An inverse problem for the
porous medium equation
with partial data and a possibly singular absorption term

Before stating the main result concering the inverse boundary value problem for equation (1), we first need to outline the sense in which we consider a function $u$ to be a solution of (1). The approach is adapted from [49].

Let $Q_{T}:=(0,T)\times\Omega$ and $S_{T}:=(0,T)\times\partial\Omega$. In order to define a suitable space of Dirichlet boundary data. let

$$C_{t}(Q_{T})=\left\{\varphi\in C^{\infty}(\overline{Q_{T}}):\mbox{supp\;}\varphi\cap\left[\{T\}\times\Omega\right]=\emptyset\right\}$$

(3)

and let $H^{1}_{t}(Q_{T})$ be the completion of this space in $H^{1}(Q_{T})$. We will denote by $H^{\frac{1}{2}}_{t}(S_{T})$ the subspace of $H^{\frac{1}{2}}(S_{T})$ that consists of traces of $H^{1}_{t}(Q_{T})$ functions. We also introduce

$$C_{\diamond}(Q_{T})=\left\{\varphi\in C^{\infty}(\overline{Q_{T}}):\mbox{supp\;}\varphi\cap\left[S_{T}\cup\{T\}\times\Omega\right]=\emptyset\right\},$$

(4)

and its completion $H^{1}_{\diamond}(Q_{T})$ in $H^{1}(Q_{T})$.

Let by $\tau_{\Omega}$ the boundary trace operator for the domain $\Omega$. It is bounded between $H^{1}(\Omega)$ and $H^{\frac{1}{2}}(\partial\Omega)$. Let $\tau_{Q_{T}}$ be the boundary trace operator for $Q_{T}$ to the side boundary $S_{T}$. It is not hard to check that $\tau_{Q_{T}}$ is bounded from $L^{2}((0,T);H^{1}(\Omega))$ to $L^{2}((0,T);H^{\frac{1}{2}}(\partial\Omega))$.

We will show in section 2 that

We have been unable to find a proof of this exact result in the available literature on the porous medium equation. For this reason, and also for the convenience of the reader, we provide our own proof. The methods needed are not new. We have followed the argument in [49], with some additional techniques from [1].

If $\psi$ is as in the above theorem and $u$ is the corresponding weak solution, the Neumann data $\gamma\partial_{\nu}u^{m}|_{S_{T}}$ can be defined by

$$\langle\gamma\partial_{\nu}u^{m}|_{S_{T}},\psi|_{S_{T}}\rangle=\int_{Q_{T}}\left(\gamma\nabla\psi\cdot\nabla(u^{m})+\lambda\psi u^{q}-\epsilon\partial_{t}\psi u\right)\,\text{d}t\,\text{d}x.$$

(9)

We can now define the Dirichlet-to-Neumann map

$$\Lambda_{\epsilon,\gamma,\lambda}^{PM}(\phi)=\gamma\partial_{\nu}u^{m}|_{S_{T}}\in\left(H^{\frac{1}{2}}_{t}(S_{T})\right)^{\prime}.$$

(10)

Suppose that we have two sets of coefficients, $\epsilon^{(i)}$, $\gamma^{(i)}$, $\lambda^{(i)}$, and $\epsilon^{(ii)}$, $\gamma^{(ii)}$, $\lambda^{(ii)}$, that are as above. The first main result of our paper is the following.

In the case when the $\gamma$ coefficient is a priori known to be constant, we also have the following partial data result.

