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Consider the viscous Burgers' equation in $\mathbb R$ with a forcing term: $$ v_t-v_{xx}=(v^2)_x +f_x,\qquad (x,t)\in \mathbb R\times [0,\infty) $$ $$ v(x,0)=v_0(x), \qquad x\in\mathbb R,$$ with kink-like boundary conditions at infinity: $v_0-\tanh(x)\in L^2(\mathbb R)$. The function $\tanh(x)$ is special because it is a stationary solution of the equation, together with all its translations. The only think to know is that $\tanh(x)\rightarrow \pm 1$ exponentially as $x\to\pm\infty$, and $\partial_x \tanh(x)=\operatorname{sech}^2(x)$. The forcing term $f$ lies in $L^2(\mathbb R\times (0,\infty)).$

Now, assume that the forcing term $f$ is small enough in $L^2(\mathbb R\times (0,\infty))$, and that the initial profile is a small perturbation of some translation of $\tanh$, i.e., $|v_0|_{L^2}$ is small enough, where $$ |z|_{L^2}:=\min_{\gamma\in \mathbb R}\|z-\tanh(x-\gamma)\|_{L^2}. $$ Then, it is possible to prove that the quantity $|v(t)|_{L^2}$ is uniformly small on $[0,\infty)$, and in particular there exists a function $\alpha:[0,\infty)\mapsto \mathbb R$ such that $v=w+\tanh(x-\alpha(t))$ with the estimates $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}.\qquad (1) $$ The function $\alpha$ is often called modulation parameter and what it is roughly meant to do is describe the "position" of the stationary solution that approximates $v$ best. In other words, we see $v$ as a perturbation of $\tanh(x-\alpha)$.

Proof (sketch). To show that, the basic argument uses some modulational analysis: call $w(x,t):=v(x,t)-\tanh(x-\alpha(t))$, where we fix $\alpha(t)$ using the orthogonality condition $$ \int_{\mathbb R}\operatorname{sech}^2(x-\alpha(t))w(x,t)\,dx=0 \quad\forall t\geq 0.\qquad\qquad(2) $$ ($\alpha(t)$ is uniquely detemined by $v(t)$ thanks to the implicit function theorem if $|v(t)|_{L^2}$ is small enough). In this case, the function $w$ satisfies $$ w_t-w_{xx}-2(\tanh(x-\alpha)w)_x=(w^2)_x+f_x+\alpha'\operatorname{sech}^2(x-\alpha).\qquad (3) $$ Multiplying by $w$ and integrating in space (as for the usual energy estimates) we obtain $$ \frac 12\frac d{dt}\|w(t)\|_{L^2}+Q_\alpha(w)\leq\|f(t)\|_{L^2_x}\|w_x\|_{L^2}, $$ where the last term vanishes thanks to the orthogonality condition $(2)$, the term $(w^2)_x$ vanishes when tested against $w$, and $Q$ is the quadratic form $$ Q_\alpha(w):=\int_{\mathbb R}w_x^2-\operatorname{sech}^2(x-\alpha)w^2\,dx. $$ Moreover (there is some work to do to show this) one can prove that, if $w$ satisfies the orthogonality condition $(2)$, then $$ Q_\alpha(w)\geq \frac{1}{2}\|w_x\|_{L^2}. $$ Thus, integrating in time, we get the nice estimate $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}. $$

Question: How to prove estimate $(1)$ for arbitrary, large data $v_0$ and $f$? Is it even possible with a continuous parameter $\alpha$? If not, what if $\alpha$ is allowed to be discontinuous?

The main problem is that $\alpha(t)$ is not uniquely determined by the orthogonality condition $(2)$ if the solution is large in the $|\cdot|_{L^2}$ norm. This leads to the fact, for instance, that the function $\alpha$ a priori is not necessarily continuous. If I knew that $\alpha$ satisfying $(2)$ exists and is continuous, then I think I can prove the statement in a similar way by approximation, but I don't know how to prove that such a function $\alpha$ is continuous. In fact, if I pick $\alpha$ arbitrarily, it is perfectly possible to have something non-continuous. In the small data case, the continuity of $\alpha$ follows from equations $(2)$ and $(3)$:

Sketch. Differentiating $(2)$ in $t$ and substituting $w_t$ using $(3)$, one obtains: $$ |\alpha'(t)| \left |\int_{\mathbb R}\operatorname{sech}^2(x-\alpha)[\operatorname{sech}^2-w_x]dx\right|\leq\dots, $$ where the right hand side has good terms.

