# Partial differential equation with infinity limit

How do I get Mathematica to solve the following partial differential equation

$$\frac{\partial u(y,t)}{\partial t} = \nu \frac{\partial^2 u(y,t)}{\partial y^2}+\frac{\partial U_0(t)}{\partial t}$$

where $U_0 = U_{0m} \cdot sech(\frac{2 \pi}{T} (t-T))$, $(T,U_{0m})$=constant

with the following boundary conditions:

$$u=0 \; {\rm for} \; y = 0$$ $$u=U_0 \; {\rm for} \;y \rightarrow \infty$$

Have tried

U0[t_] := U0m Sech[2 π/T (t - T)];
eq = D[u[y, t], t] == nu D[D[u[y, t], y], y] - D[U0[t], t];
DSolve[
{eq, u[0, t] == 0, limit[u[y, t], y -> Infinity] == U0[t]},
u[y, t],
{y, t}
]

• Do you expect there to be an analytic solution to this equation? Second: limit should be Limit, but I'm not aware that DSolve supports that as a boundary condition (but perhaps it's so). Feb 10, 2016 at 18:01
• Is it - D[U0[t], t] or + D[U0[t], t]? And where's the initial condition ($u(y,0)=?$)? Feb 11, 2016 at 3:48

I tried solving the equation with DSolve by disregarding boundary conditions, but couldn't get an analytic solution. So I think the specific equation is not solvable, which only leaves the question of how to impose a boundary conditions at infinity. This can be done by doing the transformation of variables

$$y = \tan(x)$$

All I can do with this here is to show how it's used in principle, since the actual problem doesn't yield to a symbolic treatment.

Without the boundary condition at infinity, the equation has the following general form after making the transformation of variables to the new equation eqX:

U0[t_] := U0m Sech[2 Pi/T (t - T)];
eq = D[u[y, t], t] == nu D[D[u[y, t], y], y] - D[U0[t], t];

eqX =
Simplify[eq /. u -> (ψ[ArcTan[#], #2] &) /. y -> Tan[x], Pi/2 > x > 0]


$$\nu \cos ^4(x) \psi ^{(2,0)}(x,t)+\frac{2 \pi \text{U0m} \tanh \left(\frac{2 \pi (t-T)}{T}\right) \text{sech}\left(\frac{2 \pi (t-T)}{T}\right)}{T}\\=2 \nu \sin (x) \cos ^3(x) \psi ^{(1,0)}(x,t)+\psi ^{(0,1)}(x,t)$$

The purpose of the transformation of variables is that it allows us to replace $y\to \infty$ by $x\to \pi/2$ in the next steps of the calculation. Then a boundary specification would look like ψ[Pi/2, t] == .... However, we can't go further in this case:

DSolve[eqX, ψ[x, t], {x, t}])


no solution

To see a more successful application of the same idea, see for example Solving differential equation