CUBO A Mathematical Journal
Vol.20, No¯ 02, (23–39). June 2018

http: // dx. doi. org/ 10. 4067/ S0719-06462018000200023

New approach to prove the existence of classical solutions
for a class of nonlinear parabolic equations

Svetlin G. Georgiev
1
and Khaled Zennir

2

1Sorbonne University, Paris, France,

University of Sofia, Faculty of Mathematics and Informatics,

Department of Differential Equations.
2Department Of Mathematics,

College Of Sciences and Arts, Al-Ras.

Qassim University,

Kingdom Of Saudi Arabia.

svetlingeorgiev1@gmail.com, sgg2000bg@yahoo.com, khaledzennir2@yahoo.com

ABSTRACT

In this article, we consider a class of nonlinear parabolic equations. We use an integral

representation combined with a sort of fixed point theorem to prove the existence of

classical solutions for the initial value problem (1.1), (1.2). We also obtain a result on

continuous dependence on the initial data. We propose a new approach for investigation

for existence of classical solutions of some classes nonlinear parabolic equations.

RESUMEN

En este art́ıculo, consideramos una clase de ecuaciones parabólicas nolineales. Usamos

una representación integral combinada con una especie de teorema de punto fijo para

probar la existencia de soluciones clásicas para el problema de valor inicial (1.1), (1.2).

También obtenemos un resultado sobre la dependencia continua de la data inicial.

Proponemos una estrategia nueva para la investigación de la existencia de soluciones

clásicas de algunas clases de ecuaciones parabólicas nolineales.

Keywords and Phrases: parabolic equation, existence, differentiability with respect to the initial

data

2010 AMS Mathematics Subject Classification: 35K55, 35K45.

http://dx.doi.org/10.4067/S0719-06462018000200023


24 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

1 Introduction

Here, we consider the Cauchy problem

ut − uxx = f(t, x, u, ux) in (0, ∞) × R, (1.1)

u(0, x) = φ(x) in R, (1.2)

where φ ∈ C2(R), f : [0, ∞) × R × R × R 7−→ C is a given continuous function, u : [0, ∞) × R 7−→ C

is the main unknown.

Our main results are as follows.

Theorem 1.1. Let f ∈ C([0, ∞) × R × R × R), φ ∈ C2(R). Then there exists m ∈ (0, 1) such that

the problem (1.1), (1.2) has a solution u ∈ C1([0, m], C2([0, 1])).

Theorem 1.2. Let f ∈ C([0, ∞) × R × R × R), φ ∈ C2(R). Then there exists m ∈ (0, 1) such that

the problem (1.1), (1.2) has a solution u ∈ C1([0, m], C2(R)).

For O1, O2 ⊂ R with C
1(O1, C

2(O2)) we denote the space of all continuous functions u on O1 ×O2
such that ut, ux and uxx exist and are continuous on O1 × O2.

Example 1.3. Let p > 1 and a ∈ C be chosen so that ap−1 = −
1

p − 1
. Consider the Cauchy

problem

ut − uxx = u
p in (0, ∞) × R

u(0, x) = a in R.

Then u(t, x) = a(t + 1)−
1

p−1 is its solution. Actually,

ut(t, x) = −
a

p − 1
(t + 1)

−
p

p−1 ,

and

uxx(t, x) = 0,

and

(u(t, x))p = −
a

p − 1
(t + 1)

−
p

p−1 .

Therefore

ut(t, x) − uxx(t, x) = (u(t, x))
p in (0, ∞) × R

and

u(0, x) = a in R.



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New approach to prove the existence of classical solutions . . . 25

To prove our main result we propose new integral representation of the solutions of the initial value

problem (1.1), (1.2). Many works have been devoted to the investigation of initial value problems

for parabolic equations and systems (see, for example, [13]-[16] and the references therein). We

note that in the references the IVP (1.1), (1.2) is connected with the dimension n, Fujita exponent,

Sobolev critical exponents, bounded and unbounded domain. In this article we propose new idea

which tell us that the local existence of classical solutions of the IVP is connected with the integral

representation of the solutions, it is not connected with the dimension n and if the domain is

bounded or not.

