Applications to Parabolic Equations

Here, we consider the Cauchy problem 4.1 u t − u x x = f ( t , x , u , u x ) in ( 0 , ∞ ) × R , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_1.jpg"/> 4.2 u ( 0 , x ) = ϕ ( x ) in R , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_2.jpg"/> where ϕ ∈ C 2 ( R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_1.jpg"/> , f : [ 0 , ∞ ) × R × R × R → C https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_2.jpg"/> is a given continuous function, u : [ 0 , ∞ ) × R → C https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_3.jpg"/> is the main unknown. We will start our investigations in this section with the following useful lemma. Lemma 4.1.1

Let f ∈ C ( [ a , b ] × [ c , d ] × R × R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_4.jpg"/> , g ∈ C 2 ( [ c , d ] ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_5.jpg"/> . Then the function u ∈ C 1 ( [ a , b ] , C 2 ( [ c , d ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_6.jpg"/> is a solution to the problem 4.3 u t − u x x = f ( t , x , u , u x ) in ( a , b ] × [ c , d ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_3.jpg"/> 4.4 u ( a , x ) = g ( x ) in [ c , d ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_4.jpg"/> if and only if it is a solution to the integral equation 4.5 ∫ c x ∫ c y ( u ( t , z ) − g ( z ) ) d z d y − ∫ a t ( u ( τ , x ) − u ( τ , c ) − ( x − c ) u x ( τ , c ) ) d τ = ∫ a t ∫ c x ∫ c y f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) d z d y d τ , x ∈ [ c , d ] , t ∈ [ a , b ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_5.jpg"/>

Proof 4.1.2

Let u ∈ C 1 ( [ a , b ] , C 2 ( [ c , d ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_7.jpg"/> be a solution to problem (4.3), (4.4).

We integrate equation (4.3) with respect to x and we get ∫ c x u t ( t , z ) d z − ∫ c x u x x ( t , z ) d z = ∫ c x f ( t , z , u ( t , z ) , u x ( t , z ) ) d z , x ∈ [ c , d ] , t ∈ [ a , b ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_1.jpg"/> 170or ∫ c x u t ( t , z ) d z − u x ( t , x ) + u x ( t , c ) = ∫ c x f ( t , z , u ( t , z ) , u x ( t , z ) ) d z , x ∈ [ c , d ] , t ∈ [ a , b ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_2.jpg"/> Now we integrate the last equation with respect to x and we find ∫ c x ∫ c y u t ( t , z ) d z d y − ∫ c x ( u x ( t , z ) − u x ( t , c ) ) d z = ∫ c x ∫ c y f ( t , z , u ( t , z ) , u x ( t , z ) ) d z d y , x ∈ [ c , d ] , t ∈ [ a , b ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_3.jpg"/> or ∫ c x ∫ c y u t ( t , z ) d z d y − u ( t , x ) + u ( t , c ) + ( x − c ) u x ( t , c ) = ∫ c x ∫ c y f ( t , z , u ( t , z ) , u x ( t , z ) ) d z d y , x ∈ [ c , d ] , t ∈ [ a , b ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_4.jpg"/> We integrate the last equality with respect to t and we obtain ∫ a t ∫ c x ∫ c y u t ( s , z ) d z d y d s − ∫ a t ( u ( s , x ) − u ( s , c ) − ( x − c ) u x ( s , c ) ) d s = ∫ a t ∫ c x ∫ c y f ( s , z , u ( s , z ) , u x ( s , z ) ) d z d y d s , x ∈ [ c , d ] , t ∈ [ a , b ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_5.jpg"/> or ∫ c x ∫ c y ( u ( t , z ) − g ( z ) ) d z d y − ∫ a t ( u ( s , x ) − u ( s , c ) − ( x − c ) u x ( s , c ) ) d s = ∫ a t ∫ c x ∫ c y f ( s , z , u ( s , z ) , u x ( s , z ) ) d z d y d s , x ∈ [ c , d ] , t ∈ [ a , b ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_6.jpg"/> i.e., u satisfies equation (4.5).

