Injibara University (Inu)

INJIBARA UNIVERSITY (INU)
College of Natural and Computational Science

Department of Mathematics
Course Name : PARTIAL DIFFERENTIAL EQUATIONS
course code : math 5032
GROUP TWO (INU) June 18, 2023 1 / 29

Chapter Three:- One Dimensional Wave equation;
out line
3.1 The Wave equation on the whole Line.D’Alembert Formula

3.2 The Wave Equation on the Half-Line Reflection Method
3.3 Mixed Problem For The Wave Equation
3.4 Inhomogeneous Wave Equation
3.5 Conservation of the Energy

3.1 Wave equation on the whole Line.D’Alembert Formula
The simplest hyperbolic second-order equation is the wave equation;
utt − c 2 uxx = 0,———(1)
where x signifies the spatial variable or ”position”, t the ”time”

variable, u = u(x,t) the unknown function and c is a given positive
constant. The wave equation describes vibrations of a string.
Physically u (x,t) represents the ”value” of the normal displacement
of a particle at position x and time t.The characteristic equation of
(1) is given as;
utt − c 2 uxx = 0 ⇒ (ut + cux )(ut − cux ) = 0 , since it is
reducible . Where ut + cux = 0 and ut − cux = 0 are factors of (1)
then the characteristic equations are obtained by using Lagrange
method with; p=1,q=c and R=0 and p=1 , q= -c and R=0
respectively.(i.e dtp = dx
q
= du
R
), then use as follow;
continue...

x + ct = c1
Hence, are two families of real characteristics.
x − ct = c2
Introducing
the new variables
ξ = x + ct −1 x = (ξ + η)/2
Φ: ,Φ :{ andthefunction
η = x − ct t = (ξ − η)/2c
u(ξ, η) = u((ξ + η)/2, (ξ − η)/2c), the equation (1) reduces to
uξη (ξ, η) = 0 − − − − − − − (2),
therefore, Ruξ (ξ, η) = F (η) ,
u(ξ, η) = F (ξ)dξ + g (η) = f (ξ) + g (η) and in the original
variables u (x, t) is of the form
u(x, t) = f (x + ct) + g (x − ct) − − − − − − − (3), known as the
general solution of (1). It is the sum of the function g (x - ct) which
presents a shape traveling without change to the right with speed c
and the function f (x + ct) - another shape, traveling to the left with
speed c.
Consider the Cauchy (initial value) problem for (1)
utt − c 2 uxx = 0, xεR, t > 0,
(CW): u(x, 0) = φ(x), xεR, whereφ and ψ are arbitrary
ut (x, 0) = ψ(x), xεR,
functions of x. Further we denoteR + = {t : t ≥ 0}.
Theorem 3.1 ( D’Alembert formula ):If φεC 2 (R)and, ψεC 1 (R)
the problem (CW) has a unique solution uεC 2 (RxR + ) given by the
formula R x+ct
u(x, t) = 21 (φ(x + ct) + φ(x − ct)) + 2c1 x−ct φ(s)ds − − − −(4)
Proof. We are looking for a solution of the problem in the form (3)
satisfying the initial conditions at t = 0
f (x) + g (x) = φ(x) − − − − − − − (5),
cf ′ (x) − cg ′ (x) = ψ(x).
Differentiating (5) with respect to x and solving the linear system for
f’ and g’,

continue...
we obtain
f ′ (x) = 12 φ′ (x) + 2c1 ψ(x), − − − − − − (6)
g ′ (x) = 12 φ′ (x) − 2c1 ψ(x), − − − − − − (7)
Integrating (6) and (7) R x from 0 to x we get
f (x) = 21 (φ(x) + 2c1 R0 φ(s)ds + (f (0) − 2c1 φ(0)),
x
g (x) = 21 (φ(x) − 2c1 0 φ(s)ds + (g (0) − 2c1 φ(0)).
By (5) f (0) + g (0) = φ(0).
Therefore
u(x, t) = f (x + ct) + g (x − ct) = 21 (φ(x + ct) + φ(x − ct))
R x+ct R x−ct
+ 2c1 ( x−ct ψ(s)ds − 0 ψ(s)ds)
R x+ct
1
= 2 (φ(x + ct)) + φ(x − ct) + 2c1 x−ct ψ(s)ds.

continue...
Example
3.1 Solve the problem (CW) with c = 1,φ = 0 and
cos 3 x, if xε( −π , π)
2 2
φ(x) =
0, if x̸ ε( −π , π ).
2 2
Solution:The solution of the problem is by using the general solution

R x+ct
u (x ,t) = 12 (φ(x + ct)) + φ(x − ct) + 2c1 x−ct ψ(s)ds.
From the cuachy initial condition we have
;ut (x, 0) = 0 = ψ(x), u(x, 0) = 0, so the integral part become 0.then
u (x ,t) = 21 (φ(x + ct)) + φ(x − ct)
= 12 [cos 3 (x + t) + cos 3 (x − t)]
= 21 [(cosxcost − sinxsint)3 + (cosxcost + sinxsint)3 ]
= 21 (−sin3 t + (−sin3 t)), x = −π 2
→ π2
= −sin3 t is the requaired solution.

3.2 The Wave Equation on the Half-line Reflection Method
Let us consider the problem (CW) on the half-line (0, ∞) with
Dirichlet boundary condition at the endpoint x = 0.
This is the problem:
utt − c 2 uxx = 0, 0 < x <∞, t > 0 ,
(CDW): u(x,0)=φ(x),ut (x, 0) =ψ(x) ,0 < x < ∞ ,
u(0, t) = 0,t≥ 0 ,
It can be interpreted as vibrations of a very long string with a
clamped one end.
In fact we shall reduce the problem (CDW) to a problem (CW) by
the odd reflection method. It consists in considering the odd
extensions of the initial functions φ0 (x) and ψ0 (x) where
φ(x), x > 0 ,
φo (x)= -φ(x), x < 0 ,
0 , x=0,

utt − c 2 uxx = 0, xεR, t > 0 ,
The problem (C φo ): u(x,0)=φ(x), xεR ,
ut (x, 0) = ψo (x), xεR .
, has the solution R x+ct
u(x, t) = 12 (φ0 (x + ct) + φ0 (x − ct)) + 2c1 x−ct ψ0 (s)ds.
Its restriction u(x, t) = u(x, t) : x ≥ 0 is the unique solution of the
problem (CDW) . If 0 < x < ct, then
R x+ct R x+ct R0
x−ct
ψ0 (s)ds = 0 ψ(s)ds − x−ct ψ(−s)ds
R x+ct R0 R x+ct
= 0 ψ(s)ds + ct−x ψ(s)ds = ct−x ψ(s)ds.
Therefore we have; u(x,t)= R x+ct

1/2(φ(x + ct) + φ(x − ct)) + 1/2c x−ct ψ(s)ds, ifx > ct
R x+ct
1/2(φ(x + ct) − φ(x − ct)) + 1/2c ct−x ψ(s)ds, if 0 < x < ct—(8

N
ote that u (x,t) is a continuous function if the Compatibility
condition φ(0) = 0 is satisfied.
Otherwise u (x,t) is a discontinuous solution and the jump of u (x,t)
on the characteristic x = ct is
u(ct + 0, t) − u(ct − 0, t) = φ(0).
Theorem 3.2: Let φ(x)εC 2 (R + ), ψ(x)εC 1 (R + ) and the following
compatibility conditions be satisfied:
φ(0) = φ′′ (0) = ψ(0) = 0 − − − − − (9)
Then the function u(x,t) defined by (8) is the unique solution of the
problem (CDW) of class C 2 (R + xR + )
Proof:The function u(x,t)is of class C 2 in domains
{(x, t) : x > ct > 0}, and{(x, t) : 0 < x < ct}.
We shall prove that the derivatives of u(x,t) up to order two are
continuous along the line x = ct. We have
1/2(φ′ (x + ct) + φ′ (x − ct)) + 1/2c(ψ(x + ct) − ψ(x
ux (x, t)=
1/2(φ′ (x + ct) + φ′ (ct − x)) + 1/2c(ψ(x + ct) + ψ(ct −
continue...
Therefore by (9)ux (ct + 0, t) − ux (ct − 0, t) = − c1 ψ(0) = 0.

By the same way
uxx (ct + 0, t) − uxx (ct − 0, t) = φ′′ (0) = 0,
ut (ct + 0, t) − ut (ct − 0, t) = φ(0) = 0,
utx (ct + 0, t) − utx (ct − 0, t) = cφ′′ (0) = 0,
utt (ct + 0, t) − utt (ct − 0, t) = c 2 φ′′ (0) = 0.
Moreover the function u(x , t) satisfies the equation, boundary and
initial conditions of the problem (CDW).
We can do the same for the problem with the Neumann boundary
condition ,considering even extensions of initial data.

Let us consider the problem :
utt − c 2 uxx = 0, 0 < x <, t > 0
(CNW):u(x,0)= φ(x), ut (x, 0) = ψ(x), 0 < x, ∞
ux (0, t) = 0, t ≥ 0.
In this case we reduce the problem (CNW)to (CW) with initial
φ(x), ifx ≥ 0,
functions φe (x), andψe (x) ,where φe (x) =
φ(−x), ifx ≤ 0.
As beforewe can show that the problem (CNW) has a unique solution
1/2(φ(x + ct) + φ(x − ct)) + 1/2c(ψ(x + ct) − ψ(x −
u(x, t)=
1/2(φ(x + ct) + φ(ct − x)) + 1/2c(ψ(x + ct) + ψ(ct − x
Rt
Where ψ(t) = 0 ψ(s)ds.

Example: Solve the problem
utt − uxx = 0, 0 < x < ∞, t > 0,
u(x,0)=sin3 x, ut (x, 0) = 0, 0 < x < ∞,
u(0,t)=0, t≥ 0
Solution: As before we can show that the problem (CNW) has a
unique solution

1/2(φ(x + ct) + φ(x − ct)) + 1/2c(ψ(x + ct) − ψ(x − ct
u(x, t)=
1/2(φ(x + ct) + φ(ct − x)) + 1/2c(ψ(x + ct) + ψ(ct − x
where the conditions ut (x, 0) = 0, u(0, t) = 0 by cauchy conditions.so
that 1/2c(ψ(x + ct) − ψ(x − ct) = 0
when we use the form u(x,t)=1/2(φ(x + ct) + φ(x − ct))
u(x, t) = 1/2(sin3 (x + t) + sin3 (x − t)) for c=1, is the requaired
solution of the given (CNW).

