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Daniele Tampieri
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Scalar nonlinear balancedbalance law with non-integrable source term on a bounded domain

I am considering the following PDE for $(t,x)\in\mathbb R_+\times[0,1]$:

$$\partial_t u(t,x) + \partial_x[u(1-u)]=G(x,u),$$ $$u(0,\cdot)=u_0, \hspace{5mm} u(\cdot,0)=\alpha, \hspace{5mm} u(\cdot,1)=\beta,$$

where the source term $G$ is given by

$$G=\frac{\alpha-u}{x^\gamma}+\frac{\beta-u}{(1-x)^\gamma}, \hspace{5mm} \gamma\ge1,$$$$ \begin{cases} \partial_t u(t,x) + \partial_x[u(1-u)]=G(x,u),\\ u(0,\cdot)=u_0, \quad u(\cdot,0)=\alpha, \quad u(\cdot,1)=\beta, \end{cases} $$ where

$\alpha$, $\beta$ are constants, $u_0 \in L^\infty(0,1)$ (it is fine to assume $u_0$ is smooth or BV, with any necessary compatibility).

  • the source term $G$ is given by $$ G=\frac{\alpha-u}{x^\gamma}+\frac{\beta-u}{(1-x)^\gamma}, \quad \gamma\ge1, $$
  • $\alpha$, $\beta$ are constants,
  • $u_0 \in L^\infty(0,1)$ (it is fine to assume $u_0$ is smooth or BV, with any necessary compatibility).

My question: how to define the "entropy solution" and more importantly, can we get uniqueness / stability on initial data?


I tried the vanishing viscosity method (Section 2.6 - 2.8 of [Málek, Nečas, Rokyta & Ružička, Weak and measure-valued solutions to evolutionary PDEs[1]). For $\varepsilon>0$, take the solution of the parabolic problem (let's assume $u_0$ is smooth):

$$\partial_t u^\varepsilon(t,x) + \partial_x[u^\varepsilon(1-u^\varepsilon)]=G(x,u^\varepsilon),$$ $$u^\varepsilon(0,\cdot)=u_0, \hspace{5mm} u^\varepsilon(\cdot,0)=\alpha, \hspace{5mm} u^\varepsilon(\cdot,1)=\beta.$$

Then$$ \begin{cases}\partial_t u^\varepsilon(t,x) + \partial_x[u^\varepsilon(1-u^\varepsilon)]=G(x,u^\varepsilon),\\ u^\varepsilon(0,\cdot)=u_0, \quad u^\varepsilon(\cdot,0)=\alpha, \quad u^\varepsilon(\cdot,1)=\beta. \end{cases} $$ Then take $\varepsilon\to0$ to obtain (subsequential) limit $u \in L^\infty(\mathbb R_+\times(0,1)$. For any $T<\infty$, $u$ satisfies the energy bound

$$\tag{$*$}\iint_{(0,T)\times(0,1)} \frac{(u-\alpha)^2}{x^\gamma}+\frac{(u-\beta)^2}{(1-x)^\gamma} dxdt<\infty,$$ $$ \label{1}\tag{$*$}\iint\limits_{(0,T)\times(0,1)} \frac{(u-\alpha)^2}{x^\gamma}+\frac{(u-\beta)^2}{(1-x)^\gamma} dxdt<\infty,$$

as well as the generalised entropy inequality:

$$\tag{$**$}\iint [f(u)\partial_t\psi + q(u)\partial_x\psi + G(x,u)f'(u)\psi] dxdt + \int f(u_0)\psi(0,\cdot)dx \ge 0$$$$\label{2}\tag{$**$}\iint [f(u)\partial_t\psi + q(u)\partial_x\psi + G(x,u)f'(u)\psi] dxdt + \int f(u_0)\psi(0,\cdot)dx \ge 0$$

for any Lax entropy pair $(f,q)$: $f''\ge0$, $q'(u)=(1-2u)f'(u)$ and any $\psi \in C_0^\infty((-\infty,T)\times\mathbb R)$ such that $\psi\ge0$ and

$$\iint_{(0,T)\times(0,1)} \psi^2(t,x)[x^{-\gamma}+(1-x)^{-\gamma}]dxdt<\infty.$$$$\iint\limits_{(0,T)\times(0,1)} \psi^2(t,x)[x^{-\gamma}+(1-x)^{-\gamma}]dxdt<\infty.$$

Intuitively, ($**$)\eqref{2} contains the initial data and also tells me how the solution should behave inside $(0,1)$, while ($*$)\eqref{1} controls the behaviourbehavior near the boundaries (because $\gamma$ is large). But I am not sure if they are enough to obtain uniqueness.

