Reconstruction of a nonlinear heat transfer law from uncomplete boundary data by means of infrared thermography

The theoretical background of the direct model consists of a set of results by Pao [14] in which existence and uniqueness of solutions of parabolic equations with nonlinear boundary conditions are proven under suitable hypotheses. The main theoretical statement (Theorem 1.1, chapter 4 [14]) suggests a strategy for the numerical approximation of the solution as limit of a monotone sequence of solutions of linear problems. Stability of the solution of the direct model (1)–(5) with respect to is a corollary of theorem 1.1 chapter 4 [14].

To lighten the notation, in (2)–(3) we set ${U}^{a}={U}^{0}=0$ and ${\gamma }_{a}={\gamma }_{0}=\gamma$. In the introduction, the geometry of our thin plate was described by a rectangular strip of thickness $a\gt 0$. In order to apply Pao's theorem, here we assume that Ω is the convex open set in ${{\mathbb{R}}}^{2}$

$$\begin{eqnarray*}&&{S}_{-}\cup R\cup {S}_{+},\end{eqnarray*}$$

where

$$\begin{eqnarray*}&&R=(-L,L)\times (0,a)\end{eqnarray*}$$

with ($a\ll L$),

$$\begin{eqnarray*}&&{S}_{-}={\cup }_{x\in (-L-a/2,-L)}\left(\displaystyle \frac{a}{2}-\sqrt{{a}^{2}/4-{(x+L)}^{2}},\displaystyle \frac{a}{2}+\sqrt{{a}^{2}/4-{(x+L)}^{2}}\right),\end{eqnarray*}$$

$$\begin{eqnarray*}&&{S}_{+}={\cup }_{x\in (L,L+a/2)}\left(\displaystyle \frac{a}{2}-\sqrt{{a}^{2}/4-{(x-L)}^{2}},\displaystyle \frac{a}{2}+\sqrt{{a}^{2}/4-{(x-L)}^{2}}\right).\end{eqnarray*}$$

The domain Ω looks like a finite thin strip with smoothed corners. Its boundary is a closed curve of class C¹.

Furthermore, we write down our boundary conditions in the form

$$\begin{eqnarray}&&\displaystyle \frac{\partial u}{\partial \nu }(P,t)+\gamma u=g(P(x,z),t,u),\end{eqnarray} \tag{ 6 }$$

where $P(x,z)\in \partial {\rm{\Omega }}$ and ν is the outward unit normal to the boundary $\partial {\rm{\Omega }}$. The function g is continuous and it is piecewise defined by

$$\begin{eqnarray*}&&g(P,t,u)=-\epsilon f(u(x,a,t))\ {\rm{in}}\ \{(x,a)\ {\rm{with}}\ x\in (-L,L)\}\end{eqnarray*}$$

and

$$\begin{eqnarray*}&&g(P,t,u)=\displaystyle \frac{{\rm{\Phi }}(x,t)}{\kappa }\ {\rm{in}}\ \{(x,0)\ {\rm{with}}\ x\in (-L,L)\}.\end{eqnarray*}$$

In the rest of the boundary we have $g=0=\gamma$.

To describe the essentials of this result, we must introduce the definition of upper (lower) solution: a function $\tilde{u}\in C({\bar{D}}_{\tau })\cap {C}^{2}({D}_{\tau })$ is called an upper solution of (1), homogeneous (4) and (6) if it satisfies the inequalities

$$\begin{eqnarray}&&{\tilde{u}}_{t}-\alpha {\rm{\Delta }}\tilde{u}\geqslant 0\ {\rm{in}}\ {\rm{\Omega }}\times (0,\tau ],\end{eqnarray} \tag{ 7 }$$

$$\begin{eqnarray}&&{\tilde{u}}_{\nu }+\gamma \tilde{u}\geqslant g\ {\rm{on}}\ {S}_{\tau }.\end{eqnarray} \tag{ 8 }$$

and the initial condition

$$\begin{eqnarray}&&\tilde{u}(x,z,0)\geqslant 0\ {\rm{in}}\ {\rm{\Omega }}.\end{eqnarray} \tag{ 9 }$$

