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.
y[x,0] = ...
andy(0,1)[x,0] = ...
$\endgroup$ – Bichoy May 23 '15 at 18:23