As an application of our new integral representation we deduce some results connected with the

continuous dependence on the initial data and parameters of the problem (1.1), (1.2).

Theorem 1.4. Let f ∈ C([0, ∞)×R×R×R), ∂f
∂u

, ∂f
∂ux

exist and are continuous in [0, ∞)×R×R×R,

φ ∈ C2(R). Let also, u(t, x, φ) ∈ C1([0, m], C2([c, d])) be a solution to the problem (1.1), (1.2) for

some m ∈ (0, 1) and for some [c, d] ⊂ R. Then u(t, x, φ) is differentiable with respect to φ and

v(t, x) = ∂u
∂φ

(t, x, φ) satisfies the following initial value problem

vt − vxx =
∂f
∂u

(t, x, u(t, x, φ), ux(t, x, φ))v

+ ∂f
∂ux

(t, x, u(t, x, φ), ux(t, x, φ))vx in [0, m] × [c, d],

(1.3)

v(0, x) = 1 in [c, d]. (1.4)

2 Auxiliary results

We will start with the following useful lemma.

Lemma 2.1. Let f ∈ C([a, b]×[c, d]×R×R), g ∈ C2([c, d]). Then the function u ∈ C1([a, b], C2([c, d]))

is a solution to the problem

ut − uxx = f(t, x, u, ux) in (a, b] × [c, d], (2.1)

u(a, x) = g(x) in [c, d], (2.2)

if and only if it is a solution to the integral equation

∫x
c

∫y
c
(u(t, z) − g(z)) dzdy −

∫t
a
(u(τ, x) − u(τ, c) − (x − c)ux(τ, c)) dτ

=
∫t
a

∫x
c

∫y
c
f(τ, z, u(τ, z), ux(τ, z))dzdydτ, x ∈ [c, d], t ∈ [a, b].

(2.3)

Proof. (1) Let u ∈ C1([a, b], C2([c, d])) is a solution to the problem (2.1), (2.2).

We integrate the equation (2.1) with respect to x and we get

∫x
c
ut(t, z)dz −

∫x
c
uxx(t, z)dz

=
∫x
c
f(t, z, u(t, z), ux(t, z))dz, x ∈ [c, d], t ∈ [a, b],



26 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

or ∫x
c
ut(t, z)dz − ux(t, x) + ux(t, c)

=
∫x
c
f(t, z, u(t, z), ux(t, z))dz, x ∈ [c, d], t ∈ [a, b].

Now we integrate the last equation with respect to x and we find

∫x
c

∫y
c
ut(t, z)dzdy −

∫x
c
(ux(t, z) − ux(t, c)) dz

=
∫x
c

∫y
c
f(t, z, u(t, z), ux(t, z))dzdy, x ∈ [c, d], t ∈ [a, b],

or ∫x
c

∫y
c
ut(t, z)dzdy − u(t, x) + u(t, c) + (x − c)ux(t, c)

=
∫x
c

∫y
c
f(t, z, u(t, z), ux(t, z))dzdy, x ∈ [c, d], t ∈ [a, b].

We integrate the last equality with respect to t and we obtain

∫t
a

∫x
c

∫y
c
ut(s, z)dzdyds −

∫t
a
(u(s, x) − u(s, c) − (x − c)ux(s, c)) ds

=
∫t
a

∫x
c

∫y
c
f(s, z, u(s, z), ux(s, z))dzdyds, x ∈ [c, d], t ∈ [a, b],

or ∫x
c

∫y
c
(u(t, z) − g(z)) dzdy −

∫t
a
(u(s, x) − u(s, c) − (x − c)ux(s, c)) ds

=
∫t
a

∫x
c

∫y
c
f(s, z, u(s, z), ux(s, z))dzdyds, x ∈ [c, d], t ∈ [a, b],

i.e., u satisfies the equation (2.3).

(2) Let u ∈ C1([a, b], C2([c, d])) be a solution to the integral equation (2.3).

