Analytic solution of dynamic Euler–Bernoulli beam equation with compatibility condition

The Euler–Bernoulli beam equation (also known as wave equation for beam) with pined-pined boundary has well-known solutions, but directly input the equation into Mathematica does not return them.

$$EI\cfrac{\partial^4 w}{\partial x^4} + \mu\cfrac{\partial^2 w}{\partial t^2} == 0$$ $$w(0,t) = w(L,t)=0 \\ \cfrac{\partial^2 w(0,t)}{\partial x^2}=\cfrac{\partial^2 w(L,t)}{\partial x^2}=0$$

DSolve[{EI D[y[x, t], {x, 4}] + mu D[y[x, t], {t, 2}] == 0, 
  y[0, t] == 0, y[L, t] == 0, Derivative[2, 0][y][0, t] == 0, 
  Derivative[2, 0][y][L, t] == 0}, y[x, t], {x, t}]

Can anyone give me a hint on how to solve it using DSolve?

Update:

Adding initial conditions does not help:

DSolve[{K D[y[x, t], {x, 4}] + M D[y[x, t], {t, 2}] == 0, 
  y[0, t] == 0, y[L, t] == 0, Derivative[2, 0][y][0, t] == 0, 
  Derivative[2, 0][y][L, t] == 0, y[x, 0] == Sin[x/L Pi], 
  Derivative[0, 1][y][x, 0] == 0}, y[x, t], {x, t}]

The actual situation

Here is the actual boundary conditions and compatibility conditions I am trying to solve:

1. Boundary conditions: $$w(0,t) = w(L,t)=0 \\ \cfrac{\partial^2 w(0,t)}{\partial x^2}=\cfrac{\partial^2 w(L,t)}{\partial x^2}=0$$


2. 
   

   Compatibility conditions:

   
   

   (1). compatibility condition for spring at $L/2$ $$w(x,t)\lvert_{x\to L/2^-}=w(x,t)\lvert_{x\to L/2^+}\\w'(x,t)\lvert_{x\to L/2^-}=w'(x,t)\lvert_{x\to L/2^+}\\w''(x,t)\lvert_{x\to L/2^-}=w''(x,t)\lvert_{x\to L/2^+}\\w'''(x,t)\lvert_{x\to L/2^-}=w'''(x,t)\lvert_{x\to L/2^+}+k w(L/2,t)$$ (2). compatibility condition for mass at $x_m$ $$w(x,t)\lvert_{x\to x_m^-}=w(x,t)\lvert_{x\to x_m^+}\\w'(x,t)\lvert_{x\to x_m^-}=w'(x,t)\lvert_{x\to x_m^+}\\w''(x,t)\lvert_{x\to x_m^-}=w''(x,t)\lvert_{x\to x_m^+}\\w'''(x,t)\lvert_{x\to x_m^-}=w'''(x,t)\lvert_{x\to x_m^+}-M\ddot w(x_m,t)$$

   
   

For special cases of the problem, Russian expert Filippov gave the solution in his book in 1970, but it is now very hard to find a copy of the book. And what is worse, the book is written in Russian.

Solving this problem maybe is reinvent the wheel, but the old way to manufacture the wheel is lost.

I opened a new question on how to trade compatibility condition here.

Answer

The short answer is: It's a common sense that (at least currently) DSolve is very weak on solving PDE and it simply can't handle this problem, period. However, with a little effort, you can solve it with LaplaceTransform:

eqn = ϵ D[y[x, t], {x, 4}] + μ D[y[x, t], {t, 2}] == 0;
ic = {y[x, 0] == Sin[x/L Pi], Derivative[0, 1][y][x, 0] == 0};
bc = {y[0, t] == 0, y[L, t] == 0, 
      Derivative[2, 0][y][0, t] == 0, Derivative[2, 0][y][L, t] == 0};

teqn = With[{l = LaplaceTransform}, 
            l[{eqn, bc}, t, s] /. HoldPattern@l[u_, t, s] :> u] /. Rule @@@ ic

$$\left\{\mu \left(s^2 y(x,t)-s \sin \left(\frac{\pi x}{L}\right)\right)+\epsilon y^{(4,0)}(x,t)=0,\left\{y(0,t)=0,y(L,t)=0,y^{(2,0)}(0,t)=0,y^{(2,0)}(L,t)=0\right\}\right\}$$

