# How can I solve this system of nonlinear PDEs?

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.

Edit

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)

Edit:2

I tried to solve

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

and

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

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.

• 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. Commented Dec 29, 2023 at 18:59
• 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. Commented Dec 29, 2023 at 19:21
• Why are they nonlinear? Because they contain $y^{-1}$ ? Commented Dec 29, 2023 at 19:23
• Yes, that makes the system nonlinear. Commented Dec 29, 2023 at 19:47
• @HansOlo. Hi, I thought to eliminate one of the variables z or y and make a single equation in one variable. Commented Dec 30, 2023 at 8:03

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.

• Hello @bbgodfrey. Thanks for your answer. But z[x, t] can not equal zero. Commented Dec 31, 2023 at 7:01
• 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. Commented Dec 31, 2023 at 14:24
• 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. Commented Jan 2 at 8:03
• @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. 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 $$\frac{\partial (y+z)}{\partial t}+z=\text{C1}(t)$$

• Hello @Roland F. Thanks for your answer, but how with these two equations get z or y`? Commented Dec 30, 2023 at 11:06
• Can you please look at the question's edit:2? @Roland F. Commented Dec 30, 2023 at 12:47
• 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 . Commented Dec 30, 2023 at 13:14
• 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? Commented Dec 30, 2023 at 13:55
• No, the product of two derivatives yields at best $$\int \partial_t y \partial_t z dt =y \partial_t z- \int y \partial_{t,t} z dt$$ Commented Dec 30, 2023 at 17:22