Find the solution of $u_t + bu_x = 0$

Question

let $b \in \mathbb{R} $be constant .Solve:

$$\frac{\partial u}{\partial t} (x,t) + b \frac{ \partial u}{\partial x}(x,t) =0 , \quad x \in \mathbb{R}, \quad t >0$$

$$ u(x,0) = x^2$$

My attempt : $\frac{\partial}{\partial t} u (x,t) =- b \frac{ \partial u}{\partial x}(x,t) $ , Now again differenting both l.h.S and r.h.s , we get $$\frac{\partial^2}{\partial t^2} (x,t) =- b \frac{ \partial^2 u}{\partial x^2}(x,t) $$

after that applying the D alembert formula we have $$u(x,t) = \frac{(x+ bt)^2 +(x-bt)^2}{2} + \frac{1}{2b}\int_{x-bt}^{x+bt}0\,{\rm d}s.$$

Is its true ?

Answer 1

$$\frac{\partial u}{ dt} (x,t) + b \frac{ \partial u}{dx}(x,t) =0 $$ With the method of characteristics the general solution is immediate : $$u(x,t)=F(x-bt)$$ with arbitrary function $F$.

Condition : $u(x,0)=F(x)=x^2$. Now the function is known : $F(X)=X^2$ . We put it into the above general solution where $X=(x-bt)$ . $$u(x,t)=(x-bt)^2$$

Answer 2

Hint

Let $$\hat u(\alpha ,t)=\int_{\mathbb R}u(x,t)e^{-2i\pi \alpha x}\,\mathrm d x.$$ Then, $$\partial _xu(x,t)+b\partial _tu(x,t)=0\iff \frac{\mathrm d}{\mathrm d t}\hat u(\alpha ,t)-2i\pi \alpha \cdot \hat u(\alpha ,t)=0,$$ which is just a linear ODE to solve.

Answer 3

Making the change of variables

$$ \cases{ \xi=x\\ \eta=x-b t } $$

we have

$$ u(x,t) = w(\xi,\eta) $$

and after the change of variables we have

$$ b w_{\xi}(\xi,\eta)=0 \Rightarrow w(\xi,\eta) = \phi(\eta) $$

now considering the initial conditions

$$ \phi(b x) = x^2\Rightarrow \phi(\eta) = \left(\frac{\eta}{b}\right)^2 $$

and finally

$$ u(x,t) = \left(x - b t\right)^2 $$

Answer 4

Your approach is correct provided that the solution is sufficiently smooth (which is the case here). However, you made some mistakes in the derivation of the solution. Indeed, differentiating the linear advection equation $u_t + b u_x = 0$ w.r.t. $t$, we have $u_{tt} = -b u_{xt}$. Now, we use the equality of mixed derivatives $u_{xt} = u_{tx}$, and the initial PDE itself $u_{tx} = -b u_{xx}$ after differentiation w.r.t. $x$. Thus, the wave equation $$ u_{tt} = b^2 u_{xx} $$ is obtained. D'Alembert's solution for the initial condition $u(x,0) = x^2$ and $$u_t(x,0) = -b u_x(x, 0 ) = -2b x$$ reads \begin{aligned} u(x,t) &= \tfrac{1}{2} \left[(x+b t)^2 + (x-b t)^2\right] - \int_{x-bt}^{x+bt} \xi \,\text d \xi \\ &= (x-bt)^2 . \end{aligned} This is exactly the solution obtained by the more straightforward method of characteristics (see answer by @JJacquelin).