1. ## Autonomous PDE admits a solution of exponential order?

Dear MHF members,

consider the PDE IVP
$\displaystyle \dfrac{\partial\Phi}{\partial x}+\dfrac{\partial\Phi}{\partial t}=0$ for $\displaystyle (x,t)\in\Omega:=\{(x,t)\in\mathbb{R}^{2}:\ x\geq t\geq0\}$
under the initial condition
$\displaystyle \Phi(x,0)=\Phi_{0}(x)$ for $\displaystyle x\geq0$, where $\displaystyle \Phi_{0}\in\mathrm{C}^{1}$ is a function of exponential order,
i.e., there exists $\displaystyle M,a>0$ such that $\displaystyle |\Phi_{0}(x)|\leq M\mathrm{e}^{ax}$ for all $\displaystyle x\geq0$.
Prove that for each fixed $\displaystyle x\geq0$, the solution $\displaystyle \Phi(x,\cdot)$ is of exponential order, i.e.,
there exists $\displaystyle N_{x},b_{x}>0$ such that $\displaystyle |\Phi(x,t)|\leq N_{x}\mathrm{e}^{b_{x}t}$ for all $\displaystyle t\geq0$.

Note. Do not use the solution $\displaystyle \Phi(x,t)=\Phi_{0}(x-t)$ directly to show it.

2. Expand by Taylor in $\displaystyle {x\geq t>0}$ to get $\displaystyle \Phi(x,t)=\Phi_0(x)+\Phi_x(x,0)x+R(\xi,0)x^2\leq \M{\rm e}^{ax}+\Phi_x(x,0)x+g(x)$, with the obvious choices. Invoking $\displaystyle \Phi_x=-\Phi_t$ we obtain $\displaystyle |\Phi_t|\leq \frac{1}{x}(|\Phi|+g(x))$ and the result follows by using Gronwall's inequality.

3. Originally Posted by Rebesques
Expand by Taylor in $\displaystyle {x\geq t>0}$ to get $\displaystyle \Phi(x,t)=\Phi_0(x)+\Phi_x(x,0)x+R(\xi,0)x^2\leq \M{\rm e}^{ax}+\Phi_x(x,0)x+g(x)$, with the obvious choices. Invoking $\displaystyle \Phi_x=-\Phi_t$ we obtain $\displaystyle |\Phi_t|\leq \frac{1}{x}(|\Phi|+g(x))$ and the result follows by using Gronwall's inequality.
Rebesques thanks for the reply, but it seems that there is a mistake with your Taylor's formula, it should read as follows.
$\displaystyle \Phi(x,t)=\Phi_{0}(x)+\Phi_{\mathrm{t}}(x,0)t+\dis playstyle\int_{0}^{t}(t-u)\Phi_{\mathrm{tt}}(x,u)\mathrm{d}u$.
However, as the function has first order derivatives, we can not expand it up to its second-order derivative, i.e., we can only use
$\displaystyle \Phi(x,t)=\Phi_{0}(x)+\displaystyle\int_{0}^{t}\Ph i_{\mathrm{t}}(x,u)\mathrm{d}u$
..........._$\displaystyle =\Phi_{0}(x)-\displaystyle\int_{0}^{t}\Phi_{\mathrm{x}}(x,u)\ma thrm{d}u$.
But I dont know if we can move somewhere from here...

4. ## Re: Autonomous PDE admits a solution of exponential order?

If $\displaystyle \Phi_{0}\in\mathrm{C}^{\infty}$ has the series expansion
$\displaystyle \Phi_{0}(x)=\sum_{k=0}^{\infty}a_{k}\frac{x^{k}}{k !}$
satisfying $\displaystyle |a_{k}|\leq Mr^{k}$ for all $\displaystyle k\in\mathbb{N}_{0}$,
where $\displaystyle M,r>0$, then $\displaystyle \Phi$ is of exponential order.
One can show by direct substitution that
$\displaystyle \Phi(x,t)=\sum_{k=0}^{\infty}a_{k}\frac{(x-t)^{k}}{k!}$,
which yields
$\displaystyle |\Phi(x,t)|\leq M\sum_{k=0}^{\infty}r^{k}\frac{x^{k}}{k!}=M\mathrm {e}^{rx}$ for all $\displaystyle (x,t)\in\Omega$.
Also if $\displaystyle \{a_{k}\}$ is a nonnegative sequence, then we obtain
$\displaystyle \Phi(x,t)\leq\sum_{k=0}^{\infty}a_{k}\frac{x^{k}}{ k!}=\Phi_{0}(t)$ for all $\displaystyle (x,t)\in\Omega$,
which implies $\displaystyle \Phi$ is of exponential order too.
But I dont know how we can show this only with $\displaystyle \Phi_{0}\in\mathrm{C}^{1}$.