Let u ∈ C 1 ( [ a , b ] , C 2 ( [ c , d ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_8.jpg"/> be a solution to the integral equation (4.5).

We differentiate equation (4.5) with respect to x and we get ∫ c x ( u ( t , z ) − g ( z ) ) d z − ∫ a t ( u x ( s , x ) − u x ( s , c ) ) d s = ∫ a t ∫ c x f ( s , z , u ( s , z ) , u x ( s , z ) ) d z d s , x ∈ [ c , d ] , t ∈ [ a , b ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_7.jpg"/> 171Again we differentiate with respect to x and we find 4.6 u ( t , x ) − g ( x ) − ∫ a t u x x ( s , x ) d s = ∫ a t f ( s , x , u ( s , x ) , u x ( s , x ) ) d s , x ∈ [ c , d ] , t ∈ [ a , b ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_6.jpg"/> Now we put t = a in the last equation and we find u ( a , x ) = g ( x ) , x ∈ [ c , d ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_8.jpg"/> i.e., the function u satisfies (4.4).

Now we differentiate equation (4.6) with respect to t and we find u t ( t , x ) − u x x ( t , x ) = f ( t , x , u ( t , x ) , u x ( t , x ) ) , x ∈ [ c , d ] , t ∈ [ a , b ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_9.jpg"/>

Theorem 4.1.3

Let f ∈ C ( [ 0 , ∞ ) × R × R × R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_9.jpg"/> , ϕ ∈ C 2 ( R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_10.jpg"/> . Then there exists m ∈ (0, 1) such that problem (4.1), (4.2) has a solution u ∈ C 1 ( [ 0 , m ] , C 2 ( [ 0 , 1 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_13.jpg"/> .

Proof 4.1.4

Let B > ∥ ϕ ∥ C 2 ( [ 0 , 1 ] ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_14.jpg"/> be arbitrarily chosen. Since ϕ ∈ C ( [ 0 , 1 ] ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_15.jpg"/> , f ∈ C ( [ 0 , 1 ] × [ 0 , 1 ] × [ − B , B ] × [ − B , B ] ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_16.jpg"/> we have that there exists a constant M ₁₁ > 0 such that | ϕ | ≤ M 11 in [ 0 , 1 ] , | f | ≤ M 11 in [ 0 , 1 ] × [ 0 , 1 ] × [ − B , B ] × [ − B , B ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_10.jpg"/> We take l, m ∈ (0, 1) so that 4.7 l B + l ( B + M 11 ) + 3 l B m + l M 11 m ≤ B l ( 5 B + 2 M 11 ) ≤ B . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_7.jpg"/> Let E 11 = C 1 ( [ 0 , m ] , C 2 ( [ 0 , 1 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_17.jpg"/> be endowed with the norm | | u | | = max { max ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] | u ( t , x ) | , max ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] | u t ( t , x ) | , max ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] | u x ( t , x ) | , max ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] | u x x ( t , x ) | } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_11.jpg"/> By K ~ 11 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_18.jpg"/> we denote the set of all equi-continuous families in E ₁₁, i.e., for every ε > 0 there exists δ = δ(ε) > 0 such that | u ( t 1 , x 1 ) − u ( t 2 , x 2 ) | < ϵ , | u t ( t 1 , x 1 ) − u t ( t 2 , x 2 ) | < ϵ , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_12.jpg"/> | u x ( t 1 , x 1 ) − u x ( t 2 , x 2 ) | < ϵ , | u x x ( t 1 , x 1 ) − u x x ( t 2 , x 2 ) | < ϵ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_13.jpg"/> 172whenever |t ₁ − t ₂| < δ, |x ₁ − x ₂| < δ. Let also K 11 ′ = K ~ 11 ¯ , K 11 = { u ∈ K 11 ′ : | | u | | ≤ B } https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_14.jpg"/> and L 11 = { u ∈ K 11 ′ : | | u | | ≤ ( 1 + l ) B } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_15.jpg"/> We note that K ₁₁ is a closed convex subset of L ₁₁.

