# Solution of a Transport-type equation

## Main Question or Discussion Point

Hi there,
I'm trying to find all solutions of:

$$\frac{\partial u}{\partial t}(x,t)-\frac{\partial u}{\partial x}(x,t)=-2u(x,t)$$

I know that one solution is $$u(x,t)=Ae^{x-t}$$, and any solution of $$\frac{\partial u}{\partial t}(x,t)-\frac{\partial u}{\partial x}(x,t)=0$$ is of the form $$u(x,t)=f(x+t)$$.

I tried adding these solutions together but it doesn't work...

The question says that a change of coordinates to simplify $$\frac{\partial u}{\partial t}(x,t)-\frac{\partial u}{\partial x}(x,t)$$ could be useful, but I don't know where to begin in doing that...

Any help would be much appreciated, thanks.

## Answers and Replies

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

I tried a coordinate change of $$x^{'}=t-x , t^{'}=t+x$$ and tried looking for solutions of the form $$u(x,t)=v(x^{'},t^{'})$$. This reduces the problem to $$\frac{\partial u}{\partial t}(x,t) - \frac{\partial u}{\partial x}(x,t) = 2\frac{\partial v}{\partial x^{'}}(x^{'},t^{'})$$.

Now I think I have to solve $$\frac{\partial v}{\partial x^{'}}(x^{'},t^'})+v(x^{'},t^{'}) = 0$$, all solutions of which are of the form $$v(x^{'},t^{'})=Ae^{f(t^{'})-x^{'}}$$ (I think), so our solution set are functions $$u$$ of the form $$u(x,t)=Ae^{f(x+t)+x-t}$$ for some function $$f$$.

I know that these functions solve the equation, and I don't know how to find any other kind of 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 $\xi = x+t$ and $u(x,t) = f(x+t) = f(\xi)$, then

$$u_t-u_x = 0,$$

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

$$u_t-u_x+2u= -2g'(\eta)+2g(\eta) = 0.$$

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

So I would guess that they are of the form $$u(x,t)=f(x+t)+g(x-t)$$, where if $$\eta=x-t$$, then $$g'(\eta)=g(\eta)$$, much like ODE's when you add the solutions...

But I've tried plugging that back in and I'm not sure it works... Sorry if I'm missing the point, I've only just started studying these...

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

$$u(x,t) = f(x+t) + Ae^{x-t}$$

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

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But then doesn't $$u_{t}-u_{x}+2u = 2f$$?

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 $u_t-u_x-2u=0$, we can write it as

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

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

$$\frac{d x}{d \eta} = 1,$$

$$\frac{d t}{d \eta} = -1,$$

$$\frac{d u}{d \eta} = 2 u,$$

where $x(\xi,0) = \xi,\, t(\xi,0) = \xi,\,u(\xi,0) = f(\xi)$. Hence

$$x(\xi,\eta) = \eta+\xi,$$

$$t(\xi,\eta) = -\eta+\xi,$$

$$u(\xi,\eta) = f(\xi)e^{2\eta}.$$

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

$$\xi = \frac{x+t}{2},$$

$$\eta = \frac{x-t}{2},$$

and

$$u(x,t) = f\left(\frac{x+t}{2}\right)e^{x-t}$$

is the solution of the equation.

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