Reference

[1] Jindřich Nečas, Josef Málek, Mirko Rokyta, Michael Růžička, Weak and measure-valued solutions to evolutionary PDEs, Applied Mathematics and Mathematical Computation. 13. London: Chapman & Hall, pp. vii+317 (1996), ISBN:0-412-57750-X, MR1409366, Zbl 0851.35002.

Scalar nonlinear balanced law with non-integrable source term on a bounded domain

I am considering the following PDE for $(t,x)\in\mathbb R_+\times[0,1]$:

$$\partial_t u(t,x) + \partial_x[u(1-u)]=G(x,u),$$ $$u(0,\cdot)=u_0, \hspace{5mm} u(\cdot,0)=\alpha, \hspace{5mm} u(\cdot,1)=\beta,$$

where the source term $G$ is given by

$$G=\frac{\alpha-u}{x^\gamma}+\frac{\beta-u}{(1-x)^\gamma}, \hspace{5mm} \gamma\ge1,$$

$\alpha$, $\beta$ are constants, $u_0 \in L^\infty(0,1)$ (it is fine to assume $u_0$ is smooth or BV, with any necessary compatibility).

My question: how to define the "entropy solution" and more importantly, can we get uniqueness / stability on initial data?


I tried the vanishing viscosity method (Section 2.6 - 2.8 of [Málek, Nečas, Rokyta & Ružička, Weak and measure-valued solutions to evolutionary PDEs). For $\varepsilon>0$, take the solution of the parabolic problem (let's assume $u_0$ is smooth):

$$\partial_t u^\varepsilon(t,x) + \partial_x[u^\varepsilon(1-u^\varepsilon)]=G(x,u^\varepsilon),$$ $$u^\varepsilon(0,\cdot)=u_0, \hspace{5mm} u^\varepsilon(\cdot,0)=\alpha, \hspace{5mm} u^\varepsilon(\cdot,1)=\beta.$$

Then take $\varepsilon\to0$ to obtain (subsequential) limit $u \in L^\infty(\mathbb R_+\times(0,1)$. For any $T<\infty$, $u$ satisfies the energy bound

$$\tag{$*$}\iint_{(0,T)\times(0,1)} \frac{(u-\alpha)^2}{x^\gamma}+\frac{(u-\beta)^2}{(1-x)^\gamma} dxdt<\infty,$$

as well as the generalised entropy inequality:

$$\tag{$**$}\iint [f(u)\partial_t\psi + q(u)\partial_x\psi + G(x,u)f'(u)\psi] dxdt + \int f(u_0)\psi(0,\cdot)dx \ge 0$$

for any Lax entropy pair $(f,q)$: $f''\ge0$, $q'(u)=(1-2u)f'(u)$ and any $\psi \in C_0^\infty((-\infty,T)\times\mathbb R)$ such that $\psi\ge0$ and

$$\iint_{(0,T)\times(0,1)} \psi^2(t,x)[x^{-\gamma}+(1-x)^{-\gamma}]dxdt<\infty.$$

Intuitively, ($**$) contains the initial data and also tells me how the solution should behave inside $(0,1)$, while ($*$) controls the behaviour near the boundaries (because $\gamma$ is large). But I am not sure if they are enough to obtain uniqueness.

Scalar nonlinear balance law with non-integrable source term on a bounded domain

I am considering the following PDE for $(t,x)\in\mathbb R_+\times[0,1]$: $$ \begin{cases} \partial_t u(t,x) + \partial_x[u(1-u)]=G(x,u),\\ u(0,\cdot)=u_0, \quad u(\cdot,0)=\alpha, \quad u(\cdot,1)=\beta, \end{cases} $$ where

  • the source term $G$ is given by $$ G=\frac{\alpha-u}{x^\gamma}+\frac{\beta-u}{(1-x)^\gamma}, \quad \gamma\ge1, $$
  • $\alpha$, $\beta$ are constants,
  • $u_0 \in L^\infty(0,1)$ (it is fine to assume $u_0$ is smooth or BV, with any necessary compatibility).

My question: how to define the "entropy solution" and more importantly, can we get uniqueness / stability on initial data?