For u ∈ L ₁₁, we define the operators T 11 ( u ) ( t , x ) = ( 1 + l ) u ( t , x ) , S 11 ( u ) ( t , x ) = − l u ( t , x ) + l ∫ 0 x ∫ 0 y ( u ( t , z ) − ϕ ( z ) ) d z d y − l ∫ 0 t ( u ( τ , x ) − u ( τ , 0 ) − x u x ( τ , 0 ) ) d τ − l ∫ 0 t ∫ 0 x ∫ 0 y f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) d z d y d τ , ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_16.jpg"/> We will prove that the problem 4.8 u t − u x x = f ( t , x , u x ) in [ 0 , m ] × [ 0 , 1 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_8.jpg"/> 4.9 u ( 0 , x ) = ϕ ( x ) in [ 0 , 1 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_9.jpg"/> has a solution u ∈ C 1 ( [ 0 , m ] , C 2 ( [ 0 , 1 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_19.jpg"/> .

S ₁₁:K ₁₁ → K ₁₁. Let u ∈ K ₁₁. Then S 11 ( u ) ∈ C 1 ( [ 0 , m ] , C 2 ( [ 0 , 1 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_20.jpg"/> and for (t, x) ∈ [0, m] × [0, 1], using the first inequality of (4.7), we get | S 11 ( u ) ( t , x ) | = | − l u ( t , x ) + l ∫ 0 x ∫ 0 y ( u ( t , z ) − ϕ ( z ) ) d z d y − l ∫ 0 t ( u ( τ , x ) − u ( τ , 0 ) − x u x ( τ , 0 ) ) d τ − l ∫ 0 t ∫ 0 x ∫ 0 y f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) d z d y d τ | ≤ l | u ( t , x ) | + l ∫ 0 x ∫ 0 y ( | u ( t , z ) | + | ϕ ( z ) | ) d z d y + l ∫ 0 t ( | u ( τ , x ) | + | u ( τ , 0 ) | + x | u x ( τ , 0 ) | ) d τ + l ∫ 0 t ∫ 0 x ∫ 0 y | f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) | d z d y d τ ≤ l B + l ( B + M 11 ) + 3 l B m + l M 11 m ≤ B . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_17.jpg"/> 173Note that S 11 ( u ) t ( t , x ) = − l u t ( t , x ) + l ∫ 0 x ∫ 0 y u t ( t , z ) d z d y − l ( u ( t , x ) − u ( t , 0 ) − x u x ( t , 0 ) ) − l ∫ 0 x ∫ 0 y f ( t , z , u ( t , z ) , u x ( t , z ) ) d z d y , ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_18.jpg"/> Then, using the second inequality of (4.7), we obtain | S 11 ( u ) t ( t , x ) | = | − l u t ( t , x ) + l ∫ 0 x ∫ 0 y u t ( t , z ) d z d y − l ( u ( t , x ) − u ( t , 0 ) − x u x ( t , 0 ) ) − l ∫ 0 x ∫ 0 y f ( t , z , u ( t , z ) , u x ( t , z ) ) d z d y | ≤ l | u t ( t , x ) | + l ∫ 0 x ∫ 0 y | u t ( t , z ) | d z d y + l ( | u ( t , x ) | + | u ( t , 0 ) | + x | u x ( t , 0 ) | ) + l ∫ 0 x ∫ 0 y | f ( t , z , u ( t , z ) , u x ( t , z ) ) | d z d y ≤ l B + l B + 3 l B + l M 11 = l ( 5 B + M 11 ) ≤ B , ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_19.jpg"/> 174Also, S 11 ( u ) x ( t , x ) = − l u x ( t , x ) + l ∫ 0 x ( u ( t , z ) − ϕ ( z ) ) d z − l ∫ 0 t ( u x ( τ , x ) − u x ( τ , 0 ) ) d τ − l ∫ 0 t ∫ 0 x f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) d z d τ , ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_20.jpg"/> Hence, using the first inequality of (4.7), | S 11 ( u ) x ( t , x ) | = | − l u x ( t , x ) + l ∫ 0 x ( u ( t , z ) − ϕ ( z ) ) d z − l ∫ 0 t ( u x ( τ , x ) − u x ( τ , 0 ) ) d τ − l ∫ 0 t ∫ 0 x f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) d z d τ | ≤ l | u x ( t , x ) | + l ∫ 0 x ( | u ( t , z ) | + | ϕ ( z ) | ) d z + l ∫ 0 t ( | u x ( τ , x ) | + | u x ( τ , 0 ) | ) d τ + l ∫ 0 t ∫ 0 x | f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) | d z d τ ≤ l B + l ( B + M 11 ) + 2 l B m + l M 11 m ≤ B , ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_21.jpg"/> For (t, x) ∈ [0, m] × [0, 1] we have S 11 ( u ) x x ( t , x ) = − l u x x ( t , x ) + l ( u ( t , x ) − ϕ ( x ) ) − l ∫ 0 t u x x ( τ , x ) d τ − l ∫ 0 t f ( τ , x , u ( τ , x ) , u x ( τ , x ) ) d τ , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_22.jpg"/> 175from where, using the first inequality of (4.7), | S 11 ( u ) x x ( t , x ) | = | − l u x x ( t , x ) + l ( u ( t , x ) − ϕ ( x ) ) − l ∫ 0 t u x x ( τ , x ) d τ − l ∫ 0 t f ( τ , x , u ( τ , x ) , u x ( τ , x ) ) d τ | ≤ l | u x x ( t , x ) | + l ( | u ( t , x ) | + | ϕ ( x ) | ) + l ∫ 0 t | u x x ( τ , x ) | d τ + l ∫ 0 t | f ( τ , x , u ( τ , x ) , u x ( τ , x ) ) | d τ ≤ l B + l ( B + M 11 ) + l B m + l M 11 m ≤ B . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_23.jpg"/> We note that {S ₁₁(u):u ∈ K ₁₁} is an equi-continuous family in E ₁₁. Consequently S ₁₁:K ₁₁ → K ₁₁. Also, S ₁₁(K ₁₁) ⊂ K ₁₁ ⊂ L ₁₁, i.e., S ₁₁(K ₁₁) resides in a compact subset of L ₁₁.