3.3 Mixed Problem for the Wave Equation
Let us consider the problem (CW) on a finite interval [0, ι] with
Dirichlet boundary conditions at the end-points x = 0 and x =ι .
This is the problem
utt − c 2 uxx = 0, 0 < x < ι, t > 0,
(MDW):u(x,0)= u(x,0)=φ(x), ut (x, 0) = ψ(x), 0 ≤ x ≤ ι,
u(0,t)=u(ι, t) = 0, t ≥ 0.
We can get the solution of the problem (MDW) again using the
method of reflection in this case through both ends. We extend the
initial data φ(x) and ψ(x) given on the interval (0, ι) to the whole
line using “odd”extensions φeo (x) and ψeo (x) with respect to both
sides x = 0 and x = ι, where

continue...
φ(x)if 0 < x < ι,
φeo (x) = -φ(−x)if − ι < x < 0
extended to be of period 2 ι.
Consider the problem (CWeo) :
utt − c 2 uxx = 0, xεR, t > 0,
(CWeo) : u(x,0)=φeo (x), xεR,
ut (x, 0) = ψeo (x), xεR.
By Section 3.1 it has a solution R x+ct
u(x, t) = 21 (φeo (x + ct) + φeo (x − ct)) + 2c1 x−ct ψeo (s)ds.
Its restriction,u(x, t) = v (x, t) :0≤x≤l
gives the unique solution of the problem (MDW). Note that the
solution formula is characterized by a number of reflections at each
end x = 0 and x= ι along characteristics through reflecting points.

continue...
They divide the domain R = {(x, t) : 0 < x < ι, t > 0} into

diamond-shaped domains with sides parallel to characteristics and
within each diamond the solution u (x , t) is given by a different
formula. On the data φ and ψ we impose the compatibility condition
φ(0) = φ(ι) = ψ(0) = ψ(ι) = 0. − − − − − − − (10)
In this case the solution u(x,t) is a continuous function on R.
Note that u(x, t)εC 2 (R) if
φ(0) = φ(ι) = φ′′ (x) = ψ ′′ (ι) = ψ(0) = ψ(ι) = 0. − − − − − − − (11)
We can do the same for the problem with the Neumann boundary
condition , considering even extensions of initial data.
Namely, let us consider the problem

continue...
utt − c 2 uxx = 0, 0 < x < ι, t > 0,
(MNW): u(x,0)=φ(x), ut (x, 0) = ψ(x), 0 < x < ι,
ux (0, t) = ux (ι, t) = 0, t ≥ 0.
In this case we reduce the problem (MNW) to (CWee) with initial
functions φee (x) and ψee (x) ,where
φ(x), 0 < x < ι,
φee (x)= φ(−x), −ι < x < 0,
extendedtobeofperiod2ι.
As before the problem (MNW) admits the unique solution:
u(x, t) = ω(x, t)0≤x≤ι , where ω(x, t) is the solution of the problem
ωtt − c 2 ωxx = 0, xεR, t > 0,
(CWee ): ω(x, 0) = φee (x), xεR,
ωt (x, 0) = ψee (x), xεR.

Example:Solve
the problem (MDW) with c = 1, ψ = 0 and φ(x) =
3
cos x, xε[(3π/2, 5π/2)]
0, x ε[0, 3π/2) ∪ (5π/2, 4π].

3.4 Inhomogeneous Wave Equation
Let f εC 1 (R 2 ) and consider the inhomogeneous Cauchy problem
utt − c 2 uxx = f , xεR, t > 0,
(ICW): u(x,0)=φ(x), xεR,
ut (x, 0) = ψ(x), xεR.
It can be split into two problems - one homogeneous with nonzero
initial data (CW), which we solve, and one inhomogeneous with zero
initial data
utt − c 2 uxx = f , xεR, t > 0,
u(x,0)= 0,x εR,
ut (x, 0) = 0, xεR.———–(12)
If u1 (x, t)andu2 (x, t)aresolutionsof (CW )and(12)respectively , then

continue...
Let us consider (12). Making change of variables

u(x, t) = ul (x, t) + u2 (x, t) is a solution of (ICW)
ξ = x + ct,
η = x − ct,
we transform (12) into

Uξη = − 4c1 2 F (ξ, η), − − − − − − −(13)
Uξ,ξ = Uξ (ξ, ξ) = Uη (ξ, ξ) = 0, − − − − −(14)
where
U(ξ, η) = u( ξ+η
2
, ξ−η
2c
)
ξ+η ξ−η
F(ξ, η) = f ( 2 , 2c ).

continue...
Integrating (13) with respect to η we have

Rξ
Uξ (ξ, ξ) − Uξ (ξ, η) = − 4c1 2 η F (ξ, s)ds,
Rξ
which, in view of (14), yields Uξ (ξ, η) = 4c1 2 η F (ξ, s)ds
Integrating the last equation
R ξ R zwith respect to ξ
U(ξ, η) − U(η, η) = 4c1 2 η η F (z, s)dsdz,
RξRz
U(ξ, η) = 4c1 2 η η F (z, s)dsdz. − − − − − − − −(15).
Let us make change of variables : s = σ − cτ, z = σ + cτ
∂(s,z)
which has Jacobian J = ∂(σ,τ )
= 2c. The last change transforms the
domain of integration D = {(s, z) : η ≤ s ≤ z, η ≤ z ≤ ξ} in to
D ′ = {(σ, τ ) : x − c(t − τ ) ≤ σ ≤ x + c(t − τ ), 0 ≤ τ ≤ t}.
Indeed by , η ≤ s ≤ z ≤ ξ, we havex − ct ≤ σ − cσ + cτ ≤ x + ct.

continue...
now it follows
0 ≤ 2cτ ≤ 2ct ⇔ 0 ≤ τ ≤ t, andx − c(t − τ ) ≤ σ ≤ x + c(t − τ ).
The solution of (12) is given by
R t R x+c(t−τ )
u2 (x, t) = 2c1 0 x−c(t−τ ) f (σ, τ )dσdτ = 2c1
RR
∆(x,t)
f (σ, τ )dσdτ. − −(
Where ∆(x, t) denotes the characrteristic triangle.
We prove that problem (ICW) has a solution given by the exact
formula u(x, t) = 21 φ(x + ct) + φ(x − ct)+
1
R x+ct R t R x+c(t−τ )
2c
( x−ct ψ(s)ds + 0 x−c(t−τ ) f (σ, τ )dσdτ )
The solution is.—-(17)
Note that from (17) it follows the well-posedness of (ICW)

Example: Solve the problem
utt − uxx = xt, xεR, t > 0 ,
u(x,0)= 0 ,x εR,
ut (x, 0) = 0, xεR .
1
R t R x+c(t−τ )
Solution: The solution is ; u(x, t) = 2 0 x−c(t−τ )
σ.τ dσdτ
first,we solve the inner part,
1 x+c(t−τ )
R 1
R x+c(t−τ )
2 x−c(t−τ )
σ.τ dσ = 2
τ x−c(t−τ )
σdσ
= 4 [(x + (t − τ )) − (x − (t − τ ))2 ] = τ4 [4x(t − τ )] = τ x(t − τ )
τ 2
second,
Rt by substituting
R t the first in u(x,t)R as
t Rt
0
τ x(t − τ )dτ = x 0 τ (t − τ )dτ = x[ 0 τ tdτ − 0 τ 2 dτ ]
3
= x τ6 is the required solution.

3.5 Conservation of Energy
Let R = {(x, t) : 0 < x < ι, 0 < x < ∞}, anduεC 2 (R) be a solution
of the problem
utt − c 2 uxx = 0, 0 < x < ι, t > 0
(MDW): u(x,0)= φ(x), ut (x, 0) = ψ(x), 0 < x < ι,
u(0,t)=u(ι, t) = 0, t ≥ 0 .
The quantityR ι (KE(t)) is given by
KE (t) = 12 0 ut2 (x, t)dx is known asR Kenetic Energy , where as the
ι
quantity KE(t) given by PE (t) = 21 0 c 2 ux2 (x, t)dx
is called Potential Energy.
The sum of the kinetic Rand potential energy which is given as
ι
E(t)=KE(t)+PE(t)= 12 0 (ut2 (x, t) + c 2 ux2 (x, t))dx
is the total energy of the system at the instant t.
For the above mentioned problem (MDW) we show that the total
energy E(t) is a constant independent of t.
This is the law of conservation of energy.
Theorem 3.5If uεC 2 (R) is a solution of the problem (MDW) ,the
energy E (t) is a constant E (t) = E (0).
proof: Multiplying the equation by ut , using the identities
Ut Utt = 21 ∂t∂
(ut2 ),
∂
Ut Uxx = ∂x ∂
(ux ut ) − 21 ∂t (ux2 ), R
ι
and
R ι integrating by parts we get; 0 (utt − C 2 uxx )ut dx = 0
∂
= 0 ( 21R∂t ∂
(ut2 ) − C 2 ( ∂x (ux ut )) − 21 ∂t
∂
(ux2 ))dx
1 d ι 2
= 2 dt 0 (ut )(x, t) + C 2 ux2 (x, t))dx
∼ C 2 ux2 (ι, t)ut (ι, t) + C 2 ux (0, t)ut (0, t) = dE dt
Therefore for tR > 0
ι
E(t)=E(0)= 12 0 (ut2 )(x, 0) + C 2 ux2 (x, 0))dx
ι ′
= 12 0 (ψ 2 )(x) + C 2 φ 2(x))dx so the energy is conserved.
R

Example:Show that the initial value problem

utt − c 2 uxx = f (x, t), −∞ < x < ∞,
u(x,0)= φ(x), ut (x, 0) = ψ(x),,
has a unique solution.
let as assume that the IVP has two distinct solutions, u and v. But
then their difference w = u-v will solve the homogeneous wave
equation, and will have the initial data.
w (x, 0) = u(x, 0) − v (x, 0) = φ(x) − φ(x) = 0
wt (x, 0) = ut (x, 0) − vt (x, 0) = ψ(x) − ψ(x) = 0
Hence the energy associated with the solution w at time t = 0 is
1 ∞
R
E[w](0)= 2 −∞ [(wt )(x, 0))2 + C 2 (wx (x, 0))2 ]dx = 0
This differs from the energy defined above by a constant factor of C 2
and It will subsequently be zero at any later time as well. Thus,
1 ∞
R
E[w](0)= 2 −∞ [(wt )(x, t))2 + C 2 (wx (x, t))2 ]dx = 0, ∀t.
But since the integrand in the expression of the energy is
nonnegative, the only way the integral can be zero, is if the integrand
is uniformly zero.
That is, ∇w (t, x) = (wt (x, t), wx (x, t)) = 0, ∀x, t.
This implies that w is constant for all values of x and t, but since
w(x,0)=0,the constant value must be zero. Thus,
u(x,t)-v(x,t)=w(x,t)=0
which is in contradiction with our initial assumption of distinctness of
u and v. This implies that the solution to the IVP of the given
example is unique.

