Math 124A Course Notes

124A PARTIAL DIFFERENTIAL EQUATIONS

DENIS A. LABUTIN

This is the outline for the course. We emphasize the key ideas and the structure,and often omit the details. The details will be covered on the lectures. They alsocan be found in the textbook (W. A. Strauss PDEs, an Introduction, 2nd ed.) andin the extended UCSB 124-course notes by Viktor Grigoryan. The two sourcescontain much more than will be covered during the quarter.

1. Basics

(1) A partial differential equation (in two independent variables) is a relation

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

for the unknown function u(x, y) . All laws in sciences are formulated inthe form of (system of) PDEs. A function u is a solution of the PDE in aregion if

F (x, y, u(x, y), ux(x, y), uy(x, y), uxx(x, y), uxy(x, y), . . .) 0for (x, y) . Thus it is very easy to check if a given function is a solutionof the given PDE.

(2) The general solution to the given PDE is the formula for all solutions. Forexample, we know from the calculus that the general solution u(x) of

du

dx u = 0

isu = Cex,

where C is any constant.Another example: we will learn later that the general solution u(x, y)

to the equationutt uxx = 0

is given byu(t, x) = f(x t) + g(x+ t),

where f, g are arbitrary functions of one variable. This means that allsolutions of the equation are given by this formula. It is impossible toderive this fact now, we need to develop some techniques for this.

(3) Calculus technique for solving simple PDEs.

Date: July 16, 2013.

1

2 DENIS A. LABUTIN

(4) Some simple PDEs can be integrated using just the calculus technique. Inorder to successfully solve the problems of this type one needs to understandthe following argument: according to the basic calculus the general solutionz(x) to the simplest equation

dz

dx= 0

isz(x) = C,

where C is any constant. But this means that the general solution u(x, y)to the equation

ux = 0is

u(x, y) = C(y),where C(y) is any constant with respect to x , which means that it is anyfunction of y . When writing the solutions down it is better not to use Cto denote functions, and keep it for constants only. Thus a better way ofwriting is

ux = 0 u = (y),where is any function of one variable.

(5) Similarly, the basic calculus tells us that the solution z(x) of z = f(x)is

z = f(x) z =f(x) dx+ C,

where C is any constant. This means that

ux = f(x, y) u =f(x, y) dx+ (y),

where is any function of one variable. Below are some examples ofcalculus technique solutions to PDEs.

(6) Solve (this means find the general solution to) the equation

ux = e2x+y.

There is no /y in the left hand side, and no u in the right hand side.Therefore the solution u(x, y) is given by

u =e2x+y dx+ (y)

= eye2x dx+ (y)

= ey12e2x + (y).

Thus the answer is

u =e2x+y

2+ (y),

where is any function.

(7) Here is one more example of the calculus technique for solving simple PDEs.Find all solutions u(x, y) , v(x, y) to the equations

ux + u = 0, vxx + v = 0.

124A PARTIAL DIFFERENTIAL EQUATIONS 3

Using the calculus separation of variables we see that the equation

z + z = 0

has the general solution

z(x) = Cex.

The equation for u is the same equation. There is no /y and thevariable y enters only as a parameter. Hence

ux + u = 0 u = (y)ex,

where is any function.Using the calculus separation of variables we see that the equation

z + z = 0

has the general solution

z(x) = C1 cosx+ C2 sinx,

where C1,2 are arbitrary constants. The equation for v is the sameequation. There is no /y and the variable y enters only as a parameter.Hence

vxx + v = 0 v = (y) cosx+ (y) sinx,where , are arbitrary functions.

(8) Frequently there is a need to apply several calculus steps. Here is an ex-ample of such problem. Solve

uxy = 0.

Denotev = uy.

Thenuxy = 0 vx = 0 v = (y),

where is any function of one variable. Next,

uy = (y) u =(y) dy + (x),

where , are any functions of one variable. Since is any, then(y) dy

is also any function of one variable. Thus the answer is

u = (y) + (x),

where , are any functions of one variable.Notice that , are the functions of one variable, and u(x, y) = (y)+

(x) is a function of two independent variables of a very special form. Forexample (x)(x) is not of this form. Hence x + ey is a solution, butxey is not.

(9) The integrating factor technique from calculus will also be used in thecourse. Equation for z(t)

dz

dt+ a(t)z = f(t)

4 DENIS A. LABUTIN

can be integrated using the following observation. Multiply the equationby the integrating factor

eRa(t) dt

to discover that

eRa(t) dt dz

dt+ e

Ra(t) dta(t)z = e

Ra(t) dtf(t).

The point here is that now

eRa(t) dt dz

dt+ e

Ra(t) dta(t)z =

d

dt

(e

Ra(t) dtz

).

Hence the equation becomes

d

dt

( )

= g(t)

which can be integrated.

Problem 1. Find the general solution of

uxy + a(x, y)ux = 0

in a neighborhood of (0, 0) . Here a(x, y) is a given function.Set

v = uxto discover at once that

vy + a(x, y)v = 0.

This is an ODE in the y -variable, and x enters as a parameter. Separat-ing the variables discover

dv

v= a(x, y) dy

and find after the integration

ln |v(x, y)| =eR y0 a(x,) d+C(x)

v = (x)eR y0 a(x,) d

with an arbitrary function . Returning to u discover that

ux = (x)eR y0 a(x,) d.

This is just an ODE in the x -variable, and y enters as a parameter.Integrating we find

u(x, y) = x

0

()eR y0 a(,) d d + (y),

where , are arbitrary functions of one variable.

Problem 2. Find the general solution of

uxy 1

x yux +

1x y

uy = 0.

Hint: use the observation that

(x y)vx + v =

(x y)vx, (x y)vy v =

(x y)v

y.


(10) Operators and equations. Let u be any function. A differential op-erator is an expression involving u and its derivatives. For example, theMonge-Ampere operator is

detD2u = uxxuyy u2xy,the Laplace operator is

u = uxx + uyy,

the wave operator is

u = utt uxx, cu = utt c2uxx,the heat operator is

ut uxx,the transport operator is

ux + uy,

the Schroedinger operator is11

ut uxx.

Operators a usually denoted as Au , or A[u] , or A(u) , or Lu , ... .

(11) The most important class of the operators is the class of linear operators.Operator L is called linear if the following two conditions hold:

L(u+ v) = L(u) + L(v), L(Cu) = CL(u)for all functions u, v and any constant C . To check if the operator islinear we just need to verify these conditions.

(12) A PDE is called linear homogeneous if it can be written in the form

Lu = 0with some linear operator L .

A PDE is called linear inhomogeneous if it can be written in the form

Lu = f(x, y)with some linear operator L . The important part here is that in the righthand side we have some function of x, y but the unknown u is absent.

(13) Our main objects of study will be the transport operator

(t + cx)u = ut + cux, u = u(t, x),

the wave operator

udef=(2t +

2x

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

the heat operator

(t 2x)u = ut uxx, u = u(t, x),and the Laplace operator

u def= (2x + 2y)u = uxx + uyy, u = u(x, y).

Show that they are linear operators.Linear equations can have coefficients depending on x, y . For example,

show that the equation

ux + yuy = 0

6 DENIS A. LABUTIN

is a linear (homogeneous) equation.

