# How to solve the following fourth order partial differential equation including Laplacian

I wish to solve the following fourth order partial differential equation including Laplacian

with the boundary conditions

with the initial conditions

\[Nu] = 0.34;
\[Beta] = 0.4;
solution = NDSolve[{
\!$$\*SuperscriptBox[\(w$$,
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "0", ",", "2"}], ")"}],
Derivative],
MultilineFunction->None]\)[r, \[Theta], t] +
Laplacian[
Laplacian[w[r, \[Theta], t], {r, \[Theta]},
"Polar"], {r, \[Theta]}, "Polar"] == 16,
w[\[Beta], \[Theta], t] == 0,

\!$$\*SuperscriptBox[\(w$$,
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[\[Beta], \[Theta], t] == 0,

\!$$\*SuperscriptBox[\(w$$,
TagBox[
RowBox[{"(",
RowBox[{"2", ",", "0", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, \[Theta], t] + \[Nu]/1*
\!$$\*SuperscriptBox[\(w$$,
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, \[Theta], t] + \[Nu]/1^2*
\!$$\*SuperscriptBox[\(w$$,
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, \[Theta], t] == 0,
((D[Laplacian[w[r, \[Theta], t], {r, \[Theta]}, "Polar"], r] + (
1 - \[Nu])/r^2*D[
\!$$\*SuperscriptBox[\(w$$,
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[r, \[Theta], t] - w[r, \[Theta], t]/
r, {\[Theta], 2}]) /. r -> 1) == 0,
w[r, \[Theta], 0] == 0},
w[r, \[Theta], t], {r, 0.4, 1}, {\[Theta], 0, 2*Pi}, {t, 0, 5}]


I tried to solve it with NDSolve, but failed

• Define "failed." What exactly happened? Commented Jun 30, 2015 at 12:38
• @rcollyer NDSolve::ivone: Boundary values may only be specified for one independent variable. Initial values may only be specified at one value of the other independent variable. >> Commented Jun 30, 2015 at 13:03
• @bbgodfrey I guess NDSolve cannot Solve the partial differential equations with the implicit boundaty condition Commented Jun 30, 2015 at 13:18
• Because the PDE is second-order in time, a second boundary condition in time probably is needed. Commented Jun 30, 2015 at 13:40
• You might need a periodicity boundary condition in $\theta$ as well — Mathematica doesn't know by default that $w(r,0,t) = w (r, 2\pi, t)$. Commented Jun 30, 2015 at 13:46

Solving this problem is probably going to get you in to the deep weeds of Method specifications in NDSolve. When Mathematica encounters the code as originally stated, it defaults to a finite element method. The problem is that Mathematica's finite element methods can't handle equations with higher than second-order derivatives in the spatial variables, hence the error message:

NDSolve::femcmsd: The spatial derivative order of the PDE may not exceed two. >>

We can force Mathematica to use finite-difference methods instead of finite-element methods by specifying

Method -> {"PDEDiscretization" -> {"MethodOfLines", "SpatialDiscretization" -> "TensorProductGrid"}}


but if you do this, you get the other error message

NDSolve::ivone: Boundary values may only be specified for one independent variable. Initial values may only be specified at one value of the other independent variable. >>

As has been pointed out in the comments, your problem as stated doesn't have the correct number of boundary conditions; NDSolve will throw the above error either when it's using finite-difference methods and you have either too many boundary conditions or too few. (See the page ref/message/NDSolve/ivone in the documentation.) I believe this is why the solver is defaulting to finite-element methods; you can leave some boundary conditions unspecified in these methods, and Mathematica will effectively fill them for you. (I'm not 100% clear on how it does this; I believe they're effectively Neumann boundary conditions for which the normal derivatives vanish, but I welcome correction.)

For your case, you would probably need conditions on the zeroth through third derivatives with respect to $r$ and $\theta$ at the boundaries $r = \beta$ and 1 and $\theta = 0$ and $2 \pi$. (The "zeroth" derivative is, of course, the function itself.) You would also need conditions on the zeroth and first derivatives with respect to $t$ at $t = 0$. If boundary conditions are specified on a rectangular region, then I believe that the above Method specification is no longer necessary; but there may be other wrinkles that you might yet run into once these problems are fixed....

• I know what your mean,thank you for your help.I will change my program tomorrow. Commented Jun 30, 2015 at 14:44