We differentiate the equation (2.3) with respect to x and we get

∫x
c
(u(t, z) − g(z)) dz −

∫t
a
(ux(s, x) − ux(s, c)) ds

=
∫t
a

∫x
c
f(s, z, u(s, z), ux(s, z))dzds, x ∈ [c, d], t ∈ [a, b].

Again we differentiate with respect to x and we find

u(t, x) − g(x) −
∫t
a
uxx(s, x)ds

=
∫t
a
f(s, x, u(s, x), ux(s, x))ds, x ∈ [c, d], t ∈ [a, b].

(2.4)

Now we put t = a in the last equation and we find

u(a, x) = g(x), x ∈ [c, d],



CUBO
20, 2 (2018)

New approach to prove the existence of classical solutions . . . 27

i.e., the function u satisfies (2.2).

Now we differentiate the equation (2.4) with respect to t and we find

ut(t, x) − uxx(t, x) = f(t, x, u(t, x), ux(t, x)), x ∈ [c, d], t ∈ [a, b].

The proof of the existence results are based on the following theorem.

Theorem 2.2 ([14]). Let X be a nonempty closed convex subset of a Banach space Y. Suppose

that T and S map X into Y such that

(1) S is continuous and S(X) contained in a compact subset of Y.

(2) T : X 7−→ Y is expansive and onto.

Then there exists a point x∗ ∈ X such that

Sx∗ + Tx∗ = x∗.

Definition 2.3. Let (X, d) be a metric space and M be a subset of X. The mapping T : M 7−→ X

is said to be expansive if there exists a constant h > 1 such that

d(Tx, Ty) ≥ hd(x, y)

for any x, y ∈ M.

3 Proof of Theorem 1.1

Let B > ‖φ‖C2([0,1]) be arbitrarily chosen. Since φ ∈ C([0, 1]), f ∈ C([0, 1]×[0, 1]×[−B, B]×[−B, B])

we have that there exists a constant M11 > 0 such that

|φ(x)| ≤ M11 in [0, 1],

|f(t, x, y, z)| ≤ M11 in [0, 1] × [0, 1] × [−B, B] × [−B, B].

We take l, m ∈ (0, 1) so that

lB + l(B + M11) + 3lBm + lM11m ≤ B

l(5B + 2M11) ≤ B.

(3.1)

Let E11 = C
1([0, m], C2([0, 1])) be endowed with the norm

||u|| = max
{

max
(t,x)∈[0,m]×[0,1]

|u(t, x)|, max
(t,x)∈[0,m]×[0,1]

|ut(t, x)|,

max
(t,x)∈[0,m]×[0,1]

|ux(t, x)|, max
(t,x)∈[0,m]×[0,1]

|uxx(t, x)|
}
.



28 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

By K̃11 we denote the set of all equi-continuous families in E11, i.e., for every ǫ > 0 there exists

δ = δ(ǫ) > 0 such that

|u(t1, x1) − u(t2, x2)| < ǫ, |ut(t1, x1) − ut(t2, x2)| < ǫ,

|ux(t1, x1) − ux(t2, x2)| < ǫ, |uxx(t1, x1) − uxx(t2, x2)| < ǫ

whenever |t1 − t2| < δ, |x1 − x2| < δ. Let also,

K′11 = K̃11, K11 = {u ∈ K
′
11 : ||u|| ≤ B}

and

L11 = {u ∈ K
′
11 : ||u|| ≤ (1 + l)B}.

We note that K11 is a closed convex subset of L11.

For u ∈ L11 we define the operators

T11(u)(t, x) = (1 + l)u(t, x),

S11(u)(t, x) = −lu(t, x) + l

∫x

0

∫y

0

(u(t, z) − φ(z))dzdy

−l

∫t

0

(u(τ, x) − u(τ, 0) − xux(τ, 0))dτ

−l

∫t

0

∫x

0

∫y

0

f(τ, z, u(τ, z), ux(τ, z))dzdydτ.

