General solution of Transport equation

[evinda]

Hello! (Wave)

General solution of Transport equation (homogeneous): Method of Characteristics

$$u_t+cu_x=0 (\star)$$

We know that if $f: \mathbb{R} \to \mathbb{R}$ is differentiable then $u(x,t)=f(x-ct)$ is a solution of $(\star)$.

We will show that each solution of $(\star)$ is of the form $u(x,t)=f(x-ct)$, where $f$ is an arbitrary differentiable function, $f: \mathbb{R} \to \mathbb{R}$.

We consider the lines $\epsilon_a: x-ct=a, a \in \mathbb{R}$.

We define the function $z(s)=u(x(s),t(s))$, where $x(s)=cs+a$, $t(s)=s$.

Then $(x(s), t(s)) \in \epsilon_a \forall s \in \mathbb{R}$.
Let $u$ be a solution of $(\star)$ and for this $u$ I define $z(s)$ as previously.

Then $z'(s)=c u_x+ u_t=0$, i.e. $z(s)=\beta \forall s \in \mathbb{R}$.

$u(x(s),t(s))=z(s)=z(0)=u(a,0)=u(x(s)-ct(s),0) \forall s$

Thus $u(x,t)=u(x-ct,0)=:f(x-ct)$, where $f: \mathbb{R} \to \mathbb{R}$ is differentiable.I am looking at the following exercise:

Find the general solution of the non-homogeneous Transport equation $u_t+cu_x=g(x,t)$ where $g$ is "smooth" (how smooth?) and using the above find the general solution of the wave equation.

Hint: We work as previously : $z'(s)=g(x(s),t(s))$.So in this case do we know that $u(x,t)=f(x-ct)$ is again a solution of the given equation?

Or do we have to consider an other solution and show that all the solutions are of this form? (Thinking)

 

[gtoguy97]

Yes, we can assume that $u(x,t)=f(x-ct)$ is a solution of the given equation and then show that all the other solutions are of this form.We consider the lines $\epsilon_a: x-ct=a, a \in \mathbb{R}$.We define the function $z(s)=u(x(s),t(s))$, where $x(s)=cs+a$, $t(s)=s$.Then $(x(s), t(s)) \in \epsilon_a \forall s \in \mathbb{R}$.Let $u$ be a solution of $u_t+cu_x=g(x,t)$ and for this $u$ I define $z(s)$ as previously.Then $z'(s)=c u_x+ u_t-g(x(s),t(s))=0$, i.e. $z(s)=\beta \forall s \in \mathbb{R}$.$u(x(s),t(s))=z(s)=z(0)=u(a,0)=u(x(s)-ct(s),0) + \int_0^s g(x(r),t(r))dr \forall s$Thus $u(x,t)=u(x-ct,0)+\int_0^{t} g(x-cr,r)dr=:f(x-ct)+\int_0^{t} g(x-cr,r)dr$, where $f: \mathbb{R} \to \mathbb{R}$ is differentiable.For the wave equation, we can use the same method. We consider the lines $\epsilon_a: x-ct=a, a \in \mathbb{R}$.We define the function $z(s)=u(x(s),t(s))$, where $x(s)=cs+a$, $t(s)=s$.Then $(x(s), t(s)) \in \epsilon_a \forall s \in \mathbb{R}$.