How can I solve these PDE analytically for $z$ and $y$?

\begin{eqnarray} 2 \dot{z} &=& \ddot{y}+ y^{-1}, ~~~ (1)\\ z' + \dot{z}' &=& - \dot{y}'. ~~~~~~~~~~ (2) \end{eqnarray}

Where (.) is the time derivative ($\frac{\partial}{\partial t}$ ) and (’) is the derivative with respect to x : ($\frac{\partial}{\partial x}$).

Here is my trial:

eq1[x_, t_] = 2*D[z[t, x], t] == D[y[x, t], {t, 2}] + y[x, t]^(-1); 

eq2[x_, t_] = D[z[t, x], x] + D[D[z[t, x], t], x] == -D[D[y[x, t], t], x]; 

DSolve[{eq1[x, t], eq2[x, t]}, {z[t, x], y[x, t]}, {t, x}]

But DSolve dose not give a solution.


I try to eliminate z from the second equation, using the first equation. So any help to make these steps?

  • Integrate (1) to get z(x, t): To do so, I use:

    Integrate[ D[y[x, t], {t, 2}] + y[x, t]^(-1), y[x, t], t]

But I don't understand the output:

Out[1]=Integrate[Log[y[x, t]], t] + y[x, t]*Derivative[0, 1][y][x, t]

-Then differentiate z(x, t) with respect to x to get z’(x,t).

  • Differentiate (1) with respect to x to get $\dot{z}’$

  • Then substitute by $z’$ and $\dot{z}’$ in (2)


I tried to solve

$$\frac{\partial}{\partial t} \left(\frac{\dot{y}^2}{2}+\log (y)\right) =2 \dot{y}\ \dot{z}$$


$$\frac{\partial (y+z)}{\partial t}+z=\text{C1}(t)$$

In MA by:

eq1[x, t] := D[D[y[x, t], t]^2/2 + Log[y[x, t]], t] -2*D[y[x, t], t]*D[z[x, t], t]

eq2[x, t] := D[y[x, t] + z[x, t], t] + z[x, t] - c

DSolveValue[{eq1[x, t] == eq2[x, t] == 0, y[x, 0] == 0, z[x, 0]
   == 0}, {y[x, t], z[x, t]}, {x, t}]

With or without initial conditions, or by using DSolve , there is no output solution.

  • 4
    $\begingroup$ These can't be easily solved because they are nonlinear. You also are mixing arguments. You write z[t, x] then in the call to DSolve you write z[x, t]. Strange that you did not see this error from DSolve Also your latex is confusing. This is not how partial derivatives are written in Latex. $\endgroup$
    – Nasser
    Commented Dec 29, 2023 at 18:59
  • $\begingroup$ Hello @Nasser. Yes, it’s not easy solving, also I have tried Reduce. I edit the question to make the partial derivatives clear in the latex equations. $\endgroup$
    – Dr. phy
    Commented Dec 29, 2023 at 19:21
  • $\begingroup$ Why are they nonlinear? Because they contain $y^{-1}$ ? $\endgroup$
    – Dr. phy
    Commented Dec 29, 2023 at 19:23
  • $\begingroup$ Yes, that makes the system nonlinear. $\endgroup$
    – Ghoster
    Commented Dec 29, 2023 at 19:47
  • 1
    $\begingroup$ @HansOlo. Hi, I thought to eliminate one of the variables z or y and make a single equation in one variable. $\endgroup$
    – Dr. phy
    Commented Dec 30, 2023 at 8:03

2 Answers 2


It seems doubtful to me that a symbolic general solution exists. However, a symbolic particular solution can be obtained without difficulty. Suppose that z[x, t] = 0. Then, equ2 can be dropped, and equ1 (which almost certainly is independent of x) integrated.

DSolve[D[y[t], {t, 2}] + y[t]^(-1) == 0, y[t], t] // Flatten
(* {y[t] -> E^(1/2 (C[1] - 2 InverseErf[-Sqrt[2/Pi] Sqrt[E^-C[1] (t + C[2])^2]]^2)), 
    y[t] -> E^(1/2 (C[1] - 2 InverseErf[Sqrt[2/Pi] Sqrt[E^-C[1] (t + C[2])^2]]^2))} *)

Addendum: Convert PDE pair into single ODE

Consider the two PDEs.

eq1 = 2*D[z[x, t], t] == D[y[x, t], {t, 2}] + y[x, t]^-1
eq2 = z[x, t] + D[z[x, t], t] == c[t] - D[y[x, t], t]

(eq2 here is the first-integral of eq2 in the question, as derived by RolandF in another answer to the question. It is valid, provided that y and z actually do depend on x. c[t] is an unknown function of t only, resulting from the integration.) Because z enters only linearly in both PDEs, it can be eliminated between the two as follows.