We will prove that the problem

ut − uxx = f(t, x, ux) in [0, m] × [0, 1], (3.2)

u(0, x) = φ(x) in [0, 1], (3.3)

has a solution u ∈ C1([0, m], C2([0, 1])).

a)S11 : K11 7−→ K11. Let u ∈ K11. Then S11(u) ∈ C
1([0, m], C2([0, 1])) and for (t, x) ∈ [0, m] ×



CUBO
20, 2 (2018)

New approach to prove the existence of classical solutions . . . 29

[0, 1], using the first inequality of (3.1), we get

|S11(u)(t, x)| =
∣

∣

∣
−lu(t, x) + l

∫x

0

∫y

0

(u(t, z) − φ(z))dzdy

−l

∫t

0

(u(τ, x) − u(τ, 0) − xux(τ, 0))dτ

−l

∫t

0

∫x

0

∫y

0

f(τ, z, u(τ, z), ux(τ, z))dzdydτ
∣

∣

∣

≤ l|u(t, x)| + l

∫x

0

∫y

0

(|u(t, z)| + |φ(z)|) dzdy

+l

∫t

0

(|u(τ, x)| + |u(τ, 0)| + x|ux(τ, 0)|) dτ

+l

∫t

0

∫x

0

∫y

0

|f(τ, z, u(τ, z), ux(τ, z))|dzdydτ

≤ lB + l(B + M11) + 3lBm + lM11m

≤ B.

Note that

S11(u)t(t, x) = −lut(t, x) + l

∫x

0

∫y

0

ut(t, z)dzdy

−l(u(t, x) − u(t, 0) − xux(t, 0))

−l

∫x

0

∫y

0

f(t, z, u(t, z), ux(t, z))dzdy,

(t, x) ∈ [0, m] × [0, 1].



30 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

Then, using the second inequality of (3.1), we obtain

|S11(u)t(t, x)| =
∣

∣

∣
−lut(t, x) + l

∫x

0

∫y

0

ut(t, z)dzdy

−l(u(t, x) − u(t, 0) − xux(t, 0))

−l

∫x

0

∫y

0

f(t, z, u(t, z), ux(t, z))dzdy
∣

∣

∣

≤ l|ut(t, x)| + l

∫x

0

∫y

0

|ut(t, z)|dzdy

+l (|u(t, x)| + |u(t, 0)| + x|ux(t, 0)|)

+l

∫x

0

∫y

0

|f(t, z, u(t, z), ux(t, z))|dzdy

≤ lB + lB + 3lB + lM11

= l(5B + M11)

≤ B, (t, x) ∈ [0, m] × [0, 1].

Also,

S11(u)x(t, x) = −lux(t, x) + l

∫x

0

(u(t, z) − φ(z))dz

−l

∫t

0

(ux(τ, x) − ux(τ, 0))dτ

−l

∫t

0

∫x

0

f(τ, z, u(τ, z), ux(τ, z))dzdτ,

(t, x) ∈ [0, m] × [0, 1].



CUBO
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New approach to prove the existence of classical solutions . . . 31

Hence, using the first inequality of (3.1),

|S11(u)x(t, x)| =
∣

∣

∣
−lux(t, x) + l

∫x

0

(u(t, z) − φ(z))dz

−l

∫t

0

(ux(τ, x) − ux(τ, 0))dτ

−l

∫t

0

∫x

0

f(τ, z, u(τ, z), ux(τ, z))dzdτ
∣

∣

∣

≤ l|ux(t, x)| + l

∫x

0

(|u(t, z)| + |φ(z)|) dz

+l

∫t

0

(|ux(τ, x)| + |ux(τ, 0)|) dτ

+l

∫t

0

∫x

0

|f(τ, z, u(τ, z), ux(τ, z))|dzdτ

≤ lB + l(B + M11) + 2lBm + lM11m

≤ B, (t, x) ∈ [0, m] × [0, 1].