Consider the viscous Burgers' equation in $\mathbb R$ with a forcing term: $$ \begin{cases} v_t-v_{xx}=(v^2)_x +f_x, & (x,t)\in \mathbb R\times [0,\infty) \\ v(x,0)=v_0(x), & x\in\mathbb R, \end{cases} $$ with kink-like boundary conditions at infinity: $v_0-\tanh(x)\in L^2(\mathbb R)$. The function $\tanh(x)$ is special because it is a stationary solution of the equation, together with all its translations. The only thing to take care of is the fact that $\tanh(x)\rightarrow \pm 1$ exponentially as $x\to\pm\infty$, and $\partial_x \tanh(x)=\operatorname{sech}^2(x)$. The forcing term $f$ lies in $L^2(\mathbb R\times (0,\infty)).$

Now, assume that the forcing term $f$ is small enough in $L^2(\mathbb R\times (0,\infty))$, and that the initial profile is a small perturbation of some translation of $\tanh$, i.e., $|v_0|_{L^2}$ is small enough, where $$ |z|_{L^2}:=\min_{\gamma\in \mathbb R}\|z-\tanh(x-\gamma)\|_{L^2}. $$ Then, it is possible to prove that the quantity $|v(t)|_{L^2}$ is uniformly small on $[0,\infty)$, and in particular there exists a function $\alpha:[0,\infty)\mapsto \mathbb R$ such that $v=w+\tanh(x-\alpha(t))$ with the estimates $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}.\label{1}\tag{1} $$ The function $\alpha$ is often called modulation parameter and what it is roughly meant to do is describe the "position" of the stationary solution that approximates $v$ best. In other words, we see $v$ as a perturbation of $\tanh(x-\alpha)$.

Proof (sketch). To show that, the basic argument uses some modulational analysis: call $w(x,t):=v(x,t)-\tanh(x-\alpha(t))$, where we fix $\alpha(t)$ using the orthogonality condition $$ \int_{\mathbb R}\operatorname{sech}^2(x-\alpha(t))w(x,t)\,dx=0 \quad\forall t\geq 0.\label{2}\tag{2} $$ ($\alpha(t)$ is uniquely detemined by $v(t)$ thanks to the implicit function theorem if $|v(t)|_{L^2}$ is small enough). In this case, the function $w$ satisfies $$ w_t-w_{xx}-2(\tanh(x-\alpha)w)_x=(w^2)_x+f_x+\alpha'\operatorname{sech}^2(x-\alpha).\label{3}\tag{3} $$ Multiplying by $w$ and integrating in space (as for the usual energy estimates) we obtain $$ \frac 12\frac d{dt}\|w(t)\|_{L^2}+Q_\alpha(w)\leq\|f(t)\|_{L^2_x}\|w_x\|_{L^2}, $$ where the last term vanishes thanks to the orthogonality condition \eqref{2} the term $(w^2)_x$ vanishes when tested against $w$, and $Q$ is the quadratic form $$ Q_\alpha(w):=\int_{\mathbb R}w_x^2-\operatorname{sech}^2(x-\alpha)w^2\,dx. $$ Moreover (there is some work to do to show this) one can prove that, if $w$ satisfies the orthogonality condition \eqref{2}, then $$ Q_\alpha(w)\geq \frac{1}{2}\|w_x\|_{L^2}. $$ Thus, integrating in time, we get the nice estimate $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}. $$

Question: How to prove estimate \eqref{1} for arbitrary, large data $v_0$ and $f$? Is it even possible with a continuous parameter $\alpha$? If not, what if $\alpha$ is allowed to be discontinuous?