We give a proof of Theorem 2 in section 3. The main idea for the proof is to use two sucessive transformations of equation (1). The first is the change of function $v=u^{m}$, which gives us the equation

$$\epsilon\partial_{t}v^{\frac{1}{m}}-\nabla\cdot(\gamma\nabla v)+\lambda v^{\frac{q}{m}}=0.$$

(11)

The second transformation involves the function

$$V(x)=\int_{0}^{T}(T-t)^{\alpha}v(t,x)\,\text{d}t,$$

(12)

where $\alpha,T>0$ will be chosen in the course of the proof. Since the equation satisfied by $v$ is non-linear, we do not obtain a closed form PDE for $V$. It is however possible to show that $V$ satisfies a differential inequality of the form

$$0\leq\nabla\cdot(\gamma(x)\nabla V(x))\leq C_{1}\epsilon(x)(V(x))^{\frac{1}{m}}+C_{2}\lambda(x)(V(x))^{\frac{q}{m}}.$$

(13)

If we choose Dirichlet boundary data so that $v|_{S_{T}}(t,x)=ht^{m}g(x)$, with $h$ a large parameter, we can deduce the first few terms in the asymptotic expansion of $V$ as $h\to\infty$. The Neumann data for each of these terms is determined by the Dirichlet-to-Neumann map, and we can use this information to separately show uniqueness for each of the $\epsilon$, $\gamma$, and $\lambda$ coefficients.

We prove Theorem 3 in section 4. The starting point for the argument is an integral identity derived in section 3. We reduce this integral identity to a problem similar to the linearized local Calderón problem considered in [17], but with an $L^{1}$ coefficient function rather than the $L^{\infty}$ one. The result follows by adapting the original argument to this new situation.

Finally, we would like to remark on the importance of the partial data case. Of course, one reason to consider this case is that in possible real life applications only partial data may be practically available. Here however there is an additional theoretical reason, namely that in the partial data case the equation remains in a degenerate regime for the entire time of observation, since the solution must remain zero on a part of the boundary. This shows that our method does not work by moving the equation to a non-degenerate regime, but rather that it can handle the equation even in situations that are not covered by previous works.

We will now make the above rigorous. Let $a_{k}(x,z)$, $b_{k}(x,z)$, $k=1,2,3,\ldots$, be smooth, bounded functions, with the $a_{k}$ also having strictly positive lower bounds, and such that when

$$\frac{1}{k}\leq z\leq\sup_{S_{T}}\phi+T\sup_{Q_{T}}f+\frac{1+T}{k},$$

(14)

we have

$$a_{k}(x,z)=m\gamma(x)z^{m-1},\quad b_{k}(x,z)=\lambda(x)z^{q}.$$

(15)

We also choose functions $f_{k}\in C^{\infty}(Q_{T})$ such that

$$f(t,x)\leq f_{k+1}(t,x)\leq f_{k}(t,x)\leq f(t,x)+\frac{1}{k}.$$

(16)

Let now $u_{k}$ be the solutions of

$$\left\{\begin{array}[]{l}\epsilon\partial_{t}u_{k}-\nabla\cdot\left(a_{k}(x,u_{k})\nabla u_{k})\right)+b_{k}(x,u_{k})=f_{k},\\[5.0pt] u_{k}(0,x)=\frac{1}{k},\quad u_{k}|_{[0,T)\times\partial\Omega}=\phi+\frac{1}{k}.\end{array}\right.$$

(17)

The problem (17) has a solution $u_{k}\in C^{1,2}(Q_{T})\cap C(\overline{Q_{T}})$ (see [34], or [37]), which furthermore satisfies the maximum principle

$$\frac{1}{k}\leq u_{k}(t,x)\leq\frac{1}{k}+\sup_{S_{T}}\phi+t\sup_{Q_{T}}f_{k}.$$

(18)

It follows that $u_{k}$ is also a solution to

$$\left\{\begin{array}[]{l}\epsilon\partial_{t}u_{k}-\nabla\cdot(\gamma\nabla u^{m}_{k})+\lambda u_{k}^{q}=f_{k},\\[5.0pt] u_{k}(0,x)=\frac{1}{k},\quad u_{k}|_{[0,T)\times\partial\Omega}=\phi+\frac{1}{k}.\end{array}\right.$$

(19)

By (18), $u_{k}$ is also a solution to $\epsilon\partial_{t}u_{k}-\nabla\cdot\left(a_{k+1}(x,u_{k},\nabla u_{k})\right)=f_{k}$. By the comparison principle (see [37, Theorem 9.7]), we have that

$$0\leq u_{k+1}\leq u_{k},\quad\forall k=1,2\ldots.$$

(20)

Let $u$ be the pointwise limit

$$u(t,x)=\lim_{k\to\infty}u_{k}(t,x).$$

(21)

It is clear that $u\in L^{\infty}(\Omega)$ and

$$0\leq u(t,x)\leq\sup_{S_{T}}\phi+t\sup_{Q_{T}}f.$$

(22)

A simple application of the monotone convergence theorem gives that in $L^{2}(Q_{T})$ norm we have $u_{k}\to u$, $u_{k}^{m}\to u^{m}$, and $u_{k}^{q}\to u^{q}$.

