Is there any error in the solution of the following Cauchy-Neumann problem? Let the following Cauchy-Neumann problem be:
$\partial_tn + 2xt\partial_x n = n + \frac{1}{1+t^\alpha}\partial^2_{yy}n  $, $n =n(x,y,t), (x,y,t )\in \mathbb R \times(0,1)\times (0, \infty)$
$\partial_yn(x,0,t)=0$, $\partial_yn(x,1,t)=0$ , $(x,t)\in \mathbb R \times(0, \infty)$
$n(x,y,0)=x, (x,y) \in \mathbb R \times[0,1]$
where $\alpha \in \mathbb R_+ $(that  is strictly positive). Is the following statement true or false?
$\int_{-1}^{1}\int_{0}^{1}n(x,y,t)dydx=0,$ for each $t \ge 0$
Solution:
Let $\int_0^1n(x,y,t)dy =N(x,t)$. Integrating the differential equation with respect to y
$\partial_tn + 2xt\partial_x n = n + \frac{1}{1+t^\alpha}\partial^2_{yy}n  $
$\partial_tN + 2xt\partial_x N = N + \frac{1}{1+t^\alpha}\partial_{y}n|_0^1 $.
Using the boundary conditions:$\partial_{y}n|_0^1  =0$, So
$\partial_tN + 2xt\partial_x N = N $
This equation can be solve using the method of characteristics
Let $x(t)$ be the characterstic with initial condition $x(0)=x_0$
the equation is $\frac{dx(t)}{dt}=2xt$. Solving it by separation of variables I get $x(t)=x_0 e^{t^2}$...(*)
Let $\hat N(t)=N(x(t),t)$ be the restriction to the characteristic. So the equation to be solved is $\frac{d\hat N(t)}{dt}=\hat N(t)$ with initial condition $\hat N(0) =N(x_0,t)=x_0$. The solution is $\hat N(t) = x_0 e^t$. From (*)$x_0=xe^{-t^2}$. Substituing yields $N(x,t) = xe^{-t^2+t} $
So the expresion in the given statement is  $\int_{-1}^{1}\int_{0}^{1}n(x,y,t)dydx=\int_{-1}^{1}N(x,t)dx =\int_{-1}^{1}xe^{-t^2+t}dx=2 e^{-t^2+t}$, which is clearly not 0. So the statement is FALSE
The answer I was given to this exercise was TRUE, so what am I doing wrong? Or perhaps the given answer is wrong.
Edit: Ok, don't tell me anything I already found out my mistake was the most stupid one:$\int_{-1}^{1}xe^{-t^2+t}dx=0 $. And this cost me 5 points in my final exam, otherwise I would have gotten perfect score :(.
Now I would change the question. Am I right that the $\frac{1}{1+t^\alpha}$ term does not affect the solution whatever the value of $\alpha$?
 A: $$\partial_tn + 2xt\partial_x n = n + \frac{1}{1+t^\alpha}\partial^2_{yy}n  $$
I agree with your approach but there is a mistake at the end.
With $\quad N(x,t)=\int_0^1n(x,y,t)dy$
$$ \partial_tN + 2xt\partial_x N = N \quad\text{is OK.}$$.
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dt}{1}=\frac{dx}{2xt}=\frac{dN}{N}$$
A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{2xt}$ :
$$x\:e^{-t^2}=c_1$$
A second characteristic equation comes from solving $\frac{dt}{1}=\frac{dN}{N}$ :
$$N\:e^{-t}=c_2$$
The general solution (without condition) is :
$$N\:e^{-t}=F\left(x\:e^{-t^2}\right)\quad\implies\quad N(x,t)=e^t F\left(x\:e^{-t^2}\right)$$
$F$ is an arbitrary function until some bondition be taken into account.
CONDITION : $n(x,y,0)=x$
$\begin{cases}
N(x,0)=\int_0^1 n(x,y,0)dy=\int_0^1 xdy=x\int_0^1 dy=x \\
N(x,0)=e^0 F\left(x\:e^{-0^2}\right)=F(x)
\end{cases}\quad\implies\quad F(x)=x$
Now the function $F$ is determined. We put it into the above general solution where the argument is $x\:e^{-t^2}$ :
$N(x,t)=e^t \left(x\:e^{-t^2}\right)$
$$\boxed{N(x,t)=x\:e^{t-t^2}}$$
So I agree with you up to this point.
$\int_{-1}^{1}\int_{0}^{1}n(x,y,t)dydx=\int_{-1}^{1}N(x,t)dx =\int_{-1}^{1}xe^{-t^2+t}dx=e^{-t^2+t}\int_{-1}^{1}xdx=\frac12e^{-t^2+t}\left(x^2|_{-1}^{1}\right)=\frac12e^{-t^2+t}\left(1^2-(-1)^2\right)=0$.
So the statement is TRUE.