The main problem is that $\alpha(t)$ is not uniquely determined by the orthogonality condition \eqref{2} if the solution is large in the $|\cdot|_{L^2}$ norm. This leads to the fact, for instance, that the function $\alpha$ a priori is not necessarily continuous. If I knew that $\alpha$ satisfying \eqref{2} exists and is continuous, then I think I can prove the statement in a similar way by approximation, but I don't know how to prove that such a function $\alpha$ is continuous. In fact, if I pick $\alpha$ arbitrarily, it is perfectly possible to have something non-continuous. In the small data case, the continuity of $\alpha$ follows from equations \eqref{2} and \eqref{3}:

Sketch. Differentiating \eqref{2} in $t$ and substituting $w_t$ using \eqref{3}, one obtains: $$ |\alpha'(t)| \left |\int_{\mathbb R}\operatorname{sech}^2(x-\alpha)[\operatorname{sech}^2-w_x]dx\right|\leq\dots, $$ where the right hand side has good terms.

I think my question applies to many PDE that admit stationary solutions or travelling waves. I will simply describe the setting that I am more familiar with, but the technique is standard and applies to several models. The rough question is: "How to obtain the estimates on the perturbation of the stationary solution in the case where the perturbation is large, and the modulation parameter is not uniquely defined?"

Consider the viscous Burgers' equation in $\mathbb R$ with a forcing term: $$ v_t-v_{xx}=(v^2)_x +f_x,\qquad (x,t)\in \mathbb R\times [0,\infty) $$ $$ v(x,0)=v_0(x), \qquad x\in\mathbb R,$$ with kink-like boundary conditions at infinity: $v_0-\tanh(x)\in L^2(\mathbb R)$. The function $\tanh(x)$ is special because it is a stationary solution of the equation, together with all its translations. The only think to know is that $\tanh(x)\rightarrow \pm 1$ exponentially as $x\to\pm\infty$, and $\partial_x \tanh(x)=\operatorname{sech}^2(x)$. The forcing term $f$ lies in $L^2(\mathbb R\times (0,\infty)).$

The equation is globally well-posed in the affine space $L^2(\mathbb R)+\tanh$ by standard arguments.

Now, assume that the forcing term $f$ is small enough in $L^2(\mathbb R\times (0,\infty))$, and that the initial profile is a small perturbation of some translation of $\tanh$, i.e., $|v_0|_{L^2}$ is small enough, where $$ |z|_{L^2}:=\min_{\gamma\in \mathbb R}\|z-\tanh(x-\gamma)\|_{L^2}. $$ Then, it is possible to prove that the quantity $|v(t)|_{L^2}$ is uniformly small on $[0,\infty)$, and in particular there exists a function $\alpha:[0,\infty)\mapsto \mathbb R$ such that $v=w+\tanh(x-\alpha(t))$ with the estimates $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}.\qquad (1) $$ The function $\alpha$ is often called modulation parameter and what it is roughly meant to do is describe the "position" of the stationary solution that approximates $v$ best. In other words, we see $v$ as a perturbation of $\tanh(x-\alpha)$.

Proof (sketch). To show that, the basic argument uses some modulational analysis: call $w(x,t):=v(x,t)-\tanh(x-\alpha(t))$, where we fix $\alpha(t)$ using the orthogonality condition $$ \int_{\mathbb R}\operatorname{sech}^2(x-\alpha(t))w(x,t)\,dx=0 \quad\forall t\geq 0.\qquad\qquad(2) $$ ($\alpha(t)$ is uniquely detemined by $v(t)$ thanks to the implicit function theorem if $|v(t)|_{L^2}$ is small enough). In this case, the function $w$ satisfies $$ w_t-w_{xx}-2(\tanh(x-\alpha)w)_x=(w^2)_x+f_x+\alpha'\operatorname{sech}^2(x-\alpha).\qquad (3) $$ Multiplying by $w$ and integrating in space (as for the usual energy estimates) we obtain $$ \frac 12\frac d{dt}\|w(t)\|_{L^2}+Q_\alpha(w)\leq\|f(t)\|_{L^2_x}\|w_x\|_{L^2}, $$ where the last term vanishes thanks to the orthogonality condition $(2)$, the term $(w^2)_x$ vanishes when tested against $w$, and $Q$ is the quadratic form $$ Q_\alpha(w):=\int_{\mathbb R}w_x^2-\operatorname{sech}^2(x-\alpha)w^2\,dx. $$ Moreover (there is some work to do to show this) one can prove that, if $w$ satisfies the orthogonality condition $(2)$, then $$ Q_\alpha(w)\geq \frac{1}{2}\|w_x\|_{L^2}. $$ Thus, integrating in time, we get the nice estimate $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}. $$

Now, if you made it alive until here, I really cannot thank you enough. My problem is:

Question: How to prove estimate $(1)$ for arbitrary, large data $v_0$ and $f$?