Let $\tilde{\phi}:Q_{T}\to[0,\infty)$ be a smooth extension of the Dirichlet data, such that we still have $\tilde{\phi}(0,x)=0$ for all $x\in\Omega$ and

$$||\tilde{\phi}||_{W^{1,\infty}(Q_{T})}\leq C||\phi||_{C^{0,1}(S_{T})},$$

(23)

with a constant $C>0$. Let

$$\eta_{k}=u_{k}^{m}-\left(\tilde{\phi}+\frac{1}{k}\right)^{m},$$

(24)

which is zero on $S_{T}$. We will multiply (19) by $\eta_{k}$ and integrate over $Q_{T}$. Note that

$$-\int_{Q_{T}}\eta_{k}\nabla(\gamma\nabla u_{k}^{m})\,\text{d}t\,\text{d}x\\[5.0pt] =\int_{Q_{T}}\gamma|\nabla u_{k}^{m}|^{2}\,\text{d}t\,\text{d}x-\int_{Q_{T}}\gamma\nabla\left(\tilde{\phi}+\frac{1}{k}\right)^{m}\cdot\nabla u_{k}^{m}\,\text{d}t\,\text{d}x,$$

(25)

and

$$\int_{Q_{T}}\epsilon\partial_{t}u_{k}\eta_{k}\,\text{d}t\,\text{d}x=\int_{\Omega}\frac{\epsilon}{m+1}u_{k}^{m+1}(T)\,\text{d}x\\[5.0pt] -\int_{\Omega}\epsilon u_{k}(T)\left(\tilde{\phi}(T)+\frac{1}{k}\right)^{m}\,\text{d}x+\int_{Q_{T}}\epsilon u_{k}\partial_{t}\left(\tilde{\phi}+\frac{1}{k}\right)^{m}\,\text{d}t\,\text{d}x.$$

(26)

It follows that

$$\int_{Q_{T}}\gamma|\nabla u_{k}^{m}|^{2}\,\text{d}t\,\text{d}x+\int_{Q_{T}}\lambda u_{k}^{m+q}\,\text{d}t\,\text{d}x+\int_{\Omega}\frac{\epsilon}{m+1}u_{k}^{m+1}(T)\,\text{d}x\\[5.0pt] =\int_{Q_{T}}\gamma\nabla\left(\tilde{\phi}+\frac{1}{k}\right)^{m}\cdot\nabla u_{k}^{m}\,\text{d}t\,\text{d}x+\int_{\Omega}\epsilon u_{k}(T)\left(\tilde{\phi}(T)+\frac{1}{k}\right)^{m}\,\text{d}x\\[5.0pt] -\int_{Q_{T}}\epsilon u_{k}\partial_{t}\left(\tilde{\phi}+\frac{1}{k}\right)^{m}\,\text{d}t\,\text{d}x+\int_{Q_{T}}f_{k}\left[u_{k}^{m}-\left(\tilde{\phi}+\frac{1}{k}\right)^{m}\right]\\[5.0pt] +\int_{Q_{T}}\lambda u_{k}^{q}\left(\tilde{\phi}+\frac{1}{k}\right)^{m}\,\text{d}t\,\text{d}x.$$

(27)