As for the lower solution $\hat{u}$ the definition is the same, only changing $\geqslant$ with $\leqslant$. The sector $J=\langle \hat{u},\tilde{u}\rangle$ is defined as

$$\begin{eqnarray*}&&J=\{v\in C({\bar{D}}_{T})\phantom{{aa}}s.t.\phantom{{aa}}\hat{u}\leqslant v\leqslant \tilde{u}\}.\end{eqnarray*}$$

Assume that $g\in C({S}_{\tau }\times J)$ and that g is Lipschitz in J.

The existence of a unique solution u of (1), homogeneous (4) and (6) is proved by Pao in theorem 4.1.1 [14] under the assumption that there exist a lower and an upper solution $\hat{u}$ and $\tilde{u}$ of it. The proof is based on the iterative construction of two sequences, $\{{\hat{V}}^{k}\}$ and $\{{\tilde{V}}^{k}\}$, that converge monotonically to the solution u. The decreasing sequence $\{{\tilde{V}}^{k}\}$ starts with the upper solution $\tilde{u}$ and approximates the solution u from above while $\{{\hat{V}}^{k}\}$ starts with the lower solution $\hat{u}$ and converges to u monotonically from below. Numerical implementation of $\{{\hat{V}}^{k}\}$ and $\{{\tilde{V}}^{k}\}$ seems to be very expensive though each ${\tilde{V}}^{k}$ is determined by solving a linear BVP whose boundary conditions involve ${\tilde{V}}^{k-1}$. Details about the definition of sequences $\{{\hat{V}}^{k}\}$ and $\{{\tilde{V}}^{k}\}$ are in [14] section 4.1. This construction furnishes the main tool for providing the stability estimate for u given in subsection 2.2.

Finally, we show that the sector J is not empty. Actually, consider the linear function

$$\begin{eqnarray*}&&\tilde{u}=C\parallel {\rm{\Phi }}{\parallel }_{\infty }+D\epsilon \parallel f{\parallel }_{\infty }-\parallel {\rm{\Phi }}{\parallel }_{\infty }z\end{eqnarray*}$$

with $C\geqslant \tfrac{1}{\gamma }\left(\tfrac{1}{\kappa }+1+a\right)$ and $D\geqslant \tfrac{1}{\gamma }$. Straightforward calculations show that $\tilde{u}$ is an upper solution and $\hat{u}(x,z,t)=-\tilde{u}$ is a lower solution of (1), homogeneous (4) and (6).

Theorem. Let ${u}^{\epsilon }\in C({\bar{D}}_{\tau })\cap {C}^{2}({D}_{\tau })$ be the solution of (1), homogeneous (4) and (6). We have

$$\begin{eqnarray}&&\parallel {u}^{\epsilon }-{u}^{0}{\parallel }_{\infty }\leqslant C\epsilon .\end{eqnarray} \tag{ 10 }$$

Proof. We recall that $\{{\tilde{V}}^{k}\}$ converges monotonically to ${u}^{\epsilon }$. It means that

$$\begin{eqnarray}&&\parallel u-{\tilde{V}}^{k}{\parallel }_{\infty }\leqslant {C}_{1}\epsilon \end{eqnarray} \tag{ 11 }$$

for $k\geqslant k1$. The IBVP solved by ${\tilde{V}}^{k}$ is linear. If ${\tilde{u}}^{0k}$ is an upper solution of (1), homogeneous (4) and (6) when $\epsilon =0$, it is well known (stability of linear IBVP w.r.t. parameters, see for example [16] page 507) that

$$\begin{eqnarray}&&\parallel {\tilde{V}}^{k}-{\tilde{u}}^{0k}{\parallel }_{\infty }\leqslant {C}_{2}\epsilon .\end{eqnarray} \tag{ 12 }$$

On the other hand, we have

$$\begin{eqnarray}&&\parallel {u}^{0}-{\tilde{u}}^{0k}{\parallel }_{\infty }\leqslant {C}_{3}\epsilon \end{eqnarray} \tag{ 13 }$$

for $k\geqslant k2$. Finally, (10) turns out to be true for $C={C}_{1}+{C}_{2}+{C}_{3}$ and $k\geqslant \max \{k1,k2\}$.