CHAPTER Four
One Dimentional Diffusion Equation
Injibara University
Natural and Computational Sciences
Department of Mathematics
June 21, 2023
Chapter Three
4.1 Maximum-Minimum Principle for Diffusion Equation
In this section we consider the homogeneous one dimensional
diffusion (heat)equation.
ut − α2 uxx = 0..............................4.1 which appears in the study
of heat conduction and other diffusion processes.as model for
equation (4.1) we consider a thin metal bar of length (l) whose
sides are insulated.denote byu(x, t) the temperature of the bar at
the point x at time t.the constantk = α2 is known as the thermal
conductivity.The parameter k depends only on the material from
which the bar is made .♦ Initial temperature distribution
u(x, 0) = ϕ(x), 0 ≤ x < lwhereϕ(x)is given function
♦Boundary conditions at the end of the bar,instance,we assume
that the temperature at the ends are fixed.
u(0, t) = T1 , u(l, t) = T2 , t > 0
Chapter Three
where u(x,t) satisfies the initial condition
u(x, 0) = ϕ(x), 0 < x < l.......4.3 and the boundary condition
u(0, t) = u(l, t) = 0, t > 0...........4.4
orux (0, t) = ux (l, t) = 0, t > 0....................4.5 The problem
(4.2), (4.3), (4.3) is known as the Dirichlet problem for the diffusion
equation,while (4.2), (4.3), (4.5) at the Neumann problem.
A property of diffusion equation known as the maximum-minimum
principle.
Let < = {(x, t) : 0 ≤ x ≤ l, 0 ≤ t ≤ T }be a closed rectangle and
Γ{(x, t) ∈ < : t = 0orx = 0orx = l} THEOREM
4.1(Maximum-Minimum principle)
Letu(x, t) be a continuous function in< which satisfies
equation(4.2)in<|Γ.then max<u(x, t) = maxΓu(x, t).............4.6
min<u(x, t) = minΓu(x, t)........................4.7
Chapter Three
The maximum(minimum) of u(x,t)can not be assumed any where
inside the rectangle but only on the bottom of lateral sides (unless
u is constant).
proof of theorem 4.1 denote
M = maxΓ(x, t)
we shall show that max<u(x, t) = u(x, t) + x 2 where is a
P P
positive constant.
we have for (x, t) ∈ <|ΓP
Lv (x,P t) = Lu(x, t) − 2 k < 0.........(4.8)andv (x, t) ≤
M + l 2 , if (x, t) ∈ Γ
ifv (x, t)attains its maximum at an interior point (x, t)it follows
thatLv (x, t) ≥ 0 which contradicts(4.8) there fore v (x.t)attains its
maximum at a point of
δ< = Γ ∪ γ, γ = {(x, t) ∈ < : t = Γ}supposev (x, t) has a
maximum at a point(x̃, Γ) ∈ γ, 0 < x̃ < l
thenvx (x̃, Γ) = 0, vxx (x̃, Γ) ≤ 0.
Chapter Three
corollary4.1letuj (x, t)be a solution of (HDD)with initial data
ϕj (x), j = 1, 2 thenmax0 ≤ x < l|u1 (x, t) − u2 (x, t)| ≤ max0 <
x < l|ϕ1 (x) − ϕ2 (x)|..........................(4.9)
THEOREM4.3 a,letu(x, t)be a solution of (HDD)
thenH(t1 ) ≥ H(t2 ),if0 ≤ t1 ≤ t2 ≤ T .
b,letuj (x, t)beasolutionof (HDD)correspondingtoinitialdataϕj (x), j =
Rl Rl
1, 2 then 0 (u1 (x, t) − u2 (x, t))2 dx ≤ 0 (ϕ1 (x) − ϕ2 (x)2 dx
Chapter Three
4.2 The Diffusion Equation on the Whole Line
XIn this section we give an explicit formula for the solution of the
Cauchy problem for the diffusion equation on the whole line.
(CD):ut -kux x =0,x ∈ R,0 < t < T u(x, 0) = ϕ(x); x ∈ R •The
solution of (CD) is given by the Poisson formula
1 R ∞ − (x − ξ)2
u(x, t) = √ e ϕ(ξ)dζ
2 πxt −∞ 4Kt
assuming that ϕ(ξ) is continuous and bounded on R.
•The value of u(x, t) depends on the values of the initial data
ϕ(ξ)for all xi in R.
•Conversely, the value of ϕ at a point xo has an immediate effect
everywhere for t > 0
•This effect is known as infinite speed of propagation which is in
contrast to the wave equation.
•Moreover the solution is infinitely differentiable for t > 0.
Chapter Three
cont...
•It is known that the diffusion is a smoothing process going
forward in time.
•Going backward(anti diffusion) the process becomes chaotic.
•Therefore, we would not expect-well-possessedness of the
backward-in-time problem for the diffusion equation.
•In order to prove the problem (CD) we need some preliminaries
on improper integrals.
•Recall some definitions and properties.
Let f (x, y ) be a continuous function in (x, y ) in Rx[a, b]. the
integral
R∞
I (y )= −∞ f (x, y )dx
Chapter Three
cont...
is convergent for every y ∈ [a, b]. R∞

Definition We say that integral I (y )= −∞ f (x, y )dx is uniformly
convergent for y ∈ [a, b],if for every > 0 there exists
A0 = A( ),such that if A > A0 , then
R −A R +∞
| −∞ f (x, y )dx| + | A f (x, y )dx| < ,for every y ∈ [a, b]
R∞
•If I (y )= −∞ f (x, y )dx is uniformly convergent for y ∈ [a, b],then
the function I(y) is continuous in [a, b].
∂x
•Suppose f (x, y ) and (x, y )dx are continuous functions in
∂y
R × [a, b], I (y ) is convergent for every y ∈ [a, b] and
Chapter Three
cont...
R ∞ ∂x
J(y ) = −∞ ∂y (x, y )dx is uniformly convergent for y ∈ [a, b].
Then I (y ) is a differentiable function in (a, b) and I 0 (Y ) = J(y )
•(Weierstras criterion) Suppose there exists a function g (x) such
Rthat |f (x, y )| ≤ g (x) for every y ∈ [a, b] and the integral
∞ R∞
−∞ g (x)dx is convergent.Then the integral I (y )= −∞ f (x, y )dx
is uniformly convergent for y ∈ [a, b]
Definitio A differentiable function u(x, t) is a solution of the
problem (CD) if it satisfies the equationux − Kux x =0 in
R × (0, T )and
limt ↓0 u(x, t)=ϕ(x)
Chapter Three
EXample
Solve the problem

ut -ux x =0,x ∈ R,t > 0
u(x, 0) = e −x , x ∈ R
1 R ∞ − (x − ξ)2
u(x, t)= √ e −ξ dξ
Πt −∞ 4t
using
(x − ξ)2 1
+ξ= (x 2 − 2ξx + ξ 2 + 4ξt)
4t 4t
1
= (x + ξ + 4t 2 − 2ξx + 4ξt − 4xt + 4xt − 4t 2 )
2 2
4t
(ξ + 2t − x)2
= +x − t
4t
and making change of variable
ξ + 2t − x
√ =p,we have
2 t
e t − x R ∞ − ()ξ − 2t − x)2
u(x, t)= √ e dξ
√ 2 Πt −∞ 4t
e t−x 2 t R infty
√ −∞ dp Chapter Three
Diffusion on the Half-Line
Let us consider the diffusion equation on the half-line (0, ∞) and

take the Dirichlet boundary condition at the end-point x = 0.
Using the reflection method considered for the wave equation we
shall treat the problem
{Ux − KUxx = 0, x(0, +∞), t > 0
u(x, 0) = ϕ(x), x(0, +∞), (4.23)
u(0, t) = 0, 0 ≤ t.
We are looking for a solution formula for (4.23) analogous to the
Poisson formula.
Chapter Three
Let us consider the problem (CD) with initial data ϕ0 ,which is the
old extension of ϕ(x) on the whole line
{Ux − KUxx = 0, xR, t > 0
u(x, 0) = ϕ(x), xR, (4.24)
Where
ϕ0 (x) = ϕ(x), x > 0
−ϕ(−x), x < 0,
0, x = 0
Let u0 (x, t) be the unique solution of (4.24) which by the Poisson
formula is
Chapter Three
R∞ (x−ξ)2
υo (x, t) = 2√1πkt −∞ e − 4kt ϕo (ξ)dξ

R ∞ −(x−ξ)2 R0 −(x−ξ)2
= 2√1πkt 0 e 4kt ϕ(ξ)dξ − −∞ ϕ(−ξ)e 4kt ϕ(ξ)dξ

1
R ∞ −(x−ξ)2 R∞ −(x+ξ)2
= 2√πkt 0 e 4kt ϕ(ξ)dξ − 0 ϕ(ξ)e 4kt ϕ(ξ)dξ
Chapter Three
R ∞ −(x−ξ)2 −(x+ξ)2
= 2√1πkt 0 (e 4kt − e 4kt )ϕ(ξ)dξ
(4.25) The restriction
u(x, t) = u0 (x, t)|x≥0
is the unique solution of the problem (4.23).Note that u0 (x, t)
satisfies the diffusion equation and is an odd function
u0 (−x, t) = −u0 (x, t),which easily follows from (4.25).Then
u(0, t) = u0 (0, t) = 0 and u(x, t) satisfies the diffusion equation .
More over ,u(x, t) satisfies the initial condition for x > 0
Chapter Three
Let us consider now the Neumann boundary condition at the end
point x = 0 for the diffusion equation on the half-line.Namely ,let
us consider the problem
{Ux − KUxx = 0, x(0, +∞), t > 0
u(x, 0) = ϕ(x), x(0, +∞),
u(0, t) = 0, t > 0 (4.26)
In this case we use the even reflection of ϕ(x)
ϕe (x) = ϕ(x), x ≥ 0
ϕ(−x), x ≤ 0,.
Let ue (x, t)bethesolutionoftheproblem
{Ux − KUxx = 0, xR, t > 0
u(x, 0) = ϕe (x), xR,
As before,we have
R ∞ −(x−ξ)2
ue (x, t) = 2√1πkt −∞ e 4kt ϕe (ξ)dξ
Chapter Three
R ∞ −(x−ξ)2 −(x+ξ)2
= 2√1πkt 0 (e 4kt + e 4kt )ϕ(ξ)dξ.
The restriction u(x, t) = ue (x, t)|x≥0 is the solution of (4.26).Note
that
R ∞ −(x−ξ)2 x−ξ −(x+ξ)2
∂ue
∂x (x, t) = √−1
πkt 0
(e 4kt ( 4kt ) + e 4kt (( x+ξ
4kt )ϕ(ξ)dξ.
and
∂u ∂ue
∂x (0, t) = ∂x (0, t)
2 −ξ2
∞ −ξ −ξ ξ
= √−1
R
πkt 0
(e 4kt (
4kt ) + e 4kt ((
4kt )ϕ(ξ)dξ = 0.
As before,u(x, t) satisfies the diffusion equation and the initial
condition.
Chapter Three
Example
Example 1:solve (4.23) with e −x and k = 1.

