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Willie Wong
Willie Wong
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First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}^{\lambda} = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma (applied to $\tilde{t} \tilde{\eta} = t\eta$ where $\tilde{\eta} = 2^{-\lambda(2\alpha + 1)} \eta$ and $\tilde{t} = 2^{\lambda(2\alpha + 1)} t$) we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$

First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}^{\lambda} = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$

First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}^{\lambda} = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma (applied to $\tilde{t} \tilde{\eta} = t\eta$ where $\tilde{\eta} = 2^{-\lambda(2\alpha + 1)} \eta$ and $\tilde{t} = 2^{\lambda(2\alpha + 1)} t$) we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$

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Source Link
Willie Wong
Willie Wong
  • 39k
  • 4
  • 94
  • 176

First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}(\lambda) = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$$$P_{\pm}^{\lambda} = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$

First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}(\lambda) = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$

First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}^{\lambda} = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$

Source Link
Willie Wong
Willie Wong
  • 39k
  • 4
  • 94
  • 176

First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$.

Question 1

Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate $$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi_+(2^\lambda \xi) + \chi_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}_+ + P^{-\lambda}_- $$ And we have that $$ P_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P_{\pm}^0 $$ The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space

$$ \| u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|_{B^{(1-\alpha)/2,1}_1} $$

Question 2

You don't have scaling homogeneity. But instead you can consider $$P_{\pm}(\lambda) = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi_{\pm}(\xi) ~\mathrm{d}\xi $$ for the frequency $2^{\lambda}$ piece. You see that the phase function $\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that $$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$ We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma we get exactly the same decay estimate as in Question 1.

The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that $$ \|u\|_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|_{B^{2,1}_1} $$

(a loss of one derivative compared to the $\alpha > -1/2$ case)

For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound

$$ \|u\|_{\infty} \lesssim \|f\|_{W^{1,2}}$$