For (t, x) ∈ [0, m] × [0, 1] we have

S11(u)xx(t, x) = −luxx(t, x) + l(u(t, x) − φ(x))

−l

∫t

0

uxx(τ, x)dτ

−l

∫t

0

f(τ, x, u(τ, x), ux(τ, x))dτ,



32 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

from where, using the first inequality of (3.1),

|S11(u)xx(t, x)| =
∣

∣

∣
−luxx(t, x) + l(u(t, x) − φ(x))

−l

∫t

0

uxx(τ, x)dτ

−l

∫t

0

f(τ, x, u(τ, x), ux(τ, x))dτ
∣

∣

∣

≤ l|uxx(t, x)| + l (|u(t, x)| + |φ(x)|)

+l

∫t

0

|uxx(τ, x)|dτ

+l

∫t

0

|f(τ, x, u(τ, x), ux(τ, x))|dτ

≤ lB + l(B + M11) + lBm + lM11m

≤ B.

We note that {S11(u) : u ∈ K11} is an equi-continuous family in E11. Consequently S11 :

K11 7−→ K11. Also, S11(K11) ⊂ K11 ⊂ L11, i.e., S11(K11) resides in a compact subset of L11.

b) S11 : K11 7−→ K11 is a continuous operator. We note that if {un}
∞

n=1 be a sequence of ele-

ments of K11 such that un −→ u in K11 as n −→ ∞, then S11(un) −→ S11(u) in K11 as

n −→ ∞. Therefore S11 : K11 7−→ K11 is a continuous operator.

c) T11 : K11 7−→ L11 is an expansive operator and onto. For u, v ∈ K11 we have that

||T11(u) − T11(v)|| = (1 + l)||u − v||,

i.e., T11 : K11 7−→ L11 is an expansive operator with constant 1 + l.

Let v ∈ L11. Then
v

1+l
∈ K11 and

T11

(

v

1 + l

)

= v,

i.e., T11 : K11 7−→ L11 is onto.

From a), b), c) and from Theorem 2.2, it follows that there is u11 ∈ K11 such that

T11u11 + S11u11 = u11



CUBO
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New approach to prove the existence of classical solutions . . . 33

or

(1 + l)u11(t, x) − lu11(t, x) + l

∫x

0

∫y

0

(u11(t, z) − φ(z))dzdy

−l

∫t

0

(u11(τ, x) − u11(τ, 0) − xu11x(τ, 0))dτ

−l

∫t

0

∫x

0

∫y

0

f(τ, z, u11(τ, z), u11x(τ, z))dzdydτ

= u11(t, x),

or
∫x

0

∫y

0

(u11(t, z) − φ(z))dzdy −

∫t

0

(u11(τ, x) − u11(τ, 0) − xu11x(τ, 0))dτ

−

∫t

0

∫x

0

∫y

0

f(τ, z, u11(τ, z), u11x(τ, z))dzdydτ

= 0, (t, x) ∈ [0, m] × [0, 1],

whereupon, using Lemma 2.1, we conclude that u11 ∈ C
1([0, 1], C2([0, 1])) is a solution to the

problem (3.2), (3.3).

4 Proof of Theorem 1.2

Now we consider the problem

ut − uxx = f(t, x, u(t, x), ux(t, x)) in (0, m] × [1, 2], (4.1)

u(0, x) = φ(x) in [1, 2]. (4.2)

Let E12 = C
1([0, m], C2([1, 2])) be endowed with the norm

||u|| = max
{

max
(t,x)∈[0,m]×[1,2]

|u(t, x)|, max
(t,x)∈[0,m]×[1,2]

|ut(t, x)|,

max
(t,x)∈[0,m]×[1,2]

|ux(t, x)|, max
(t,x)∈[0,m]×[1,2]

|uxx(t, x)|
}
.

By K̃12 we denote the set of all equi-continuous families in E12.

Let K′12 = K̃12,

K12 = {u ∈ K
′
12 : ||u|| ≤ B}.



34 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

Since φ ∈ C([1, 2]), f ∈ C([0, m] × [1, 2] × [−B, B] × [−B, B]) we have that there exists a constant

M12 > 0 such that

|φ(x)| ≤ M12 in [1, 2],

|f(t, x, y, z)| ≤ M12 in [0, m] × [1, 2] × [−B, B] × [−B, B].

Let l1 > 0 be chosen so that

l1(5B + 2M12) ≤ B

l1B + l1(B + M12) + 3l1Bm + l1M12m ≤ B

Let also,

L12 = {u ∈ K
′
12 : ||u|| ≤ (1 + l1)B}.