Notice that $y(x,t)$ actually represents $\mathcal{L}_t[y(x,t)](x)$ in teqn. I made this replacement because DSolve has some difficulty in understanding $\mathcal{L}_t[y(x,t)](x)$. Now we just need to solve teqn with DSolve:

tsol = DSolve[teqn, y[x, t], x][[1, 1, -1]]

$$\frac{\mu L^4 s \sin \left(\frac{\pi x}{L}\right)}{\left(\pi ^2 \sqrt{\epsilon }-i \sqrt{\mu } L^2 s\right) \left(\pi ^2 \sqrt{\epsilon }+i \sqrt{\mu } L^2 s\right)}$$

and change the transformed solution back:

sol = InverseLaplaceTransform[tsol, s, t]

$$\frac{1}{2} \sin \left(\frac{\pi x}{L}\right) e^{-\frac{i \pi ^2 t \sqrt{\epsilon }}{\sqrt{\mu } L^2}} \left(1+e^{\frac{2 i \pi ^2 t \sqrt{\epsilon }}{\sqrt{\mu } L^2}}\right)$$

When dealing with an initial boundary value problem, the above approach is more automatic than Jens' method of separation of variables. You can wrap the procedure into a function:

pdeSolveWithLaplaceTransform[eqn_, ic_, func : _[__], t_, nott_] := 
 With[{l = LaplaceTransform}, 
  Module[{s}, 

   InverseLaplaceTransform[
    func /. First@
      DSolve[l[eqn, t, s] /. HoldPattern@l[u_, t, s] :> u /. Rule @@@ Flatten@{ic}, 
     func, nott], s, t]]]

This function will probably fail in more complex cases, but does have a certain generality, for example, it can handle the problem in this post like this:

eqn = D[p[x, t], {t, 2}] == c^2 (D[p[x, t], {x, 2}]);
ic = {p[x, 0] == Exp[x], D[p[x, t], t] == Sin[x] /. t -> 0};

pdeSolveWithLaplaceTransform[eqn, ic, p[x, t], t, x]

$$c_1 \delta \left(t+\frac{x}{c}\right)+c_2 \delta \left(t-\frac{x}{c}\right)+\frac{c \left(e^{2 c t}+1\right) e^{x-c t}-i e^{-i c t} \left(-1+e^{2 i c t}\right) \sin (x)}{2 c}$$

---

Update: solution to the actual situation

OK, since a solution containing InverseLaplaceTransform is acceptable for you, I'd like to make this complement. Still, I'll use LaplaceTransform for your actual situation. For brevity, let's define a helper function, a pdeSolveWithLaplaceTransform without inverse transform:

helper[eqn_, ic_, func : _[__], t_, s_, nott_, const_: C] := 
 func /. First@
   DSolve[With[{l = LaplaceTransform}, l[eqn, t, s] /. HoldPattern@l[u_, t, s] :> u] /. 

     Rule @@@ ic, func, nott, GeneratedParameters -> const]

First find the transformed solutions with boundary conditions at only one side respectively:

eqn = ϵ D[y[x, t], {x, 4}] + μ D[y[x, t], {t, 2}] == 0;
ic = {y[x, 0] == Sin[x/L Pi], Derivative[0, 1][y][x, 0] == 0};
bcL = {y[0, t] == 0, Derivative[2, 0][y][0, t] == 0};
bcR = {y[L, t] == 0, Derivative[2, 0][y][L, t] == 0};

tsolL = helper[{eqn, bcL}, ic, y[x, t], t, s, x, cL]
tsolR = helper[{eqn, bcR}, ic, y[x, t], t, s, x, cR]

Needless to say, tsolL and tsolR involve constants. (To be more specific, cL[1], cL[2], cR[1], cR[2].) How to eliminate them? We still have compatibility conditions unused:

(1) compatibility condition for spring at $L/2$

cond1 = Solve[{# == #2, D[#, x] == D[#2, x], D[#, {x, 2}] == D[#2, {x, 2}], 
               D[#, {x, 3}] == D[#2, {x, 3}] + k #} &[tsolL, tsolR] /. x -> L/2, 
              {cL[1], cL[2], cR[1], cR[2]}][[1]];

tsolLcond1 = tsolL /. cond1 (*// Simplify*)
tsolRcond1 = tsolR /. cond1 (*// Simplify*)

(2) compatibility condition for mass at $x_m$

cond2 = Solve[{# == #2, D[#, x] == D[#2, x], D[#, {x, 2}] == D[#2, {x, 2}], 
               D[#, {x, 3}] == D[#2, {x, 3}] + s^2 # - s ic[[1, -1]] - ic[[2, -1]]} &
              [tsolL, tsolR] /. x -> xm, {cL[1], cL[2], cR[1], cR[2]}][[1]];

tsolLcond2 = tsolL /. cond2 (*// Simplify*)
tsolRcond2 = tsolR /. cond2 (*// Simplify*)

The result is quite lengthy so I'd like to omit them here. The final step is to make the inverse transform. As mentioned above, InverseLaplaceTransform will remain unevaluated. If you want to calculate the transform numerically in the future work, have a look at this package.