After straightforward estimates we obtain that

$$\int_{Q_{T}}|\nabla u_{k}^{m}|^{2}\,\text{d}t\,\text{d}x<C^{\prime}(1+T)\Bigg{(}\left\|\left(\phi+\frac{1}{k}\right)^{m}\right\|_{C^{0,1}(S_{T})}^{2}+\left\|\left(\phi+\frac{1}{k}\right)^{m}\right\|_{C(S_{T})}\\[5.0pt] +\left\|\left(\phi+\frac{1}{k}\right)^{m+q}\right\|_{C(S_{T})}+\left\|\phi+\frac{1}{k}\right\|_{C(S_{T})}^{2}+\left\|f+\frac{1}{k}\right\|_{L^{\infty}(Q_{T})}^{2}\Bigg{)},$$

(28)

where $C^{\prime}>0$ is a constant independent of $k$ and $T$.

It follows that (a subsequence of) $\nabla u_{k}^{m}$ converges weakly in $L^{2}(Q_{T})$ to a limit $U$. It is easy to see that $U=\nabla u^{m}$ in the sense of distributions. Since $\tau_{Q_{T}}(u_{k}^{m})(t)\to\phi^{m}$ in $L^{2}((0,T);H^{\frac{1}{2}}(\partial\Omega))$ by construction, and at the same time $\tau_{Q_{T}}(u_{k}^{m})\rightharpoonup\tau_{Q_{T}}(u^{m})$, it follows that

$$\tau_{Q_{T}}(u^{m})=\phi^{m}.$$

(29)

Therefore $u$ is a weak solution of (1) in $Q_{T}$.

Before proving the energy estimate and the comparison principle part of Theorem 1, we first need to prove the uniqueness of weak solutions. Once uniqueness has been established, we can deduce the desired properties from the corresponding ones that hold for $u_{k}$, since we would then know that each weak solution can be obtained as a limit of such approximate soultions. The following argument uses ideas from the proofs of [49, Theorem 6.5, Theorem 6.6] and from [1, Section 3].

Suppose $u_{1}$, $u_{2}$ are both weak solutions of (1) in $Q_{T}$, with possibly different boundary data $\phi_{1}\leq\phi_{2}$ and source terms $f_{1}$ and $f_{2}$. Then for any $\varphi\in C_{\diamond}(Q_{T})\cap C^{\infty}(Q_{T})$ we have

$$-\int_{Q_{T}}\epsilon\partial_{t}\varphi(u_{1}-u_{2})\,\text{d}t\,\text{d}x+\int_{Q_{T}}\lambda\varphi(u_{1}^{q}-u_{2}^{q})\,\text{d}t\,\text{d}x\\[5.0pt] \leq\int_{Q_{T}}\nabla\cdot(\gamma\nabla\varphi)(u_{1}^{m}-u_{2}^{m})\,\text{d}t\,\text{d}x+\int_{Q_{T}}\varphi(f_{1}-f_{2})\,\text{d}t\,\text{d}x.$$

(30)

Let

$$a(t,x)=\left\{\begin{array}[]{l}\frac{u_{1}^{m}(t,x)-u_{2}^{m}(t,x)}{u_{1}-u_{2}},\quad\text{if }u_{1}(t,x)\neq u_{2}(t,x),\\[5.0pt] 0,\quad\text{if }u_{1}(t,x)=u_{2}(t,x).\end{array}\right.$$

(31)

The function $a$ is continuous, but may not be smooth. For $k=(k_{1},k_{2})\in\mathbb{N}^{2}$, let $a_{k}\in C^{\infty}(Q_{T})$ be such that

$$\frac{1}{k_{1}}\leq a_{k}\leq k_{2},$$

(32)

and $a_{k}\to a$.

