# Have I obtained the proper solution to this PDE?

I'm a little stuck on this.

Consider $u_t -(1+t^2)u_x = \phi(x,t) \quad u(x,0)=u_0(x)$

Via the method of characteristics, the total derivative of $u(x,t)$ is

$$\frac{du}{dt} = \dfrac{\partial u}{\partial t} + \frac{dx}{dt}\frac{\partial u}{\partial x}\>.$$

Therefore, the characteristics satisfy

$$\frac{dx}{dt} = -(1+t^2) \implies x(t)=-\left(t+\frac{t^3}{3}\right) +x_0\>,$$

And the value of $u$ satisfies

$$\frac{du}{dt} = \phi(t)\>.$$

Here, the $x$ dependance has been dropped, as $x$ is a function of $t$.

Now, I could write the solution to this DE as $$u(x(t),t) = u_0(t_0) + \int_{t_0}^{t} \phi(z) dz = u_0(t_0) -\Phi(t_0) + \Phi(t)$$

But that doesn't seem to get me anywhere. The solution should have the initial profile moving along the characteristic, with the magnitude of the profile changing according to the solution to the differential equation for $u$ w.r.t $t$.

So how can I rectify this?

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=-(1+t^2)=-1-s^2$ , letting $x(0)=x_0$ , we have $x=x_0-s-\dfrac{s^3}{3}=x_0-t-\dfrac{t^3}{3}$

$\dfrac{du}{ds}=\phi(x,t)=\phi\left(x_0-s-\dfrac{s^3}{3},s\right)$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)+\int_0^s\phi\left(x_0-r-\dfrac{r^3}{3},r\right)dr=f\left(x+t+\dfrac{t^3}{3}\right)+\int_0^t\phi\left(x+t-r+\dfrac{t^3-r^3}{3},r\right)dr$

$u(x,0)=u_0(x)$ :

$f(x)=u_0(x)$

$\therefore u(x,t)=u_0\left(x+t+\dfrac{t^3}{3}\right)+\int_0^t\phi\left(x+t-r+\dfrac{t^3-r^3}{3},r\right)dr$

It is a matter of your repeated use of symbols. You cannot use the same $t$ again when you try to make $u$ a function on characteristics. Since the idea of using characteristics is based on change of coordinates $(x,t) \leftrightarrow (r,s)$

Define $r$ such that $u(t,x) = u( t(r), x(r))$ $$\frac{du}{dr} = \frac{dt}{dr}u_t+\frac{dx}{dr}u_x$$ before going on, you need to multiply both sides of your eqn by a good integrating factor, say, $$\frac{-1}{1+t^2}u_t+u_x = \psi(x,t)$$

then you may obtain $t+\frac{1}{3}t^3=r$ and $x=r+s$, where $s$ is an integration constant and just added to one of these differential equation only since $r$ is not uniquely defined. You can now change the coordinates to $(r,s)$ and solve a single variable PDE.

By doing so, you just have to change back to coordinate $(t,x)$ at the end.