We note that K12 is a closed convex subset of L12.

For u ∈ L12 we define the operators

T12(u)(t, x) = (1 + l1)u(t, x),

S12(u)(t, x) = −l1u(t, x) + l1

∫x

1

∫y

1

(u(t, z) − φ(z))dzdy

−l1

∫t

0

(u(τ, x) − u11(τ, 1) − (x − 1)u11x(τ, 1))dτ

−l1

∫t

0

∫x

1

∫y

1

f(τ, z, u(τ, z), ux(τ, z))dzdydτ.

As in the previous section one can prove that there is u12 ∈ C
1([0, 1], C2([1, 2])) which is a solution

to the problem (4.1), (4.2). This solution u12 satisfies the integral equation

∫x

1

∫y

1

(u12(t, z) − φ(z))dzdy

−
∫t
0
(u12(τ, x) − u11(τ, 1) − (x − 1)u11x(τ, 1))dτ

−
∫t
0

∫x
1

∫y
1
f(τ, z, u12(τ, z), u12x(τ, z))dzdydτ

= 0, (t, x) ∈ [0, m] × [1, 2].

(4.3)

Now we put x = 1 in (4.3) and we find

∫t

0

(u12(τ, 1) − u11(τ, 1))dτ = 0,



CUBO
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New approach to prove the existence of classical solutions . . . 35

which we differentiate with respect to t and we get

u12(t, 1) = u11(t, 1) in [0, m]. (4.4)

Now we differentiate (4.3) with respect to x and we find

∫x

1

(u12(t, z) − φ(z))dz −

∫t

0

(u12x(τ, x) − u11x(τ, 1))dτ

−

∫t

0

∫x

1

f(τ, z, u12(τ, z), u12x(τ, z))dzdτ = 0, (t, x) ∈ [0, m] × [1, 2].

In the last equation we put x = 1 and we become

∫t

0

(u12x(τ, x) − u11x(τ, 1))dτ = 0, (t, x) ∈ [0, m] × [1, 2],

which we differentiate with respect to t and we find

u12x(t, 1) = u11x(t, 1) in [0, m]. (4.5)

Now we differentiate (4.4) with respect to t and we get

u12t(t, 1) = u11t(t, 1) in [0, m].

Hence, (4.4), (4.5) and

f(t, 1, u11(t, 1), u11x(t, 1)) = f(t, 1, u12(t, 1), u12x(t, 1)),

we find

u12xx(t, 1) = u12t(t, 1) − f(t, 1, u12(t, 1), u12x(t, 1))

= u11t(t, 1) − f(t, 1, u11(t, 1), u11x(t, 1))

= u11xx(t, 1) in [0, m].

Consequently the function

u(t, x) =






u11(t, x) in [0, m] × [0, 1]

u12(t, x) in [0, m] × [1, 2],

is a C1([0, m], C2([0, 2]))-solution to the problem

ut − uxx = f(t, x, u(t, x), ux(t, x)) in (0, m] × [0, 2],

u(0, x) = φ(x) in [0, 2].



36 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

Then we consider the problem

ut − uxx = f(t, x, u(t, x), ux(t, x)) in (0, m] × [2, 3]

u(0, x) = φ(x) in [2, 3].

(4.6)

As in above there is u13 ∈ C
1([0, m], C2([2, 3])) which is a solution to the problem (4.6) and satisfies

the integral equation

∫x

2

∫y

2

(u13(t, z) − φ(z))dzdy

−

∫t

0

(u13(τ, x) − u12(τ, 2) − (x − 2)u12x(τ, 2))dτ

−

∫t

0

∫x

2

∫y

2

f(τ, z, u13(τ, z), u13x(τ, z))dzdydτ

= 0, t ∈ [0, m], x ∈ [2, 3].

The function

u(t, x) =






u11(t, x) in [0, m] × [0, 1]

u12(t, x) in [0, m] × [1, 2]

u13(t, x) in [0, m] × [2, 3]

is a C1([0, m], C2([0, 3]))-solution to the problem

ut − uxx = f(t, x, u(t, x), ux(t, x)) in [0, m] × [0, 3],

u(0, x) = φ(x) in [0, 3].