The main problem is that $\alpha(t)$ is not uniquely determined by the orthogonality condition $(2)$ if the solution is large in the $|\cdot|_{L^2}$ norm. This leads to the fact, for instance, that the function $\alpha$ a priori is not necessarily continuous. If I knew that $\alpha$ satisfying $(2)$ exists and is continuous, then I think I can prove the statement in a similar way by approximation, but I don't know how to prove that such a function $\alpha$ is continuous. In fact, if I pick $\alpha$ arbitrarily, it is perfectly possible to have something non-continuous. In the small data case, the continuity of $\alpha$ follows from equations $(2)$ and $(3)$:

Sketch. Differentiating $(2)$ in $t$ and substituting $w_t$ using $(3)$, one obtains: $$ |\alpha'(t)| \left |\int_{\mathbb R}\operatorname{sech}^2(x-\alpha)[\operatorname{sech}^2-w_x]dx\right|\leq\dots, $$ where the right hand side has good terms.

I feel like this issue is typical of modulational analysis for large data, but I have no idea how to treat this situation.

I think my question applies to many PDE that admit stationary solutions or travelling waves. I will simply describe the setting that I am more familiar with, but the technique is standard and applies to several models. The rough question is: "How to obtain the estimates on the perturbation of the stationary solution in the case where the perturbation is large, and the modulation parameter is not uniquely defined?"

Consider the viscous Burgers' equation in $\mathbb R$ with a forcing term: $$ v_t-v_{xx}=(v^2)_x +f_x,\qquad (x,t)\in \mathbb R\times [0,\infty) $$ $$ v(x,0)=v_0(x), \qquad x\in\mathbb R,$$ with kink-like boundary conditions at infinity: $v_0-\tanh(x)\in L^2(\mathbb R)$. The function $\tanh(x)$ is special because it is a stationary solution of the equation, together with all its translations. The only think to know is that $\tanh(x)\rightarrow \pm 1$ exponentially as $x\to\pm\infty$, and $\partial_x \tanh(x)=\operatorname{sech}^2(x)$. The forcing term $f$ lies in $L^2(\mathbb R\times (0,\infty)).$

The equation is globally well-posed in the affine space $L^2(\mathbb R)+\tanh$ by standard arguments.

Now, assume that the forcing term $f$ is small enough in $L^2(\mathbb R\times (0,\infty))$, and that the initial profile is a small perturbation of some translation of $\tanh$, i.e., $|v_0|_{L^2}$ is small enough, where $$ |z|_{L^2}:=\min_{\gamma\in \mathbb R}\|z-\tanh(x-\gamma)\|_{L^2}. $$ Then, it is possible to prove that the quantity $|v(t)|_{L^2}$ is uniformly small on $[0,\infty)$, and in particular there exists a function $\alpha:[0,\infty)\mapsto \mathbb R$ such that $v=w+\tanh(x-\alpha(t))$ with the estimates $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}.\qquad (1) $$ The function $\alpha$ is often called modulation parameter and what it is roughly meant to do is describe the "position" of the stationary solution that approximates $v$ best. In other words, we see $v$ as a perturbation of $\tanh(x-\alpha)$.

