I was able to get full credit for part a, but my attempts for part b and c proved miserable since I received no points. Can someone please explain how to do this?
Let \begin{eqnarray*} u(x,y) = 3x^2y-y^3 \end{eqnarray*}
b) Translate u to polar coordinates.
c) Show that your polar version of u satisfies the polar Laplace equation:
\begin{eqnarray*}
u_{rr} + u_r /r + u_{θθ} /r^2 = 0
\end{eqnarray*}
I suppose the first thing to do would be to change to polar coordinates via:
Part B:
$$x=r \cos(\theta)$$$$y=r \sin(\theta)$$
Substituting that back into the function you get:
$$ u(r,\theta) =3r^3\cos^2(\theta)\sin(\theta)-r^3\sin^3(\theta) $$
Part C:
For part c you will take the respective partial derivatives. The partial derivatives for $u_{\theta\theta}$ will be a bit messy. Once you have $u_{\theta\theta}$, $u_{rr}$, and $u_{r}$ you can plug those back into the polar Laplace equation directly. Assuming the derivatives are correct, you "should" compute that polar Laplacian of $ u(r,\theta)$ is 0. Thus, showing that the given $u(x,y)$ satisfies the equation.
Here is what part C looks like in mathematica:
First we tell mathematica that $u(r,\theta)=3r^3\cos^2(\theta)\sin(\theta)-r^3\sin^3(\theta))$
u[r_, t_] = 3 r^3 Cos[t]^2 Sin[t] - r^3 Sin[t]^3;Then we take the derivatives $u_r$,$u_{rr}$, and $u_{\theta\theta}$
ur = D[u[r, t], r]
urr = D[u[r, t], r, r]
utt = D[u[r, t], t, t]the output:
9 r^2 Cos[t]^2 Sin[t] - 3 r^2 Sin[t]^3
18 r Cos[t]^2 Sin[t] - 6 r Sin[t]^3
-27 r^3 Cos[t]^2 Sin[t] + 9 r^3 Sin[t]^3So we have
$$\begin{aligned}
u_r(r,\theta)=r^2(9\cos^2(\theta)\sin(\theta)-&3\sin^3(\theta))
\\u_{rr}(r,\theta)=r(18\cos^2(\theta)\sin(\theta)-&6\sin^3(\theta))
\\u_{\theta\theta}=r^3(-27\cos^2(\theta)\sin(\theta)+&9\sin(\theta))
\end{aligned}$$
Its actually pretty easy to do this by hand at this point since the trig terms are all the same, but Mathematica isn't as error prone as we are.
urr + ur/r + utt/r^2 // FullSimplifyWhich gives
0What a surpirise, the Laplacian is satisfied! The worst part in this whole thing is by far the angular derivative, but its pretty easy to see how all this should go together to zero once we have the derivatives.
I am confused as to why r = 1? Are we arbitrarily setting r to 1 or is there a reason?
What I don't understand is how the r's in the equation are eliminated? Because if we were just looking at the r's then I would think the urr+ur/r+utt/r^2=r+r+r. Unless we assume that r=1 and I do not understand that either, just like Becca.
@dwillia2 has it right, the r's don't go away and his work up top is all correct there.
Yea, I am not sure why I included $r=1$ that was a mistake. I have edited my previous post to correct. Looking at my test I didn't even do that. Regardless of the value for $r$, the flow the problem is the same even if you just have $r=1$ for your Laplacian. To do this problem correctly $r$ is left as a variable.
If we look at just the $r$ cancellation after this step:
the cancellation of r goes the following way(ignoring the trig components and factoring out a 9, 18, and 27 respectively):
$$9r+18(1/r) r^2 -27 (1/r^2)=0$$