Let us consider now the one-dimensional case of Problem 1.1 for the temperature defined by
Problem 3.1
Find the temperature u at \((x, t)\) such that it satisfies the following conditions:
$$\begin{aligned} & u_{t} - u_{xx} = -F \biggl( \int_{0}^{t} u_{x}(0, s)\,ds \biggr), \quad x>0, t>0, \\ & u(0, t)= 0, \quad t>0, \\ & u(x, 0) = h(x), \quad x>0. \end{aligned}$$
Taking into account that
$$\begin{aligned} \int_{0}^{t} G(x, t, \xi, \tau) \,d\xi= \operatorname{erf} \biggl( {x\over 2\sqrt{t-\tau}} \biggr), \end{aligned}$$
(3.1)
thus the solution of Problem 3.1 is given by
$$\begin{aligned} u(x, t) = u_{0}(x, t) - \int_{0}^{t} \operatorname{erf} \biggl( {x\over 2\sqrt{t-\tau}} \biggr) F \biggl( \int_{0}^{\tau} W(\sigma)\,d\sigma \biggr) \,d\tau \end{aligned}$$
(3.2)
with
$$\begin{aligned} u_{0}(x, t) = \int_{0}^{t} G(x, t, \xi, 0) h(\xi) \,d\xi, \end{aligned}$$
(3.3)
and \(W(t)= u_{x}(0, t)\) is the solution of the following Volterra integral equation:
$$\begin{aligned} W(t)= V_{0}(t) - \int_{0}^{t} {F ( \int_{0}^{\tau} W(\sigma)\,d\sigma) \over \sqrt{\pi(t-\tau)}} \,d\tau, \end{aligned}$$
(3.4)
where
$$\begin{aligned} V_{0}(t)= {1\over 2\sqrt{\pi}t^{3/2}} \int_{0}^{+\infty} \xi e^{-\xi^{2}/4t}h(\xi) \,d\xi= {2\over \sqrt{\pi t}} \int_{0}^{+\infty} \eta e^{-\eta^{2}}h(2\sqrt{t} \eta) \,d\eta. \end{aligned}$$
(3.5)
For the particular case
$$\begin{aligned} h(x)=h_{0} >0 \mbox{ pour } x>0, \mbox{ and } F(W)= \lambda W \mbox{ for } \lambda\in\mathbb {R}, \end{aligned}$$
(3.6)
we have
$$\begin{aligned} u_{0}(t, x)= h_{0} \operatorname{erf} \biggl( {x\over 2\sqrt{t}} \biggr), \end{aligned}$$
(3.7)
and integral equation (3.4) becomes
$$\begin{aligned} W(t) = {h_{0}\over \sqrt{\pi t}} - \lambda \int_{0}^{t}{ \int_{0}^{\tau} W(\sigma) \,d\sigma\over \sqrt{\pi (t-\tau)}}\,d\tau. \end{aligned}$$
(3.8)
Lemma 3.1
Assume (3.6) holds. The solution of Problem
3.1
is given by
$$\begin{aligned} u(x, t) = h_{0} \operatorname{erf} \biggl( {x\over 2\sqrt{t}} \biggr) - \lambda \int_{0}^{t} \operatorname{erf} \biggl( {x\over 2\sqrt{t-\tau}} \biggr) U(\tau)\,d\tau, \end{aligned}$$
(3.9)
where
U
is given by
$$\begin{aligned} U(t)= {h_{0}\over \sqrt{\pi}} \int_{0}^{t} {g(\tau)\over \sqrt{t-\tau}}\,d\tau, \end{aligned}$$
(3.10)
and
g
is the solution of the Volterra integral equation
$$\begin{aligned} g(t)= 1- {2\lambda\over \sqrt{\pi}} \int_{0}^{t} g(\tau) \sqrt{t-\tau} \,d\tau. \end{aligned}$$
(3.11)
Moreover, the heat flux on
\(x=0\)
is given by
$$\begin{aligned} u_{x}(0, t) = U'(t)= {h_{0}\over \sqrt{\pi t}} - h_{0}\lambda \int_{0}^{t} g(\tau)\,d\tau, \quad t>0. \end{aligned}$$
(3.12)
Proof
We set
$$\begin{aligned} U(t) = \int_{0}^{t}W(\tau) \,d\tau, \end{aligned}$$
(3.13)
thus the function U satisfies the following new Volterra integral equation:
$$\begin{aligned} U(t) =& 2h_{0} \sqrt{{t\over \pi}} - {\lambda\over \sqrt{\pi}} \int_{0}^{t} \int_{0}^{\tau}{U(\sigma)\over \sqrt{\tau-\sigma}}\,d\sigma \,d \tau \\ =& 2h_{0} \sqrt{{t\over \pi}} - {2\lambda\over \sqrt{\pi}} \int_{0}^{t} U(\tau) \sqrt{t-\tau} \,d\tau, \quad t> 0 \end{aligned}$$
(3.14)
by using the following equality:
$$\begin{aligned} \int_{\sigma}^{t} {d\tau\over \sqrt{\tau-\sigma}} = 2\sqrt{t- \sigma}, \quad0< \sigma< t. \end{aligned}$$
(3.15)
From [21], p.229, the solution \(t\mapsto U(t)\) of integral equation (3.14) is given by (3.10), where g is the solution of Volterra equation (3.11).