Let $\theta\in C^{\infty}_{0}(Q_{T})$, $\theta\geq 0$ be an arbitrary function and $\varphi_{k}$ be the unique smooth solution of the backwards in time linear parabolic problem in $Q_{T}$

$$\left\{\begin{array}[]{l}\epsilon\partial_{t}\varphi_{k}+a_{k}\nabla\cdot(\gamma\nabla\varphi_{k})+\theta=0,\\[5.0pt] \varphi_{k}(T,x)=0,\;\forall x\in\Omega,\quad\varphi_{k}|_{S_{T}}=0.\end{array}\right.$$

(33)

By the maximum principle we have that $\varphi_{k}\geq 0$. Using $\varphi_{k}$ as a test function in (30) we get

$$\int_{Q_{T}}\theta(u_{1}-u_{2})\,\text{d}t\,\text{d}x+\int_{Q_{T}}\lambda\varphi_{k}(u_{1}^{q}-u_{2}^{q})\,\text{d}t\,\text{d}x\\[5.0pt] \leq\int_{Q_{T}}(a-a_{k})\nabla\cdot(\gamma\nabla\varphi_{k})(u_{1}-u_{2})\,\text{d}t\,\text{d}x+\int_{Q_{T}}\varphi_{k}(f_{1}-f_{2})\,\text{d}t\,\text{d}x.$$

(34)

Let

$$J_{k}=\int_{Q_{T}}|u_{1}-u_{2}|\,|a-a_{k}|\,|\nabla\cdot(\gamma\nabla\varphi_{k})|\,\text{d}t\,\text{d}x$$

(35)

and note immediately that

$$J_{k}\leq\left(\int_{Q_{T}}a_{k}[\nabla\cdot(\gamma\nabla\varphi_{k})]^{2}\,\text{d}t\,\text{d}x\right)^{\frac{1}{2}}\left(\int_{Q_{T}}|u_{1}-u_{2}|^{2}\frac{|a-a_{k}|^{2}}{a_{k}}\,\text{d}t\,\text{d}x\right)^{\frac{1}{2}}.$$

(36)

We will control each of the two factors on the right hand side separately.

Let $\zeta:[0,T]\to[\frac{1}{2},1]$ be a smooth function such that $\zeta^{\prime}(t)\geq c>0$. We multiply (33) by $\frac{\zeta}{\epsilon}\nabla\cdot(\gamma\nabla\varphi_{k})$ and integrate to obtain

$$\int_{Q_{T}}\partial_{t}\varphi_{k}\zeta\nabla\cdot(\gamma\nabla\varphi_{k})\,\text{d}t\,\text{d}x+\int_{Q_{T}}\frac{\zeta}{\epsilon}a_{k}[\nabla\cdot(\gamma\nabla\varphi_{k})]^{2}\,\text{d}t\,\text{d}x+\int_{Q_{T}}\zeta\frac{\theta}{\epsilon}\nabla\cdot(\gamma\nabla\varphi_{k})\,\text{d}t\,\text{d}x=0.$$

(37)

We have that

$$\int_{Q_{T}}\partial_{t}\varphi_{k}\zeta\nabla\cdot(\gamma\nabla\varphi_{k})\,\text{d}t\,\text{d}x=-\int_{Q_{T}}\zeta\gamma\nabla\varphi_{k}\cdot\nabla(\partial_{t}\varphi_{k})\,\text{d}t\,\text{d}x\\[5.0pt] =-\frac{1}{2}\int_{Q_{T}}\zeta\gamma\partial_{t}(\nabla\varphi_{k})^{2}\,\text{d}t\,\text{d}x\geq\frac{1}{2}\int_{Q_{T}}\zeta^{\prime}\gamma(\nabla\varphi_{k})^{2}\,\text{d}t\,\text{d}x.$$

(38)

Then

$$\frac{1}{2}\int_{Q_{T}}\zeta^{\prime}\gamma|\nabla\varphi_{k}|^{2}\,\text{d}t\,\text{d}x+\int_{Q_{T}}\frac{\zeta}{\epsilon}a_{k}[\nabla\cdot(\gamma\nabla\varphi_{k})]^{2}\,\text{d}t\,\text{d}x\leq\int_{Q_{T}}\zeta\gamma\nabla\frac{\theta}{\epsilon}\cdot\nabla\varphi_{k}\,\text{d}t\,\text{d}x.$$

(39)