Proof (sketch). To show that, the basic argument uses some modulational analysis: call $w(x,t):=v(x,t)-\tanh(x-\alpha(t))$, where we fix $\alpha(t)$ using the orthogonality condition $$ \int_{\mathbb R}\operatorname{sech}^2(x-\alpha(t))w(x,t)\,dx=0 \quad\forall t\geq 0.\qquad\qquad(2) $$ ($\alpha(t)$ is uniquely detemined by $v(t)$ thanks to the implicit function theorem if $|v(t)|_{L^2}$ is small enough). In this case, the function $w$ satisfies $$ w_t-w_{xx}-2(\tanh(x-\alpha)w)_x=(w^2)_x+f_x+\alpha'\operatorname{sech}^2(x-\alpha).\qquad (3) $$ Multiplying by $w$ and integrating in space (as for the usual energy estimates) we obtain $$ \frac 12\frac d{dt}\|w(t)\|_{L^2}+Q_\alpha(w)\leq\|f(t)\|_{L^2_x}\|w_x\|_{L^2}, $$ where the last term vanishes thanks to the orthogonality condition $(2)$, the term $(w^2)_x$ vanishes when tested against $w$, and $Q$ is the quadratic form $$ Q_\alpha(w):=\int_{\mathbb R}w_x^2-\operatorname{sech}^2(x-\alpha)w^2\,dx. $$ Moreover (there is some work to do to show this) one can prove that, if $w$ satisfies the orthogonality condition $(2)$, then $$ Q_\alpha(w)\geq \frac{1}{2}\|w_x\|_{L^2}. $$ Thus, integrating in time, we get the nice estimate $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}. $$

Now, if you made it alive until here, I really cannot thank you enough. My problem is:

Question: How to prove estimate $(1)$ for arbitrary, large data $v_0$ and $f$? Is it even possible with a continuous parameter $\alpha$? If not, what if $\alpha$ is allowed to be discontinuous?

The main problem is that $\alpha(t)$ is not uniquely determined by the orthogonality condition $(2)$ if the solution is large in the $|\cdot|_{L^2}$ norm. This leads to the fact, for instance, that the function $\alpha$ a priori is not necessarily continuous. If I knew that $\alpha$ satisfying $(2)$ exists and is continuous, then I think I can prove the statement in a similar way by approximation, but I don't know how to prove that such a function $\alpha$ is continuous. In fact, if I pick $\alpha$ arbitrarily, it is perfectly possible to have something non-continuous. In the small data case, the continuity of $\alpha$ follows from equations $(2)$ and $(3)$:

Sketch. Differentiating $(2)$ in $t$ and substituting $w_t$ using $(3)$, one obtains: $$ |\alpha'(t)| \left |\int_{\mathbb R}\operatorname{sech}^2(x-\alpha)[\operatorname{sech}^2-w_x]dx\right|\leq\dots, $$ where the right hand side has good terms.

I feel like this issue is typical of modulational analysis for large data, but I have no idea how to treat this situation.

I think my question applies to many PDE that admit stationary solutions or travelling waves. I will simply describe the setting that I am more familiar with, but the technique is standard and applies to several models.

Consider the viscous Burgers' equation in $\mathbb R$ with a forcing term: $$ v_t-v_{xx}=(v^2)_x +f_x,\qquad (x,t)\in \mathbb R\times [0,\infty) $$ $$ v(x,0)=v_0(x), \qquad x\in\mathbb R,$$ with kink-like boundary conditions at infinity: $v_0-\tanh(x)\in L^2(\mathbb R)$. The function $\tanh(x)$ is special because it is a stationary solution of the equation, together with all its translations. The only think to know is that $\tanh(x)\rightarrow \pm 1$ exponentially as $x\to\pm\infty$, and $\partial_x \tanh(x)=\operatorname{sech}^2(x)$. The forcing term $f$ lies in $L^2(\mathbb R\times (0,\infty)).$

The equation is globally well-posed in the affine space $L^2(\mathbb R)+\tanh$ by standard arguments.

Now, assume that the forcing term $f$ is small enough in $L^2(\mathbb R\times (0,\infty))$, and that the initial profile is a small perturbation of some translation of $\tanh$, i.e., $|v_0|_{L^2}$ is small enough, where $$ |z|_{L^2}:=\min_{\gamma\in \mathbb R}\|z-\tanh(x-\gamma)\|_{L^2}. $$ Then, it is possible to prove that the quantity $|v(t)|_{L^2}$ is uniformly small on $[0,\infty)$, and in particular there exists a function $\alpha:[0,\infty)\mapsto \mathbb R$ such that $v=w+\tanh(x-\alpha(t))$ with the estimates $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}.\qquad (1) $$ The function $\alpha$ is often called modulation parameter and what it is roughly meant to do is describe the "position" of the stationary solution that approximates $v$ best. In other words, we see $v$ as a perturbation of $\tanh(x-\alpha)$.