(14) Equationux + uuy = 0

is nonlinear. Indeed, every term in it contains u . Hence the equation islinear if and only if the operator

Au = ux + uuy

is linear. But A is not linear because the two conditions of the linearityare not satisfied. For example if C is a constant, then

A(Cu) = (Cu)x + (Cu)(Cu)y = Cux + C2uuy.

At the same timeCAu = Cux + Cuuy.

HenceA(Cu) 6= CA(u)

for all constants C .

(15) Superposition principle for linear equations. The key feature of lin-ear equations is the superposition principle. It is also called the linearityprinciple. It consists of two parts.

The first part states that if u1 , . . . , uN are solutions to the linearhomogeneous equation

Lu = 0,then for any constants C1 , . . . , CN the function C1u1 + +CNuN isalso a solution:

L(C1u1 + + CNuN ) = 0.

This follows at once from the definition of the linearity of L .The second part states that if u0 solves the linear inhomogeneous equa-

tionLu = f,

and if v is the general solution to the homogeneous equation

Lv = 0

then the general solution to the inhomogeneous equation is given by

u = u0 + v.

(16) The second part is the most important and useful for problem solving.Indeed, in plain language is states that

{all solutions to Lu = f} = {some solution to Lu = f}+{all solutions to Lu = 0}.

Thus to find all solution to Lu = f it is enough to find all solutions onlyto Lu = 0 and somehow guess just one particular solution to Lu = f .

(17) Let us illustrate the power of the superposition principle by solving a prob-lem. Solve (that is, find all solutions to)

uxy = x2y.


This is a linear inhomogeneous equation. Hence we can apply the super-position principle. We have computed earlier the general solution to thehomogeneous equation:

vxy = 0 v = f(x) + g(y),

where f, g any functions. Hence we just need to guess one solution to theinhomogeneous equation. This is easy for our equation. For example

u0 =x3

3y2

2=x3y2

6will do the job. Therefore, by the superposition principle the answer is:

u(x, y) =x3y2

6+ f(x) + g(y),

where f, g any functions.

(18) The superposition principle will be one of the main tools in this course bothfor the theory and the problem solving. Superposition principle does nothold for nonlinear equations.

2. Changing variables in PDEs

(1) The changing variables technique is fundamental for PDEs. It is based onvarious tools from calculus such as chain rule, implicit differentiation, andso on. Many problems in this course require the ability to change variablescorrectly. First we explain the generalities, underline the difficult part inthe change of variables procedure, and then solve some problems.

(2) To abbreviate we write x = (x1, . . . , xn) , = (1, . . . , n) , ux =(x1u, . . . , xnu) , and so on. In the differential equation

F (x, u, ux, . . .) = 0

we would like to change the variables x . The change of variables

x = H()

at once givesv() = u(H()).

and

(2.1) F (H(), v(), (ux)(H()), . . .) = 0.

We know F and H . Consequently we almost have the differential equa-tion for the new unknown v() .

It is just left to relate (ux)(H()) with v . This is the somewhatcomplicated part. To accomplish that we need the inverse change

= (x).

The inverse implies

(H()) = , H((x)) = x.

8 DENIS A. LABUTIN

Therefore together with v() = u(H()) we also have u(x) = v((x)) .By the chain rule

ux(x) = v()=(x)

x(x),

and we need to express this in variables:

(ux)(H()) =(v()

=(x)

x(x))

x=H()

= v() (x) (H()).

We see that the main computation when changing variables is finding

(x) (H())

and the higher derivatives needed for uxx , uxxx , . . . . That is, we needto express the derivative x = x(x) as a function of , x = H() .

When this is done, equation (2.1) becomes

0 = F (H(), v(), v() (x) (H()), . . .)

m0 = G(, v, v, . . .) .

(3) Thus to change the independent variables x to in the equation

F (x, u, ux, . . .) = 0

we need:Step 1. Use x = H() to introduce v() = u(H()) , and write the inverse

change = (x) so that u(x) = v((x)) .

Step 2. Compute by the chain rule

xu(x) = v((x)) x(x)

substituting x = H()(xu)(H()) = v() (x)(H()) ,

and higher derivatives, if necessary

2xu(x) = x(v((x)) x(x)

)= 2v((x)) (x(x))

2 + v((x)) 2x(x)

substituting x = H()

(xxu)(H()) = 2v()(

(x)(H()))2

+ ;

Step 3. Plug the results of steps 1 and 2 into the equation to write it in thenew variables

0 = F (H(), u(H()), ux(H()) . . .)

= G(, v, v, . . .).

(4) The successful strategy for changing variables is not to memorize the generalformula above, but to repeat the three steps for individual problem. Let usshow this by examples.


(5) In the eiconal equation

(ux)2 + (uy)2 = 1

pass to the polar coordinates

x = r cos , y = r sin .

(When dealing with polar coordinates one always assumes (x, y) 6= (0, 0)and hence r 6= 0 .)

First, changing the variables from (x, y) we obtain the new unkown

v(r, ) = u(x, y)x=r cos , y=r sin

.

Second, to obtain the equation for v we need the inverse change

r = r(x, y), = (x, y).

Then u(x, y) = v(r(x, y), (x, y)) and

ux = vrrx + vxuy = vrry + vy.

We need to express the right hand sides in terms of (r, ) and plug theresult into the eiconal equation.

With polar coordinates the inverse change does not have a convenientform. The trick is to use the implicit differentiation of polar coordinates:

x r(x, y) cos (x, y), y r(x, y) sin (x, y),

therefore

1 = rx cos r sin x0 = ry cos r sin y1 = ry sin + r cos y0 = rx sin + r cos x

and

rx = cos ry = sin

x =sin r

y =cos r

Combining the results discover

1 = u2x + u2y

= (vrrx + vx)2 + (vrry + vy)2

=(vr cos + v

( sin

r

))2+(vr sin + v

(cos r

))2= v2r +

1r2v2 .

10 DENIS A. LABUTIN

3. Derivation of equations and boundary conditions

(1) The Gauss divergence theorem is crucial for deriving PDEs. It says that if is the boundary of the domain , and if ~n is the outer normal on , then

div ~F dvol =

~F ~n dA.

Here dvol is the volume lement in and dA is the area element on .

(2) Derivation of the transport equation. Let be the density of thetransported component, ~V be the velocity of the transporting flow. Letalso f be the density of the external sources. That is the external produc-tion of the component at the point x during the time interval (t, t+ t)is

M = f(t, x) dvol(x) t.

The key observation: let m be the mass transported by the flow ~Vthrough the flat plate S in the direction of the unit normal ~n , |~n| = 1 ,during the time interval t . Then is

|m| = vol(R)= S height= S |~V |t cos.

We must think that m > 0 if ~V points in the same direction as ~n ,and m < 0 if the directions are opposite. Hence

m = St (~V ~n).

The mass balance condition must always hold for the component. There-fore for any spacial volume

(t+ t, x) dvol(x)

(t, x) dvol(x) ={production inside }

{flux through in direction ~n}

=t

f dvol

t

(t, x) ~V ~n dA(x)

=t

f div((t, x)~V

)dvol(x).