S ₁₁:K ₁₁ → K ₁₁ is a continuous operator. We note that if { u n } n = 1 ∞ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_21.jpg"/> is a sequence of elements of K ₁₁ such that u n ⟶ u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_22.jpg"/> in K ₁₁ as n ⟶ ∞ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_23.jpg"/> , then S 11 ( u n ) ⟶ S 11 ( u ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_24.jpg"/> in K ₁₁ as n ⟶ ∞ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_25.jpg"/> . Therefore S ₁₁:K ₁₁ → K ₁₁ is a continuous operator.

T ₁₁:K ₁₁ → L ₁₁ is an expansive operator and onto. For u, v ∈ K ₁₁ we have that | | T 11 ( u ) − T 11 ( v ) | | = ( 1 + l ) | | u − v | | , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_24.jpg"/> i.e., T ₁₁:K ₁₁ → L ₁₁ is an expansive operator with constant 1 + l.

Let v ∈ L ₁₁. Then v 1 + l ∈ K 11 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_26.jpg"/> and T 11 ( v 1 + l ) = v , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_25.jpg"/> i.e., T ₁₁:K ₁₁ → L ₁₁ is onto.

176From items a, b, and c, and from Theorem 1.9.12, it follows that there is u ₁₁ ∈ K ₁₁ such that T 11 u 11 + S 11 u 11 = u 11 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_26.jpg"/> or ( 1 + l ) u 11 ( t , x ) − l u 11 ( t , x ) + l ∫ 0 x ∫ 0 y ( u 11 ( t , z ) − ϕ ( z ) ) d z d y − l ∫ 0 t ( u 11 ( τ , x ) − u 11 ( τ , 0 ) − x u 11 x ( τ , 0 ) ) d τ − l ∫ 0 t ∫ 0 x ∫ 0 y f ( τ , z , u 11 ( τ , z ) , u 11 x ( τ , z ) ) d z d y d τ = u 11 ( t , x ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_27.jpg"/> or ∫ 0 x ∫ 0 y ( u 11 ( t , z ) − ϕ ( z ) ) d z d y − ∫ 0 t ( u 11 ( τ , x ) − u 11 ( τ , 0 ) − x u 11 x ( τ , 0 ) ) d τ − ∫ 0 t ∫ 0 x ∫ 0 y f ( τ , z , u 11 ( τ , z ) , u 11 x ( τ , z ) ) d z d y d τ = 0 , ( t , x ) ∈ [ 0 , m ] × [ 0 , 1 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_28.jpg"/> whereupon, using Lemma 4.1.1, we conclude that u 11 ∈ C 1 ( [ 0 , 1 ] , C 2 ( [ 0 , 1 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_27.jpg"/> is a solution to problem (4.8), (4.9). This completes the proof.

Theorem 4.1.5

Let f ∈ C ( [ 0 , ∞ ) × R × R × R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_28.jpg"/> , ϕ ∈ C 2 ( R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_29.jpg"/> . Then there exists m ∈ (0, 1) such that problem (4.1), (4.2) has a solution u ∈ C 1 ( [ 0 , m ] , C 2 ( R ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_30.jpg"/> .

Example 4.1.6

Let p > 1 and a ∈ C https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_31.jpg"/> be chosen so that a p − 1 = − 1 p − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_32.jpg"/> . Consider the Cauchy problem u t − u x x = u p in ( 0 , ∞ ) × R u ( 0 , x ) = a in R . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_29.jpg"/>

177Then u ( t , x ) = a ( t + 1 ) − 1 p − 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_33.jpg"/> is its solution. Actually, u t ( t , x ) = − a p − 1 ( t + 1 ) − p p − 1 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_30.jpg"/> and u x x ( t , x ) = 0 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_31.jpg"/> and ( u ( t , x ) ) p = − a p − 1 ( t + 1 ) − p p − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_32.jpg"/> Therefore u t ( t , x ) − u x x ( t , x ) = ( u ( t , x ) ) p in ( 0 , ∞ ) × R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_33.jpg"/> and u ( 0 , x ) = a in R . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_34.jpg"/>

Proof 4.1.7

Consider 4.10 u t − u x x = f ( t , x , u ( t , x ) , u x ( t , x ) ) in ( 0 , m ] × [ 1 , 2 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_10.jpg"/> 4.11 u ( 0 , x ) = ϕ ( x ) in [ 1 , 2 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_11.jpg"/> Let E 12 = C 1 ( [ 0 , m ] , C 2 ( [ 1 , 2 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_34.jpg"/> be endowed with the norm | | u | | = max { max ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] | u ( t , x ) | , max ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] | u t ( t , x ) | , max ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] | u x ( t , x ) | , max ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] | u x x ( t , x ) | } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_35.jpg"/> By K ~ 12 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_35.jpg"/> we denote the set of all equi-continuous families in E ₁₂.

Let K 12 ′ = K ~ 12 ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_36.jpg"/> , K 12 = { u ∈ K 12 ′ : | | u | | ≤ B } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_36.jpg"/> Since ϕ ∈ C ( [ 1 , 2 ] ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_37.jpg"/> , f ∈ C ( [ 0 , m ] × [ 1 , 2 ] × [ − B , B ] × [ − B , B ] ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_38.jpg"/> we have that there exists a constant M ₁₂ > 0 such that | ϕ ( x ) | ≤ M 12 in [ 1 , 2 ] , | f ( t , x , y , z ) | ≤ M 12 in [ 0 , m ] × [ 1 , 2 ] × [ − B , B ] × [ − B , B ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_37.jpg"/> Let l ₁ > 0 be chosen so that l 1 ( 5 B + 2 M 12 ) ≤ B , l 1 B + l 1 ( B + M 12 ) + 3 l 1 B m + l 1 M 12 m ≤ B . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_38.jpg"/> 178Let also L 12 = { u ∈ K 12 ′ : | | u | | ≤ ( 1 + l 1 ) B } . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_39.jpg"/> We note that K ₁₂ is a closed convex subset of L ₁₂.

