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Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x_1}(u^2+f)& &\text{in } \mathbb R^n,\\ & \,\,\,u(x)\to 0\,,& &x\to\infty. \end{aligned}\right. $$$$ \left\{\begin{aligned} &-\Delta u=\partial_{x}(u^2+f)& &\text{in } \mathbb R^3,\\ & \,\,\,u(\mathbf x)\to 0\,,& &\mathbf x\to\infty, \end{aligned}\right. $$ where $x=(x_1,...,x_n)$$\mathbf x=(x,y,z)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{n}(\mathbb R^n), f\in L^{n/2}(\mathbb R^n). $$$$ u\in L^{3}(\mathbb R^3), f\in L^{3/2}(\mathbb R^3). $$ Assume $n=3$ to fix ideas. II know two ways of finding solutions.

  1. Strong solutions. Assume $n=3$ to fix ideas. ByBy using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

  2. Weak solutions. If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound $$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$ Using this a-priori bound it is possible to prove* the existence of weak solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$). In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x_1}(u^2+f)& &\text{in } \mathbb R^n,\\ & \,\,\,u(x)\to 0\,,& &x\to\infty. \end{aligned}\right. $$ where $x=(x_1,...,x_n)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{n}(\mathbb R^n), f\in L^{n/2}(\mathbb R^n). $$ Assume $n=3$ to fix ideas. I know two ways of finding solutions.

  1. Strong solutions. Assume $n=3$ to fix ideas. By using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

  2. Weak solutions. If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound $$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$ Using this a-priori bound it is possible to prove* the existence of weak solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$). In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x}(u^2+f)& &\text{in } \mathbb R^3,\\ & \,\,\,u(\mathbf x)\to 0\,,& &\mathbf x\to\infty, \end{aligned}\right. $$ where $\mathbf x=(x,y,z)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{3}(\mathbb R^3), f\in L^{3/2}(\mathbb R^3). $$ I know two ways of finding solutions.

  1. Strong solutions. By using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

  2. Weak solutions. If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound $$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$ Using this a-priori bound it is possible to prove* the existence of weak solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$). In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

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The PDE

Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x_1}(u^2+f)& &\text{in } \mathbb R^n,\\ & \,\,\,u(x)\to 0\,,& &x\to\infty. \end{aligned}\right. $$ where $x=(x_1,...,x_n)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{n}(\mathbb R^n), f\in L^{n/2}(\mathbb R^n). $

Strong solutions

Assume$$ u\in L^{n}(\mathbb R^n), f\in L^{n/2}(\mathbb R^n). $$ Assume $n=3$ to fix ideas. By using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

Weak solutions

If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound

$$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$

Using this a-priori bound (please correct me if I am wrong*) it is possible to prove the existenceknow two ways of weakfinding solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$).

In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

  1. Strong solutions. Assume $n=3$ to fix ideas. By using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

  2. Weak solutions. If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound $$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$ Using this a-priori bound it is possible to prove* the existence of weak solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$). In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

RemarksFurther remarks.

*I think there should beis more than one way of doing that. The way I thought would be to approximate the Cauchy problem with that of the same PDE on a half plane (with Dirichlet boundary conditions), where the "edge" of the plane goes to infinity. On a half plane, you can look at the PDE as an evolution equation via the Poisson kernel, so you can show that a solution exists, and then you can use the uniform bound and weak compactness to show the existence of the solution of the original problem... Probably not the easiest way. I have to say I am not an elliptic person, I am mostly dispersive, sometimes parabolic... In any case, feel free to comment on this point.

The PDE

Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x_1}(u^2+f)& &\text{in } \mathbb R^n,\\ & \,\,\,u(x)\to 0\,,& &x\to\infty. \end{aligned}\right. $$ where $x=(x_1,...,x_n)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{n}(\mathbb R^n), f\in L^{n/2}(\mathbb R^n). $

Strong solutions

Assume $n=3$ to fix ideas. By using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

Weak solutions

If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound

$$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$

Using this a-priori bound (please correct me if I am wrong*) it is possible to prove the existence of weak solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$).

In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

Remarks.

*I think there should be more than one way of doing that. The way I thought would be to approximate the Cauchy problem with that of the same PDE on a half plane (with Dirichlet boundary conditions), where the "edge" of the plane goes to infinity. On a half plane, you can look at the PDE as an evolution equation via the Poisson kernel, so you can show that a solution exists, and then you can use the uniform bound and weak compactness to show the existence of the solution of the original problem... Probably not the easiest way. I have to say I am not an elliptic person, I am mostly dispersive, sometimes parabolic... In any case, feel free to comment on this point.

Consider a very simple semilinear elliptic PDE: $$ \left\{\begin{aligned} &-\Delta u=\partial_{x_1}(u^2+f)& &\text{in } \mathbb R^n,\\ & \,\,\,u(x)\to 0\,,& &x\to\infty. \end{aligned}\right. $$ where $x=(x_1,...,x_n)$. The boundary conditions are not rigorous, it is just to describe the kind of solutions we want. The critical Lebesgue spaces with respect to the natural scaling symmetry are: $$ u\in L^{n}(\mathbb R^n), f\in L^{n/2}(\mathbb R^n). $$ Assume $n=3$ to fix ideas. I know two ways of finding solutions.

  1. Strong solutions. Assume $n=3$ to fix ideas. By using the fixed point theorem, it is possible to show that if $f\in L^{3/2}(\mathbb R^3)$ is small enough, then there exists a unique small solution $u\in L^3(\mathbb R^3)$, which moreover satisfies $$ \|\nabla u\|_{L^{3/2}(\mathbb R^3)}\lesssim \|f\|_{L^{3/2}(\mathbb R^3)} $$ (note that $\dot W^{1,3/2}(\mathbb R^3)\hookrightarrow L^3(\mathbb R^3)$).

  2. Weak solutions. If I instead consider a scaling-subcritical datum, like $f\in L^2(\mathbb R^3)$, I cannot use the fixed point theorem. However, notice that multiplying the equation by $u$ and integrating by parts, we obtain the a priori bound $$ \|\nabla u\|_{L^2(\mathbb R^3)}\lesssim \|f\|_{L^2(\mathbb R^3)}. $$ Using this a-priori bound it is possible to prove* the existence of weak solutions $u\in \dot H^1(\mathbb R^3)$ for any $f\in L^2(\mathbb R^3)$ (we have $\dot H^1(\mathbb R^3)\hookrightarrow L^6(\mathbb R^3)$). In this case, estimates for the difference of two solutions do not seem to exist, so the uniqueness cannot be proved by standard arguments (this is the only reason why I am calling them “weak solutions”; you can see that they are more regular than strong solutions).

Further remarks.

*I think there is more than one way of doing that. The way I thought would be to approximate the Cauchy problem with that of the same PDE on a half plane (with Dirichlet boundary conditions), where the "edge" of the plane goes to infinity. On a half plane, you can look at the PDE as an evolution equation via the Poisson kernel, so you can show that a solution exists, and then you can use the uniform bound and weak compactness to show the existence of the solution of the original problem... Probably not the easiest way. I have to say I am not an elliptic person, I am mostly dispersive, sometimes parabolic... In any case, feel free to comment on this point.

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