Use NDSolve to solve PDE's with different domains

Question

Good morning,

I want to find a numerical solution to the following equation using NDSOLVE:

$$ (1)\hspace{1cm}EJ\frac{\partial^4 v}{\partial x^4} +kv+\rho A \frac{\partial^2 v}{\partial t^2} = q \hspace{1cm} \forall x \in [0;L_o] $$

$$ (2)\hspace{1,7cm}EJ\frac{\partial^4 w}{\partial x^4} +\rho A \frac{\partial^2 w}{\partial t^2} = q \hspace{1cm} \forall x \in [L_o;L] $$

$$ BC_{(1)}:\hspace{1cm} \left|\frac{\partial^3 w}{\partial x^3}\right|_{x=0} =0\hspace{1cm}\left|\frac{\partial^2 w}{\partial x^2}\right|_{x=0} =0\hspace{1cm} \forall t $$

$$ BC_{(2)}:\hspace{1cm} \left|\frac{\partial^3 v}{\partial x^3}\right|_{x=L} =0\hspace{1cm}\left|\frac{\partial^2 v}{\partial x^2}\right|_{x=L} =0\hspace{1cm} \forall t $$

$$ BC_{(1\space and\space2)}:\hspace{1cm} \left|\frac{\partial^i v}{\partial x^i}\right|_{x=L_o} = \left|\frac{\partial^i w}{\partial x^i}\right|_{x=L_o} \hspace{1cm} i=0,...,3 \hspace{0,9cm}\forall t $$

$$IC_{(1\space and\space2)}:\hspace{1cm} v(x,t)=w(x,t)=0\hspace{1cm}\forall x, \space t=0$$

$$\hspace{3,1cm} \frac{\partial v}{\partial t}=\frac{\partial w}{\partial t}=0\hspace{1cm}\forall x, \space t=0 $$

How can I use 'NDSOLVE' and imposing two-domain integration? As suggested let's think that Eq. (2) is valid in y domain so: $$ (2')\hspace{1,7cm}EJ\frac{\partial^4 w}{\partial y^4} +\rho A \frac{\partial^2 w}{\partial t^2} = q \hspace{1cm} \forall y \in [L_o;L] $$ And (as suggested in a comment) let's assume $$ y=y(x),\hspace{1,7cm} y=L-\frac{x}{L_o}(L-L_o)$$ if it's the case we have: $$ \frac{\partial w}{\partial x} =\frac{\partial w}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x} \hspace{0,2cm}\rightarrow \hspace{0,2cm} \frac{\partial^2 w}{\partial x^2}=\frac{\partial^2 w}{\partial y^2}\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2 \hspace{0,2cm} \rightarrow \space...\space\rightarrow \hspace{0,2cm} \frac{\partial^4 w}{\partial y^4}=\frac{\partial^4 w}{\partial x^4}\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^{-4} \hspace{0,2cm} $$ At the end EQ(2) is replaced with: $$ (2'')\hspace{1,7cm}EJ\Big(\frac{L_o}{L_o-L}\Big)^4\frac{\partial^4 w}{\partial x^4} +\rho A \frac{\partial^2 w}{\partial t^2} = q \hspace{1cm} \forall x \in [0;L_o] $$ I wrote this code but is not working.

Clear[b, h, Emod, m, Lo, Lf, q, k, J, y, u, v];
b = 0.20; (*m*)
h = 0.20; (*m*)
Emod =10000000000;(*modulo el N/m^2*)
Lo = 2;(*l ini m*)
L = 4;(*l fin m*)
rho = 10;(*rho kN/m^3*)
m = b*h*rho;(*N/m*)
q = 1000;(*N/m*)
k = 50000;(*N*)
P = 0.00000; 
J = (b*h^3)/12;
tmax = 30;
sol = NDSolve[{
     (*1st*)
+Emod*J*D[u[t, x], {x, 4}] + k*u[t, x] == q - m*(D[u[t, x], {t, 2}]),
     (*2nd*)
+Emod*J*(Lo/(Lo - L))^4*D[v[t, x], {x, 4}] + k*v[t, x] == q - m*(D[v[t, x],{t, 2}]),