From (3.11) we obtain that
$$\begin{aligned} \int_{0}^{t} {g(\tau)\over \sqrt{t-\tau}} \,d\tau= 2 \sqrt{t} - \lambda\sqrt{\pi} \int_{0}^{t}g(\tau)\sqrt{t-\tau}\,d\tau, \end{aligned}$$
(3.16)
using the following equality:
$$\begin{aligned} \int_{\sigma}^{t} {\sqrt{\tau-\sigma}\over \sqrt{t-\tau}}\,d\tau =&(t- \sigma) \int_{0}^{1}{\sqrt{\xi}\over \sqrt{1-\xi}} \,d\xi=(t- \sigma)B \biggl({3\over 2}, {1\over 2} \biggr) \\ =&(t-\sigma) {\Gamma({3\over 2}) \Gamma( {1\over 2})\over \Gamma (2)} ={\pi\over 2}(t-\sigma), \end{aligned}$$
(3.17)
where B and Γ are the classical beta and gamma functions defined below.
Therefore, we have that
$$\begin{aligned} U(t) = 2h_{0}\sqrt{{t\over \pi}} - \lambda h_{0} \int_{0}^{t} g(\tau) (t-\tau) \,d\tau, \end{aligned}$$
(3.18)
and then the heat flux on \(x=0\) is given by \(u_{x}(0, t) = W(t)= U'(t)\), that is, (3.12) holds. □
We recall here the well-known beta and gamma functions defined respectively by
$$\begin{aligned} &B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \,dt, \quad x>0, y>0, \\ & \Gamma(x) = \int_{0}^{+\infty} t^{x-1} e^{-t} \,dt, \quad x>0. \end{aligned}$$
We will use in the next theorem the well-known relations
$$\begin{aligned} & B(x, y) ={\Gamma(x)\Gamma(y)\over \Gamma(x+y)}, \qquad\Gamma (x+1)=x\Gamma(x) \quad\forall x>0, \\ & \Gamma \biggl({1\over 2} \biggr)= \sqrt{\pi}, \qquad \Gamma(n+1)=n! \quad\forall n\in\mathbb {N}, \end{aligned}$$
and in particular the following one.
Lemma 3.2
For all integers
\(n \geq1\), we have
$$\begin{aligned} \Gamma \biggl(n+ {1\over 2} \biggr)= {(2n-1)!!\over 2^{n}} \sqrt{ \pi}, \end{aligned}$$
and we use the definition
$$(2n-1)!!= (2n-1) (2n-3) (2n-5) \cdots5\cdot3\cdot1 $$
for compactness expression.
Proof
For \(n=1\), we get \(\Gamma({3\over 2})= {\sqrt{\pi}\over 2}\), which is true. By induction we obtain that
$$\begin{aligned} \Gamma \biggl(n+1+{1\over 2} \biggr) =& \Gamma \biggl( \biggl(n+ {1\over 2} \biggr)+1 \biggr) = \biggl(n +{1\over 2} \biggr) \Gamma \biggl(n +{1\over 2} \biggr) \\ =& \biggl(n +{1\over 2} \biggr) {(2n-1)!!\over 2^{n} }\sqrt{ \pi} ={(2n+1)!!\over 2^{n+1}} \sqrt{\pi}, \end{aligned}$$
thus the lemma is true. □
Corollary 3.3
For all integers
\(n \geq0\), we have also
$$\begin{aligned} & \Gamma \biggl(3n+5 +{1\over 2} \biggr)= {(6n+9)!!\over 2^{3n+5}} \sqrt{\pi}, \\ & B \biggl({3\over 2}, 3n+4 \biggr) = {\Gamma({3\over 2} )\Gamma (3(n+1)+1)\over \Gamma(3n+5+{1\over 2} )} = {(3(n+1))! 2^{3(n+1)+1}\over (6n+9)!!}, \\ & B \biggl({3\over 2}, 3n+{5\over 2} \biggr) = {\Gamma({3\over 2} )\Gamma(3n+ {5\over 2} ) \over \Gamma(3(n+1)+1)} ={\pi(6n+3)!! \over (3(n+1))! 2^{3(n+1)} }, \end{aligned}$$
which will be useful in the next lemma.