solution: By the solution formula for (4.23)
R ∞ −(x−ξ)2 −(x+ξ)2
u(x, t) = 2√1πt 0 (e 4t − e 4t )e −ξ dξ
R∞ 2 2
= 2√1πt 0 (e − ((x−ξ)
4t + ξ) − e − (x+ξ)
4t + ξ)dξ
Using
(x−ξ)2 2
4t + ξ = (ξ+2t−x)
4t +x −t
(x+ξ)2 2
4t + ξ = (ξ+2t+x)
4t − (x + t)
we obtain
R ∞ (ξ+2t−x)2 R ∞ (ξ+2t+x)2
u(x, t) = 2√1πt [e t−x 0 e − 4t dξ − e t+x 0 e − 4t dξ]
t−x R ∞ 2 t+x R ∞ 2
= e√π 2t−x√
e −p dp − e√π 2t+x √
e −p dp
2 t 2 t
= 12 (e t−x − e t+x ) + 12 (e t+x erf ( x+2t
√ ) + e t−x erf ( x−2t
2 t
√ )).
2 t
√ √
Note that u(0, t) = 12 e t (erf ( t) + erf (− t)) = 0
Chapter Three
Inhomogeneous Diffusion Equation on the whole line
The problem of finding the a function u(x,t) in the inhomogeneous

diffusion equation is known as Cauchy problem.
This is given as

ut − kuxx = f (x, t), xR, t > o, (1)
v (x, 0) = ϕ(x)xR
By linear properties of operator Lu ≡ ut − kuxx , the solution of (1)

u(x, t)is the sum of the solutionυ(x, t)andω(x, t) of the problems

υt − kυxx = 0 xR, t>0 (2)
υ(x, 0) = ϕ(x) xR,
Chapter Three
One Dimensional Diffusion Equation
Dimensional Diffusion Equation and

ωt − kωxx = f (x, t) xR, t>0, (3)
ω(x, 0) = 0 xR,
respectively (4)
x2
1 −
Can be formulated or written as G (x, t) = √ e 4kt , and it is
2 πkt
known as Green, s function or fundamental solution of the diffusion
operator L.It is clear that
L(G (x, t)) = 0, (x, t)R × (0, ∞), (5)

L(G (x − ξ, t − τ )) = 0, (x, t) × R,(0, ∞)
Chapter Three
Gree , sfunction
From (5)
it follows that the solution of (3) is υ(x, t) =

R∞ R1R∞
−∞ G (x − ξ, t)ϕ(ξ)dξ + 0 −∞ G (x − ξ, t − τ )f (ξ, τ )dξdτ .(4.33)
Assuming that f(x,t) is bounded and continuous on R × (0.∞),We
prove that the function ω(x, t)givenby (4.31) satisfies the problem
(4.29).By the maximum principle for the diffusion equation on the
whole line it follows that the line υ(x, t) given by (4.33)is the
unique solution of(1).
Chapter Three
Theorem:
Let f (x, t)C (R × (0, ∞)) be a bounded function and

Z ∞
υ(x, t, τ ) = G (x − ξ, t − τ )f (ξ, τ )dξ.
−∞
then the function
Z t
ω(x, t) = υ(x, t, τ )dτ
0
is asolution of the problem (3)

Proof. By virtue of the estimates of section 4.2 and the
assumption of the theorem the integrals
Chapter Three
Z t Z t Z t
υ(x, t, τ )dτ, υt (x, t, τ )dτ, υxx (x, t, τ )dτ
0 0 0
are uniformly convergent over bounded closed intervals of R.By
the formula for differentiation of integrals depending on parameters
Z t Z t
∂
f (x, t)dx = ft (x, t)dx + f (t, t)
∂t 0 0
and we have
Z t
∂ω ∂
= υ(x, t, τ )dτ
∂t ∂t 0
∂ R t−ε
= limε→0 0 υ(x, t, τ )dτ
∂t
∂ R t−ε
= limε→0 υ(x, t, τ )dτ
∂t 0
Chapter Three
Rt
= 0 υ(tx, t, τ )dτ + limε→0 υ(x, t, t − ε)
Rt R∞
= 0 K υXX (X , t, τ )dτ + limε→ 0 −∞ G(X − ξ, ε)f (ξ, t − ε)dξ =
∂ 2 R t R∞ 2 √
k 2 0 υ(x, t, τ )dτ +limε→0 √1π −∞ e −p f (x −2p kε, t −ε)dp
∂x
∂x 2 ω
= k 2 + f (x, t.)
∂
To show that the initial condition is satisfied observe that
Z t
|υ(x, t)| ≤ |υ(x, t, τ )|dτ
0
Z t Z ∞
1 2
e −p |f (x − 2p
p
≤ √ k(t − τ ), τ )|dpdτ
0 π −∞
≤ tsup|f (x, t)|

∴ limt↓0 ω(x, t) = 0
Chapter Three
One Dimensional Diffusional Equation
Examples:
consider inhomogeneous problem on the half -line with the
Dirichlet boundary condition

υt − kυxx = f (x, t), x > 0, t > 0
ω(0, t) = h(t)t > 0,

ω(x, 0) = ϕ(x)x > 0
We reduce the above inhomogeneous problems to a simpler
problem υ(x, t) = ω(x, t) − h(t).Then υ(x, t) satisfies the problem

0
υt − kυxx = f (x, t) − h (t), x > 0, t > 0,
υ(0, t) t¿0,
υ1 (x, 0) = ϕ(x) − h(0) x > 0.
Chapter Three
One Dimensional Diffusional
continue...

ωt − kωxx = FO (x, t), xR t> 0
ω(x, 0) = 0 xR R R
t ∞
Given byω(x, t) = 0 −∞ G (X − ξ, t − τ )Fo (ξ, τ )dξdτ then
υ2 (x, t) = ω(x, t) | x ≥ 0 is the solution of 9(4.37)
Let as show that υ2 (0, t) = 0 for t¿0.We have
Z tZ 0
υ2 (0, t) = − G (x − ξ, t − τ )F (−ξ, τ )dξdτ
0 −∞
Z tZ ∞
+ G (x − ξ, t − τ )F (ξ, τ )dξdτ
0
Z tZ ∞
= (G (X − ξ, t − τ ) − G (x + ξ, t − τ ))F (ξ, τ )dξdτ
0 0
Chapter Three
Z tZ ∞
= (G (X − ξ, t − τ ) − G (x + ξ, t − τ ))F (ξ, τ )dξdτ
0 0
Z tZ ∞
υ2 (0, t) = (G (−ξ, t − τ ) − G (ξ, t − τ ))F (ξ, τ )dξdτ = 0
0 0
Chapter Three
CHAPTER FIVE
Weak Solution,Shock Waves and Conservation Laws
June 21, 2023
Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

IUWassie June 21, 2023 1 / 36
1 Weak Derivative Weak solution
2 Conservation Laws
3 Burger’s Equation
4 Weak Solution,Riemann Problem

Weak derivative and Weak Solution
To define a weak (generalized) solutions of the problem by passing to the

limit in L2 space of corresponding solutions uk
Let L2 (Ω) be the usual Lebesgue space of a square integrable functions
U : Ω 7−→ R where Ω is the measurable domain in R.n .
L2 (Ω) is a Banach space R
with a norm ||U||2 L2 (Ω) = Ω U 2 (x)dx
Next L2 (Ω) is the space of functions U : Ω 7−→R such that for every
compact sub set k ⊂ Ω it holds U|k ∈ L2 (k)

Definition
a
A function vL2loc (Ω) is said to be the weak axj derivative of the given
2
Rfunction uLloc (Ω) iffR ap
Ω v (x)p(x)dx = − Ω u(x) axj dx,
For every test function p(x)C0∞ (Ω)
a au
If uC 1 (Ω) it is easy to see that the weak axj derivative of u(x) equal to ax j
The Sobolev 1 space W 2,1 (Ω) is introduce
au 2
U : Ω → R for which there exist a weak derivatives axj L (Ω)
j = 1, 2, 3, ..., n
W 2,1 is a BanachR space with the norm
n ( au )2 )dx
||U||2 w 2,1 (Ω) = Ω (u 2 + Σj=1 axj

2,1
By Wloc (Ω) we denote the space of functions
U:Ω→R
such that for every compact subset k ⊂ Ω it holds U|k W 2,1 (k)
Every function of W 2,1 (Ω) can be approximated by a sequence of smooth
functions w.r.tW 2,1 norm on compact sub-domains of Ω
Consider the so-called Dirac2 Kernels or Mollifying Kernels
or the Friedrichs 3 Mollifiers
Let ε > 0 and p0(x)c0∞ (Ω) R
where the constant c0 is such that R n p0(x)dx = 1
Define the sequence of Mollifiers
x
pε(x) = ε1n p0( )
ε
For which pε (x)C ∞ (Ω) and R n pε (x)dx = 1
R

Mollifcations Regularizations Jε u of a function u ∈ L1loc (Ω)
are definedR as
Jε u(x) = Ω pε (x − y )u(y )dy
If u ∈ L1loc (Ω) ,then Jε u ∈ C ∞ (R n ).
Theorem
Theorem 5.1 For a given function u ∈ W 2,1 (Ω) the regularizations Jε u
tends to u in W 2,1 (k)
for every compact k ⊂ Ω i.e
||Jε u||w 2,1 (k) → 0 asε → 0.

Proof.
Let ε < dist(Ka, Ω)
By changeRof variables y = x − εz and Cauchy Schwarz4 inequality we have
Jεu(x) = Ω pε(x − y )u(y )dy
nR
= 1ε |y −x|≤1 p0( x−yε )u(y )dy
R
= |z|≤1 p0(z)u(x − εz)dz
R 1 1
|Jεu(x)| ≤ |z|≤1 p0 2 (z)p0 2 (z)dz
1 R 1
≤ ( |z|≤1 p0(z)dz) 2 ( |z|≤1 p0(z)u 2 (x − εz)dz) 2
R
1
= ( |z|≤1 p0(z)u 2 (x − εz)dz) 2 ,
R
| k Jεu(x)|2 dx ≤ |z|≤1 p0(z)dz( k u 2 (x − εz)dx)dz

R R R
1
= ( |z|≤1 p0(z)u 2 (x − εz)dz) 2
R
|Jεu(x)|2 dx ≤ |z|≤1 p0(z)( k u 2 u 2 (x − εz)dx)dz

R R R
kR
≤ |z|≤1 p0(z)dz kε u 2 (x)dx
R
= kε u 2 (x)dx
R

Definition
1,2
Definition 5.2. A function u ∈ Wloc (R 2 ) is said to be a weak solution of
the
R wave equation utt − c 2 uxx = 0 iff
2
R 2 u(ptt − c pxx )dxdt = 0
for every test function p ∈ co∞ (R 2 )
Definition
2,1
Definition 5.3. A function u ∈ Wloc (R 2 ) is said to be a weak solution of
2
the wave equation utt − c uxx = 0 iff there exists a sequence of smooth
solutions uk (x, t) ∈ c 2 (R 2 ) of the wave eqation such that for every
compact set k ⊂ R 2 ,||uk − u||w 2,1 (k) → 0 as k → ∞