The nonlinear term $\epsilon \,f$ in (2) is unknown. We will use the equations (1)–(5)of the direct model and the knowledge of thermal contrast at z = 0 to recover $\epsilon \,f$.

First, we plug the formal expansion ${u}^{\epsilon }=u0+\epsilon u1+O({\epsilon }^{2})$ in the IBVP (1)–(5) and transform it in a perturbative hierarchy of linear problems. We consider only order zero and order one of the scheme.

The term u0 satisfies the heat equation

$$\begin{eqnarray}&&{u}_{t}=\alpha {\rm{\Delta }}u\ {\rm{in}}\ {D}_{\tau }\end{eqnarray} \tag{ 14 }$$

with boundary-initial conditions

$$\begin{eqnarray}&&-{u}_{z}(x,0,t)=\displaystyle \frac{{\rm{\Phi }}(x,t)}{\kappa }\ {\rm{for}}\ x\in (-L,L)\phantom{{aa}}t\in (0,\tau ],\end{eqnarray} \tag{ 15 }$$

$$\begin{eqnarray}&&{u}_{z}(x,a,t)+\gamma (u(x,a,t)-{U}^{a})=0\ {\rm{for}}\ x\in (-L,L)\phantom{{aa}}t\in (0,\tau ],\end{eqnarray} \tag{ 16 }$$

$$\begin{eqnarray}&&u(x,z,0)={T}_{{\rm{in}}}(x,z)\ {\rm{for}}\ (x,z)\ {\rm{in}}\ {\rm{\Omega }},\end{eqnarray} \tag{ 17 }$$

$$\begin{eqnarray}&&-{u}_{x}(-L,z,t)={u}_{x}(L,z,t)=0\ {\rm{for}}\ (z,t)\in (0,a)\times (0,\tau ].\end{eqnarray} \tag{ 18 }$$

The solution u0 is just the background temperature u⁰. Observe that (14)–(34) is well-posed and, in particular, once the physical parameters T_in, α, γ, κ and Φ are known, u⁰ is uniquely determined. On the other hand, the function $W=\epsilon u1$ fulfills the heat equation

$$\begin{eqnarray}&&{W}_{t}=\alpha {\rm{\Delta }}W\ {\rm{in}}\ {D}_{\tau }\end{eqnarray} \tag{ 19 }$$

with linear boundary conditions

$$\begin{eqnarray}&&{W}_{z}(x,a,t)+\gamma W(x,a,t)+\tilde{f}({u}^{0})=0,\phantom{{aa}}x\in (-L,L),\phantom{{aa}}t\in (0,\tau ]\end{eqnarray} \tag{ 20 }$$

with $\tilde{f}=\epsilon f$

$$\begin{eqnarray}&&-{W}_{x}(-L,z,t)={W}_{x}(L,z,t)=0\ {\rm{for}}\ (z,t)\in (0,a)\times (0,\tau ],\end{eqnarray} \tag{ 21 }$$

$$\begin{eqnarray}&&{W}_{z}(x,0,t)=0,x\in (-L,L),t\in (0,\tau ],\end{eqnarray} \tag{ 22 }$$

and initial condition

$$\begin{eqnarray}&&W(x,z,0)=0\ {\rm{for}}\ (x,z)\in {\rm{\Omega }}.\end{eqnarray} \tag{ 23 }$$

The thermal contrast (measured in real cases by means of an infrared camera)

$$\begin{eqnarray}&&G(x,t)={u}^{\epsilon }(x,0,t)-{u}^{0}(x,0,t),x\in (-L,L),t\in (0,\tau ]\end{eqnarray} \tag{ 24 }$$

gives us the following noisy additional boundary condition that will be used in section 3.2 to recover $\tilde{f}$:

$$\begin{eqnarray}&&W(x,0,t)\approx G(x,t),x\in (-L,L),t\in (0,\tau ].\end{eqnarray} \tag{ 25 }$$

