Show that IVP has a solution which remains bounded as $t\to \infty$.

Question

Find the solution of the Cauchy problem for the pde

$\dfrac{\partial u }{\partial t}+x\dfrac{\partial u}{\partial x}=x$

$u(x,0)=2x$ and show that it has a solution which remains bounded as $t\to \infty$.

By Lagrange's Equations

$\dfrac{dt}{1}=\dfrac{dx}{x}=\dfrac{du}{x}$

If I solve the above I get ;

$t=\ln x+\ln c\implies t=\ln(cx )\implies cx=e^t$

Also, $x=u+k$.Hence the solution becomes $u-x=f(\dfrac{e^t}{x})$ where $f$ is an arbitrary function.

Now applying the boundary conditions on $ u=x+f(\dfrac{e^t}{x})$; I get $2x=x+f(\dfrac{1}{x})\implies f(\dfrac{1}{x})=x=\dfrac{1}{\dfrac{1}{x}}$

So $f(t)=\dfrac{1}{t}$. So $u-x=\dfrac{x}{e^t}\implies u=x+\dfrac{x}{e^t}\to_{t\to \infty} 2x$

But how to show that the solution is bounded from here?Please help.

Answer 1

We have $$ \lim_{t \to \infty}\left(\frac1{e^t}\right)=0 $$ thus, there exists a $t_0>0$ such that $$ \left|\frac1{e^t}\right|\leq \color{red}{1}, \quad \text{for all}\quad t\geq t_0, $$ giving, for any fixed $x$, as $t \to \infty$:

$$ |u(x,t)|=\left|x+\dfrac{x}{e^t}\right| \leq \left|x\right|+\left|\dfrac{x}{e^t}\right|\leq\left|x\right|+|x|\times \color{red}{1}\leq2|x|. $$

The preceding inequality shows that $u(x,\cdot)$ is bounded as $t \to \infty$.