Theorem
Definition 5.2 is equivalent to Definition 5.3
proof Definition 5.3 ⇒ Definition 5.2

Let ρc0∞ (R 2 ),suppρ ⊂ k and (u)k be the sequence of C 2 smooth solution
such that ||uk − u||w 2,1 (k) → 0 as k → ∞
Integrating by parts, we have
0 =R R 2 (uktt − c 2 ukxx )ρdxdt
R
= R 2 uk ρtt − c 2 ρxx dxdt

By
R Definition 5.3 and the Cauchy Schwarz inequality it follows
| kR u(ρtt − c 2 ρxx )dxdt| = | k (u − uk )(ρtt − c 2 ρxx )dxdt|
R
≤ |u − uk ||ρtt − c 2 ρxx |dxdt

≤ ||u − uk ||L2 (k)||ρtt − c 2 ρxx ||L2 (K )
≤ ||u − uk ||w 2,1 (k)||ρtt − c 2 ρxx ||L2 (k) → 0
as k → ∞

Therefore
2 2
R R
k u(ρtt − c ρxx )dxdt = R 2 (ρtt − c ρxx )dxdt = 0

Conservation Laws
We consider the solution of the hyperbolic systems of conservation
laws.These are systems of PDEs of the form
ut (x, t) + (f (u(x, t)))x = 0 (1)
where u : R × R −→ R m is a vector function

 
u1 (x, t)
u(x, t) = 
 .. 
. 
um (x, t)
and f : R m −→ R m is a mapping
 
f1 (u1 (x, t), ..., um (x, t))
f (u(x, t)) = 
 .. 
. 
fm (u1 (x, t), ..., um (x, t))
The function u describes physical quantities as mass,momentum,energy in
fluid dynamical problems.The mapping f(u) is called a flux function.The
system (1) is hyperbolic iff the Jacobian matrix
 ∂f1 ∂f1 
∂u1 , ..., ∂um
A = Jf (u) = 
 .. 
. 
∂fm ∂fm
∂u1 , ..., ∂um
has only real eigenvalue and is diagonalizable,i.e. there exists a complete

set of m linearly independent eigenvectors.
The Euler 5 system in gas dynamics is a system of conservation laws.In one
space dimension these equations are

   
ρ ρv
∂  ∂ 
ρv  + ρv 2 + p  = 0 (2)
∂t ∂x
E v (E + p)
Where ρ = ρ(x, t) is the density, v is the velocity ,ρv is the momentom ,ρ
is the pressure and E is the energy.The equation (2) are called
1 ρt + (ρv )x = 0 , conservation of mass
2 (ρv )t + (ρv 2 + p)x = 0 , conservation of momentum
,
3 Et + (v (E + p))x = 0 ,conservation of energy
Introduce new variables

   
ρ(x, t) u1 (x, t)
u(x, t) =  ρ(x, t)v (x, t)  =  u2 (x, t) 
E (x, t) u3 (x, t)
the system can be written in the form (1) with
 
u2
2
f (u) =  uu2 + p(u)
 
1

u2
u1 (u3 + p(u))
Example Assume p(u) = u1 .Then the system (2) is hyperbolic in

R 3 \ {u : u1 = 0}.
Solution The Jacobian of
 
u2
2
f (u) =  uu2 + p(u)
 
1

u2
(u
u1 3 + p(u))
is the matrix
 
0 1 0
−u22
A= +1 2 uu12 0 
 
u12
−u2 u3 +u1 u3 +u1 u2
u1 ( u1 − 1 u1 u1
with eigenvalues
λ1 = uu12 , λ2 = uu21 + 1, λ3 = uu21 − 1
corresponding eigenvectors are
 
0
v1 =  0 
1
1
 
v 2 =  uu21 + 1 
u3 +u2
u1 + 1
1
 
v 3 =  uu21 − 1 
u3 −u2
u1 + 1

0 1 1
u2 u2
0
1
+1 u1 − 1 = −2
u3 −u2
u3 +u2

1
u1 + 1 u1 + 1

The simple initial value problem for the system (1) is the Cauchy problem
in which (1)holds x ∈ R,t > 0 and
u(x, 0) = u0 (x) (3)

where u0 (x) is a prescribed function.
ut + aux = 0, x ∈ R, t > 0 (4)
u(x, 0) = u0 (x), x ∈ R,
where a is a constant,known as the linear advection equation or one

side wave equation.
The problem (4) has the unique solution
u(x, t) = u0 (x − at) (5)
if the initial function u0 (x) ∈ C 1 (R)

Definition Assume that u0 (x) ∈ L1loc (R). A function
u(x, t) ∈ L1loc (R × [0, ∞)) is a weak solution of (4) iff
Z∞ Z∞ Z∞
u(ρt + aρx)dxdt + u0 (x)ρ(x, 0)dx = 0
0 −∞ −∞
for every test function ρ ∈ C01 (R × [0, ∞)).

one side wave equations we consider linear strictly hyperbolic systems
ut + Aux = 0, (6)
u(x, 0) = u0 (x)
u : R × R → R, A ∈ R m is a constant matrix.
The system is strictly hyperbolic iff the matrix A is diagonalizable and
has m distinct real eigenvalues.Let
A = RΛR −1 (7)
where
Λ = diag (λ1 , λ2 , ..., λm ) ,R = [r1 |r2 |...|rm ]
is the matrix of right eigenvectors
Ark = λk rk , k = 1, 2, 3, ..., m
changing the variables
v = R −1 u,
by R −1 ut + ΛR −1 ux = 0 we obtain
vt + Λvx = 0, (8)
or componetwise
(vk )t + λk (vk )x = 0, k = 1, 2, ..., m

The initial conditions to (8)are
v (x, 0) = v0 (x) = R −1 u0 (x) (9)
vk (x, 0) = vk0 = (R −1 u0 (x))k

Then

vk (x, t) = vk0 (x − λk t) (10)
is the solution of (8),(9)
The solution of (6)is
m
X
u(x, t) = vk0 (x − λk t)rk .
k=1

Example solve the problem
ut + Aux = 0,
u(x, 0) = u0 (x),
where u : R × R → R 3 ,
 
3 1 −2
A =  −1 2 1 
4 1 −3
f1 (x) − 27 f2 (x) + f3 (x)

 
u0 (x) =  f1 (x) + f2 (x) 

13
f1 (x) − 2 f2 (x) + f3 (x),

−sin3 x x ∈ [−3π, −2π]

f1 (x) =
0 x∈
/ [−3π, −2π
−sin3 x x ∈ [−3π, 0]

f2 (x) =
0 x∈/ [−3π, 0]
−sin3 x x ∈ [π, 2π]

f3 (x) =
0 x∈ / [π, 2π]
solution.The matrix
 
3 1 −2
A =  −1 2 1 
4 1 −3
has eigenvalues
λ1 = 2 , λ2 = −1 , λ3 = 1
and corresponding eigenvectors

   −7   
1 2 1
r1 =  1  , r2 =  1  , r3 =  0 
−13
1 2 1
Then
1 −7
 
2 1
R = [r1 |r2 |r3 ] =  1 ! 0 
−13
1 2 1
 −1 1
  
3 1 3 2 0 0
R −1 =  31 0 −1 
3 , Λ = R −1 AR =  0 −1 0 
5 −3
2 −1 2 0 0 1
 
f1 (x − 2t)
v = f2 (x + t)  ,

f3 (x − t)
f1 (x − 2t) − 27 f2 (x + t) + f3 (x − t)
 
u = Rv  f1 (x − 2t) + f2 (x + t) 
13
f1 (x − 2t) − 2 f2 (x + t) + f3 (x − t)
Burger equation
Definition The simples equation combining both nonlinear propagation and

diffusion effects is Burgers’equation.
ut + uux = cuxx
This way of the quasi linear PDE form then.
a(x, y , u)ux + b(x, y , u)uy = f (x, y , u)
can be treated similar manner to the semi-linear case.if we can determine
dy
two(linear-ind)first integrals if. dx du
dt /a = dt /b = dt /f .

.and also
dx dy du
a(x,y ,u) = b(x,y ,u) = f (x,y ,u) .of the form h(x, y , u) = c1 and j(x, y , u) = c2
then the general solution to the PDE
may be written in implicit form

J(x, y , u) = F (h(x, y , u)) where F is an arbitrary function.
Example-1
uux + uy = 0, u(x, 0) = φ(x)-homogeneous problem is.aux + buy = f
quasi-linear.

dy dy
solution consider dx du dx du
a = b = f i.e u = 1 = 0 then the first and the
three ratio dx du
u = 0 and also. we haveu = c1 wherec1 is constant and the
dy du
R R
second and three ratio 1 = 0 then 0dy = 1du which
dy
giveu = c1 .also dx
u = 1 , dx = udy dived both side by dy which
dx
givenu = dy =u = c1 and for x = c1 y + c2 , (c1 constant function.

solution x = uy + c2 thus,x − uy = c2 combine to from u = F (x − uy ) for
arbitrary differential function(c1 = F (c2 )) Applying are initiation condition
we obtain u(x, 0) = φ(x) = F (x)phiF = φ Thus the solution to our
Cauchy problem associated with Burgers equation is u(x, y ) = φ(x − uy )
implicit form.
Example-2
find the instant of gradient catastrophe for the problem
ut + uux = 0, x ∈ R, t > 0,
u(x, 0) = −tanh( x ), x ∈ R
solution
consider dx dt
a = b = u
du
aut + bux = 0whena = 1, b = u, f = 0also dx dt du

u = 1 = 0 then
dt du
1 = 0 , 0 = 1du,both side integral then u = c1 to choose the first and
second ratio.

ratio dx dt dx
u = 1 , dx = udt, u = dt , u = c1 and
forx = c1 t + c2 , x = ut + c2 thusx − ut = c2 ,combine to
fromu = F (x − ut)for arbitrary differential F,(c1 = F (c2 ))
Applying are initiation condition we obtainu(x, 0) = −tanh( x ),u(x, 0) =
c1 (x − u(x, 0), 0) = −tanh( x ),c1 (x) = −tanh( x )
thusu = x − ut the solution of the problem in the implicit form
isu(x, t) = −tanh( 1 (x − u(x, t)t)).
for the function u0 (x) = −tanh( x )
1
u00 (x) = − cosh 2x

which has a minimum at x = 0
1
min u00 x = −max cosh 2x

1
= - mincosh 2x

=− 10
because x
(e +e − x )2
cosh2 x = 2 ≥1

Weak Solutions. Riemann Problem
Consider the Cauchy Problem for quasilinear equation

ut + uux = 0, x ∈ R, t > 0 ..................(5.28)
u(x, 0) = u0 (x) ,x ∈ R
Which is a limit case of Burgers’ equation as ε → 0. If u0 (x) is none
smooth we introduce,as in section 5.2,a notation of weak solution.
Definition
Definition 5.6. Assume u0 (x) ∈ L1loc (R). A function u(x, t) ∈
L2loc (R)x[0, ∞)) is a weak solution of (5.28) . iff
R∞R∞ u2
R∞
0 − ∞ (uρt + 2 ρx )dxdt + −∞ u0 (x)ρ(x, 0)dx = 0 ................. (5.29)
For every test function ρc01 (Rx[0, ∞))

Proposition 5.2 Let u C 1 (Rx[0, ∞)) be a smooth solution of the
equation ut + uux = 0 and a weak solution of the problem (5.28). If u0 (x)
is continuous at a point x0 ,then u(x0 , 0) = u0 (x0 ).
1
R ∞ Let ρ(x, t)C0 (Rx[0, ∞))
Proof
−∞ (u(x, 0) − u0 (x))ρ(x, 0)dx = 0
Suppose u(x0 , 0) > u0 (x0 ). By continuity there exists a neighborhood U
such that u(x, 0) > u0 (x), xU.
Take ρ(x, t) ∈ C01 (Rx[0, ∞)) such that supp
ρ(x, 0) = [a, b] ⊂ U, ρ(x) > 0, x(a, b)
ρ(a) = ρ(b) = 0, then
R∞ Rb
−∞ (u(x, 0) − u0 (x))ρ(x, 0)dx = a (U(x, 0) − u0 (x))ρ(x, 0) > 0
Which is a contradiction then
U(x0 , 0) = u0 (x0 ).