eq1 /. Solve[eq2, D[z[x, t], t]] // Flatten;
fr = Solve[%, z[x, t]] // Flatten

(* {z[x, t] -> (-1 + 2*c[t]*y[x, t] - 2*y[x, t]*D[y[x, t], t] - 
                y[x, t]*D[y[x, t], t, t]) / (2*y[x, t])} *)

With z now given in terms of y, it can be eliminated from eq2.

eq2 /. z -> Function[{x, t}, Evaluate[fr[[1, 2]]]] // Simplify;
eq3 = MultiplySides[%, y[x, t], Assumptions -> y[x, t] != 0] // Simplify

(* D[y[x, t], t] + y[x, t]^2*(2*c'[t] - 3*D[y[x, t], t, t] 
   - D[y[x, t], t, t, t]) == y[x, t] *)

Since x enters only as a parameter, the pair of PDEs has indeed been reduced to a single ODE. Unfortunately, DSolve cannot integrate this ODE either. As a final attempt, I converted eq3 to a second-order ODE, which is possible for constant c, because eq3 is autonomous (i.e., t does not enter explicitly. The resulting second-order ODE (for c[t] = 0), obtained by replacing y'[t] by w[y] and ignoring the parameter x, is

w[y] + y^2 (-3 w'[y] w[y] - w''[y] w[y]^2 - w'[y]^2 w[y]) == y

It too cannot be integrated by DSolve.

  • $\begingroup$ Hello @bbgodfrey. Thanks for your answer. But z[x, t] can not equal zero. $\endgroup$
    – Dr. phy
    Commented Dec 31, 2023 at 7:01
  • 1
    $\begingroup$ Could z be a constant or dependent only on t? Could your C1[t] be a constant? Any information that could narrow the range of possible solutions would be helpful. $\endgroup$
    – bbgodfrey
    Commented Dec 31, 2023 at 14:24
  • $\begingroup$ Thank you very much @bbgodfrey for your solution. This is exactly what I thought about a single DE in y. But how MA can solve this? . C1[t] can vanish, since the initial conditions can be chosen so that the integration constants vanish. But still z and y functions of [x, t] . I think this term D[y[x, t], t, t, t] is the problem, also I wonder can DSolve or AsympotiticDSolveValue solve something like DSolve[eq, {z[x, t] ,y[x, t]}, {x, t}] , if not, I think one can find the numeric solution first, then find the analytic expressions by fitting. $\endgroup$
    – Dr. phy
    Commented Jan 2 at 8:03
  • $\begingroup$ @Dr.phy, I am confident that DSolve cannot solve your equations, even if` C1[t] vanishes. In fact, I doubt that a symbolic solution even exists. A symbolic asymptotic or power series solution probably does exist, at least for most starting points. A numerical solution requires three boundary conditions in t. Provide them, and I can see what I can do. $\endgroup$
    – bbgodfrey
    Commented Jan 2 at 14:18

Most what one can do is using the energy trick for the first equation

$$\frac{\partial}{\partial t} \left(\frac{\dot{y}^2}{2}+\log (y)\right) =2 \dot{y}\ \dot{z}$$ and finding a first integral of the second $$\frac{\partial (y+z)}{\partial t}+z=\text{C1}(t)$$

  • $\begingroup$ Hello @Roland F. Thanks for your answer, but how with these two equations get z or y? $\endgroup$
    – Dr. phy
    Commented Dec 30, 2023 at 11:06
  • $\begingroup$ Can you please look at the question's edit:2? @Roland F. $\endgroup$
    – Dr. phy
    Commented Dec 30, 2023 at 12:47
  • 1
    $\begingroup$ For such an algebraically transparent equation, I don't see a solution. Partial differential equations represent examples from wave field theories, harmonic function theories and diffusion of matter. Quadratic time derivative sugests a wave equation, two components suggest a change of the components of a separated wave equation into a radial part y and a circualr wave in z. Give the origin of the equation, perhaps it comes with a Solve strategy. System PDE's without boundary and start conditions, solution spaces predefined by function norms don't form solvable problem classes . $\endgroup$
    – Roland F
    Commented Dec 30, 2023 at 13:14
  • $\begingroup$ Even without Mathematica, can't the first equation in your answer be integrated as: $ \int \frac{\partial}{\partial t} \left(\frac{\dot{y}^2}{2}+\log (y)\right) =2 \int \dot{y}\ \dot{z} = \left(\frac{\dot{y}^2}{2}+\log (y)\right) =2 z ~\dot{y} $. Is that correct? $\endgroup$
    – Dr. phy
    Commented Dec 30, 2023 at 13:55
  • $\begingroup$ No, the product of two derivatives yields at best $$\int \partial_t y \partial_t z dt =y \partial_t z- \int y \partial_{t,t} z dt$$ $\endgroup$
    – Roland F
    Commented Dec 30, 2023 at 17:22

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