There is a problem I have much difficulties to solve. It is about a temperature difference induced by pressure variation. It is this equation solved by kieransquared, but this time with specific boundary conditions. Here is the full problem:
$$ \left[ \partial_x^2 - \alpha ( \partial_t + v(t) \partial_x ) \right] \tau(x,t) = f(x,t) $$
with $\alpha > 0$, $v(t)$ real function of $t$ (oscillating function) and $f(x,t)$ real function of $x$ and $t$ (also oscillating). $0 < x < l$ is position and $t>0$ is time.
This my boundary conditions:
$$ \partial_x \tau(x,t) \vert_{x = 0} = P \\ \tau(l,t) = \tau_c $$.
With $P$ and $\tau_c$ constants. My initial condition is not important so maybe the PDE is solvable, but ideally to include the initial condition it is:
$$\tau(x,0) = \tau_c$$.
Do you have the solution or a solution without the initial value (i.e for the oscillating steady states)?
What I know
So thanks to kieransquared we can simplify the heat-convection equation into a heat equation like this (not so sure of my notation):
by the transformation $x \rightarrow x + \nu(t)$, with $\nu(t)$ to be determined, we arrive to the equation:
$$ \left[ \partial_x^2 - \alpha ( \partial_t + v(t) \partial_x ) \right] \tau(x+ \nu(t),t) = f(x+ \nu(t),t) $$
Then some partial derivatives can cancel each other by posing $ \partial_t \nu = - v(t) $ so we obtain:
$$ \left( \partial_x^2 - \alpha \partial_t \right) \tau(x + \nu(t),t) = f(x+ \nu(t),t) $$
we can pose $g(x,t) = \tau(x + \nu(t),t)$ and solve after that:
$$ \left( \partial_x^2 - \alpha \partial_t \right) g(x,t) = f(x+ \nu(t),t) $$
We notice that $g(x-\nu(t),t) = \tau(x,t)$, so the BC for $g$ are:
$$ g(l-\nu(t),t) = \tau_c \\ \partial_x g(x-\nu(t),t)\vert_{x = 0} = P $$
I am stuck on this one. I know that, in principle, I need a Fourier series on $x$ with sine/cos but I cannot fit the BC since the sine and cos will depend on $\nu(t)$ on each side ($0$ and $l$).