Divide by t 0 to discover

t(t, x) dvol(x) =

f div((t, x)~V

)dvol(x).

Then divide by vol() 0 to find that the equation for the density

= (t, x)

of the component transported by the flow ~V with the sources density fis

t = div(~V ) + f.


In real processes the presence of the component influences the flow ~V =~V (t, x, ) and the sources density f = f(t, x, ) . So the transport equationis nonlinear:

t+ divx((t, x)~V (t, x, (t, x))) = f(t, x, ).

If the flow and the source are not influenced by the component the equationbecomes

(3.2) t+ divx((t, x)~V (t, x)) = f(t, x).

Prove that (3.2) is a linear (non-homogeneous) equation for the unknown = (t, x) . If the flow has a constant velocity

~V (t, x) ~C,the transport equation is

t+ ~C = f.If the constant velocity flow is one dimensional then the transport equationbecomes

= (t, x), t+ cx = f(t, x).

(3) Derivation of the heat equation. The heat equation (also called theheat transport equation) shares some features with the transport equation.The unknown is the temperature T = T (t, x) . The key fact is the Fourierslaw of the heat conduction: let Q be the amount of the heat energytransported through the flat plate S in the direction of the unit normal~n , |~n| = 1 , during the time interval t . Then

Q = ktS gradT ~n.Here k is the (positive) materials conductivity coefficient. Notice thesimilarities with the transport. What is the meaning of the sign inthe formula?

Let f be the intensity of the external heat sources. That is the amountof energy produced externally at the point x during the time interval(t, t+ t) is

E = f(t, x) dvol(x) t.The heat energy E of the volume dvol is given by

E = C T dvol,

where C is the thermal capacity coefficient of the material. Using theFourier law we write the conservation of energy for any spacial volume

CT (t+ t, x) dvol(x)

CT (t, x) dvol(x) ={heat production inside }

{heat flux through in direction ~n}

=t

f dvol

+ t

k gradT ~n dA

=t

f + div(k gradT

)dvol.

Divide by t 0 to discover

CtT (t, x) dvol(x) =

f + div(k gradT (t, x)

)dvol(x).

12 DENIS A. LABUTIN

Then divide by vol() 0 to find that the equation for the temperature

T = T (t, x, y, z)

with the sources density f is

tT =k

Cdiv(gradT ) + f.

(4) Laplace operator. The operator

div(gradu) = (2x + 2y +

2z )u

is called the Laplace operator or Laplacian and is denoted by ,

= div(grad ) = 2x + 2y +

2z .

Thus the heat equation for the unknown T = T (t, x, y, z) with the givensources of the density f is

tT = aT + f.

In the one dimensional case the heat equation becomes

T = T (t, x), tT = a2xT + f, a > 0.

Suppose that the temperature regime is stationary, that is T = T (x, y, z)and

tT = 0.

Then the temperature distribution is described by the Laplace (or Poisson)equation

aT = f.Solutions u of

u = 0

are called the harmonic functions.(5) Diffusion equation. This is the same equation as the heat equation. Let

N = N(t, x) be the concentration of the component which is being diffusedin medium. The key for the derivation is the Ficks diffusion law: let be the amount of the component passed through the flat plate S in thedirection of the unit normal ~n , |~n| = 1 , during the time interval tunder the natural diffusion. Then

= dtS gradN ~n.

Here k is the (positive) materials diffusion coefficient. This is exactly thesame formula as in the Fouriers heat conduction law.

(6) Wave equation. We derive the equation for small vibrations of a onedimensional horisontal string. Small means that if the length of thestring at rest is l , and if the change in length during the vibration is l ,then

ll 1.

This implies that the change in the tension force value

TT0v

ll 1.

Therefore the tension force has the constant value T0 during the vibra-tions. Small also means tag the pieces of the string move only in thevertical direction. Accepting these observations we derive the equation.


Let u(t, x) be the vertical coordinate of a piece (x, x + x) of thestring. The Newtons law of the motion projected on the vertical axisstates

mutt = F + Tvert(t, x) + Tvert(t, x+ x).Here F is an external vertical force on the piece (x, x+ x) , and m =0x . For the tension force we have

~T (x) = T0 ({tangent vector at x})and

~T (x+ x) = T0 {tangent vector at x+ x}.Here the tangent vector to the graph of u is

(1, ux)1 + u2x

.

The smallness of vibrations allows to writeux

1 + u2x= ux(1 +

12u2x + ) ' ux.

Thus

Tvert(t, x) + Tvert(t, x+ x) =T0(ux(t, x+ x) ux(t, x)).Therefore the Newtons law becomes

utt =T00

ux(t, x+ x) ux(t, x)x

+F

0 x.

Letting x 0 discover

utt = c2uxx + f(t, x),

where f is the given external force density and c =T0/0 . The oper-

atorcu =

(2t c22x

)u

is called the wave (or Dalamberts) operator.(7) Initial vaue problem. For the equation of the type

tu+ L(x)u = f

the initial value problem is to find u(t, x) satisfying{ut + Lu = f for t > 0, x (,+)u|t=0 = (x).

Here f(t, x) and (x) are given functions.For the wave equation the initial value problem is to find u(t, x) satis-

fying utt + uxx = f for t > 0, x (,+)u|t=0 = (x)ut|t=0 = (x).

Here f(t, x) , (x) , and (x) are given functions.(8) Boundary conditions. In one dimension u = u(t, x) , t > 0 , a < x 0;Neumann, functions f(t) , g(t) are given, u satsifies

ux|x=a = f(t), ux|x=b = g(t) for all t > 0;

14 DENIS A. LABUTIN

Robin, positive constants C1, C2 > 0 and functions f(t) , g(t) are given,u satsifies

(ux C1u)|x=a = f(t), (ux + C2u)|x=b = g(t) for all t > 0;Conditions at , u and all its derivatives tend to 0 as x fast enough for all

t > 0 .(9) The meaning of the Dirichlet conditions is obvious (prescribing the tem-

perature of the ends of the rod or the motion of the ends of the string).The least obvious conditions are the Robins. For the heat propagation

they are derived by the following argument based on Newtons law of theheat energy conduction and on Fouriers heat transport law. Suppose thatthe right end of the rod, 0 < x < l , exchanges the energy with the massivereservoir of the given temperature T (t) . The Newtons law states that

the energy flux Q from the rod into the reservoir in time t isproportional to the temperature difference,

Q = C(u(t, l) T (t))t.At the same time

the amount of heat transported through the right end of the rodalso outside the rod in time t by Fouriers law is

Q = ux(t, l)t.Hence we must have ux(t, l) = C(u(t, l) T (t)) , which is equivalent to(

ux +C

u

)x=l

=C

T (t).

Similarly, using the Fourier law at the left end of the rod we derive

Q = ux(t, 0)

and (ux

C

u

)x=0

= CT (t).

For the string vibration consider the situation when, say, the left end ofthe string is attached to an elastic spring acting with the returning verticalforce

F = ku(t, 0).Then the Newtons law for the element x projected on the vertical axisstates that

(x)utt(t, 0) = T + F= T0ux(t, 0) ku(t, 0).

