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

Dear MHF members,

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

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

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

3. Originally Posted by Rebesques
Expand by Taylor in ${x\geq t>0}$ to get $\Phi(x,t)=\Phi_0(x)+\Phi_x(x,0)x+R(\xi,0)x^2\leq \M{\rm e}^{ax}+\Phi_x(x,0)x+g(x)$, with the obvious choices. Invoking $\Phi_x=-\Phi_t$ we obtain $|\Phi_t|\leq \frac{1}{x}(|\Phi|+g(x))$ 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.
$\Phi(x,t)=\Phi_{0}(x)+\Phi_{\mathrm{t}}(x,0)t+\dis playstyle\int_{0}^{t}(t-u)\Phi_{\mathrm{tt}}(x,u)\mathrm{d}u$.
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
$\Phi(x,t)=\Phi_{0}(x)+\displaystyle\int_{0}^{t}\Ph i_{\mathrm{t}}(x,u)\mathrm{d}u$
..........._ $=\Phi_{0}(x)-\displaystyle\int_{0}^{t}\Phi_{\mathrm{x}}(x,u)\ma thrm{d}u$.
But I dont know if we can move somewhere from here...

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

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