For u ∈ L ₁₂, we define the operators T 12 ( u ) ( t , x ) = ( 1 + l 1 ) u ( t , x ) , S 12 ( u ) ( t , x ) = − l 1 u ( t , x ) + l 1 ∫ 1 x ∫ 1 y ( u ( t , z ) − ϕ ( z ) ) d z d y − l 1 ∫ 0 t ( u ( τ , x ) − u 11 ( τ , 1 ) − ( x − 1 ) u 11 x ( τ , 1 ) ) d τ − l 1 ∫ 0 t ∫ 1 x ∫ 1 y f ( τ , z , u ( τ , z ) , u x ( τ , z ) ) d z d y d τ , ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_40.jpg"/> As in the previous theorem, one can prove that there is u 12 ∈ C 1 ( [ 0 , 1 ] , C 2 ( [ 1 , 2 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_39.jpg"/> which is a solution to problem (4.10), (4.11). This solution u ₁₂ satisfies the integral equation 4.12 ∫ 1 x ∫ 1 y ( u 12 ( t , z ) − ϕ ( z ) ) d z d y − ∫ 0 t ( u 12 ( τ , x ) − u 11 ( τ , 1 ) − ( x − 1 ) u 11 x ( τ , 1 ) ) d τ − ∫ 0 t ∫ 1 x ∫ 1 y f ( τ , z , u 12 ( τ , z ) , u 12 x ( τ , z ) ) d z d y d τ = 0 , ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_12.jpg"/> Now we put x = 1 in (4.12) and we find ∫ 0 t ( u 12 ( τ , 1 ) − u 11 ( τ , 1 ) ) d τ = 0 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_41.jpg"/> which we differentiate with respect to t and we get 4.13 u 12 ( t , 1 ) = u 11 ( t , 1 ) in [ 0 , m ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_13.jpg"/> Now we differentiate (4.12) with respect to x and we find ∫ 1 x ( u 12 ( t , z ) − ϕ ( z ) ) d z − ∫ 0 t ( u 12 x ( τ , x ) − u 11 x ( τ , 1 ) ) d τ − ∫ 0 t ∫ 1 x f ( τ , z , u 12 ( τ , z ) , u 12 x ( τ , z ) ) d z d τ = 0 , ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_42.jpg"/> 179In the last equation we put x = 1 and we obtain ∫ 0 t ( u 12 x ( τ , x ) − u 11 x ( τ , 1 ) ) d τ = 0 , ( t , x ) ∈ [ 0 , m ] × [ 1 , 2 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_43.jpg"/> which we differentiate with respect to t and we find 4.14 u 12 x ( t , 1 ) = u 11 x ( t , 1 ) in [ 0 , m ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_14.jpg"/> Now we differentiate (4.13) with respect to t and we get u 12 t ( t , 1 ) = u 11 t ( t , 1 ) in [ 0 , m ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_44.jpg"/> Hence, (4.13), (4.14) and f ( t , 1 , u 11 ( t , 1 ) , u 11 x ( t , 1 ) ) = f ( t , 1 , u 12 ( t , 1 ) , u 12 x ( t , 1 ) ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_45.jpg"/> we find u 12 x x ( t , 1 ) = u 12 t ( t , 1 ) − f ( t , 1 , u 12 ( t , 1 ) , u 12 x ( t , 1 ) ) = u 