(Here to get the formula T = T0ux we assumed that the vibrations aresmall, hence t ux is small, and hence

sin ' tan = ux,where is the angle at which the end of the string slides along the verticalaxis.) Letting x 0 deduce that(

ux k

T0u

)x=0

= 0.

The tension force on the right end of the string is

T = T0ux(t, l),


and the similar argument gives(ux +

k

T0u

)x=l

= 0.

4. Transport and the first order equations

(1) Main tool. The important calculus notion for the transport equation isthe directional derivative of a function. Let u be a multivariable function,

u = u(~r),

~r = (r1, . . . , rn) . Let ~V be a fixed vector. Consider the line through thepoint ~r0 in the direction ~V :

~r(t) = ~r0 + t~V , < t

16 DENIS A. LABUTIN

How to express the last property of u(x, y) as a formula?One way to argue is as follows. Any line parallel to (a, b) has the

equationbx+ ay = C

with some constant C . Along this line the value of u is (some other)constant. Therefore the level sets of functions u(x, y) and bx + aycoincide! This is possible if and only if

u(x, y) = f(bx+ ay)with some function f of one variable. Thus

aux + buy = 0 u(x, y) = f(bx+ ay)where f is an arbitrary function of one variable.

Here is another way to derive this formula. Namely, fix a point (x0, y0) . The linethrough (x0, y0) parallel to (a, b) has the equation

bx+ ay = bx0 + ay0.

At least one coefficients has to be nonzero, suppose a 6= 0 . Since u is constant alongthis line

u(x0, y0) = u

x,bx bx0 + ay0

a

x=0

= u

0,bx0 + ay0

a

.

But u(0, z) is just a function of one variable z . Denote it by f(z) . Then

u(x0, y0) = f

bx0 + ay0

a

= g(bx0 + ay0)

with the one variable function g , g(z) = f(z/a) .

(3) As we just saw, the lines bx+ ay = c which are parallel to (a, b) , playan important role for the transport equation aux + buy = 0 . They arecalled the characteristics of the equation. Above it is explained, that theequation is equivalent to the vanishing of the derivative of u along thecharacteristic.

Using the formula for the general solution we discover at once that thesolution to the initial value problem{

ut + cux = 0 for t > 0, x (,+)u|t=0 = (x)

is given by the wave propagating to the right

u = (x ct).

Problem 3. Why this formula indeed gives a wave propagating to theright? Draw pictures.

(4) Using the property of the characteristics it is easy to solve the initial valueproblem for the inhomogeneous transport equation in the general form{

ut + cux = f(t, x) for t > 0, x (,+)u|t=0 = (x) for x (,+) .

However, it is not very convenient for problem solving. For that it is almostalways easier to compute the solution using the superposition principle.

Since the equation is linear by the superposition principle we write

u = v + w


where {vt + cvx = f for t > 0, x (,+)v|t=0 = 0 for x (,+) .

and {wt + cwx = 0 for t > 0, x (,+)w|t=0 = (x) for x (,+) .

We already know that

w(t, x) = (x ct).

It is left to find v .The key observation for that is that any parametrization of a straight line ~r =

~r0 + ~V , according to the chain rule gives

v(~r)

~V= v(~r) ~V =

d

dv(~r0 + ~V ).

That is, our PDE is equivalent to the ODE along the characteristic with

~V = (1, c).

Now, fix a point (t0, x0) and consider the characteristic line

x ct = x0 ct0 x = ct+ (x0 ct0)

passing through it. Take t as the parameter along the line. Then according to our

observation, the restriction z of u to the characteristic,

z(t) = v(t, x0 ct0 + ct),

satisfies 8 0

z(0) = 0,

and the integration gives

v(t0, x0) = z(t0)

= z(0) +

Z t00

dz(t)

dtdt

= 0 +

Z t00

f(t, x0 ct0 + ct) dt.

Hence the answer is

u(t, x) = (x ct) +Z t

0f(, x c(t ) d.

(5) Variable coefficients first order equation. The equation

a(x, y)ux + b(x, y)uy = 0

similarly to the constant coefficients transport equation, can be rewrittenwith the vector field

~A(x, y) =(a(x, y)b(x, y)

)as

a(x, y)ux + b(x, y)uy = 0u

~A(x, y)= 0.

Unlike the constant coefficients case, the vector field now depends on(x, y) . Instead of the characteristic lines we need the characteristic curvestangent to ~A . These are the solutions of the system(

x(t)y(t)

)= ~A(x(t), y(t)).

18 DENIS A. LABUTIN

Along any such curved

dtu(x(t), y(t)) = aux + buy

and thereforeu

~A= 0

{u is constant along any curve

tangent to ~A

}.

How to express the last condition as a formula?The easiest way to argue is as follows. Suppose we have the equation of

the characteristic curves not in the form

t 7(x(t)y(t)

),

but in the form of a level curve

F (x, y) = const.

The gradient F is orthogonal to the level curve, whereas ~A is tangentto the level curve. Hence

F (x, y) ~A(x, y)(that is aFx + bFy = 0 ). At the same time for the unknown u we alsohave

aux + buy = 0 u ~A = 0,and hence

u(x, y) ~A(x, y).But u is orthogonal to the level curve of u . Therefore the tangent lineto the level curve of u coincides with the tangent line to the level curveof F . But this means that the level curves of F and u coincide! Thisis equivalent to

u(x, y) = f(F (x, y))with an arbitrary function f of one variable.

(6) Thus for the problem solving we need to have an effective way of findingthe characteristic curve in the form

F (x, y) = const.

This is easy: for any curve (x(t), y(t)) expressed as a graph y = y(x) wehave

dy

dx=

dy

dtdx

dt

=y

x.

But the equation for the characteristics is

x = a(x, y)y = b(x, y) .

Hencedy

dx=b(x, y)a(x, y)

.

It is better to write this equation in more symmetric form

dy

b(x, y)=

dx

a(x, y)integration F (x, y) = const.

This is the formula for the characteristic curves for the equation

aux + buy = 0.


Then the general solution to the equation is

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

where f is an arbitrary function of one variable.(7) Let us give a rigorous proofs of (4.3). The equation

(4.4) a(x, y)ux + b(x, y)uy = 0

has the characteristic system

x = a(x, y)

y = b(x, y).

Suppose we know that F is the first integral of the system. That is, it is constant alongany solution (x(t), y(t)) . This is equivalent to

0 = tF (x(t), y(t))

= Fxx+ Fy y

= Fxa+ Fyb .

That is, suppose that

aFx + bFy = 0.

Suppose Fy 6= 0 (if Fx 6= 0 the argument is similar). Introduce the the map m ,xy

7

x

F (x, y)

.

That is, make the change of variables (, ) = m(x, y) ,

= x

= F (x, y).

Geometrically m maps vertical lines into vertical

{x = const} m { = const},

and the level curves of F into horisontal

{(x, y) | F (x, y) = const} m { = const}.

The map m is invertible since

m =

1 0Fx Fy

, det(m) = 1 Fy 6= 0.

Thus

v(, ) = u(m1(, )), u(x, y) = v(m(x, y)).

By this change of variables we derive

ux = vx + vx

= v1 + vFx,

uy = vy + vy

= vFy .

Consequently

0 = aux + buy

= a(v + vFx) + bvFy

= av + v(aFx + bFy)

= av + 0 .

