It is necessary for me to find $\Delta^2 u=u_{xxxx}+u_{yyyy}+2u_{xxyy}$ where $u=u(r,t).$
I tried by the following but got an error in Maple.
$$x:=rcos(t);~y:=rsin(t);$$ then tried to find $$diff(u,x$4)+diff(u,y$4)+2diff(diff(u,x$2),y$2).$$
but there was an error, what is the problem? how can I find this?
First the Laplacian can be written as $$ \Delta = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} =\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial t^2}. $$ Then the biharmonic opeartor $\Delta^2$ is the square of the Laplacian, $$ \Delta^2 u = \left(\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial t^2}\right)^2 u= \frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r} \frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r} u +\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r} \frac{1}{r^2}\frac{\partial^2}{\partial t^2}u +\frac{1}{r^2}\frac{\partial^2}{\partial t^2} \frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r} u +\frac{1}{r^4}\frac{\partial^4}{\partial t^4}u. $$
You have to use the package PDEtools in MAPLE to do change of variables.
restart;
with(PDEtools)
PDE := diff(u(x, y),$(x, 4))+diff(u(x, y),$(y, 4))+2*(diff(diff(u(x, y),$(x, 2)),$(y, 2)));
tr := {x = r*cos(t), y = r*sin(t)}
eq := dchange(tr, PDE)
simplify(eq)
A simpler way is
L:= D[1,1]+D[2,2]:
PDEtools:-dchange({x= r*cos(t), y= r*sin(t)}, (L@@2)(u)(x,y), simplify);