Estimating

$$\int_{Q_{T}}\zeta\gamma\nabla\frac{\theta}{\epsilon}\cdot\nabla\varphi_{k}\,\text{d}t\,\text{d}x\leq\frac{c}{4}\int_{Q_{T}}\gamma|\nabla\varphi_{k}|^{2}\,\text{d}t\,\text{d}x+\frac{1}{c}\int_{Q_{T}}\gamma|\nabla\frac{\theta}{\epsilon}|^{2}\,\text{d}t\,\text{d}x,$$

(40)

we conclude that

$$\int_{Q_{T}}\gamma|\nabla\varphi_{k}|^{2}\,\text{d}t\,\text{d}x+\int_{Q_{T}}a_{k}[\nabla\cdot(\gamma\nabla\varphi_{k})]^{2}\,\text{d}t\,\text{d}x\leq C\int_{Q_{T}}\gamma|\nabla\frac{\theta}{\epsilon}|^{2}\,\text{d}t\,\text{d}x,$$

(41)

with a constant $C>0$ which do not depend on $k$.

The estimate for the second factor on the right hand side of (36) relies only on the strategy for approximating $a$ by $a_{k}$. There is no difference at all between our case here and the argument in [49], so we quote the result

$$\int_{Q_{T}}|u_{1}-u_{2}|^{2}\frac{|a-a_{k}|^{2}}{a_{k}}\,\text{d}t\,\text{d}x\leq\frac{C}{k_{1}}.$$

(42)

It follows that in the limit $J_{k}\to 0$.

Let $t_{0}\in(0,T)$, $0\leq\psi(x)\leq 1$ be a smooth function, and $\rho_{\eta}$, $\eta>0$, be a standard mollifier. We set

$$\theta(t,x)=\epsilon(x)\psi(x)\rho_{\eta}(t-t_{0}).$$

(43)

Then

$$\varphi(t,x)=\int_{t}^{T}\rho_{\eta}(s-t_{0})\,\text{d}s$$

(44)

is a supersolution for (33), so for all $k$ we have that $\varphi_{k}\leq\varphi$.

Taking the limit in (34), we obtain

$$\int_{Q_{T}}\epsilon\psi\rho_{\eta}(u_{1}-u_{2})\,\text{d}t\,\text{d}x\leq\int_{Q_{T}}\lambda(u_{2}^{q}-u_{1}^{q})_{+}\left(\int_{t}^{T}\rho_{\eta}(s-t_{0})\,\text{d}s\right)\,\text{d}t\,\text{d}x\\[5.0pt] +\int_{Q_{T}}(f_{1}-f_{2})_{+}\left(\int_{t}^{T}\rho_{\eta}(s-t_{0})\,\text{d}s\right)\,\text{d}t\,\text{d}x.$$

(45)

Taking the $\eta\to 0$ limit, we get

$$\int_{\Omega}\epsilon\psi(u_{1}(t_{0})-u_{2}(t_{0}))_{+}\,\text{d}x\leq\int_{0}^{t_{0}}\int_{\Omega}\lambda(u_{2}^{q}-u_{1}^{q})_{+}\,\text{d}t\,\text{d}x+\int_{0}^{t_{0}}\int_{\Omega}(f_{1}-f_{2})_{+}\,\text{d}t\,\text{d}x,$$

(46)

which easily gives that

$$\int_{\Omega}(u_{1}(t_{0})-u_{2}(t_{0}))_{+}\,\text{d}x\leq C\left(\int_{0}^{t_{0}}\int_{\Omega}(u_{2}^{q}-u_{1}^{q})_{+}\,\text{d}t\,\text{d}x+\int_{0}^{t_{0}}\int_{\Omega}(f_{1}-f_{2})_{+}\,\text{d}t\,\text{d}x\right).$$

(47)

We need to distinguish two cases. The first is when $q\geq 1$. In this case we can from the start assume that $\phi_{1}=\phi_{2}$ and $f_{1}=f_{2}$. Note that both $u_{1}$ and $u_{2}$ are bounded, so there exists $M>0$ such that

$$(u_{2}^{q}-u_{1}^{q})_{+}\leq M(u_{2}-u_{1})_{+}.$$

(48)

