# Undetermined PDE Error

I'm trying to solve the Fokker-Planck equation

$$\frac{\partial p}{\partial L}(L, \eta)= \frac{\partial}{\partial \eta}\left[\left(\eta^{2}-1\right) \frac{\partial p}{\partial \eta}(L, \eta)\right] = (\eta^2-1)\frac{\partial^2p}{\partial\eta^2}(L,\eta)+2\eta\frac{\partial p}{\partial\eta}(L,\eta), \quad \eta>1$$

with initial condition

$$p(L=0, \eta)=\delta(\eta-1).$$

Ignoring the initial condition my code is

FPE = D[P[L, eta],
L] == (1/Lloc)*((eta^2 - 1)*D[P[L, eta], {eta, 2}] +
2*eta*D[P[L, eta], eta])

s = NDSolve[{FPE, P[0, eta] == DiracDelta[eta-1]}, P, {L, 0, 30}, {eta, 1, 30}]


to which I get the error "Encountered non-numerical value for a derivative at L == 0." Is there something I have done wrong? I know this FPE has an analytical solution in terms of Legendre polynomials. Any help would be much appreciated. Thanks.

• The meaning of the error message seems pretty straightforward to me: you are trying to solve for three unknowns (i.e. the function and its derivatives), but only provided one equation. You need more. Notice also that NDSolve is a numerical solver: you also need information on the range for L and eta, and initial conditions. Everything should have a numerical value. If you want solutions depending on parameters, look at ParametricNDSolve. Perhaps start from the docs and try to reformulate your problem. – MarcoB Dec 6 '19 at 17:21
• Thanks for your comment. I have updated my question. – rami_salazar Dec 6 '19 at 17:32
• @rodger_kicks Don't you need two boundary conditions? You only have an initial condition. Without the boundary conditions, you cannot solve. – dearN Dec 8 '19 at 16:37

Try

n = 20;
FPE = D[P[L, eta], L] - D[(eta^2 - 1) D[P[L, eta], eta], eta]
s = Quiet @ NDSolve[{FPE == 0, P[0, eta] == D[Tanh[n (eta - 1)], eta]}, P, {L, 0, 30}, {eta, 1, 30}][[1]]
Plot3D[Evaluate[P[L, eta] /. s], {L, 0, 30}, {eta, 1, 30}]


Here the problem is with DiracDelta[]. Use instead an approximation as

DiracDelta[eta] ~ D[Tanh[n eta], eta]


for $$n \gg 1$$ and $$\eta \ge 0$$

• I can't thank you enough. That approximation is amazing. I didn't think of that. – rami_salazar Dec 7 '19 at 19:04