While trying to solve: $$ z \frac{\partial}{\partial z} (z G(s,z))=z \cosh (s) \frac{\partial}{\partial z} G(s,z)-\frac{1}{2} \sinh (s) \frac{\partial}{\partial s} G(s,z) $$ using DSolve it gives a family of solutions:

 DSolve[-z D[z G[s, z], 
     z] == - z Cosh[s] D[G[s, z], z] + (1/2) Sinh[s] D[G[s, z], s],
   G, {s, z}]]
(* {{G -> 
     z}, -(((-2 Csch[s]^2 - 2 z Log[Tanh[s/2]] - 
        z Csch[s]^2 Log[Tanh[s/2]] + 
        z Cosh[2 s] Csch[s]^2 Log[Tanh[s/2]]) C[1][(
       Csch[s]^2 (2 - 2 z Cosh[s] + z Log[Tanh[s/2]] - 
          z Cosh[2 s] Log[Tanh[s/2]]))/(2 z)])/z)]}} *)

However I want a specific solution which satisfies:

$$ G(0,z)=\frac{1}{z-1} \;\; \mbox{and} \;\;\frac{\partial}{\partial s}G(s,z) \Big \vert_{s=0}=\frac{\sqrt{z}}{(z-1)^\frac{3}{2}} \;. $$ I do not know how to implement the conditions. Any help is appreciated.

  • 1
    $\begingroup$ As you see from the solution, the function is not defined for s==0because of the Csch term. Therefore you can not specify G[0,z]. The same pertains to the second condition. $\endgroup$ Commented Jan 26, 2021 at 15:22

1 Answer 1


Let us suppose that we are interested only in the solution for small s, which seems reasonable, because the question addresses applying boundary conditions at s == 0. Then the PDE becomes

eq = -z D[z G[s, z], z] == -z D[G[s, z], z] + (1/2) s D[G[s, z], s]

and solving it with DSolve yields

G[s, z] -> C[1][Log[-((-1 + z)/(s^2 z))]]/(s^2 z)

i.e., an arbitrary function of Log[-((-1 + z)/(s^2 z))] divided by s^2 z. The problem with this solution is that DSolve does not concern itself with such niceties as branch cuts. If, in the same spirit, I do not worry about branch cuts either, then just a few minutes of thought suggests the solution

 sol = G -> Function[{s, z}, 1/(z - 1) + s Sqrt[z/(z - 1)^3]]

Clearly, it satisfies the the two conditions listed in the question. And, it actually satisfies the PDE, as can be seen from

(* True *)

Probably this solution is not unique, more general solutions being power series in s Sqrt[z/(z - 1)].

Although I do not have time now to solve the actual PDE posed in the question, I can say with confidence that it does have a solution, and that solution reduces to the one here in the limit of small s.

Addendum: Solution for original PDE

The full PDE in the question,

eq1 = -z D[z G[s, z], z] == -z Cosh[s] D[G[s, z], z] + (1/2) Sinh[s] D[G[s, z], s]

has the solution

DSolve[eq1, G[s, z], {s, z}][[1, 1]]
(* G[s, z] -> (2 Csch[s]^2 C[1][(Csch[s]^2 (2 - 2 z Cosh[s] + 
   z Log[Tanh[s/2]] - z Cosh[2 s] Log[Tanh[s/2]]))/(2 z)])/z *)

obtained using Mathematica "12.2.0 for Microsoft Windows (64-bit) (December 12, 2020)". From this general solution, construct the specific solution

sol1 = G -> Function[{s, z}, 
    1/(z Cosh[s] - 1 + 1/2 z (Cosh[2 s] - 1) Log[Tanh[s/2]]) + 
    Sinh[s] Sqrt[z/(z Cosh[s] - 1 + z/2 (Cosh[2 s] - 1) Log[Tanh[s/2]])^3]];

It satisfies the two constraints,

Series[G[s, z] /. sol1, {s, 0, 0}] // Normal
(* 1/(-1 + z) *)

Series[D[G[s, z] /. sol1, s], {s, 0, 0}] // Normal
(* Sqrt[z/(-1 + z)^3] *)

and, of course, satisfies the PDE.

FullSimplify[eq1 /. sol1]
(* True *)

Note that the solution given is not necessarily the most general one, which would be a power series in Sinh[s] with the two lowest order terms given in sol1. Note also that sol1 may not be analytical at s == 0, but it is continuously differentiable there, which is sufficient for our purposes. For completeness, the characteristic curves of the PDE can be illustrated by

StreamPlot[{Sinh[s]/2, z^2 - z Cosh[s]}, {s, -1, 1}, {z, -1, 1}, 
    StreamScale -> None, StreamColorFunction -> Blue, ImageSize -> Large,
    FrameLabel -> {s, z}, LabelStyle -> {15, Bold, Black}]

enter image description here

A more detailed image can be obtained using the approach illustrated in my answer to 238547.

  • $\begingroup$ Excellent starting point. Thank you so much. $\endgroup$
    – Ferhat
    Commented Jan 27, 2021 at 7:30
  • 1
    $\begingroup$ @Farhad I have just added the full solution. Best wishes $\endgroup$
    – bbgodfrey
    Commented Jan 27, 2021 at 12:55
  • $\begingroup$ This is fantastic.Thank you for such a wonderful contribution. @bbgodfrey $\endgroup$
    – Ferhat
    Commented Jan 27, 2021 at 19:03

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