First, we need some preliminary simple results in order to obtain the solution of integral equation (2.1).
$$\begin{aligned} & \int_{0}^{t} \sqrt{t-\tau} \,d\tau= {2\over 3}t^{3/2}, \qquad \int_{0}^{t}\tau^{3/2} \sqrt{t-\tau} \,d \tau = {\pi\over 2^{4}}t^{3}, \end{aligned}$$
(3.19)
$$\begin{aligned} & \int_{0}^{t}\tau^{3} \sqrt{t-\tau} \,d \tau ={2^{4} 3!\over 9!!}t^{9/2}, \qquad \int_{0}^{t}\tau^{9/2} \sqrt{t-\tau} \,d \tau ={\pi 9!! \over 2^{6} 6!}t^{6}, \end{aligned}$$
(3.20)
$$\begin{aligned} & \int_{0}^{t}\tau^{6} \sqrt{t-\tau} \,d \tau ={2^{7} 6!\over 15!!}t^{15/2}, \qquad \int_{0}^{t}\tau^{15/2} \sqrt{t-\tau} \,d \tau ={\pi 15!! \over 2^{9} 9!}t^{9}, \end{aligned}$$
(3.21)
which can be generalized by the following ones.
Lemma 3.4
For all integers
\(n \geq0\), we have
$$\begin{aligned} & \int_{0}^{t}\tau^{2n+3} \sqrt{t-\tau} \,d \tau ={2^{3n+4} (3(n+1))!\over (6n+9)!!}t^{3(2n+3)/2}, \end{aligned}$$
(3.22)
$$\begin{aligned} & \int_{0}^{t}\tau^{3(2n+1)\over 2} \sqrt{t-\tau} \,d \tau ={\pi (6n+3)!! \over 2^{3(n+1)} (3(n+1))!}t^{3(n+1)}. \end{aligned}$$
(3.23)
Proof
Taking the change of variable \(\tau= t\xi\) in (3.22) and using Corollary 3.3, we get
$$\begin{aligned} \int_{0}^{t}\tau^{2n+3} \sqrt{t-\tau} \,d \tau =& t^{3(2n+3)\over 2} \int_{0}^{1}\xi^{3n+3} (1- \xi)^{1\over 2} \,d\xi \\ =&t^{3(2n+3)\over 2} \int_{0}^{1}\xi^{(3n+4)-1}(1- \xi)^{{3\over 2} -1}\,d\xi \\ =& t^{3(2n+3)\over 2} B \biggl({3\over 2}, 3n+4 \biggr) = {(3(n+1))! 2^{3(n+1) +1}\over (6(n+9))!!} \end{aligned}$$
and
$$\begin{aligned} \int_{0}^{t}\tau^{3(2n+1)\over 2} \sqrt{t-\tau} \,d \tau =& t^{3(2n+1)\over 2} \int_{0}^{1} \xi^{(3n+{5\over 2})-1}(1- \xi)^{{3\over 2}-1}\,d\xi \\ =&t^{3(2n+1)\over 2} B \biggl({3\over 2}, 3n+{5\over 2} \biggr) = {\pi (6n+3)!! \over (3(n+1))! 2^{3(n+1)}} t^{3(n+1)}, \end{aligned}$$
thus (3.22)-(3.23) hold. □
Now, we will obtain the explicit solution of the integral equation
$$\begin{aligned} y(t)= 1- {2\lambda\over \sqrt{\pi}} \int_{0}^{t} y(\tau) \sqrt{t-\tau} \,d\tau, \quad t>0, \end{aligned}$$
(3.24)
by using the Adomian decomposition method [22–24] through a series expansion.
Theorem 3.5
The solution of integral equation (3.24) is given by the following expression:
$$\begin{aligned} y(t) = I(t) - \sqrt{{2\over \pi}} J(t), \quad t>0, \end{aligned}$$
(3.25)
with
$$\begin{aligned} I(t) = \sum_{n=0}^{+\infty} {(\lambda^{2/3}t)^{3n}\over (3n)!} \end{aligned}$$
(3.26)
and
$$\begin{aligned} J(t) = \sum_{n=0}^{+\infty} {(\lambda^{2/3}t)^{{3(2n+1)\over 2}}\over (3(2n+1))!!} \end{aligned}$$
(3.27)
are series with infinite radii of convergence.