Actually, we prove that $\tfrac{\parallel u-{u}^{0}-W\parallel }{\epsilon }=O(\epsilon )$. We set

$$\begin{eqnarray*}&&v=\displaystyle \frac{u-{u}^{0}-W}{\epsilon }.\end{eqnarray*}$$

We have

$$\begin{eqnarray}&&{v}_{t}=\alpha {\rm{\Delta }}v\ {\rm{in}}\ {D}_{\tau }\end{eqnarray} \tag{ 26 }$$

with boundary and initial conditions

$$\begin{eqnarray}&&{v}_{z}(x,a,t)+\gamma v(x,a,t)-(f(u(x,a,t))-f({u}^{0}(x,a,t)))=0,x\in (-L,L),t\in (0,\tau ]\end{eqnarray} \tag{ 27 }$$

$$\begin{eqnarray}&&{v}_{z}(x,0,t)=0,x\in (-L,L),t\in (0,\tau ]\end{eqnarray} \tag{ 28 }$$

$$\begin{eqnarray}&&-{v}_{x}(-L,z,t)={v}_{x}(L,z,t)=0\ {\rm{for}}\ (z,t)\in (0,a)\times (0,\tau ]\end{eqnarray} \tag{ 29 }$$

and

$$\begin{eqnarray}&&v(x,z,0)=0,(x,z)\in {\rm{\Omega }}.\end{eqnarray} \tag{ 30 }$$

Since $f\in C(J)$ is supposed Lipschitz in J, we have from (10) that

$$\begin{eqnarray*}&&\parallel f(u)-f({u}^{0})\parallel \leqslant \parallel f{\parallel }_{\infty }\parallel u-{u}^{0}\parallel \leqslant C\parallel f{\parallel }_{\infty }\epsilon .\end{eqnarray*}$$

It comes from classical estimates (see [16] page 507) that

$$\begin{eqnarray*}&&\parallel v\parallel \leqslant A\epsilon .\end{eqnarray*}$$

Let ${\{{{\rm{\Psi }}}_{k}\ :J\to {\mathbb{R}}\}}_{k=0}^{\infty }$ be a sequence of linearly independent functions that span $C(J)$ so that $\tilde{f}(v)={\sum }_{k}{\beta }_{k}{{\rm{\Psi }}}_{k}(v)$. Furthermore, let ${W}^{(k)}$ solve the linear IBVP

$$\begin{eqnarray}&&{W}_{t}=\alpha {\rm{\Delta }}W\ {\rm{in}}\ {D}_{\tau }\end{eqnarray} \tag{ 31 }$$

with boundary conditions

$$\begin{eqnarray}&&{W}_{z}(x,0,t)=0\end{eqnarray} \tag{ 32 }$$

$$\begin{eqnarray}&&{W}_{z}(x,a,t)+\gamma W(x,a,t)-{{\rm{\Psi }}}_{k}({u}^{0})=0\end{eqnarray} \tag{ 33 }$$

$$\begin{eqnarray}&&-{W}_{x}(-L,z,t)={W}_{x}(L,z,t)=0\ {\rm{for}}\ (z,t)\in (0,a)\times (0,\tau ]\end{eqnarray} \tag{ 34 }$$

and initial data

$$\begin{eqnarray}&&W(x,z,0)=0.\end{eqnarray} \tag{ 35 }$$

It comes from linearity that the solution of (19)–(23) takes the form of

$$\begin{eqnarray*}&&W(x,z,t;\beta )=\displaystyle \sum _{k}{\beta }_{k}{W}^{(k)}({u}^{0}(x,z,t)).\end{eqnarray*}$$

