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I am attempting solve a system of three partial differential equations in 1D, to obtain values of G(x,t), X(x,t) and Y(x,t). I was using the guidance of this post: Solving system of differential equations.

In 3D these equations look like this:

enter image description here

I have simplified to 1D and left out the $\chi$ function. Most of the values in the 3 equations are constants. Here is my attempt to solve them:

GEqns = {G'[x, t] == DG*G''[x] - (kn1 + kp2)*G[x] + kn2*X[x] + kp1*A, 
  Constants -> {DG, kn1, kp2, kn2, kp1, A}}

XEqns = {X'[x, t] == 
   DX*X''[x] + kp2*G[x] - (kn2 + kp3*B + kp5)*X[x] + kn3*Z*Y[x] - 
    kn4*X^3 + kp4*X[x]^2*Y[x] + kn5*Omega, 
  Constants -> {DX, kp2, kn2, kp3, B, kp5, kn3, Z, kn4, kp4, kp5, 
    Omega}}

YEqns = {Y'[x, t] == 
   DY*Y''[x] + kp3*B*X[x] - kn3*Z*Y[x] + kn4*X[x]^3 - 
    kp4*X[x]^2*Y[x], Constants -> {DY, kp3, B, kn3, Z, kn4, kp4}}

ResultXYG = 
 DSolve[Flatten[{GEqns, XEqns, YEqns}], {G[x, t], Y[x, t], X[x, t]}, 
    t] // FullSimplify // Flatten

I get the error message:

enter image description here

Can anyone give advise where I am going wrong?

Thank you.

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    $\begingroup$ for starters, dont use underscores. $\endgroup$ Jun 17, 2023 at 14:51
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    $\begingroup$ What is your intention using Constants->... in the definition of equations?? $\endgroup$ Jun 17, 2023 at 17:28
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    $\begingroup$ Eliminate the "Constants->.." . Further, all function must have the same arguments. And all arguments (x,t) must be specified as last element of DSolve. $\endgroup$ Jun 17, 2023 at 18:39
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    $\begingroup$ I strongly doubt that Mathematica will be able to find a general solution for these equations, even in 1D. You will probably need to use NDSolve, which will require you to provide a set of initial conditions and boundary conditions for your functions, as well as specific numerical values for all of your constants. $\endgroup$ Jun 19, 2023 at 20:50
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    $\begingroup$ Further to @DanielHuber's point, here's how to correctly provide your first equation to Mathematica: D[G[x, t], t] == DG*D[G[x, t], {x, 2}] - (kn1 + kp2)*G[x, t] + kn2*X[x, t] + kp1*A Note the use of the D operator to specify the derivatives, and the fact that all functions of x and t are specified to have those arguments. $\endgroup$ Jun 19, 2023 at 20:52

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