An so on. We construct a solution u1 ∈ C
1([0, m], C2(R)) which is a solution to the problem

ut − uxx = f(t, x, u(t, x), ux(t, x)) in (0, m] × R,

u(0, x) = φ(x) in R.



CUBO
20, 2 (2018)

New approach to prove the existence of classical solutions . . . 37

5 Proof of Theorem 1.4

We have that the solution u(t, x, φ) satisfies the following integral equation

Q(φ) =

∫x

c

∫y

c

(u(t, z, φ(z)) − φ(z))dzdy

−

∫t

0

(u(τ, x, φ(x)) − u(τ, c, φ(c)) − (x − c)ux(τ, c, φ(c)))dτ

−

∫t

0

∫x

c

∫y

c

f(τ, z, u(τ, z, φ(z)), ux(τ, z, φ(z)))dz

= 0, t ∈ [0, m], x ∈ [c, d].

Then

Q(φ) − Q(φ1) =

∫x

c

∫y

c

(u(t, z, φ(z)) − u(t, z, φ1(z)) − (φ(z) − φ1(z)))dzdy

−

∫t

0

(u(τ, x, φ(x)) − u(τ, x, φ1(x)))dτ

+

∫t

0

(u(τ, c, φ(c)) − u(τ, c, φ1(c)))dτ

+

∫t

0

(x − c)(ux(τ, c, φ(c)) − ux(τ, c, φ1(c)))dτ

−

∫t

0

∫x

c

∫y

c

(

f(τ, z, u(τ, z, φ(z)), ux(τ, z, φ(z)))

−f(τ, z, u(τ, z, φ1(z)), ux(τ, z, φ1(z)))
)

dzdydτ

=

∫x

c

∫y

c

(

∂u

∂φ
(t, z, φ(z)) − 1

)

dzdy

−

∫t

0

∂u

∂φ
(τ, x, φ(x))dτ +

∫t

0

∂u

∂φ
(τ, c, φ(c))dτ +

∫t

0

(x − c)

(

∂u

∂φ

)

x

(τ, c, φ(c))dτ

−

∫t

0

∫x

c

∫y

c

∂f

∂u
(τ, z, u(τ, z, φ(z)), ux(τ, z, φ(z)))

∂u

∂φ
(τ, z, φ(z))dzdydτ

−

∫t

0

∫x

c

∫y

c

∂f

∂ux
(τ, z, u(τ, z, φ(z)), ux(τ, z, φ(z)))

(

∂u

∂φ

)

x

(τ, z, φ(z))dzdydτ

+δ{φ, φ1},



38 Svetlin G. Georgiev and Khaled Zennir CUBO
20, 2 (2018)

where δ{φ, φ1} −→ 0 as φ(x) −→ φ1(x) for every x ∈ [c, d]. Hence, when φ(x) −→ φ1(x) for every

x ∈ [c, d], we get

0 =
∫x
c

∫y
c
(v(t, z) − 1)dzdy −

∫t
0
v(τ, x)dτ

+
∫t
0
v(τ, c)dτ +

∫t
0
xvx(τ, c)dτ

−
∫t
0

∫x
c

∫y
c

∂f
∂u

(τ, z, u(τ, z, φ(z)), ux(τ, z, φ(z)))v(τ, z)dzdydτ

−
∫t
0

∫x
c

∫y
c

∂f
∂ux

(τ, z, u(τ, z, φ(z)), ux(τ, z, φ(z)))vx(τ, z)dzdydτ,

(5.1)

which we differentiate twice in x and once in t and we get that v satisfies (1.3). Now we put t = 0

in (5.1) and then we differentiate twice in x, and we find that v satisfies (1.4).

Acknowledgments

The authors would like to thank the anonymous referees for their helpful comments and suggestions.

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	Introduction
	Auxiliary results
	Proof of Theorem 1.1
	Proof of Theorem 1.2
	Proof of Theorem 1.4