Since $\tilde{f}$ (or, equivalently, the vector β) is not known, we can try to approximate it from the knowledge of the thermal contrast $G(x,t)={u}^{\epsilon }(x,0,t)-{u}^{0}(x,0,t)$. To do this, we set the following minimum problem in finite dimension

$$\begin{eqnarray}&&\mathop{\min }\limits_{\beta =({\beta }_{0},\ldots ,{\beta }_{N})}\{\parallel W(x,0,t;\beta )-G(x,t){\parallel }_{2}^{2}+\lambda B(\beta )\}.\end{eqnarray} \tag{ 36 }$$

Since the ${W}^{(k)}{\rm{s}}$ are linearly independent, any finite Gram matrix ${H}^{(N)}={\left({\int }_{[-L,L]\times [0,\tau ]}{W}^{(j)}{W}^{(k)}{\rm{d}}{x}{\rm{d}}{t}\right)}_{j,k=0,\ldots ,N}$ is expected to be nonsingular, even if it could be severely ill-conditioned. For this reason, regularization is required to handle truncation errors and the effects of noise affecting our data.

As for the penalty $B(\beta )$, it must be chosen using a priori information if available. In fact, we assume that $\tilde{f}$ is smooth and increasing. This assumption is supported by a number of known examples of nonlinear heat transfer coefficients ([1, 4, 5, 12] and many others) and by private discussions. The idea is that the higher the temperature, the greater the rate at which heat passes through an interface. Moreover, sudden jumps related to the temperature increase are not expected. A good choice of B could be the L² norm of the first derivative of $\tilde{f}$.

Hence, we assume that $\tilde{f}(\xi )$ ($\xi \in J$) is a non decreasing function which can be approximated by means of low-order polynomials. At this stage, we assume ${{\rm{\Psi }}}_{k}(\xi )={\xi }^{k}$ so that we will work with the finite expansion

$$\begin{eqnarray}&&{\tilde{f}}^{(N)}(\xi )=\displaystyle \sum _{k=0}^{N}{\beta }_{k}{\xi }^{k}.\end{eqnarray} \tag{ 37 }$$

In our tests, we take N = 4. This value of N comes from a compromise between the accuracy of the approximation and the stability of the solutions. We write the L² norm of $\tilde{f}^{\prime} (\xi )={\sum }_{k=1}^{N}k{\beta }_{k}{\xi }^{k-1}$ as the quadratic form in β

$$\begin{eqnarray*}&&({B}^{(N)}\beta ,\beta )=\displaystyle \sum _{k,j=1}^{N}k{\beta }_{k}j{\beta }_{j}{\int }_{[-L,L]\times [0,\tau ]}{u}^{0}{(x,0,t)}^{j+k-2}{\rm{d}}{x}{\rm{d}}{t}.\end{eqnarray*}$$

Finally, the Euler equations of our minimum problem are

$$\begin{eqnarray}&&\displaystyle \sum _{k=0}^{N}\left({\displaystyle \int }_{[-L,L]\times [0,\tau ]}\ {W}^{(j)}{W}^{(k)}{\rm{d}}{x}{\rm{d}}{t}\right.+\left.\lambda {kj}{\displaystyle \int }_{[-L,L]\times [0,\tau ]}{u}^{0}{(x,0,t)}^{j+k-2}{\rm{d}}{x}{\rm{d}}{t}\right){\beta }_{k}\\ &&\quad ={\displaystyle \int }_{[-L,L]\times [0,\tau ]}{W}^{(j)}{G}{\rm{d}}{x}{\rm{d}}{y}{\rm{d}}{t}\end{eqnarray} \tag{ 38 }$$

for $j=0,\ldots ,N$.

We suppose that the heat flux Φ is constant in x and that the unknown nonlinear function is $\tilde{f}=\epsilon \cdot \eta \cdot {(u(a,t)-{T}_{a})}^{2}$ with $\epsilon =0.001$ and $\eta =100$. Hence, it is natural to reduce ourselves to a one-dimensional problem (see also [3]), where $z\in [0,a]$. We provide heat to the boundary z = 0 through a flux of density ${\rm{\Phi }}(t)={{Qe}}^{-\tfrac{t}{\theta }}$ while nonlinear heat exchange takes place on the opposite boundary z = a.