Proof
Following the idea of [25–33], we propose, for the solution of integral equation (3.24), the series of expansion functions given by
$$\begin{aligned} y(t) = \sum_{n=0}^{+\infty} y_{n}(t), \end{aligned}$$
(3.28)
and we obtain the following recurrence expressions:
$$\begin{aligned} y_{0}(t)=1, \qquad y_{n}(t)= - {2\lambda\over \sqrt{\pi}} \int_{0}^{t} y_{n-1}(\tau) \sqrt{t-\tau} \,d \tau, \quad\forall n \geq1. \end{aligned}$$
(3.29)
Then we get
$$\begin{aligned} & y_{1}(t)=-{2\lambda\over \sqrt{\pi}} \int_{0}^{t} \sqrt{t-\tau} \,d\tau= - {4\lambda\over 3 \sqrt{\pi}}t^{3/2} = -\sqrt{{2\over \pi}} { (2\lambda^{2/3}t )^{3/2}\over 3!!}, \end{aligned}$$
(3.30)
$$\begin{aligned} & y_{2}(t)=-{2\lambda\over \sqrt{\pi}} \int_{0}^{t} \biggl(-{4\lambda\over 3 \sqrt{\pi}} \tau^{3/2} \biggr) \sqrt{t-\tau} \,d\tau= {8\lambda^{2}\over 3 \pi} \int_{0}^{t} \tau^{3/2}\sqrt{t-\tau} \,d \tau = {\lambda^{2}t^{3}\over 3!}. \end{aligned}$$
(3.31)
The first step of the double induction principle is just verified by (3.25) taking into account (3.30), (3.31). For the second step, we suppose by induction hypothesis that we have
$$\begin{aligned} J_{2n}(t) = {\lambda^{2n}\over (3n)!}t^{3n}, \qquad J_{2n+1}(t) = - {2^{3n+2}\over (3(2n+1))!!}{\lambda^{2n+1}\over \sqrt{\pi}} t^{{3(2n+1)\over 2}}. \end{aligned}$$
(3.32)
Therefore, we obtain
$$\begin{aligned} J_{2n+2}(t) =& -{2\lambda\over \sqrt{\pi}} \int_{0}^{t} y_{2n+1}(\tau)\sqrt{t-\tau} \,d \tau= {\lambda^{2n+2}\over \pi}{2^{3n+3}\over (6n+3)!!} \int_{0}^{t} \tau^{3(2n+1)\over 2}\sqrt{t-\tau} \,d \tau \\ =& {\lambda^{2n+2}\over \pi} {2^{3(n+1)}\over (6n+3)!!} {\pi\over 2^{3(n+1)}} {(6n+3)!!\over (3(n+1))!} t^{3(n+1)} \\ =& {\lambda^{2n+2}\over (3(n+1))!} t^{3(n+1)} \end{aligned}$$
(3.33)
and
$$\begin{aligned} Y_{2n+3}(t) =& -{2\lambda\over \sqrt{\pi}} \int_{0}^{t} y_{2n+2}(\tau)\sqrt{t-\tau}\,d \tau= -{2\lambda^{2n+3}\over (3(n+1))! \sqrt{\pi}} \int_{0}^{t} \tau^{3n+3}\sqrt{t-\tau} \,d \tau \\ =& -{2\lambda^{2n+3}\over (3(n+1))! \sqrt{\pi}} {2^{3n+4}(3(n+1))!\over (6n+9)!!} t^{3(2n+3)\over 2} \\ =& -{2^{3(n+1)+2} \lambda^{2(n+1)+1}\over \sqrt{\pi} (3(2n+1)+1)!!} t^{{3(n+1)+1}\over 2}. \end{aligned}$$
(3.34)
This ends the proof. □
Remark 3.6
Taking \(t\to0^{+}\) in (3.14), (3.12), and (3.11), we obtain
$$\begin{aligned} & W \bigl( 0^{+} \bigr)= +\infty, \qquad W' \bigl( 0^{+} \bigr)= -\infty, \\ & U \bigl( 0^{+} \bigr) = 0, \qquad U' \bigl( 0^{+} \bigr) =+\infty, \\ & g \bigl( 0^{+} \bigr) = 1, \qquad g' \bigl(0^{+} \bigr) = 0. \end{aligned}$$
So we deduce that the heat flux W and the total heat flux U, and also g, are positive functions in a neighborhood of \(t=0\).