11 t ( t , 1 ) − f ( t , 1 , u 11 ( t , 1 ) , u 11 x ( t , 1 ) ) = u 11 x x ( t , 1 ) in [ 0 , m ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_46.jpg"/> Consequently, the function u ( t , x ) = { u 11 ( t , x ) in [ 0 , m ] × [ 0 , 1 ] u 12 ( t , x ) in [ 0 , m ] × [ 1 , 2 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_47.jpg"/> is a C 1 ( [ 0 , m ] , C 2 ( [ 0 , 2 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_45.jpg"/> -solution to the problem u t − u x x = f ( t , x , u ( t , x ) , u x ( t , x ) ) in ( 0 , m ] × [ 0 , 2 ] , u ( 0 , x ) = ϕ ( x ) in [ 0 , 2 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_48.jpg"/> Then we consider the problem 4.15 u t − u x x = f ( t , x , u ( t , x ) , u x ( t , x ) ) in ( 0 , m ] × [ 2 , 3 ] u ( 0 , x ) = ϕ ( x ) in [ 2 , 3 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_15.jpg"/> As in the above there is u 13 ∈ C 1 ( [ 0 , m ] , C 2 ( [ 2 , 3 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_46.jpg"/> which is a solution to problem (4.15) and satisfies the integral equation ∫ 2 x ∫ 2 y ( u 13 ( t , z ) − ϕ ( z ) ) d z d y − ∫ 0 t ( u 13 ( τ , x ) − u 12 ( τ , 2 ) − ( x − 2 ) u 12 x ( τ , 2 ) ) d τ − ∫ 0 t ∫ 2 x ∫ 2 y f ( τ , z , u 13 ( τ , z ) , u 13 x ( τ , z ) ) d z d y d τ = 0 , t ∈ [ 0 , m ] , x ∈ [ 2 , 3 ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_49.jpg"/> 180The function u ( t , x ) = { u 11 ( t , x ) in [ 0 , m ] × [ 0 , 1 ] u 12 ( t , x ) in [ 0 , m ] × [ 1 , 2 ] u 13 ( t , x ) in [ 0 , m ] × [ 2 , 3 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_50.jpg"/> is a C 1 ( [ 0 , m ] , C 2 ( [ 0 , 3 ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_47.jpg"/> -solution to the problem u t − u x x = f ( t , x , u ( t , x ) , u x ( t , x ) ) in [ 0 , m ] × [ 0 , 3 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_51.jpg"/> u ( 0 , x ) = ϕ ( x ) in [ 0 , 3 ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_52.jpg"/> and so on. We construct a solution u 1 ∈ C 1 ( [ 0 , m ] , C 2 ( R ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_48.jpg"/> which is a solution to the problem u t − u x x = f ( t , x , u ( t , x ) , u x ( t , x ) ) in ( 0 , m ] × R , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_53.jpg"/> u ( 0 , x ) = ϕ ( x ) in R . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_54.jpg"/> This completes the proof.