The temperature of our one-dimensional sample is a function ${u}^{\epsilon }(z,t)$ that solves the IBVP

$$\begin{eqnarray}&&{u}_{t}=\alpha {u}_{{zz}},\end{eqnarray} \tag{ 39 }$$

$$\begin{eqnarray}&&{u}_{z}(0,t)={\gamma }_{0}\cdot (u(0,t)-{T}_{a})-\displaystyle \frac{Q}{\kappa }{e}^{-\displaystyle \frac{t}{\theta }},\end{eqnarray} \tag{ 40 }$$

$$\begin{eqnarray}&&{u}_{z}(a,t)=-{\gamma }_{a}\cdot (u(a,t)-{T}_{a})-\epsilon \cdot \eta \cdot {(u(a,t)-{T}_{a})}^{2},\end{eqnarray} \tag{ 41 }$$

$$\begin{eqnarray}&&u(z,0)={T}_{a};\end{eqnarray} \tag{ 42 }$$

where $\alpha =6\times {10}^{-5}\,{\rm{s}}\,{{\rm{m}}}^{-2}$, ${T}_{a}=20\,^\circ {\rm{C}}$, $\gamma ={\gamma }_{0}={\gamma }_{a}=10\,{{\rm{m}}}^{-1}$, $\tfrac{Q}{\kappa }=10000\,^\circ {\rm{C}}\,{{\rm{m}}}^{-1}$, $\theta =1\,{\rm{s}}$.

We solve numerically equations (39)–(42) by means of a finite difference scheme (for the explicit and implicit-explicit numerical schemes adopted here, see [2]). The trace ${u}^{\epsilon }(0,t)$ of the solution is assumed to play the role of real temperature data. In experiments, these values are taken by means of an infrared camera.

${u}^{0}(x,t)$ is the background temperature corresponding to $\epsilon =0$, i.e. the situation in which heat exchange through the boundary z = a follows Newton's cooling law.

Consider the expansion ${u}^{\epsilon }={u}^{0}+\epsilon u1+O({\epsilon }^{2})$. It is easy to check that $\epsilon u1(0,t)+O({\epsilon }^{2})$ is just the thermal contrast considered in the definition of the inverse problem IP. Since is actually not known, we set $W=\epsilon u1$ and $\tilde{\eta }=\epsilon \eta$. We plug this expansion in the IBVP above and carry out a perturbative analysis.

4.1.1. Order zero

The function u⁰ fulfills

$$\begin{eqnarray}&&{u}_{t}=\alpha {u}_{{zz}},\end{eqnarray} \tag{ 43 }$$

$$\begin{eqnarray}&&{u}_{z}(0,t)=\gamma \cdot (u(0,t)-{T}_{a})-\displaystyle \frac{Q}{\kappa }{e}^{-t},\end{eqnarray} \tag{ 44 }$$

$$\begin{eqnarray}&&{u}_{z}(a,t)=-\gamma \cdot (u(a,t)-{T}_{a}).\end{eqnarray} \tag{ 45 }$$

Because of the hierarchic structure of perturbations, u⁰ will enter as a given quantity in the following IBVP corresponding to the order one.

4.1.2. Order one

The scaled first order solution W solves

$$\begin{eqnarray}&&{W}_{t}=\alpha {W}_{{zz}},\end{eqnarray} \tag{ 46 }$$

$$\begin{eqnarray}&&{W}_{z}(0,t)=\gamma W(0,t),\end{eqnarray} \tag{ 47 }$$

$$\begin{eqnarray}&&{W}_{z}(a,t)=-\gamma \cdot W(a,t)-\tilde{\eta }{({u}^{0}(a,t)-{T}_{a})}^{2},\end{eqnarray} \tag{ 48 }$$

