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differential equations - Use NDSolve to solve a PDE


I have the following PDE (a master equation, and $P$ is probability density, $0\le x\le1$ and $0\le y\le1$):


$$ \partial_t P(x,y,t)=x\partial_xP(x,y,t)+(1-y)\partial_yP(x,y,t)+2P(x,y,t) $$


The initial conditions are $P(x,y,t=0)=\delta(x-0.05,y-0.85)$. For the boundary conditions, if the probability density is going to flow through the boundary, I won't do anything to stop it. (How to express the BC in Mathematica?)


I used the following command:


NDSolve[{D[P[x, y, t], t] == 
x D[P[x, y, t], x] + (1 - y) D[P[x, y, t], y] + 2 P[x, y, t], 
P[x, y, 0] == DiracDelta[x - 0.05, y - 0.85]}, P, 
{t, 0, 2}, {x, 0, 1}, {y, 0, 1}]


First, Mathematica gave the following warning:


Warning: an insufficient number of boundary conditions have been specified for the direction of 
independent variable y. Artificial boundary effects may be present in the solution. >>

Moreover, the result it gave obviously is wrong:


Plot3D[Evaluate[(P[x, y, t] /. %) /. {t -> 0}], {x, 0, 1}, {y, 0, 1}]

enter image description here


My question is: 1) How can I express the boundary conditions I mentioned 2) How can I make the NDSolve work?




Answer



Your specific example can be solved by DSolve. First let's get the general solution of the PDE:


gsol = DSolve[{D[p[x, y, t], t] == 
               x D[p[x, y, t], x] + (1 - y) D[p[x, y, t], y] + 2 p[x, y, t]}, p, {x, y, t}]


{{p -> Function[{x, y, t}, C[1][x (-1 + y), t + Log[x]]/x^2]}}

Then substitute it into the boundary condition:


eqn = p[x, y, 0] == DiracDelta[x - 0.05, y - 0.85] /. gsol[[1]]



C[1][x (-1 + y), Log[x]]/x^2 == DiracDelta[-0.05 + x, -0.85 + y]

Solve for C[1][_, _]:


Solve[Cases[eqn, C[1][a__] -> a, Infinity] == {p1, p2}, {x, y}]
Solve[Simplify[eqn /. %, {p1, p2} ∈ Reals][[1]], C[1][p1, p2]]
bc = % /. (a_[b__] -> c_) -> (a -> ({b} \[Function] c))



{{x -> E^p2, y -> E^-p2 (E^p2 + p1)}}
{{C[1][p1, p2] -> E^(2 p2) DiracDelta[-0.05 + E^p2, 0.15 + E^-p2 p1]}}
{{C[1] -> Function[{p1, p2}, E^(2 p2) DiracDelta[-0.05 + E^p2, 0.15 + E^-p2 p1]]}}

Substitute it back to the general solution:


 sol[x_, y_, t_] = (p[x, y, t] /. gsol /. bc)[[1, 1]]


(E^(2 (t + Log[x])) DiracDelta[-0.05 + E^t x, 0.15 + E^-t (-1 + y)])/x^2


BTW, when exploring this I encountered another problem, see here for details.


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