Then

$$\int_{\Omega}(u_{1}(t_{0})-u_{2}(t_{0}))_{+}\,\text{d}x\leq C\int_{0}^{t_{0}}\int_{\Omega}(u_{2}(t)-u_{1}(t))_{+}\,\text{d}t\,\text{d}x\\[5.0pt] \leq C\int_{0}^{t_{0}}\int_{0}^{t}\int_{\Omega}(u_{1}(s)-u_{2}(s))_{+}\,\text{d}s\,\text{d}t\,\text{d}x\\[5.0pt] \leq Ct_{0}\int_{0}^{t_{0}}\int_{\Omega}(u_{1}(s)-u_{2}(s))_{+}\,\text{d}s\,\text{d}x.$$

(49)

Gronwall’s inequality implies that

$$\int_{\Omega}(u_{1}(t_{0})-u_{2}(t_{0}))_{+}\,\text{d}x=0,\quad\forall t_{0}\in(0,T),$$

(50)

and by exchanging the roles of $u_{1}$ and $u_{2}$ we conclude that $u_{1}=u_{2}$ everywhere.

The second case is when $0<q<1$. Here we would like to take $u_{1}=\hat{u}$ to be any weak solution of (1) in $Q_{T}$, with boundary data $\phi$ and source term $f$, and $u_{2}$ to be the approximate solution $u_{k}$ constructed above in the existence step. There is a slight difference to the assumtions above since $u_{k}$ has positive initial data $\frac{1}{k}$. Going over the argument with this in mind quickly shows that an additional term of the form

$$\int_{\Omega}(-u_{k}(0,x))_{+}\,\text{d}x$$

(51)

should appear on the right hand side of (47). Since this term is actually zero, we may continue without modifying anything.

There is another modification we wish to make. Let $\omega>0$ and let $u_{1}=e^{\omega t}\hat{u}$, $u_{2}=e^{\omega t}u_{k}$. It is easy to see that $u_{1}$ satisfies the equation

$$\partial_{t}u_{1}-\nabla\cdot(\gamma\nabla u_{1})+\lambda u_{1}^{q}-\omega u_{1}=e^{\omega t}f$$

(52)

in the weak sense, with a similar situation holding for $u_{2}$. If we apply (47) to these choices of $u_{1}$ and $u_{2}$ we have

$$e^{\omega t_{0}}\int_{\Omega}(\hat{u}(t_{0})-u_{k}(t_{0}))_{+}\leq C\left(\int_{0}^{t_{0}}\int_{\Omega}e^{\omega t}(u_{k}^{q}-\hat{u}^{q}-\omega(u_{k}-\hat{u}))_{+}\,\text{d}t\,\text{d}x\right.\\[5.0pt] \left.+\int_{0}^{t_{0}}\int_{\Omega}e^{\omega t}(f-f_{k})_{+}\,\text{d}t\,\text{d}x\right).$$

(53)

The last term is zero. If

$$u_{k}^{q}(t,x)-\hat{u}^{q}(t,x)-\omega(u_{k}(t,x)-\hat{u}(t,x))\geq 0,$$

(54)

and $u_{k}(t,x)\geq\hat{u}(t,x)$, then we must have that

$$\omega\leq\frac{u_{k}^{q}(t,x)-\hat{u}^{q}(t,x)}{u_{k}(t,x)-\hat{u}(t,x)}\leq(u_{k}(t,x)-\hat{u}(t,x))^{q-1}\leq k^{1-q}.$$

(55)

If we choose $\omega>k^{1-q}$, then we see that (54) can only hold when $\hat{u}(t,x)\geq u_{k}(t,x)$. In this case we have

$$e^{\omega t_{0}}\int_{\Omega}(\hat{u}(t_{0})-u_{k}(t_{0}))_{+}\leq C\omega\int_{0}^{t_{0}}\int_{\Omega}e^{\omega t}(\hat{u}-u_{k})_{+}\,\text{d}t\,\text{d}x.$$

(56)

By Gronwall’s inequality we conclude that $\hat{u}\leq u_{k}$, and therefore $\hat{u}\leq u$, where $u$ is the solution constructed in the previous subsection.