I tried the vanishing viscosity method (Section 2.6 - 2.8 of [1]). For $\varepsilon>0$, take the solution of the parabolic problem (let's assume $u_0$ is smooth): $$ \begin{cases}\partial_t u^\varepsilon(t,x) + \partial_x[u^\varepsilon(1-u^\varepsilon)]=G(x,u^\varepsilon),\\ u^\varepsilon(0,\cdot)=u_0, \quad u^\varepsilon(\cdot,0)=\alpha, \quad u^\varepsilon(\cdot,1)=\beta. \end{cases} $$ Then take $\varepsilon\to0$ to obtain (subsequential) limit $u \in L^\infty(\mathbb R_+\times(0,1)$. For any $T<\infty$, $u$ satisfies the energy bound $$ \label{1}\tag{$*$}\iint\limits_{(0,T)\times(0,1)} \frac{(u-\alpha)^2}{x^\gamma}+\frac{(u-\beta)^2}{(1-x)^\gamma} dxdt<\infty,$$

as well as the generalised entropy inequality:

$$\label{2}\tag{$**$}\iint [f(u)\partial_t\psi + q(u)\partial_x\psi + G(x,u)f'(u)\psi] dxdt + \int f(u_0)\psi(0,\cdot)dx \ge 0$$

for any Lax entropy pair $(f,q)$: $f''\ge0$, $q'(u)=(1-2u)f'(u)$ and any $\psi \in C_0^\infty((-\infty,T)\times\mathbb R)$ such that $\psi\ge0$ and

$$\iint\limits_{(0,T)\times(0,1)} \psi^2(t,x)[x^{-\gamma}+(1-x)^{-\gamma}]dxdt<\infty.$$

Intuitively, \eqref{2} contains the initial data and also tells me how the solution should behave inside $(0,1)$, while \eqref{1} controls the behavior near the boundaries (because $\gamma$ is large). But I am not sure if they are enough to obtain uniqueness.

Reference

[1] Jindřich Nečas, Josef Málek, Mirko Rokyta, Michael Růžička, Weak and measure-valued solutions to evolutionary PDEs, Applied Mathematics and Mathematical Computation. 13. London: Chapman & Hall, pp. vii+317 (1996), ISBN:0-412-57750-X, MR1409366, Zbl 0851.35002.

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Scalar nonlinear balanced law with non-integrable source term on a bounded domain

I am considering the following PDE for $(t,x)\in\mathbb R_+\times[0,1]$:

$$\partial_t u(t,x) + \partial_x[u(1-u)]=G(x,u),$$ $$u(0,\cdot)=u_0, \hspace{5mm} u(\cdot,0)=\alpha, \hspace{5mm} u(\cdot,1)=\beta,$$

where the source term $G$ is given by

$$G=\frac{\alpha-u}{x^\gamma}+\frac{\beta-u}{(1-x)^\gamma}, \hspace{5mm} \gamma\ge1,$$

$\alpha$, $\beta$ are constants, $u_0 \in L^\infty(0,1)$ (it is fine to assume $u_0$ is smooth or BV, with any necessary compatibility).

My question: how to define the "entropy solution" and more importantly, can we get uniqueness / stability on initial data?


I tried the vanishing viscosity method (Section 2.6 - 2.8 of [Málek, Nečas, Rokyta & Ružička, Weak and measure-valued solutions to evolutionary PDEs). For $\varepsilon>0$, take the solution of the parabolic problem (let's assume $u_0$ is smooth):

$$\partial_t u^\varepsilon(t,x) + \partial_x[u^\varepsilon(1-u^\varepsilon)]=G(x,u^\varepsilon),$$ $$u^\varepsilon(0,\cdot)=u_0, \hspace{5mm} u^\varepsilon(\cdot,0)=\alpha, \hspace{5mm} u^\varepsilon(\cdot,1)=\beta.$$

Then take $\varepsilon\to0$ to obtain (subsequential) limit $u \in L^\infty(\mathbb R_+\times(0,1)$. For any $T<\infty$, $u$ satisfies the energy bound

$$\tag{$*$}\iint_{(0,T)\times(0,1)} \frac{(u-\alpha)^2}{x^\gamma}+\frac{(u-\beta)^2}{(1-x)^\gamma} dxdt<\infty,$$

as well as the generalised entropy inequality:

$$\tag{$**$}\iint [f(u)\partial_t\psi + q(u)\partial_x\psi + G(x,u)f'(u)\psi] dxdt + \int f(u_0)\psi(0,\cdot)dx \ge 0$$

for any Lax entropy pair $(f,q)$: $f''\ge0$, $q'(u)=(1-2u)f'(u)$ and any $\psi \in C_0^\infty((-\infty,T)\times\mathbb R)$ such that $\psi\ge0$ and

$$\iint_{(0,T)\times(0,1)} \psi^2(t,x)[x^{-\gamma}+(1-x)^{-\gamma}]dxdt<\infty.$$

Intuitively, ($**$) contains the initial data and also tells me how the solution should behave inside $(0,1)$, while ($*$) controls the behaviour near the boundaries (because $\gamma$ is large). But I am not sure if they are enough to obtain uniqueness.