Proof (sketch). To show that, the basic argument uses some modulational analysis: call $w(x,t):=v(x,t)-\tanh(x-\alpha(t))$, where we fix $\alpha(t)$ using the orthogonality condition $$ \int_{\mathbb R}\operatorname{sech}^2(x-\alpha(t))w(x,t)\,dx=0 \quad\forall t\geq 0.\qquad\qquad(2) $$ ($\alpha(t)$ is uniquely detemined by $v(t)$ thanks to the implicit function theorem if $|v(t)|_{L^2}$ is small enough). In this case, the function $w$ satisfies $$ w_t-w_{xx}-2(\tanh(x-\alpha)w)_x=(w^2)_x+f_x+\alpha'\operatorname{sech}^2(x-\alpha).\qquad (3) $$ Multiplying by $w$ and integrating in space (as for the usual energy estimates) we obtain $$ \frac 12\frac d{dt}\|w(t)\|_{L^2}+Q_\alpha(w)\leq\|f(t)\|_{L^2_x}\|w_x\|_{L^2}, $$ where the last term vanishes thanks to the orthogonality condition $(2)$, the term $(w^2)_x$ vanishes when tested against $w$, and $Q$ is the quadratic form $$ Q_\alpha(w):=\int_{\mathbb R}w_x^2-\operatorname{sech}^2(x-\alpha)w^2\,dx. $$ Moreover (there is some work to do to show this) one can prove that, if $w$ satisfies the orthogonality condition $(2)$, then $$ Q_\alpha(w)\geq \frac{1}{2}\|w_x\|_{L^2}. $$ Thus, integrating in time, we get the nice estimate $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}. $$

Now, if you made it alive until here, I really cannot thank you enough. My problem now is:

Question: How to prove estimate $(1)$ for arbitrary, large data $v_0$ and $f$?

The main problem is that $\alpha(t)$ is not uniquely determined by the orthogonality condition $(2)$ if the solution is large in the $|\cdot|_{L^2}$ norm. This leads to the fact, for instance, that the function $\alpha$ a priori is not necessarily continuous. If I knew that $\alpha$ satisfying $(2)$ exists and is continuous, then I think I can prove the statement in a similar way by approximation, but I don't know how to prove that such a function $\alpha$ is continuous. In fact, if I pick $\alpha$ arbitrarily, it is perfectly possible to have something non-continuous. In the small data case, the continuity of $\alpha$ follows from equations $(2)$ and $(3)$:

Sketch. Differentiating $(2)$ in $t$ and substituting $w_t$ using $(3)$, one obtains: $$ |\alpha'(t)| \left |\int_{\mathbb R}\operatorname{sech}^2(x-\alpha)[\operatorname{sech}^2-w_x]dx\right|\leq\dots, $$ where the right hand side has good terms.

I feel like this issue is typical of modulational analysis for large data, but I have no idea how to treat this situation.

I think my question applies to many PDE that admit stationary solutions or travelling waves. I will simply describe the setting that I am more familiar with, but the technique is standard and applies to several models. The rough question is: "How to obtain the estimates on the perturbation of the stationary solution in the case where the perturbation is large, and the modulation parameter is not uniquely defined?"

Consider the viscous Burgers' equation in $\mathbb R$ with a forcing term: $$ v_t-v_{xx}=(v^2)_x +f_x,\qquad (x,t)\in \mathbb R\times [0,\infty) $$ $$ v(x,0)=v_0(x), \qquad x\in\mathbb R,$$ with kink-like boundary conditions at infinity: $v_0-\tanh(x)\in L^2(\mathbb R)$. The function $\tanh(x)$ is special because it is a stationary solution of the equation, together with all its translations. The only think to know is that $\tanh(x)\rightarrow \pm 1$ exponentially as $x\to\pm\infty$, and $\partial_x \tanh(x)=\operatorname{sech}^2(x)$. The forcing term $f$ lies in $L^2(\mathbb R\times (0,\infty)).$

The equation is globally well-posed in the affine space $L^2(\mathbb R)+\tanh$ by standard arguments.