$$\begin{eqnarray}&&W(z,0)=0.\end{eqnarray} \tag{ 49 }$$

It comes from the linearity of (46)–(49) that

$$\begin{eqnarray*}&&W(z,t)=\displaystyle \sum _{k=0}^{N}\beta (k){W}^{(k)}(z,t),\end{eqnarray*}$$

where ${W}^{(k)}$ is the solution of

$$\begin{eqnarray}&&{W}_{t}=\alpha {W}_{{zz}},\end{eqnarray} \tag{ 50 }$$

$$\begin{eqnarray}&&{W}_{z}(0,t)=\gamma \cdot W(0,t),\end{eqnarray} \tag{ 51 }$$

$$\begin{eqnarray}&&{W}_{z}(a,t)=-\gamma \cdot W(a,t)-{({u}^{0}(a,t)-{T}_{a})}^{k},\end{eqnarray} \tag{ 52 }$$

$$\begin{eqnarray}&&W(z,0)=0,\end{eqnarray} \tag{ 53 }$$

for $k=0,\mathrm{..},N$.

We recall that $W(0,t)$ is the thermal contrast defined in (20). To make contrast data more realistic, we add white Gaussian noise ${g}_{\sigma }(t)$ of average zero and standard deviation $\sigma =0.3$ °C.

Since the matrix ${\int }_{0}^{\tau }{W}^{(j)}(t){W}^{(k)}(t){\rm{d}}{t}$ is ill-conditioned, to determine the values of $\beta (0),\ldots ,\beta (4)$ we solve the linear system (38) for N = 4 with an optimal choice of the regularization parameter λ. Recall that the penalty function B is the Euclidean norm of the first derivative of the nonlinear term f.

The parameter λ is determined by means of Hansen's L-curve method [9]. In figure 1 we summarize the essentials of the numerical example:

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Figure 1(a). We plot the graphs of $U(t)={u}^{\epsilon }(0,t)+{g}_{\sigma }(t)$ (that plays the role of experimental temperature data at the boundary z = 0) and ${u}^{0}(0,t)$ (the solution of order zero).

Figure 1(b). We plot the curve $(\parallel \sum {\beta }_{k}(\lambda ){W}^{(k)}-U{\parallel }_{2}^{2},\parallel B(\beta (\lambda )){\parallel }_{2}^{2})$ parametrized by λ. The corner suggests a choice for the numerical value of λ.

Figure 1(c). Finally, we compare the unknown $\tilde{\eta }$ with the polynomial reconstructions with $\lambda =0$ and ${\lambda }_{{\rm{opt}}}=1.4\times {10}^{-10}$.

We simulate cooling of a one-dimensional specimen represented by the interval $(0,a)$ by means of the following IBVP (see [5]):

$$\begin{eqnarray}&&{u}_{t}={10}^{-6}{u}_{{zz}}\ {\rm{in}}\ {D}_{\tau },\end{eqnarray} \tag{ 54 }$$

where ${D}_{\tau }=(0,a)\times (0,\tau )$

$$\begin{eqnarray}&&{u}_{z}(0,t)=0,\end{eqnarray} \tag{ 55 }$$

$$\begin{eqnarray}&&{u}_{z}(a,t)+F(u)=0,\end{eqnarray} \tag{ 56 }$$

$$\begin{eqnarray}&&u(z,0)=1500\,^\circ {\rm{C}}.\end{eqnarray} \tag{ 57 }$$

Let us assume

$$\begin{eqnarray}&&F(u)=\displaystyle \frac{A}{2000-u}(u-500),\end{eqnarray} \tag{ 58 }$$

where A = 367 500. We recall that there is no controlled heat flux here (${\rm{\Phi }}=0$).

The function $T(t)=u(a,t)$ simulates the collection of data by means of the infrared camera. Simulated temperature for z = a become more realistic by adding for all $t\in (0,\tau )$ a random variable g with uniform distribution on the interval $[-5\,^\circ {\rm{C}},5\,^\circ {\rm{C}}]$ (the same noise used in [5]).

