + bhy = 0.

Note that

x ¡ at = x0 (Char. Line)

y ¡ bt = y0

is a parametric representation of the line passing through (x0, y0) withdirection ha, bi . Along this line, the solution remains constant

u = h (x ¡ at, y ¡ bt) = h (x0, y0) .

We call line (Char. Line) a characteristic line. General solutions fortransport equations are functions that remain constant along any char-acteristic line.

16

² General First-order linear Partial Di¤erential Equations of two vari-ables

a (x, y)ux + b (x, y)uy + c (x, y) u = f (x, y) .

To …nd general solutions, we introduce characteristic curves by solv-ing characteristic equation

dy

dx=

b (x, y)

a (x, y). (Char. Equ.)

The general solution takes the form

ψ(x, y) = constant

for any constant, where ψ is called the integral of Equ (Char. Equ)that satis…es

ψx (x, y) +ψy (x, y)dy

dx= 0

orb (x, y)

a (x, y)= ¡ψx (x, y)

ψy (x, y).

Indeed, we can verify that if y = y (x) represents an implicit functionde…ned by ψ(x, y) =constant, then

dψ(x, y)

dx= ψx +ψy

dy

dx= 0.

We now change variables

ξ= x

η= ψ(x, y)

and introduce new function

w (ξ, η) = u (x, y) .

Then

ux = wξξx + wηηx = wξ+ wηψx

uy = wξξy + wηηy = wηψy

17

and the original PDE

a (x, y)ux + b (x, y)uy + c (x, y) u = f (x, y) .

reduces to

a (wξ+ wηψx) + b (wηψy) + cu = f,

awξ+ (awηψx + bwηψy) + cu = f,

awξ+ wη(aψx + bψy) + cu = f

orawξ+ cw = f (03.5)

sinceb (x, y)

a (x, y)= ¡ψx (x, y)

ψy (x, y)=) aψx + bψy = 0.

Equ (03.5) is a linear ODE (with …xed η), if a 6= 0, and can be solvedusing formula (Sol Lin ODE) as

w = exp

µ¡

Z ξ

0

P (s) ds

¶Z ·Q (ξ) exp

µZ ξ

0

P (s) ds

¶¸dξ,

with

P =c

a, Q =

f

a, ξ is the variable with …xed η.

Note that the general solution w (ξ, η) involves an arbitrary functionc (η) , since we treated η like a constant during integrations. As a gen-eral principle, whenever a variable is temporarily frozen to a constant,this variable must be incorporated into any consequent "constant".

Example 0.3.4. Find the general solution of

2ux ¡ 3uy + 2u = 2x.

Solution: We …rst solve the characteristic equation

dy

dx=

b

a= ¡3

2.

The general solution, i.e., the characteristic curves, is

y = ¡32x+ C () 3x+ 2y = 2C.

18

Thus ψ= 3x+ 2y, and new variables

ξ= x, η= 3x+ 2y.

Note that one can solve for x and y as

x = ξ, y =1

2(η¡ 3x) = 1

2(η¡ 3ξ) (03.6)

The corresponding ODE for the new function

w (ξ, η) = u (x, y) = u

µξ,1

2(η¡ 3ξ)

¶,

oru (x, y) = w (x, 3x+ 2y)

satis…es Equ (03.5). This can also be directly calculated as follows:

ux = wξ+ 3wη, uy = 2wη.

Substituting these into the original PDE to arrive at

2 (wξ+ 3wη)¡ 3 (2wη) + 2w = 2x.

Note that x and y must be substituted by ξ and η through relation(03.6). We thus obtain

2wξ+ 2w = 2x

wξ+ w = ξ.

To solve this last ODE, instead of plugging directly into the formula(Sol Lin ODE), we observe that¡

weξ¢

ξ= wξe

ξ+ weξ= (wξ+ w) eξ= ξeξ.

So, by integration by parts,

weξ=

Zξeξdξ=

Zξd

¡eξ

¢= ξeξ ¡

Zeξdξ= ξeξ ¡ eξ+ C.

However, the "constant" C is constant only relative to ξ. It actually isa function of η. Therefore,

w = e¡ξ¡ξeξ ¡ eξ+ C

¢= ξ¡ 1 + e¡ξC (η) .

19

Returning back to the original variables x and y, we …nd the generalsolution

u (x, y) = w (x, 3x+ 2y)

= x ¡ 1 + e¡xC (3x+ 2y) .