Now, assume that the forcing term $f$ is small enough in $L^2(\mathbb R\times (0,\infty))$, and that the initial profile is a small perturbation of some translation of $\tanh$, i.e., $|v_0|_{L^2}$ is small enough, where $$ |z|_{L^2}:=\min_{\gamma\in \mathbb R}\|z-\tanh(x-\gamma)\|_{L^2}. $$ Then, it is possible to prove that the quantity $|v(t)|_{L^2}$ is uniformly small on $[0,\infty)$, and in particular there exists a function $\alpha:[0,\infty)\mapsto \mathbb R$ such that $v=w+\tanh(x-\alpha(t))$ with the estimates $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}.\qquad (1) $$ The function $\alpha$ is often called modulation parameter and what it is roughly meant to do is describe the "position" of the stationary solution that approximates $v$ best. In other words, we see $v$ as a perturbation of $\tanh(x-\alpha)$.

Proof (sketch). To show that, the basic argument uses some modulational analysis: call $w(x,t):=v(x,t)-\tanh(x-\alpha(t))$, where we fix $\alpha(t)$ using the orthogonality condition $$ \int_{\mathbb R}\operatorname{sech}^2(x-\alpha(t))w(x,t)\,dx=0 \quad\forall t\geq 0.\qquad\qquad(2) $$ ($\alpha(t)$ is uniquely detemined by $v(t)$ thanks to the implicit function theorem if $|v(t)|_{L^2}$ is small enough). In this case, the function $w$ satisfies $$ w_t-w_{xx}-2(\tanh(x-\alpha)w)_x=(w^2)_x+f_x+\alpha'\operatorname{sech}^2(x-\alpha).\qquad (3) $$ Multiplying by $w$ and integrating in space (as for the usual energy estimates) we obtain $$ \frac 12\frac d{dt}\|w(t)\|_{L^2}+Q_\alpha(w)\leq\|f(t)\|_{L^2_x}\|w_x\|_{L^2}, $$ where the last term vanishes thanks to the orthogonality condition $(2)$, the term $(w^2)_x$ vanishes when tested against $w$, and $Q$ is the quadratic form $$ Q_\alpha(w):=\int_{\mathbb R}w_x^2-\operatorname{sech}^2(x-\alpha)w^2\,dx. $$ Moreover (there is some work to do to show this) one can prove that, if $w$ satisfies the orthogonality condition $(2)$, then $$ Q_\alpha(w)\geq \frac{1}{2}\|w_x\|_{L^2}. $$ Thus, integrating in time, we get the nice estimate $$ \|w\|_{L^\infty_tL^2_x}+\|w_x\|_{L^2_tL^2_x}\lesssim \|w_0\|_{L^2}+\|f\|_{L^2_tL^2_x}. $$

Now, if you made it alive until here, I really cannot thank you enough. My problem is:

Question: How to prove estimate $(1)$ for arbitrary, large data $v_0$ and $f$?

The main problem is that $\alpha(t)$ is not uniquely determined by the orthogonality condition $(2)$ if the solution is large in the $|\cdot|_{L^2}$ norm. This leads to the fact, for instance, that the function $\alpha$ a priori is not necessarily continuous. If I knew that $\alpha$ satisfying $(2)$ exists and is continuous, then I think I can prove the statement in a similar way by approximation, but I don't know how to prove that such a function $\alpha$ is continuous. In fact, if I pick $\alpha$ arbitrarily, it is perfectly possible to have something non-continuous. In the small data case, the continuity of $\alpha$ follows from equations $(2)$ and $(3)$:

Sketch. Differentiating $(2)$ in $t$ and substituting $w_t$ using $(3)$, one obtains: $$ |\alpha'(t)| \left |\int_{\mathbb R}\operatorname{sech}^2(x-\alpha)[\operatorname{sech}^2-w_x]dx\right|\leq\dots, $$ where the right hand side has good terms.

I feel like this issue is typical of modulational analysis for large data, but I have no idea how to treat this situation.