Hence

v = 0 v(, ) = f()for an arbitrary function f of one variable. Actually, this is to be expected since u isconstant along the level curves of F and, as we saw above, the latter are mapped intothe horisontal lines = const . Thus v must depend only on the vertical direction .Going back to the (x, y) -variables discover

(4.4) u(x, y) = v(m(x, y)) = f(F (x, y)).

20 DENIS A. LABUTIN

(8) Hence solving (4.4) is equivalent to finding the first integral of the characteristic system

x = a(x, y)

y = b(x, y).

In the the derivation of (4.3) we were saying that F (x, y) = C was the expression

of the solution to the system as a level curve. Let us show that (in the plane) this isthe same as saying that F is the first integral of the system. That is, finding the first

integral of the plane system of ODEs is the same as solving it.

On the one hand if F (x, y) = C is the solution formula for the system. Then everysolution (x(t), y(t)) after the substitution gives

F (x(t), y(t)) C,

so F is the first integral.

On the other hand, suppose that F is the first integral. Assume, for example,

Fy 6= 0 . Then by the implicit function theorem

F (x, y) = C y = (x)

and

(x) = Fx(x, y)

Fy(x, y), F (x, y) = C.

Solve the simplest equation

x = a(x, (x))

to obtain x(t) , and define

y(t) = (x(t)).

We have F (x(t), y(t)) C , and consequently

y(t) = (x(t)) x(t)

= Fx(x, y)

Fy(x, y)a(x, y)

= b(x, y).

Thus (x(t), y(t)) is the solution to the characteristic system.(9) The map m allows to prove the solvability of the nonlinear (more precisely the semi-

linear) equation

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

following the same strategy. We can prove the existence result for a suitably posed IVP.To formulate the IVP we argue as follows. The characteristics for the nonlinear

equation are the same as above, they are given by the level curves of F . Suppose thecondition

Fy 6= 0holds in a neighbourhood of (0, 0) . The tangent vector to a characteristic is ~A =(Fy ,Fx) . Therefore the characteristics are not tangential to the line

{x = 0}

in the neighborhood of (0, 0) . Consider the following problem: for a given smooth find u satisfying(

a(x, y)ux + b(x, y)uy = H(x, y, u) near (0, 0)

u(0, y) = (y) on the line {x = 0}.

We now prove the existence of such u .The (, ) -variables computation, we have done for (4.4), gives for v(, ) the

equation

av + v(aFx + bFy) = H(x, y, v), (x, y) = m1(, ).

Introducing

a(, ) = a m1(, ), H(, , v) = H(m1(, ), v),

we rewrite the equation as

a(, )v = H(, , v) v = G(, , v),

with

G(, , v) =H(, , v)

a(, ).


The neighborhood of (0, 0) is mapped by m onto some neighborhood N of (0, 0) ,

where 0 = F (0, 0) . The line {x = 0} is mapped onto = 0 . Thus m1 pullsback our problem to finding v(, ) satisfying(

v = G(, , v) in N

v(0, ) = () on the line { = 0} N,

with

() = (y)y=(m1(0,))2

.

This is an IVP for the ODE for v , for which the standard existence-uniqueness theoremproduces a smooth solution v(, ) . Hence we obtain the desired solution

u(x, y) = v(m(x, y)) = v(x, F (x, y)).

(10) Solving problems. For the problem solving the new difficulties areintroduced by the right hand side

aux + buy = f

and by the lower order terms

aux + buy + cu = f.

The right strategy is to use the superposition principle for non homogeneousequations and integrating factors for the lower order terms.

Problem 4. Find u = u(t, x) solving the initial value problem

2ut = ux + xu, u(0, x) = 1.

We cannot use any formulas given above due to the term xu . First wehave to get rid of it. Compute

x dx =x2

2+ C.

The constant is not needed for the integrating factor technique: namely,multiplying the equation by ex

2/2 ,

2ex2/2ut = ex

2/2ux + xex2/2u,

find that

ex2/2ut =

(ex

2/2u)t, ex

2/2ux + xex2/2u =

(ex

2/2u)x.

Hence we can substitute

v = ex2/2u

to find that2vt = vx, v(0, x) = ex

2/2.

This is the simplest homogeneous transport equation. We find at once that

v = f(x+

t

2

), f(x) = ex

2/2.

Thus

u(t, x) = ex2/2v

= ex2/2e(x+

t2 )

2/2

= ext2 +

t28 .

Problem 5. Find u = u(t, x) solving the initial value problem

ut + (1 + t2)ux + u = 1 u(0, x) = ex.

22 DENIS A. LABUTIN

This equation has the variable coefficients, the lower order term, and thenontrivial right hand side. We need to make several steps to reduce it tothe homogeneous equation aut + bux = 0 .

First, the easiest way is to combine ut + u . The integrating factor canbe taken et . Multiplying deduce

etut + etu+ et(1 + t2)ux = et,

so that(etu)t + (1 + t2)(etu)x = et.

Substitutev = etu

and findvt + (1 + t2)vx = et, v(0, x) = ex.

Next, we easily guess that et is solution to the non homogeneous equation.Therefore

w = v et

satisfies

wt + (1 + t2)wx = 0, w(0, x) = ex 1.

This is a simple homogeneous first order equation. The characteristics are

dt

1=

dx

1 + t2

(1 + t2) dt =

dx t+ t3

3+ C = x.

Thus the general solution is

w = f(x t t

3

3

).

The initial condition gives

f(x+ 0) = ex 1.

Consequently

w = ex+t+t33 1,

thenv = et + w = et + ex+t+

t33 1,

and thenu = etv = 1 + ex+

t33 et.

(11) Initial value problem for nonlinear transport equations.

5. Initial value problem on the real line

(1) Initial value problem for the wave equation. For the wave equation

cu = 0

we can find the general solution

u(t, x) = f(x+ ct) + g(x ct),


f, g are arbitrary functions of one variable. This is done using the re-duction to the transport equations. Then the solution to the initial valueproblem on the line

cu = 0, t > 0, x (,+)u|t=0 = (x)ut|t=0 = (x)

can be derived in the form of dAlamberts formula:

u(t, x) =(x+ ct) + (x ct)

2

+12c

x+tcxtc

(s) ds.

(2) The important issue is to visualise in the (t, x) -plane the calculations withthe dAlamberts formula.

(3) Initial value problem for the heat equation. Solutions for the initialvalue problem both for the transport and the wave equation were obtainedby the following procedure. First, we derived the formula for the generalsolution. Second, we specified the general solution to satisfy the initialconditions.

Such approach does not work for the heat equation on the line. Theredoes not exist a simple formula for the general solution to the heat equation.Instead we have to treat the initial value problem directly.

6. Energy method

(1) In this course the energy method is used only to prove the uniqueness of theboundary value problems. In fact the energy method is a truly fundamentaltool for analysis of linear and nonlinear PDEs.

(2) Ifut = c2uxx, t > 0, x (a, b),

then the convenient energy is

E [u](t) = ba

u2

2dx =

ba

u(t, x)2

2dx.

Ifutt = c2uxx, t > 0, x (a, b),

then the convenient energy is

E [u](t) = ba

12(u2t + c

2u2x)dx =

ba

12(ut(t, x)2 + c2ux(t, x)2

)dx.