Theorem 4.1.8

Let f ∈ C ( [ 0 , ∞ ) × R × R × R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_49.jpg"/> , ∂ f ∂ u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_50.jpg"/> , ∂ f ∂ u x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_51.jpg"/> exist and are continuous in [ 0 , ∞ ) × R × R × R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_52.jpg"/> , ϕ ∈ C 2 ( R ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_53.jpg"/> . Let also u ( t , x , ϕ ) ∈ C 1 ( [ 0 , m ] , C 2 ( [ c , d ] ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_54.jpg"/> be a solution to problem (4.1), (4.2) for some m ∈ (0, 1) and for some [ c , d ] ⊂ R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_55.jpg"/> . Then u(t, x, ϕ) is differentiable with respect to ϕ and v ( t , x ) = ∂ u ∂ ϕ ( t , x , ϕ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_56.jpg"/> satisfies the following initial value problem 4.16 v t − v x x = ∂ f ∂ u ( t , x , u ( t , x , ϕ ) , u x ( t , x , ϕ ) ) v + ∂ f ∂ u x ( t , x , u ( t , x , ϕ ) , u x ( t , x , ϕ ) ) v x in [ 0 , m ] × [ c , d ] , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_16.jpg"/> 4.17 v ( 0 , x ) = 1 in [ c , d ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_17.jpg"/>

