Toronto Math Forum

Hopeful solution
By letting $u(x,t)=X(x)T(t)$ and plug into the wave equation we can get
$$\frac{X''(x)}{X(x)}-\frac{T''(t)}{4T(t)}=0 \Rightarrow \frac{X''(x)}{X(x)}=\frac{T''(t)}{4T(T)}=-\lambda$$
We know that $\lambda\geq 0$, so let of first develop the the case where $\lambda =0$. Hence,
$$X''(x)=0 \Rightarrow X(x)=Ax+B \Rightarrow X'(0)=A=0 \Rightarrow X(x)=B$$
Now if $\lambda =\omega ^2,\,\omega >0$ we get the following solution
$$X(x)=C\sin(\omega X)+D\cos(\omega X)$$
Applying the boundry conditions on the solution we see that
$$X'(0)=C\omega\cos(\omega (0))+D\omega\sin(\omega (0))=C=0\,\mathrm{and}\, X(1)=D\omega\sin(\omega (1))=0\Rightarrow \omega=n\pi$$
$$\Rightarrow X_n (x)=D_n \cos(n\pi x)$$
As for the solution for $T(t)$, we again consider first when $\lambda=0$
$$T''(t)=0 \Rightarrow T(t)=Et+F \Rightarrow T'(0)=E=0 \Rightarrow T(t)=F$$
Now when $\lambda =(n\pi)^2$
$$T(t)=G\cos(2n\pi x)+H\sin(2n\pi x)$$
Applying the boundary conditions we see that
$$T'(0)=2n\pi G\sin(n(0))+2n\pi H\cos(n(0))=2n\pi F=0$$
Hence,
$$T_n (t)=G_n \cos(2n\pi t)$$
So the full solution can be expressed as
$$u(x,y)=\alpha _0 +\sum\limits_{n=1}^\infty \alpha _n \cos(n\pi x)\cos(2n\pi t)$$
where we let $\alpha _0 =B+F$ and $\alpha _n = D_n G_n$. We have one last boundary conditions to apply to get our final answer:
$$u(x,0)=x(1-x)=\alpha _0 +\sum\limits_{n=1}^\infty \alpha _n \cos(n\pi x)$$
Fourier tells us that the coefficiants will be given by
$$\alpha _0 =\int_0^1 x-x^2\mathrm{d}x=[\frac{x^2}{2}-\frac{x^3}{3}]_0^1=\frac{1}{6}$$
$$\alpha _n =2\int_0^1 (x-x^2)\cos(n\pi x)\mathrm{d}x=2[(-n\pi(-1+2 x)\cos(n\pi x)+(2-n^2\pi^2 (-1+x)x)\sin(n\pi x))/(n^3\pi^3)]_0^1=\frac{(1+(-1)^n)}{\pi^2 n^2}$$
Therefore, our final solution is:
$$u(x,y)=\frac{1}{6} +\sum\limits_{n=1}^\infty \frac{2(1+(-1)^n)}{\pi^2 n^2} \cos(n\pi x)\cos(2n\pi t)$$

Hopeful solution

Find function $u$ harmonic in $\{x^2+y^2+z^2\le 1\}$ and coinciding with $g=z^4$ as $x^2+y^2+z^2=1$.

Hint. According to [Subsection 28.1] solution must be a harmonic polynomial of degree $4$ and it should depend only on $x^2+y^2+z^2$ and $z$ (Explain why). The only way to achive it (and still coincide with $g$ on $\{x^2+y^2+z^2=1\}$) is to find
\begin{equation*}
u= z^4 + az^2(1-x^2-y^2-z^2)+b(1-x^2-y^2-z^2)^2
\end{equation*}
with unknown coefficients $a,b$.

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It seems to me that we should have a harmonic polynomial of degree 4 if we want the function to coincide with $g(x)=z^4$ on ${x^2+y^2+z^2=1}$, I'm I missing something?