(3) The energy method computations always involve evaluating of

d

dtE [u](t)

by differentiation under the integral sign, using the equation for replacingsome derivatives by the other, integration by parts relying on the givenboundary conditions.

24 DENIS A. LABUTIN

7. Reflection method for problems on the half line

(1) Method of reflections allows to derive the solutions to some (but not all!)problems with the boundary conditions (some, not all!) from the solutionsto the initial value problem on (,+) . Thus the reflection methodcan be summarized as the reduction

I.V.P. for Lu = 0 on x (0,+)with boundary conditions at x = 0

I.V.P. for Lu = 0 on x (,+)without the boundary

Below we introduce the reflections approach through some examples.(2) Problems with homogeneous boundary conditions. Consider the

wave equation on the half-line with the zero Dirichlet boundary condition:u =0, t > 0, x (0,+)

u|t=0 =(x), x > 0ut|t=0 =(x), x > 0u|x=0 =0, t > 0.

Here is the argument allowing to write down the solution to the problemusing the dAlamberts formula. Take the odd reflection of , from(0,+) to (, 0) :

, (x) =

, (x) for x > 0

0 for x = 0

,(x) for x < 0.

Consider the initial value problem on (,+) with the reflected data, :

v = 0, t > 0, x (,+)

v|t=0 = (x)

vt|t=0 = (x).

The data , are the odd functions. Hence by the property of the waveequation the solution v is odd in x for any t . But an odd function mustbe equal to zero at the origin, thus

v(t, 0) = 0 for all t.

Hence v for x 0 is the solution to the problem for u since it satisfiesboth the initial and the boundary conditions there. Using the dAlambertsformula

u(t, x) =(x+ t) + (x t)

2

+12

x+txt

(s) ds .

Using the visualisation of the dAlamberts formula in the (t, x) -plane andthe oddness of the data it is easy to express the answer using only , (not , ). The formula will be somewhat lengthy but it admits a simplevisualisation on the (t, x) -plane.

(3) The method of reflection allows to solve some boundary value problems onthe finite interval (0, `) . For example, one can solve the Dirichlet problem


for the wave equationu = 0, t > 0, x (0, `)

u|t=0 = (x), x (0, `)ut|t=0 = (x), x (0, `)

u|x=0 = u|x=` = 0, t > 0.The solution will have an incredibly lengthy formal description, but it hasa clean visualisation on the (t, x) -plane.

(4) The homogeneous Neumann boundary conditions are also manageable bythe reflection method. For example, consider the heat equation on thehalf-line

ut uxx = 0, t > 0, x (0,+)u|t=0 = (x), x > 0ux|x=0 = 0, t > 0.

Take the even reflection of from (0,+) to (, 0) :

(x) =

{(x) for x 0

(x) for x 0.

Consider the initial value problem on (,+) with the reflected datum : {

vt vxx = 0, t > 0, x (,+)v|t=0 = (x) .

The datum is an even function. Hence by the property of the heatequation the solution v is even in x for any t . But an even (smooth)function must have zero derivative at the origin, thus

vx(t, 0) = 0 for all t.

Hence v is the solution to the problem for u since satisfies both the initialand the boundary conditions there. Using the heat kernel

u(t, x) = +

S(t, x y)(y) dy .

Using the evenness property of the heat kernel and the evenness of it iseasy to express the answer using alone (instead of ).

(5) Reflections for the non-homogeneous boundary conditions.Method of reflections allows also to reduce some problems with the nonho-mogeneous boundary conditions to the IVP on (,+) . For example,consider the general Dirichlet problem for the wave equation on (0,+) :

utt c2uxx = 0, t > 0, x (0,+)u|t=0 = (x), x > 0ut|t=0 = (x), x > 0u|x=0 = h(t), t > 0.

The Dirichlet boundary condition is now given by h(t) , h 6= 0 . Here isthe reduction argument. Recall that

cf(x ct) = 0for an arbitrary function f of one variable. Choose f() = h ( /c) , sothat

f(x ct) = h(t x

c

).

26 DENIS A. LABUTIN

Definev(t, x) = u(t, x) h

(t x

c

),

so that v(t, 0) = h(t) h(t) = 0 for all t > 0 . Then v solvesvtt c2vxx = 0, t > 0, x (0,+)v|t=0 = 1(x), x > 0vt|t=0 = 1(x), x > 0

v|x=0 = 0, t > 0,

with 1(x) = (x) h(x/c) , 1(x) = (x) h(x/c) . This is theproblem with the zero Dirichlet condition. We solved it above using theodd reflection. Hence the answer is

for x > ct :

u(t, x) =1(x+ ct) + 1(x ct)

2

+12c

x+tcxtc

1(s) ds

+ h(t x

c

)=(x+ ct) + (x ct)

2+

12c

x+tcxtc

(s) ds

h((x+ ct)/c) + h((x ct)/c)2

12c

x+tcxtc

h(s/c) ds

+ h(t x

c

)=(x+ ct) + (x ct)

2+

12c

x+tcxtc

(s) ds

h((x+ ct)/c) + h((x ct)/c)2

+12

(x+tc)/c(xtc)/c

h(r) dr

+ h(t x

c

)=(x+ ct) + (x ct)

2+

12c

x+tcxtc

(s) ds

h((x ct)/c)

+ h(t x

c

)=(x+ ct) + (x ct)

2+

12c

x+tcxtc

(s) ds;

for ct x > 0 using the similar computations:

u(t, x) =(x+ ct) + (x ct)

2+

12c

x+tcxtc

(s) ds

+ h(t x

c

).


The meaning of the solutions is simple. For x > ct the wave launched bythe motion h(t) at the origin does not reach the point x , and the solutionis given by the usual dAlamberts formula. For ct > x the solution is thesuperposition of the free wave coming from the dAlambert and the wavelaunched from the origin by the prescribed motion h(t) .

8. Fundamental solution and Duhamels principle

(1) Delta function. Delta function is not a function! The meaning willbe assigned not to itself, but to expressions involving it. Therefore forus formulas as +

(x y) dy = 1, (x) = 0 x 6= 0, . . .

will have a meaning, but itself will not. The formulas involving -function (such as above) express the limiting results of the correspondingprocesses.

(2) Intuitively -function expresses such physical notions as the density of thepoint electrostatic charge, the energy of the point heat source, and similar.This is done using the limit of -shaped functions in the same way as inphysics one deals with the point-concentrated objects as limits of smearedones, as smearing gets smaller and smaller.

A -shaped function (x) is the function satisfying:

(a) +

(x) dx = 1 ;

(b) (x) 0 for all x ;(c) (x) = 0 for all x with |x| > (or lim0 (x) = 0 for every

x 6= 0 .)The following two examples of -shaped sequences are of fundamental

importance for this course.First, the characteristic function approximation of is

(8.5)1(x) =

{0 , x / (0, )

1/ , x [0, ],where

(x) ={

0 , x / (0, )1 , x [0, ].

Second, the heat kernel approximation is1G(x/

) =

14

ex24 .

Now, -function is the limit of a -shaped sequence as 0 . Thus

(8.6) (x),1(x),

14

ex24 (x), 0.