The problem (5.28) with discontinuous initial data is known as A Riemann
problem.
Let as consider the initial data


ul x < 0,
u0 (x) =
ur x > 0,
where ul and ur are constant
The two cases ul > ur and ul < ur are quite different with respect to the
solvability of the problem.It can be proved that if ul > ur ,then the weak
solution is unique,while iful < ur there exists infinitely many solutions.
case I ul > ur
consider the problem
ut + uux = εuxx, x ∈ R, t > 0
u(x, 0) = u0 (x), x ∈ R.
If ul > ur we are in a situation to apply theorem.
Letx > 0 be fixed and
ul + ur
s=
2
The instant T of theorem is determined by the slope k of the straight line
through the point (x, 0) and (0, s)
−1
Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale −s
as ε → 0
The unique solution of the above formula is known as a shock wave ,while
s = ul +u
2
T
is a shock speed , the speed at which the discontinuity of the
solution travels.
proportion the function

ul x < st,
u(x, t) =
ur x > st,
is a weak solution of the problem with initial data where

u − l > uT ,s = ul +u
2
T

Proof Let ρ(x, t)C01 RX [0, ∞)). Denote for simplicity
R∞R∞ 2
A := 0 −∞ (Uρt + u2 ρx )dxdt
R∞
B := − −∞ u(x, 0)ρ(x, 0)dx. We have
R ∞ R st u2 R∞R∞ 2
A = 0 −∞ (ρt ul + ρx 2l )dxdt + 0 st (ρt ur + ρx u2r )dxdt
R ∞ R st u2
A1 := 0 −∞ (ρt ul + ρx 2l )dxdt
R ∞ R st u 2 R ∞ R st
= ul 0 ( −∞ ρt )dt + 2l 0 ( −∞ ρx dx)dt
By
R st R st
d
−∞ ρt (x, t)dx = dt −∞ ρ(x, t)dx − ρ(st, t)s
and
R st
−∞ ρx (x, t)dx = ρ(s, t) − ρ(−∞, t) = ρ(st, t)
It follows
R ∞ R st R ∞ d R st
0 −∞ ρt (x, t)dxdt = 0 ( dt ( −∞ ρ(x, t)dx) − ρ(s, t)s)dt
R ∞ d R st R∞ R0
0 R ( dt −∞ ρ(x, t)dx)dt = −∞ ρ(x, ∞)dx − −∞ ρ(x, 0)dx =
0
− −∞ ρ(x, 0)dx
R ∞ R st R0 R∞
Then 0 −∞ ρt (x, t)dxdt = − −∞ ρ(x, 0)dx − s 0 ρ(st, t)dt
R0 R∞ u2 R ∞
A1 = −ul ( −∞ ρ(x, 0)dx + s 0 ρ(st, t)dt + 2l 0 ρ(st, t)dt
Similarly
R∞R∞ 2
A2 := 0 st (ρt ur + ρx u2r )dxdt
R∞ R∞ 2 R∞R∞
= ur ( 0 ( st ρt dx)dt) + u2r ( 0 st ρx dxdt)
R∞ R∞ 2 R∞
= −ur ( 0 ρ(x, 0)dx − s 0 ρ(st, t)dt − u2r 0 ρ(st, t)dt
Because
R∞ R∞
d
st
R∞ ρ t (x, t)dx = dt st ρ(x, t)dx + sρ(st, t)
st ρx (x, t)dx = −ρ(st, t).
Then
A = A1 + A2
R0 R∞ u2 R ∞
= −ul ( −∞ ρ(x, 0)dx + s 0 ρ(st, t)dt) + 2l 0 ρ(st, t)dt
R∞ R∞ 2 R∞
−ur ( 0 ρ(x, 0)dx − s 0 ρ(st, t)dt) − u2r 0 ρ(st, t)dt

One theR other hand
∞
B = − −∞ u(x, 0)ρ(x, 0)dx
R0 R∞
= − −∞ u(x, 0)ρ(x, 0)dx − 0 u(x, 0)ρ(x, 0)dx
R0 R∞
= −ul −∞ ρ(x, 0)dx − ur 0 ρ(x, 0)
Finally
u 2 −u 2 R ∞
A = B + (s(ur − ul ) + l 2 r ) 0 ρ(x, t)dt
And Since
u 2 −u 2 ul2 −ur2
s(ur − ul ) + l 2 r = ur +u
2 (ur − ul ) +
l
2 =0
We obtain A=B
Case2
ul < ur
In this case there exist more than one weak solutions

Chapter:The Laplace Equation
Harmonic Function
INJIBARA UNIVERSITY
June 21, 2023
Group3Alemayehu Simneh ID 265/14Belay Fentahun

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The Laplace Equation

Green’s Identity
Green’s Function
Green’s Functions for a Half-space and Sphere
Harnack’s Identities and Theorem

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The Laplace equation or potential equation is
∆u = 0,
where ∆u is the Laplace equation of the function u
∆u = ∇2 u = uxx + uyy in two dimensions,

∆u = ∇2 u = uxx + uyy + uzz in three dimensions
A function u ∈ c 2 (Ω) which satisfies the Laplace equation is called a

harmonic function. The in-homogeneous laplace equation
∆u = f
where f is a given function is known as the Poisson equation.

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The Laplace equation is very important in applications. It appears in
physical phenomena such as
1 steady-state heat condition in a homogeneous body with constant
heat capacity and constant conductivity.

2 steady-state in-compressible fluid flow.
3 Electrical potential of a stationary electrical fluid in a region without
charge.
The basic mathematical problem is to solve the Laplace or Poisson
equation in a given domain possibly with a condition on its boundary
∂Ω = Ω̄ − Ω.Let ψ and ϕ be continuous function on ∂Ω. the problem of
finding u ∈ C 2 (Ω) ∩ C (Ω̄) such that
(
∆u = 0, in Ω
(DL) :
u = ϕ on ∂Ω
is called the Dirichlet or first boundary value problem (BVP) for the
Laplace equation.
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Historically, the name boundary value problem was attributed to only

problems for which the PDE was of the elliptic type.
The Neumann or second (BVP) is
(
∆u = 0, in Ω
(NL) : ∂u
∂n = ψ on ∂Ω
∂u
where n~ denotes the outward unit normal to ∂Ω and ∂n = Ou.~
n is the
normal derivative.
The Robin or third BVP is
(
∆u = 0, in Ω
(RL) :
δu + ∂u
∂n = ψ on ∂Ω
where δ is continuous function on ∂Ω.

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A boundary value problem for the Laplace equation is well posed in sense
of Hadamard with respect to a class of boundary data if
1 A solution of the problem exists;
2 the solution is unique;
3 small variations of the boundary data yield small variation on the
corresponding solutions.
Example 1.
Show thatT (x, y ) = e −y sinx; 0 < x < π, y > 0 is Harmonic function.
solution: from the given function,
T (x, y ) = e −y sinx,
Tx = e −y cosx; Txx = −e −y sinx
Ty = −e −y sinx; Tyy = e −y sinx
then by harmonic function
Txx + Tyy = −e −y sinx + e −y sinx = 0
Thus T is Harmonic function
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Example 2.
f (x, y ) = 3x 2 y − y 3 + y
Show this is Harmonic function. solution
f (x) = 6xy , fxx = 6y

fy = 3x 2 − 3y 2 fyy = −6y
Therefor
fxx + fyy = 6y − 6y = 0
Thus f is Harmonic.

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Maximum and minimum principle
Suppose that u(x, y ) is harmonic in a bounded domain D and contain in
D̄ = DUB. Then prove that U attains its maximum as well as minimum
on the boundary B of D proof
Let the maximum of U on B be M.
Let us now suppose that the maximum of U on D̄ is not attained at any
point on B. Then it must be attained at some point p(x0 , y0 ) in D.
If M0 = p(x0 , y0 ), then M0 > M
M0 −M
Consider D(x, y ) = U(x, y ) + 4R 2
[(x − x0 )2 + (y − y0 )2 ]
Where (x, y ) ∈ D and R is radius of the circle with center x0 , y0
containing D.We know that,
D(x0 , y0 ) = U(x0 , y0 ) = M0
on the boundary B we have
|M0 −M|
|U(x, y )| ≤ |U(x, y )| + 4R 2
|(x − x0 )2 + (y − y0 )2 |
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|M0 −M|
< |U(x, y )| + 4R 2
.R 2
|M0 −M| M0 M M0 3M
≤M+ 4 =M+ 4 − 4 = 4 + 4
M0 3M0
< 4 + 4 = M0 .
Thus D(x, y ), attains its maximum at point in D like U(x, y ) has.
Therefor
Dxx ≤ 0 and Dyy ≤ 0 at a point in D
⇒ Dxx + Dyy ≤ 0 at some point D
M0 −M M0 −M
But, Dxx = Uxx + 4R 2
.2 = uxx + 2R 2
M0 −M M0 −M
Dyy = Uyy + 4R 2
.2 = uyy + 2R 2
M0 −M M0 −M
Dxx + Dyy = uxx + uyy + 2R 2
+ 2R 2

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M0 −M
= uxx + uyy + R2
= uxx + uyy = 0, because of Harmonic

M0 −M
So, = uxx + uyy + R2
M0 −M
=0+ R2
>0 while contradiction
Thus U must attains maximum value on U.