In summary, in order to solve,

a (x, y)ux + b (x, y)uy + c (x, y) u = f (x, y)

we follow three steps:Step 1: solve characteristic equation

dy

dx=

b (x, y)

a (x, y),

and represent general solution in the form (i.e., solve the constant)

ψ(x, y) = C

Step 2: change variable

ξ= x

η= ψ(x, y)

and solve from these two equations for x, y

x = x(ξ, η)

y = y(ξ, η)

Step 3: Solveawξ+ cw = f

The solution of PDE is

u (x, y) = w(ξ,η) = w (x, ψ(x, y))

In the above example, we can easily follow these three steps: Step 1:

dy

dx=

b

a= ¡3

2.

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The general solution is

y = ¡32x+ C () 3x+ 2y = 2C.

ψ= 3x+ 2y

Step 2:

ξ= x

η= 3x+ 2y.

and solve for x and y as

x = ξ

y =1

2(η¡ 3x) = 1

2(η¡ 3ξ)

Step 3:wξ+ w = ξ

and the solution isw = ξ¡ 1 + e¡ξC (η) .

Thus, the solution of PDE is

u = w (ξ, η) = w (x, 3x+ 2y)

= x ¡ 1 + e¡xC (3x+ 2y)

² Nonlinear PDEs

Burger’s equationut = u2x

This is a typical example from ‡uid dynamics which exhibits non-uniqueness of solutions, or shock waves.

² Auxiliary Conditions

In the case of ODEs, general solutions involve only constants. So initialconditions are needed and are su¢cient to guarantee uniqueness. Under

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appropriate conditions (e.g., f has continuous partial derivatives), theinitial-value problem

y0 = f (y, t)

y (0) = y0

admits a unique solution. For PDE, as we have discussed, generalsolutions involves arbitrary functions. Therefore, more conditions areneeded. In general, in addition to initial conditions, we also prescribeBoundary Conditions. For instance, the wave equation

utt (x, t) = uxx (x, t)

have general solution (we shall discuss later on)

u (x, t) = f (x+ t) + g (x ¡ t) .

If we consider the PDE for 0 < x < 1, i.e.,

utt (x, t) = uxx (x, t) for t > 0, 0 < x < 1,

then we may have issues determining function f (x) and g (x) . We willsee that under the initial condition

u (x, 0) = a (x)

together with Boundary Condition

u (0, t) = g (t) , u (1, t) = h (t)

where a (x) , g (t) , h (t) are given functions, one may …nd a uniquesolution. In applications, t is often time and x is location. The term"boundary" means the boundary of the spatial domain or the range oflocations. We call

utt (x, t) = uxx (x, t) for t > 0, 0 < x < 1

u (x, 0) = a (x)

u (0, t) = g (t) , u (1, t) = h (t)

initial-boundary- value problem. There are other auxiliary conditions.

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Example 0.3.5. Find the solution of

2ux ¡ 3uy + 2u = 2x

which assumes the value

u = x2 on the line y = ¡x

2or

u³x, ¡x

2

´= x2.

Solution: From Example 0.3.4, we have the general solution

u (x, y) = x ¡ 1 + e¡xC (3x+ 2y)

for any function C (ξ) . Set y = ¡x/2. We see that, in order to …nd thedesired solution, we must choose C (ξ) such that

x2 = u³x, ¡x

2

´= x ¡ 1 + e¡xC

³3x+ 2

³¡x

2

´´or

x2 = x ¡ 1 + e¡xC (2x) .

ThusC (2x) =

¡x2 ¡ x+ 1

¢ex.

Let ξ= 2x.Then

C (ξ) =

µξ2

4¡ ξ

2+ 1

¶eξ/2,

and

C (3x+ 2y) =

Ã(3x+ 2y)2

4¡ (3x+ 2y)

2+ 1

!e(3x+2y)/2.

We …nd the solution to be

u (x, y) = x ¡ 1 + e¡xC (3x+ 2y)

= x ¡ 1 + e¡x

Ã(3x+ 2y)2

4¡ (3x+ 2y)

2+ 1

!e(3x+2y)/2

= x ¡ 1 +Ã(3x+ 2y)2

4¡ (3x+ 2y)

2+ 1

!e(x/2+y).

Homework #1.

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1. Determine the order of each of the following PDEs and …nd which arelinear.

(a) uxy + x2ux = 1

(b) uux + uyy ¡ 2uy = sinxy

(c) xux + yuy + (x2 + y2)u = 1

(d) yux + (uy)2 + x = 0

2. Show that the given function is a solution.

(a) u (x, y) = ax2 + by2; xux + yuy = 2u.

(b) u (x, y) = (x ¡ a)2 + (y ¡ b)2 ; (ux)2 + (uy)

2 = 4u.

(c) u (x, y) = ln (x+ t) + (x ¡ t)2 ; utt = uxx.

3. Find general solutions.

(a) uxy = x+ 2y.

(b) ux + yu = 2xy.

(c) ux + uy ¡ u = 0.

(d) xux + yuy = 1.

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