Solution of a Transport-type equation

Take [itex]\xi = x+t[/itex] and [itex]u(x,t)=f(x+t)=f(\xi)[/itex]. What you are looking for is traveling wave solutions.

 

You are missing the point. First of all, your ecuation is a first order linear equation, wich means that given an initial condition (well behaved and non-characteristic of course), your solution is unique. This tells you that, since you don't have initial condition, you'll find a family of solutions.
Now, lets backtrack to the first suggestion. If [itex]\xi = x+t [/itex] and [itex]u(x,t) = f(x+t) = f(\xi)[/itex], then

[tex] u_t-u_x = 0,[/tex]

so [itex]f(x+t)[/itex] is a solution of the homogeneous equation. Now, for the particular solution, let [itex]\eta = x-t[/itex] and [itex]u(x,t) = g(x-t) = g(\eta)[/itex], then

[tex] u_t-u_x+2u= -2g'(\eta)+2g(\eta) = 0.[/tex]

Now, what is the family of solutions for your PDE?

 

Yup, that's correct,, but there is a more explicit form of the solution, given the ordinary differential equation for the particular solution, this means that

[tex]u(x,t) = f(x+t) + Ae^{x-t}[/tex]

is the family of solutions you are looking for (and given an initial condition, you can easily determine A and f).

 

True, my mistake. There is no inhomogeneous equation. The whole thing is wrong. Let's start over. We are going to use the method of characteristics. Given the equation [itex]u_t-u_x-2u=0[/itex], we can write it as

[tex]\left(\begin{array}{c} 1 \\ -1 \\ 2u\end{array}\right) \cdot \left( \begin{array}{c} u_x \\ u_t \\ -1 \end{array}\right)=0.[/tex]

This equation geometrically states that the vector field (1,-1,2u) is tangent to the surface [itex]\bigl(x(\xi,\eta),t(\xi,\eta),u(\xi,\eta)\bigr)[/itex], so given an initial condition [itex]u(\xi,0) = f(\xi)[/itex], we want to continue the surface trough the parameter [itex]\eta[/itex] using the ortogonality condition. So

[tex]\frac{d x}{d \eta} = 1,[/tex]

[tex]\frac{d t}{d \eta} = -1,[/tex]

[tex]\frac{d u}{d \eta} = 2 u,[/tex]

where [itex]x(\xi,0) = \xi,\, t(\xi,0) = \xi,\,u(\xi,0) = f(\xi)[/itex]. Hence

[tex]x(\xi,\eta) = \eta+\xi,[/tex]

[tex]t(\xi,\eta) = -\eta+\xi,[/tex]

[tex]u(\xi,\eta) = f(\xi)e^{2\eta}.[/tex]

Now you have a surface that satisfies the initial condition plus the PDE, but you need a function, wich means invertin the transformation [itex](\xi,\eta) \rightarrow (x,t)[/itex], so

[tex]\xi = \frac{x+t}{2},[/tex]

[tex]\eta = \frac{x-t}{2},[/tex]

and

[tex]u(x,t) = f\left(\frac{x+t}{2}\right)e^{x-t}[/tex]

is the solution of the equation.