The limit in (8.6) is understood in the following sense: the validity of anygiven formula

F() = fmeans that after the substitution F() and letting 0 we find

lim0

F() = f.

28 DENIS A. LABUTIN

(3) Let us illustrate the technique of dealing with .

Problem 6. Let us show that +

(x) dx = 1.

Indeed, take any -shaped sequence (for example / ). We have +

(x) dx = 1 > 0,

hence trivially

lim0

( +

(x) dx)

= 1.

Thus according to our definition we may write +

(x) dx = 1.

Problem 7. Show that

(x) = 0 x 6= 0.

Problem 8. Show that for any continuous function f +

f(x) (x) dx = f(0).

To prove this formula it is convenient to use the -shaped sequence1(x) .

Problem 9. Fix x0 . For the shifted (x x0) show that(x x0) = 0 x 6= x0, +

(x x0) dx = 1,

and +

f(x) (x x0) dx = f(x0).

Problem 10. Prove that for a continuous function f

(8.7) +

f(y) (x0 y) dy = f(x0).

The formula can be approximately rewritten as

f(x0) ' +

f(y)(x0 y) dy

'j

(x0 yj) f(yj) yj

'j

(x0 yj) f(yj) yj

=j

f(yj)yj (x yj)x=x0

.

This form emphasizes the important meaning of (8.7). Namely, an arbitraryforce (or signal, or field, or ) f(x) at a point x is approximately equal to


the value at x of the sum of the point-impulse forces (or signals, or fields,or ) shifted to yj :

(8.8) f(x) 'j

f(yj) (yj+1 yj) (x yj).

To derive (8.7) just use +

f(y)(x0 y) dy =

+f(x0 z)(z) (dz) since x0 y = z,dy = dz

= +

f(x0 z)(z) dz

f(x0 z)z=0

.

Problem 11. For the Heaviside step-function

(x) = H(x) ={ 0, x < 0

1, x 0

show that(x) = (x).

For that smear the step-function (x) on scale . Then show that(x) = (x) is a -shaped sequence.

(4) Fundamental solution. Consider the initial value problem for a verygeneral evolution equation with the unknown u(t, x){

tu = Lu t > 0, x (,+)u|t=0 = (x).

Think thatL = L(x)

is, say, a constant coefficient operator involving only x , 2x , . . . . Manyimportant problems from sciences have such form. For example, we havealready encountered the heat diffusion L = k22x , and transport L =cx .

(5) The fundamental solution (or propagator) is the evolution of the initialpoint-impulse. That is, the solution of{

tu = Lu, t > 0, x (,+)u|t=0 = (x)

This means that

tK(t, x) L(x)K(t, x) = 0, t > 0, x (,+)

and K(t, x) is -shaped in the same sense as in (8.6) with relaced byt . That is

K(t, x) (x), t 0.Hence the fundamental solution shows how the equation smears for t > 0the initial (at time t = 0 ) datum .

As we will see below, knowing the propagator for L is the key for solv-ing the initial value problem with arbitrary data. This explains the namefundamental. We will denote the propagator by K(t, x) or KL(t, x) .

30 DENIS A. LABUTIN

(6) For example, the Gaussian (heat kernel)

G(t, x) =14t

ex24t

satisfies (t 2x

)G(t, x) = 0, t > 0.

Moreover, it forms a -shaped sequence,

G(t, x) (x), t 0.

Hence the Gaussian is the fundamental solution to the heat equation.(7) Somewhat unusual example is the fundamental solution to the transport equation

tu+ cxu = 0.

One easily makes the formal computations for K(t, x) = (x ct) :

x(x ct) = xct

1,

t(x ct) = xct

(c),

so that

(t + cx)(x ct) = c+ c = 0.

We also have (x ct) (x) as t 0 (without any smearing). Hence (x ct) isthe fundamental solution of the transport equation.

This answer has a clear physical meaning. It says that the initial point-mass with

the density is transported by the one dimensional flow with the velocity c along

the x -access without smearing or spreading. That is indeed what the perfect transport

(without difusion) does!

(8) We verified that two suggested functions are the fundamental solutions ofthe heat and transport equation. There is a general method based on theFourier transform for the computation of the fundamental solution for agiven equation from the scratch. It will be studied in the second quarter.

(9) Fundamental solution and the formula for IVP. Let us show that thesuperposition principle implies that the solution of{

tu = Lu, t > 0, x (,+)u|t=0 = (x)

is given by

(8.9) u(t, x) = +

K(t, x y)(y) dy.

Hence, knowing the fundamental solution K(t, x) allows us to calculatethe solution with any initial data by the simple integration!

Verifying that (8.9) solves the IVP is easy. Indeed, firstly for t > 0 andx (,+) we have

(t L(x))( +

K(t, x y)(y) dy)

= +

(t L(x))K(t, x y)(y) dy

= +

(sK(s, z) L(z)K(s, z))(s,z)=(t,xy)

(y) dy

= +

0(y) dy

= 0.


Secondly, for every x (,+)

limt0

( +

K(t, x y)(y) dy)

= +

(x y)(y) dy

= (x).

Instead of verifying the formula given by somebody, one can derive it asshown below.

(10) The basic building block for the derivation of (8.9) is the following simple observation.

Consider the IVP with the shifted :

C(x a).

That is (tu = Lu, t > 0, x (,+)

u|t=0 = C(x a).Use the linearity to prove that the solution is

u = CK(t, x a),

where K is the fundamental solution. This is the formula we will need.(11) Now fix a small > 0 , discretize the x variable

xj = j, j = 0, 1, 2 . . . ,

and approximate our datum

(x) 'Xj

(xj)(x xj).

The key observation is to write

(xj)(x xj) = (xj)1

(x xj)

with the characteristic function approximation of from (8.5). Therefore

(xj)1

(x xj) ' (xj) (x xj).

The summation then gives

(8.10) (x) 'Xj

(xj) (x xj),

which is nothing but the physical approximation of a general charge distribution by asum of point charges. That is we approximated by a sum of discrete impulses located

at points xj .

Notice that we could skip the details of the discretization and at once deduce (8.10)from (8.8)!

(12) Next, by the linearity

u(t, x) 'Xj

uj(t, x),

where uj solves (tuj = Luj , t > 0, x (,+)

uj |t=0 = (xj) (x xj).

By our building block solution we already know that

uj(t, x) = (xj)K(t, x xj).

Consequently

u(t, x) 'Xj

uj(t, x)

=Xj

(xj)K(t, x xj)

=Xj

K(t, x xj)(xj)(xj+1 xj).

Finally, notice that the last sum approximates an integral over x . Letting the discreti-sation

xj+1 xj = 0

32 DENIS A. LABUTIN

we discover

u(t, x) = lim0

0@Xj

K(t, x xj)(xj)(xj+1 xj)

1A=

Z +

K(t, x y)(y) dy,

which is desired (8.9).

(13) To make the formulae shorter, it is convenient to introduce the solutionoperator etL acting on the functions F on R . Define

(etLF )(x) = +

K(t, x y)F (y) dy.