Proof 4.1.9

181We have that the solution u(t, x, ϕ) satisfies the following integral equation: Q ( ϕ ) = ∫ c x ∫ c y ( u ( t , z , ϕ ( z ) ) − ϕ ( z ) ) d z d y − ∫ 0 t ( u ( τ , x , ϕ ( x ) ) − u ( τ , c , ϕ ( c ) ) − ( x − c ) u x ( τ , c , ϕ ( c ) ) ) d τ − ∫ 0 t ∫ c x ∫ c y f ( τ , z , u ( τ , z , ϕ ( z ) ) , u x ( τ , z , ϕ ( z ) ) ) d z = 0 , t ∈ [ 0 , m ] , x ∈ [ c , d ] . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_55.jpg"/> Then Q ( ϕ ) − Q ( ϕ 1 ) = ∫ c x ∫ c y ( u ( t , z , ϕ ( z ) ) − u ( t , z , ϕ 1 ( z ) ) − ( ϕ ( z ) − ϕ 1 ( z ) ) ) d z d y − ∫ 0 t ( u ( τ , x , ϕ ( x ) ) − u ( τ , x , ϕ 1 ( x ) ) ) d τ + ∫ 0 t ( u ( τ , c , ϕ ( c ) ) − u ( τ , c , ϕ 1 ( c ) ) ) d τ + ∫ 0 t ( x − c ) ( u x ( τ , c , ϕ ( c ) ) − u x ( τ , c , ϕ 1 ( c ) ) ) d τ − ∫ 0 t ∫ c x ∫ c y ( f ( τ , z , u ( τ , z , ϕ ( z ) ) , u x ( τ , z , ϕ ( z ) ) ) − f ( τ , z , u ( τ , z , ϕ 1 ( z ) ) , u x ( τ , z , ϕ 1 ( z ) ) ) ) d z d y d τ = ∫ c x ∫ c y ( ∂ u ∂ ϕ ( t , z , ϕ ( z ) ) − 1 ) d z d y https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_56.jpg"/> − ∫ 0 t ∂ u ∂ ϕ ( τ , x , ϕ ( x ) ) d τ + ∫ 0 t ∂ u ∂ ϕ ( τ , c , ϕ ( c ) ) d τ + ∫ 0 t ( x − c ) ( ∂ u ∂ ϕ ) x ( τ , c , ϕ ( c ) ) d τ − ∫ 0 t ∫ c x ∫ c y ∂ f ∂ u ( τ , z , u ( τ , z , ϕ ( z ) ) , u x ( τ , z , ϕ ( z ) ) ) ∂ u ∂ ϕ ( τ , z , ϕ ( z ) ) d z d y d τ − ∫ 0 t ∫ c x ∫ c y ∂ f ∂ u x ( τ , z , u ( τ , z , ϕ ( z ) ) , u x ( τ , z , ϕ ( z ) ) ) ( ∂ u ∂ ϕ ) x ( τ , z , ϕ ( z ) ) d z d y d τ + δ { ϕ , ϕ 1 } , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/umath4_57.jpg"/> 182where δ { ϕ , ϕ 1 } ⟶ 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_57.jpg"/> as ϕ ( x ) ⟶ ϕ 1 ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_58.jpg"/> for every x ∈ [c, d]. Hence, when ϕ ( x ) ⟶ ϕ 1 ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/inline-math4_59.jpg"/> for every x ∈ [c, d], we get 4.18 0 = ∫ c x ∫ c y ( v ( t , z ) − 1 ) d z d y − ∫ 0 t v ( τ , x ) d τ + ∫ 0 t v ( τ , c ) d τ + ∫ 0 t ( x − c ) v x ( τ , c ) d τ − ∫ 0 t ∫ c x ∫ c y ∂ f ∂ u ( τ , z , u ( τ , z , ϕ ( z ) ) , u x ( τ , z , ϕ ( z ) ) ) v ( τ , z ) d z d y d τ − ∫ 0 t ∫ c x ∫ c y ∂ f ∂ u x ( τ , z , u ( τ , z , ϕ ( z ) ) , u x ( τ , z , ϕ ( z ) ) ) v x ( τ , z ) d z d y d τ , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003028727/1d426271-33df-4258-80cb-5188d2ed19c2/content/math4_18.jpg"/> which we differentiate twice in x and once in t and we get that v satisfies (4.16). Now we put t = 0 in (4.18) and then we differentiate twice in x, and we find that v satisfies (4.17). This completes the proof.