(*1st boundary conditions*)
Derivative[0, 2][u][t, 0] == 0,
Derivative[0, 3][u][t, 0] == P*2/Pi*ArcTan[t],
(*1st initial conditions*)
u[0, x] == 0,
Derivative[1, 0][u][0, x] == 0,

(*2nd boundary conditions*)
(Lo/(Lo - L))^2*Derivative[0, 2][v][t, 0] == 0,
(Lo/(Lo - L))^3*Derivative[0, 3][v][t, 0] == 0,
(*2nd initial conditions*)
v[0, x] == 0,
Derivative[1, 0][v][0, x] == 0,

(*commonn*)
u[t, Lo] == -v[t, Lo],
Derivative[0, 1][u][t, Lo] == (Lo/(Lo - L))*
Derivative[0, 1][v][t, Lo],
Derivative[0, 2][u][t, Lo] == (Lo/(Lo - L))*
Derivative[0, 2][v][t, Lo],
Derivative[0, 3][u][t, Lo] == (Lo/(Lo - L))*
Derivative[0, 3][v][t, Lo]
}, {u, v}, {t, 0, tmax}, {x, 0, Lo}, PrecisionGoal -> 2] 

---

ADDITIONAL INTERESTING QUESTION: As suggested in a comment I used "UnitStep" function, but now I want to add an ODE that prescribes the "lenght of UniStep". What I want to do is easy understandable by the following code (but it doesn't work):

Lo = 1; L = 3; J = 1; m = 100; Emod = 1; k = 0.5; q = 0.001; tmax = 20;
sol = NDSolve[{
m*D[u1[t, x], {t, 2}] + Emod*J*D[u1[t, x], {x, 4}] + 
UnitStep[Lo + u2[t] - x] k*u1[t, x] == q,
Derivative[0, 2][u1][t, 0] == 0,
Derivative[0, 3][u1][t, 0] == 0,
Derivative[0, 2][u1][t, L] == 0,
Derivative[0, 3][u1][t, L] == 0,
u1[0, x] == -0.001*x,
Derivative[1, 0][u1][0, x] == 0,
D[u2[t], {t,1}] == -NeumannValue[D[u1[t, x], {t, 1}], x == Lo + u2[t]]/
 NeumannValue[D[u1[t, x], {x, 1}], x == Lo + u2[t]],
u2[0] == 0},
{u1, u2}, {t, 0, tmax}, {x, 0, L}, MaxSteps -> 100000]
Plot3D[Evaluate[u1[t, x] /. sol[[1, 1]]], {t, 0, tmax}, {x, 0, L}]

Answer 1

As the equations are almost the same in the two domains, and all your conditions at the interface are just continuity, you can easily just combine into one equation by using UnitStep to only include the $k$ term in the $v$ region. Here for $L=3$:

Lo = 1;
L = 3;
J = 1;
m = 1;
Emod = 1;
k = 0.5;
q = 0.001;
tmax = 100;
sol = NDSolve[{m*D[u[t, x], {t, 2}] + Emod*J*D[u[t, x], {x, 4}] + 
     UnitStep[Lo - x] k*u[t, x] == q, Derivative[0, 2][u][t, 0] == 0, 
   Derivative[0, 3][u][t, 0] == 0, Derivative[0, 2][u][t, L] == 0, 
   Derivative[0, 3][u][t, L] == 0, u[0, x] == 0, 
   Derivative[1, 0][u][0, x] == 0}, u, {t, 0, tmax}, {x, 0, L}, 
  MaxSteps -> 100000]

Plot3D[Evaluate[u[t, x] /. sol[[1, 1]]], {t, 0, tmax}, {x, 0, L}]