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Green’s Identity
Let u, v ∈ C 2 (Ω), Ω be a domain with smooth boundary ∂Ω, n̄ be the
outward unit normal vector to ∂Ω. Recall the following notations of field
theory
gradu = Ou = (ux , uy , uz ),
div F~ = O.F~ = fx + gy + hz ,
rot F~ = Ox F~ = hy − gz , fz − hx , gx − fy ,
M u = div (Ou) = O2 u = uxx + uyy + uzz ,
We have the divergence theorem

Divergence Theorem states that the total divergence of a vector field in a
solid region V equals the total flow across the boundary surface S.
V must be a solid region bounded by S oriented by a unit external vector
directed out ward .The partial derivatives must be continuous on V
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R R RR R
S Fnds = V divFdv
RR R ∂f ∂g ∂h
= V ( ∂x + ∂y + ∂z )dv
Example.
Determine the flux of vector field F~ =(x 2 , y 2 , z 3 ) across the surface
x = 0, x = 2; y = 0, y = 2; z = 0, z = 4
RR RR R R R R ∂f ∂g ∂h
S F .nds = V divFdV = V ( ∂x + ∂y + ∂z )dxdydz
(2x + 2y + 3z 2 )dxdydz
RR R
V
R4R2
0 0 (x 2 + 2xy + 3xz 2 )|20 dydz
R4R2 R4
0 0 (4 + 4y + 6z 2 )dydz = 04y + 2y 2 + 6yz 2 |20 dz = 8 + 8 + 12Z 2 =
16 + 12Z 2
R4
0 (16 + 12Z 2 )dz = 16z + 4z 2 |40 = 64 + 256 = 320

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How to drive green’s first and second identity
where F~ (f , g , h) is a vector field.Denoted dV = dxdydz, dSp a surface
element at p on ∂Ω.
The divergence theorem or the Gauss-Ostrogradskii formula
~ ~ ndSp
RR R RR
Ω div F dV = ∂Ω F .~
If F = Ou, we have
∂u
RR R RR
Ω M udV = ∂Ω ∂n dSp known as a Gauss formula
By the product rule
(vux )x = vx ux + vuxx , it follows
div (v Ou) = Ov .Ou + v M u and
by divergent theorem,
∂u
RR R RR R RR
Ω v M udV + Ω Ov .Oudv = ∂Ω v ∂n dSp,
Known as Green’s first identity.

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Green’s Identity
Changing the role of v and u we have

∂v
RR R RR R RR
Ω u M vdV + Ω Ou.Ovdv = ∂Ω u ∂n dSp,
Subtracting Green’s first identity from the above we obtain

∂v ∂u
RR R RR
Ω (u M v − v M u)dV = ∂Ω (u ∂n − v ∂n )dSp,
Known as Green’s second identity.

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Mean Value property
Theorem
The average value of any Harmonic function over any sphere is equal to its
value at the center. proof. Let u(p)be a harmonic function on B, where
B = Ba (P0 ) = {p ∈ R 3 : |P − P0 | ≤ a}
S = Sa (P0 ) = {p ∈ R 3 : |P − P0 | = a}
by green’s first identity it follows

∂u
RR R RR
0= B M udV = S ∂n dSp
For the sphere S the unit normal vector at P ∈ S is

P−P0 0 y −y0 z−z0
n~= a = ( x−x
a , a , a )

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Mean Value property
Let us make the change of variables


x = x0 + ρcoseθsinψ,

y = y0 + ρsinθsinψ,

z = z0 + ρcosψ

Then for
u(ρ, θ, ψ) = u(x0 + ρcoseθsinψ, y0 + ρsinθsinψ, z0 + ρcosψ)
we have
y −y0
∂u
= x−x
∂n |S a ux + a uy + a uz
0 z−z0
= cosθsinψux + sinθ sin ψuy + cosψuz

∂
= ∂ρ u(ρ, θ, ψ)|ρ=a

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Mean Value property
Therefore ,. . . becomes
RR ∂
0 = s ∂p u(ρ, θ, ψ)|p=a dSp
R 2π R π
= 0 0 up (ρ, θ, ψ)|ρ=a a2 sinψdψdθ
and as a > 0
R 2π R π
= 0 0 up (ρ, θ, ψ)|ρ=a sinψdψdθ
the last identity is valued for every a > 0. so that we can consider a as a
variable r and we have
d
R 2π R π
dr ( 0 0 u(r , θ, ψ)sinψdψdθ) = 0
then
R 2π R π
I (r ) = ( 0 0 u(r , θ, ψ)sinψdψdθ)
is independent of r.
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Mean Value property
Letting r → 0, we get
R 2π R π
lim I (r ) = 0 0 u(0, θ, ψ)sinψdψdθ
r →0
R 2π R π
= 0 0 u(P0 )sinψdψdθ
= 4πu(P0 )
Then it follows
R 2π R π
4πu(P0 ) = 0 0 u(a, θ, ψ)sinψdψdθ,
2π R π
4πa2 u(P0 ) = 0 RR0 u(a, θ, ψ)a2 sinψdψdθ,
R
= s u(P)dSp
or
1
RR
u(p0 ) = 4πa2 s u(P)dSp.

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Dirichlet principle
Theorem:
Let Ω ⊂ R 3 be a domain with boundary Γ = ∂Ω.
2
T
Among all functions v (P) ∈ C C C (Ω̃) that satisfy the Dirichlet
boundary condition
v (P) = ψ(P) on Γ
where ψ(P) ∈ C (Γ), the lowest energy
E (v ) = 12 2
RR R
Ω |Ov | dV ,
is attained by a harmonic function satisfying v (P) = ψ(P) on Γ Proof: we

prove that if u is the unique harmonic function, such that u(P) = ψ(P) on
Γ, then for every v ∈ C 2 (Ω) C (Ω̃) with v (P) = ψ(P) on Γ, we have
T
E (v ) ≥ E (u).
We can represent v = u − w ,where w (P) = 0 on Γ

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Dirichlet principle
By the Green’s Identity
E (v ) = E (u − w ) = 12 (|Ou|2 − 2OuOw + |Ow |2 )dV

RRR
R R RΩ RR ∂u
= E (u) + E (w ) + Ω w M udV − Γ w ∂n dSp
= E (u) + E (w ) ≥ E (u)
⇒ E (v ) ≥ E (u)
Which completes the proof.

CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
21, 2023
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20 Arega
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Representation Formula
Theorem:6.5
Let u ∈ C 2 (Ω) ∩ C 1 (Ω̄) be such that ∆u ∈ L1 (Ω).Then for every
P ∈ Ω, if N ∈= 3
1
RR 1 ∂u(Q) ∂ 1 1
RRR ∆u(Q)
u(P) = 4π ∂Ω ( |Q−P| ∂n − u(Q) ∂n ( |Q−P| ))dSQ − 4π Ω |Q−P| dVQ
and if N = 2
1
H ∂ ∂u(Q)
u(P) = 2π (u(Q) ∂n ln|Q − P| − ln|Q − P| ∂n )dsQ
1
RR
+ 2π Ω ln|Q − P|∆u(Q)dxdy
proof
consider the three dimensional case.Fix u ∈ Ω and let ε be sufficiently
small such that Bε (P) ⊂ Ω.Let us apply the green’s second identity in
1
ΩwithoutBε (P)for the function u(Q) and v (Q) = |Q−P| which is harmonic
for Q 6= P.Denote for simplicity
Ωε = Ω without Bε (P), Bε = Bε (P), Sε = ∂Bε (P)
On Sε ,
CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
21, 2023
Habtamu
21 Arega
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∂ ∂
∂n = − ∂r r = |Q − P|.........(6.15)
it follows by the Green’s second identity that
RRR ∆uQ
− Ωε |Q−P| dVQ
∂ 1 1 ∂u
RR
= ∂Ω (u ∂n ( |Q−P| ) − |Q−P| ∂n )dSQ
∂ 1 1 ∂u
+(u ∂n ( |Q−P| )− |Q−P| ∂n )dSQ..........(6.16)
BY(6.15)
∂ 1 1 ∂u
RR
Aε =
Sε (u (
∂n |Q−P| ) − |Q−P| ∂n )dSQ................(6.17)
= Sε (u( −∂ ( 1 )) + 1∂u )ds
RR
RR ∂r r r1=RR ∂r
1
= 2 Sε uds + Sε ∂u ∂r ds
= 4π |Sε| Sε uds + 4π |s | Sε ∂u
1 1
RR RR
∂r ds
1
RR ∂u
= 4πM (u) + 4πM |s | Sε ∂r ,
whereM (u) denotes the mean value of u over s and |s | the area of s .
CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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22 Arega
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As u ∈ c 1 (Ω̄),letting
R
→0 ,we have
lim→0 A = 4πu(P)
Then, by (6.16),we obtain as → 0

RRR ∆(Q) RR ∂ 1 1 ∂u
Ω |Q−P| dVQ = ∂Ω (u ∂n ( |Q−P| ) − |Q−P| ∂n )dSQ + 4πu(P),
Or
1
RR 1 ∂u ∂ 1 1
RRR ∆u(Q)dVQ
u(P) = 4π ∂Ω ( |Q−P| ∂n − u ∂n ( |Q−P| ))dSQ − 4π Ω |Q−P|
motivated by the representation formula we set
(
1
if N = 3
F (Q, P) = 4π|Q−P| 1
− 2πln|Q−P| if N = 2
The function F (Q, P) is called a fundamental solution of the Laplace with
pole at P.
CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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Green’s Function
Definition
Green’s Functions in PDEs are used to solve homogeneous PDEs as well as
non-homogeneous of PDEs.
Now we use Green’s identity to study Dirichlet problem.
Consider the problem of finding a function
Φ(Q, P) ∈ C 2 (Ω) ∩ C 1 (Ω̄) such that
(
∆Q Φ(Q, P) = 0, Q ∈ Ω,
Φ(Q, P) = F (Q, P), Q ∈ ∂Ω wherep ∈ Ω is fixed and
(
1
4π|Q−P| if N = 3,
F(Q,P)= 1
− 2π ln|Q − P| if N =2
Suppose that the above equation has a solution and u ∈ C 2 (Ω) ∩ C 1 (Ω̄) is
Harmonic function.
CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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Green’s Function
By the Green’s second identity we have
RR ∂u(Q) ∂Φ
∂Ω (Φ(Q, P) ∂nQ − u(Q) ∂nQ (Q, P))dSQ = 0.
Consider the Dirichlet problem of finding a function

u(P) ∈ C 2 (Ω) ∩ C 1 (Ω̄) such that
(
∆u(P) = 0, p ∈ Ω,
u(P) = ϕ(P), p ∈ ∂Ω,
where ϕ(P) ∈ C (∂Ω). By the representation formula
RR ∂u(Q) ∂F (Q,P)
u(P) = ∂Ω (F (Q, P) ∂nQ − u(Q) ∂nQ )dSQ
When we subtracting these two equation like as,

∂u(Q) ∂F (Q,P)
( RR
u(P) = ∂Ω (F (Q, P) ∂nQ − u(Q) ∂nQ )dSQ
- RR ∂u(Q) ∂Φ
0= ∂Ω (Φ(Q, P) ∂nQ − u(Q) ∂nQ (Q, P))dSQ .