In order to apply again the perturbative scheme of the previous example we introduce a linear part ${\gamma }_{0}(u-q)$ so that $\tilde{f}(u)=F(u)-{\gamma }_{0}(u-q)$. A natural choice for q is the surrounding temperature $q=500\,^\circ {\rm{C}}$. We leave ${\gamma }_{0}$ as a free parameter that will be identified in the next subsection.

4.2.1. Order zero. Choice of the linear part

The order zero solution ${u}^{0,{\gamma }_{0}}$ (we are stressing the dependence on ${\gamma }_{0}$) solves the IBVP

$$\begin{eqnarray}&&{u}_{t}={10}^{-6}{u}_{{zz}},\end{eqnarray} \tag{ 59 }$$

$$\begin{eqnarray}&&{u}_{z}(0,t)=0,\end{eqnarray} \tag{ 60 }$$

$$\begin{eqnarray}&&{u}_{z}(a,t)+{\gamma }_{0}(u(a,t)-500)=0.\end{eqnarray} \tag{ 61 }$$

The parameter ${\gamma }_{0}$ must be chosen to minimize the integral distance ${\int }_{1000}^{1500}| {u}^{0,{\gamma }_{0}}(a,t)-T(t){| }^{2}{\rm{d}}{t}$. The best value is ${\gamma }_{0}\approx 443$ ${{\rm{m}}}^{-1}$.

4.2.2. Order one

We compute a finite basis $\{{W}^{(k)}\}$ solving the IBVP

$$\begin{eqnarray}&&{W}_{t}={10}^{-6}{W}_{{zz}}\ {\rm{in}}\ {D}_{\tau },\end{eqnarray} \tag{ 62 }$$

$$\begin{eqnarray}&&{W}_{z}(0,t)=0,\end{eqnarray} \tag{ 63 }$$

$$\begin{eqnarray}&&{W}_{z}(a,t)+{\gamma }_{0}W+{({u}^{0}(a,t)-500)}^{k}=0,\end{eqnarray} \tag{ 64 }$$

for $k=0,\,..,4$. Then, using equations (37) and (38), we estimate the vector coefficient β to approximate the nonlinear part $\tilde{f}(u)=F(u)-{\gamma }_{0}(u-500)$.

Remark. We observe numerically that the temperature approximation of the first order ${u}^{0}(a,t)+W(a,t)$ is a very good approximation of the solution of (54)–(57) (see figure 2).

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Since the matrix ${\int }_{0}^{\tau }{W}^{(j)}(t){W}^{(k)}(t){\rm{d}}{t}$ is severely ill-conditioned, the values of $\beta (0),\ldots ,\beta (0)$ are obtained by solving the regularized linear system (38) for N = 4. The regularization parameter λ is determined by means of Hansen's L-curve method [9].

In figure 3, we summarize the essentials of the numerical example:

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Figure 3(a). Here, we plot the functions $u(a,t)+g(t)$ (that plays the role of experimental temperaure data at the boundary z = 0) and ${u}^{0}(a,t)$ (the solution of order zero).

Figure 3(b). We plot of the curve $(\parallel \sum {\beta }_{k}(\lambda ){W}^{(k)}-U{\parallel }_{2}^{2},\parallel B(\beta (\lambda )){\parallel }_{2}^{2})$ parametrized by λ. The corner suggests the choice for λ.

Figure 3(c). We compare the unknown $F(u)$ with the polynomial reconstructions with $\lambda =0$ and ${\lambda }_{{\rm{opt}}}=3.3\times {10}^{-16}$.

Finally, the regularized curve shown in figure 3(c) is a monotone approximation of $F(u)$ obtained by smoothing the input data and regularizing the solutions. In [5], the additional condition

$$\begin{eqnarray*}&&F(T(0.005))=\displaystyle \frac{A}{2000-u(0.005)}(u(0.005)-500)\end{eqnarray*}$$

is used to obtain a monotone approximation of F. In our approximation, we don't use any assumption on the values of F.