<s>[INST] Solution for $\Delta u = 0$ for which $u(\cos(t), \sin(t)) = t(2\pi - t) \ , \ 0 \leq t \leq 2\pi$ I have the following problem: Find the solution for $\Delta u = 0$ for which $u(\cos(t), \sin(t)) = t(2\pi - t), \\ 0 \leq t \leq 2\pi$.
My attempt:
I rewrite $u(\cos(t), \sin(t)) = u(x(t), y(t))$ with $x(t) = \cos(t), \ y(t) = \sin(t)$. The Laplace equation is $\frac{d^2 u}{dx^2} + \frac{d^2 u}{dy^2} = 0$, where $\frac{d^2 u}{dx^2} = \frac{d}{dx} (\frac{du}{dx})$ and $\frac{d^2 u}{dy^2} = \frac{d}{dy} (\frac{du}{dy})$. Since $x(t) = \cos(t), \ y(t) = \sin(t)$, $u$ is basically a function of $t$, so I use the chain rule to get $\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}$.
So we have $\frac{dx}{dt} = \frac{d}{dt} \cos(t) = -\sin(t)$ and $\frac{dy}{dt} = \frac{d}{dt} \sin(t) = \cos(t)$. Here I run into some problems.

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*How do I calculate $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ ?

Because, as far as I understand it, I should rewrite the part $t(2\pi - t)$ in terms of $x$ and $y$ (or maybe in terms of $\cos(t)$ and $\sin(t)$ ?). But I don't know how to do it (if it's even necessary)


*How would the Laplace equation then look in terms of $u$ being dependent on $t$?

 [/INST] I’ve made some mistakes in the original answer and is easier to redo everything. I can’t delete the answer so I’m compelled to at another one.
As I understand the problem is this: solve $\Delta u=0$ for $u(\cos t, \sin t)=t(2\pi-t), 0\le t\lt 2\pi$
There is a lot of differentiating from now on.
$$\frac{du}{dx}=\frac{\frac{du}{dt}}{\frac{dx}{dt}}=\frac{2\pi-2t}{-\sin t}=2\frac{t-\pi}{\sin t}$$
$$\frac{d^2u}{dx^2}=\frac{d}{dx}\left(\frac{du}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{du}{dx}\right)}{\frac{dx}{dt}}=\frac{2\frac{\sin t-(t-\pi)\cos t}{\sin ^2 t}}{-\sin t}=-2\frac{\sin t-(t-\pi)\cos t}{\sin ^3 t}$$
In similar manner,
$$\frac{du}{dy}=\frac{2\pi-2t}{\cos t}=-2\frac{t-\pi}{\cos t}$$
$$\frac{d^2u}{dy^2}=\frac{-2\frac{\cos t-(t-\pi)(-\sin t)}{\cos ^2 t}}{\cos t}=-2\frac{\cos t+(t-\pi)\sin t}{\cos^3 t}$$
Now replacing $\sin t =y, \cos t=x$:
$$\Delta u =0\Leftrightarrow \frac{y-(t-\pi)x}{y^3}+\frac{x+(t-\pi)y}{x^3}=0, x^2+y^2=1$$
Solving for t:$$t=\pi+\frac{xy}{x^2-y^2}$$
Replacing this expression of t in the definition of u:
$$u(\cos t, \sin t)=t(2\pi-t)\Leftrightarrow u(x,y)= \pi^2-\left(\frac{xy}{x^2-y^2}\right)^2, x^2+y^2=1$$
Now the function u could be differentiated for x and y:
$$\frac{du}{dx}=-2\frac{xy}{x^2-y^2}\cdot \frac{y(x^2-y^2)-xy\cdot 2x}{(x^2-y^2)^2}=2\frac{xy^2}{(x^2-y^2)^3}$$
$$\frac{d^2u}{dx^2}=2\frac{y^2(x^2-y^2)^3-xy^2\cdot 3(x^2-y^2)^2\cdot 2x}{(x^2-y^2)^6}=-2y^2\frac{1+4x^2}{(x^2-y^2)^4}$$
$$\frac{du}{dy}= -2\frac{xy}{x^2-y^2}\cdot \frac{x(x^2-y^2)-xy\cdot (-2y)}{(x^2-y^2)^2}=-2\frac{x^2y}{(x^2-y^2)^3} $$
$$\frac{d^2u}{dy^2}=-2\frac{x^2(x^2-y^2)^3-x^2y\cdot 3(x^2-y^2)^2\cdot (-2y)}{(x^2-y^2)^6}=-2x^2\frac{1+4y^2}{(x^2-y^2)^4}$$
Solving $\Delta u=0$:
$$\Delta u=0\Leftrightarrow-y^2(1+4x^2)-x^2(1+4y^2)=0\Leftrightarrow 1+8x^2y^2=0$$
This equation has no solutions.
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