CHAPTER
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Zeleke ID
Equation
ID 273/14
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Green’s Function Half Space
we obtained
∂
RR
u(P) = − ∂Ω (Φ(Q) ∂nQ G (Q, P))dSQ ,
where G (Q, P) = F (Q, P) − Φ(Q, P) is known as Green’s Function for the

Laplacian inΩ
Green’s Function Half Space
LetD ⊂ R 3 be the half space of points P(x, y , z), z > 0. Each point
P(x, y , z) ∈ D has a reflected pointP ∗ (x, y , −z) ∈
/ D.
Suppose Q(ξ, η, 0) ∈ ∂D = {z = 0}. By symmetry of P and P ∗ with
respect to ∂D
|Q − P| = |Q − P ∗ |.
as D is an infinite region, all the properties on Green’s function are still

valid if we impose the so-called boundary condition at infinity, that is the
function and its derivatives are tending to 0 as |P| → ∞.
CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
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Sphere
Let BR be the open ball of radius R centered at the origin O,SR = ∂BR .
Let P ∈ BR and P ∗ be the inverse point of p with respect to SR defined as
~∗= R2 ~
OP |P|∗ OP.
which implies
|P||P ∗ | = R 2 .
If Q ∈ SR by the above it follows that the two triangles QOP ∗ and POQ
are similar, because they have a common angle
∠QOP = ∠QOP ∗ , |Q| = R and
|P| |Q|
|Q| = |P ∗ | .
Then it follows
|Q−P| |P|
|Q−P ∗ | = R .

CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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Sphere
Or
1 R |P ∗ |
|Q−P| = |P||Q−P ∗ | = R|Q−P ∗ |
If Q ∈ SR , P ∈ BR without {0} the solution of the above is given by

(
R
∗ | if P 6= 0,
Φ(Q, P) = 4π|P||Q−P
1
4πR if P = 0.
The functionΦ(Q, P) is harmonic with respect to P in BR without {P}

1 1
because |Q−P| and |Q−P ∗ | are harmonic.By the above it follows that the
1
boundary condition Φ(Q, P) = 4π|Q−P| , Q ∈ SR is satisfied.
In the two dimensional case we have
(
1
− 2π ln|Q − P ∗ | |P|
R if P 6= 0,
Φ(Q, P) = 1
− 2π lnR if P = 0.

CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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Sphere
Thus the Green’s function for the sphere SR in R 3 is
n
1 1 R
G (Q, P) = 4π ( Q−P − |P||Q−P ∗| )
In order to solve the Dirichlet problem according to the formula,let us

compute the normal derivative ∂n∂Q G (Q, P), which for the sphere SR is
∂
∂|Q| G (Q, P)|Q∈SR
Applying the cosine theorem to the triangles OQP and OQP ∗ we have
|Q − P|2 = |P|2 + |Q|2 − 2|P||Q|cosγ
2
|Q − P ∗ |2 = |PR∗ |2 + |Q|2 − 2 R|P|Q|
4
∗ | cosγ
where γ = ∠QOP. Therefore Q ∈ SR , γ fixed.

∂|Q−P| |Q|−|P|cosγ
∂|Q| = |Q−P|
2
∂|Q−P ∗ | |Q|− |PR∗ | cosγ
∂|Q| = |Q−P ∗ |

CHAPTER
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SIX The Laplace
Zeleke ID
Equation
ID 273/14
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Sphere
− ∂n∂Q G (Q, P)|Q=R = − ∂|Q|

∂
G (Q, P)|Q=R
R 2
1 |Q|−|P|cosγ R(|Q|− |P| cosγ)
= 4π ( |Q−P|3 − |P||Q−P|3
)
1 |Q|−|P|cosγ P(|Q||P|−R 2 cosγ)

= 4π ( |Q−P|3 − R 2 |Q−P|3
)
1 R 3 −R|P|2
4π R 2 |Q−P|3
1 R 2 −|P|2
4π R|Q−P|3

CHAPTER
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SIX The Laplace
Zeleke ID
Equation
ID 273/14
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Harnack’s Inequality and Theorems
Harmonic Function :- A function u(x, y ) is called Harmonic if it is twice

contineously differentiable and satisfies the following partial differential
equation.
i.e
uxx + uyy = 0
,so it is called Laplace equation.
Harnack’s Inequality :-is an inequality relating the value of a positive
harmonic function at two points ,introduced by A.Harnack.
Theorems:-let u be a nonnegative harmonic function in Ω .Then for
every PO ∈ Ω, BR (PO ) ⊂ Ω and every ρ ∈ (0, R), P ∈ Sρ (PO )
R(R − ρ) R(R + ρ)
u(PO ) ≤ u(P) ≤ u(PO )
(R + ρ)2 (R − ρ)2

CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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Proof
using a translation of the argument, poission formula and the mean value
property. let PO = O
R 2 − |P|2
Z Z
u(Q)
u(P) = 3
dSQ
4πR SR |Q − P|
R 2 − |P|2
Z Z
u(Q)
≤ 3
dSQ
4πR SR (|Q| − |P|)
,By absolute value property
R 2 − |P|2
Z Z
= u(Q)dSQ
4πR(R − P)3 SR
But |Q| = R and P = ρ

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Zeleke ID
Equation
ID 273/14
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R 2 − ρ2
= 4πR 2 u(O)
4πR(R − ρ)3
R(R + ρ)
= u(O).
(R − ρ)2
By the same way,
R 2 − |P|2
Z Z
u(Q)
u(P) = 3
dSQ
4πR SR |Q − P|
R 2 − |P|2
Z Z
u(Q)
≥ dSQ
4πR SR (|Q| + |P|)3
,By absolute value property
R 2 − |P|2
Z Z
= u(Q)dSQ
4πR(R + P)3 SR
But |Q| =
Group3Alemayehu R ID
Simneh and P = ρFentahun
265/14Belay CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
21, 2023
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33 Arega
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R 2 − ρ2
= 4πR 2 u(O)
4πR(R + ρ)3
R(R − ρ)
= u(O).
(R + ρ)2
R(R − ρ) R(R + ρ)
⇔ 2
u(PO ) ≤ u(P) ≤ u(PO )
(R + ρ) (R − ρ)2
But PO = O
Theorem(Harnack’s first theorem):- Let un (P) be a sequence of
harmonic function in a domain Ω ,uniformly convergent on every compact
K ⊂ Ω .Then the limit function u(P) is harmonic in Ω.

CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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34 Arega
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Proof
It is clear that the limit function u(P) is a continuous function .Let B be
an open ball,S = ∂B and P ∈ B.passing to the limit in
Z Z
un (P) = un (Q)Π(Q, P)dSQ
S
,
R 2 −|P|2
But Π(Q, P) = 4πR|Q−P|3
,Then it is harmonic.that means ∆PΠ(Q, P) = 0
We obtain ,
Z Z
u(P) = u(Q)Π(Q, P)dSQ
S
As Π(Q, P) is a harmonic function in B ,By
Z Z
∆Pu(Q, P) = u(Q)∆PΠ(Q, P)dSQ = 0
S
Hence:- it follows that u(P) is harmonic in P. As B is arbitrary, u(P) is

harmonic in Ω.
CHAPTER
ID 269/14Belay
SIX The Laplace
Zeleke ID
Equation
ID 273/14
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35 Arega
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INJIBARA UNIVERSITY (INU)

College of Natural and Computational Science

GROUP TWO (INU) June 18, 2023 1 / 29

3.1 The Wave equation on the whole Line.D’Alembert Formula

GROUP TWO (INU) June 18, 2023 2 / 29

utt − c 2 uxx = 0,———(1)

where x signifies the spatial variable or ”position”, t the ”time”

GROUP TWO (INU) June 18, 2023 5 / 29

GROUP TWO (INU) June 18, 2023 6 / 29

Solution:The solution of the problem is by using the general solution

GROUP TWO (INU) June 18, 2023 7 / 29

GROUP TWO (INU) June 18, 2023 8 / 29

GROUP TWO (INU) June 18, 2023 9 / 29

Therefore by (9)ux (ct + 0, t) − ux (ct − 0, t) = − c1 ψ(0) = 0.

GROUP TWO (INU) June 18, 2023 11 / 29

GROUP TWO (INU) June 18, 2023 12 / 29

GROUP TWO (INU) June 18, 2023 13 / 29

GROUP TWO (INU) June 18, 2023 14 / 29

GROUP TWO (INU) June 18, 2023 15 / 29

They divide the domain R = {(x, t) : 0 < x < ι, t > 0} into

GROUP TWO (INU) June 18, 2023 16 / 29

GROUP TWO (INU) June 18, 2023 17 / 29

GROUP TWO (INU) June 18, 2023 18 / 29

GROUP TWO (INU) June 18, 2023 19 / 29

Let us consider (12). Making change of variables

we transform (12) into

GROUP TWO (INU) June 18, 2023 20 / 29

Integrating (13) with respect to η we have

GROUP TWO (INU) June 18, 2023 21 / 29

GROUP TWO (INU) June 18, 2023 22 / 29

GROUP TWO (INU) June 18, 2023 23 / 29

GROUP TWO (INU) June 18, 2023 25 / 29

GROUP TWO (INU) June 18, 2023 27 / 29

is convergent for every y ∈ [a, b]. R∞

Solve the problem

Let us consider the diffusion equation on the half-line (0, ∞) and

Example 1:solve (4.23) with e −x and k = 1.

The problem of finding the a function u(x,t) in the inhomogeneous

By linear properties of operator Lu ≡ ut − kuxx , the solution of (1)

υ(x, 0) = ϕ(x) xR,

Dimensional Diffusion Equation and

L(G (x, t)) = 0, (x, t)R × (0, ∞), (5)

it follows that the solution of (3) is υ(x, t) =

Let f (x, t)C (R × (0, ∞)) be a bounded function and

is asolution of the problem (3)

≤ tsup|f (x, t)|

ω(0, t) = h(t)t > 0,

June 21, 2023

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

To define a weak (generalized) solutions of the problem by passing to the

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

| k Jεu(x)|2 dx ≤ |z|≤1 p0(z)dz( k u 2 (x − εz)dx)dz

|Jεu(x)|2 dx ≤ |z|≤1 p0(z)( k u 2 u 2 (x − εz)dx)dz

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

Ali Muhammed Gezahegn GatinetHabtamu AlemayehuTeshale

υ(x, 0) = ϕ(x) xR,

L(G (x, t)) = 0, (x, t)R × (0, ∞), (5)

Let f (x, t)C (R × (0, ∞)) be a bounded function and

For every test function ρc01 (Rx[0, ∞))