In the case L is a number etL is the usual exponential. In these notationswe have{

tu = Lu, t > 0, x Ru|t=0 = (x)

u(t, x) = etL(x)

(14) The propagator approach gives the same formula for any constant coeffi-cients linear operator L . Notice, however, that the wave equation doesnot fit immediately in this form we have 2t instead of t in the initialvalue problem for the wave equation. Later we will show how to adapt thearguments for the wave equation.

(15) Duhamels principle. Knowing the propagator also allows us to solvethe non-homogeneous problem

(8.11)

{tu =Lu+ f(t, x), t > 0, x R

u|t=0 =0

via the Duhamels formula:

u(t, x) = t

0

e(ts)Lf(s) ds

= t

0

+

K(t s, x y)f(s, y) dy ds.

Again, once the formula is given, its verification is not hard. Firstly,

u(0, x) = 0

0

( )dt = 0,

therefore the initial condition holds. Secondly, equality

L(x)u(t, x) = t

0

+

L(x)K(t s, x y)f(s, y) dy ds

= t

0

+

L(z)K(r, z)(r,z)=(ts,xy)

f(s, y) dy ds


together with

tu(t, x) = +

K(t s, x y)f(s, y) dys=t

+ t

0

+

tK(t s, x y)f(s, y) dy ds

= +

(x y)f(t, y) dy

+ t

0

+

rK(r, z)(r,z)=(ts,xy)

f(s, y) dy ds

imply

(t L)u = +

(x y)f(t, y) dy

+ t

0

+

(rK(r, z) L(z)K(r, z)

)(r,z)=(ts,xy)

f(s, y) dy ds

=f(t, x) + t

0

+

0 f(s, y) dy ds

=f(t, x).

(16) We now derive Duhamels formula on the physical level of rigor. For that we shall usethe following basic building block problem. Consider the delayed IVP:(

tu = Lu, t > s, x (,+)u|t=s = (x).

Using the invariance in time, one can easily see that the value of the solution to the

delayed problem is

u(t, x) =

Z +

K(t s, x y)(y) dy

where K is the fundamental solution. This is the formula we will need.

(17) Fix T > 0 . To compute u(T, x) solving (8.11) we first split the time interval [0, T ]into N segments of the length t = T/N ,

0 = t0 < t1 = t

< t2 = 2t = t1 + t

< tj = jt = tj1 + t

< tN = T.

Next, in (8.11) discretise f in time by splitting it into

f(t, x) 'N1Xj=0

fj(t, x), fj(t, x) = f(tj+1, x)(tj ,tj+t)(t).

By the linearity

u(T, x) 'N1Xj=0

uj(T, x),

where uj(t, x) solves(tuj = Luj + fj , t > 0, x (,+)

uj |t=0 = 0.

The point here is that the right hand side fj(t, x) is identical zero if 0 < t < tj and

tj + t < t < T . Only on the short time interval (tj , tj + t) it acts on the equation.Let us evaluate uj(T, x) for every j . By (8.9) we have

uj(t, x) = 0 for 0 < t tj .

34 DENIS A. LABUTIN

Ignoring the tiny time interval (tj , tj+1) of length t we use the first order approxi-

mation to evaluate for t = tj+1

uj(tj+1, x) ' uj(tj , x) + t tuj(tj , x)

= 0 + tL(x)uj + fj

(tj ,x)

= t

0 + f(tj+1, x).

Consequently for tj+1 < t < T our uj solves(tv = Lv, tj+1 < t < T, x (,+)

v|t=tj+1 = f(tj+1, x)t.

But this is just the delayed IVP. Hence, as we shown above

uj(T, x) =

Z +

K(T tj+1, x y)f(tj+1, y)t dy.

Therefore

u(T, x) 'N1Xj=0

Z +

K(T tj+1, x y)f(tj+1, y) dy

t.

Recognising the Riemann integral sum here we discover letting t 0 that

u(T, x) =

Z T0

Z +

K(T s, x y)f(s, y) dy ds.

(18) Notice that the Duhamels formula gives the following statement for theheat equation. If (x) is odd (even) and for every t the function f(t, x)is odd (even) in x , then the solution of{

tu = Lu+ f(t, x), t > 0, x (,+)u|t=0 = (x),

u = u(t, x) is odd (even) in x for any fixed t .This follows at once from the representation

u(t, x) = +

K(t, x y)(y) dy + t

0

+

K(t s, x y)f(s, y) dy ds

with

K(t, z) =1tG

(zt

),

and the evenness of the Gaussian

G(x) =14e

x24 .

9. Boundary value problems and separation of variables

(1) Separation of variables method for the heat and wave equations with eitherDirichlet or Neumann boundary conditions is familiar from calculus. Actu-ally the method works for more general equations and boundary conditions.

(2) Suppose f is a given function on a finite segment [a, b] .

Boundary conditions

1f(a) + 1f(b) + 1f (a) + 1f (b) = 0

2f(a) + 2f(b) + 2f (a) + 2f (b) = 0


are called symmetric if for any functions f , g satisfyingthem one has

f (x)g(x) f(x)g(x)x=bx=a

= 0.

This definition is not very inspiring. The true meaning of the symmetryis the symmetry of the operator d2/dx2 with respect to the L2 innerproduct. Indeed,

the boundary conditions are symmetric if and only for any smoothfunctions f , g satisfying them one has(

d2f

dx2, g

)L2

=(f,d2g

dx2

)L2

(3) The theoretical basis for the separation of variables method is provided bythe following theorem. Its most important part (iii) states that the sepa-ration of variables method always works provided the boundary conditionsare symmetric.

Theorem. Consider the eigenvalue problem

d2

dx2f = f on (a, b)

with some symmetric boundary conditions. Then:(i) all eigenvalues are real and form an infinite discrete sequence

1 2 ,n + as n ;

(ii) if n 6= m then the corresponding eigenfunctions are L2 -orthogonal(fn, fm)L2 = 0 ;

(iii) any L2 can be expanded as

=n=1

Anfn on [a, b]

with the coefficients given by

An =(, fn)L2(fn, fn)L2

.

The convergence of the series in the theorem is in the L2 sense. Thetheorem unites several theorems in Strauss.

(4) We know from the calculus that any can be expanded either into thefull Fourier series, or into the cos -half range Fourier series, or into thesin -half range series. The theorem provides us with the expansion withrespect to the general system {fn} of the eigenfunctions not covered bythe three calculus cases.

For example, the mixed boundary conditions

u|x=0 = 0, ux|x=l = 0on (0, l) lead through the separation to the eigenvalue problem

f = f, f(0) = f (l) = 0.The eigenvalues turn out to be

n =(

(n+ 12 )l

)2, n = 0, 1, 2, . . . ,

36 DENIS A. LABUTIN

and the eigenfunctions

fn(x) = sin(

(n+ 12 )xl

).

This is neither the full Fourier nor a half range Fourier system. Hence thecalculus expansions do not work in this case. However, the theorem tellsthat any initial datum on [0, l] can be expanded with respect to {fn} .Indeed, by the theorem

(x) =n=0

An sin(

(n+ 12 )xl

), x [0, l],

with the coefficients computed by the formula

An =

l0

(x) sin(

(n+ 12 )xl

)dx l

0

sin(

(n+ 12 )xl

)sin(

(n+ 12 )xl

)dx

.

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Math 124A Course Notes