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Messages - Zacharie Leger

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Final Exam / Re: FE Problem 3
« on: April 16, 2015, 02:16:09 AM »
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
$$T'(0)=2n\pi G\sin(n(0))+2n\pi H\cos(n(0))=2n\pi F=0$$
$$T_n (t)=G_n \cos(2n\pi t)$$
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:
$$u(x,y)=\frac{1}{6} +\sum\limits_{n=1}^\infty \frac{2(1+(-1)^n)}{\pi^2 n^2} \cos(n\pi x)\cos(2n\pi t)$$

Final Exam / Re: FE Problem 5
« on: April 15, 2015, 07:22:27 PM »
Hopeful solution
a. The eigenvalue problem can be found using separation of variables, hence if we let $u(x,y)=X(x)Y(y)$ and plug into the Laplace equation we can get
$$\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=0 \Rightarrow -\frac{X''(x)}{X(x)}=\frac{Y''(y)}{Y(y)}=-\lambda$$

b. The boundry conditions the problem tell us that $\lambda >0$, since we have periodic boundary conditions for $Y(y)$. Furthermore, if we let $\lambda =\omega ^2,\,\omega >0$ we get the following solution
$$Y(y)=A\sin(\omega y)+B\cos(\omega y)$$
Applying the boundry conditions on the solution we see that
$$Y(0)=A\sin(\omega (0))+B\cos(\omega (0))=B=0\,\mathrm{and}\, Y(\pi)=A\sin(\omega (\pi))=0\Rightarrow \omega =n$$
$$\Rightarrow Y_n (y)=A_n \sin(ny)$$
as for the $X(x)$ solution we get
Applying the boundry conditions all we can say is that
$$X_n (x)=C_n\cosh(nx)+\sinh(nx)$$
The full solution can be expressed as
$$u(x,y)=\sum\limits_{n=1}^\infty [(C_n \cosh(nx)+\sinh(nx))A_n\sin(ny)]$$

c. From this we get
$$u_x (0,y)=1=\sum\limits_{n=1}^\infty [(nC_n \sinh(n(0))+n\cosh(n(0)))A_n \sin(ny)]=\sum\limits_{n=1}^\infty [A_n \sin(ny)]$$
The coefficiants are given by
$$A_n=\frac{2}{pi} \int_0^\pi (1)\sin(ny)\mathrm{d}x=\frac{2}{pi}\frac{-cos(nx)}{n}|_0^\pi = \frac{2}{\pi}\frac{1+(-1)^2}{n}$$
Finally, we get
$$u(x,y)=\sum\limits_{n=1}^\infty \frac{2}{\pi}\frac{1+(-1)^2}{n}[(C_n \cosh(nx)+\sinh(nx))\sin(ny)]$$

HA3 / Re: HA3 problem 5
« on: April 14, 2015, 10:40:46 PM »
Hopeful solution

Web Bonus Problems / Re: Web Bonus Problem 5
« on: April 14, 2015, 01:35:22 PM »
Following hint (a) we let $u(x,y)=X(x)Y(y)$ to obtain
from equation (1). For $\lambda=\omega^2, \omega>0$ we obtain solutions of the form
$$X(x)=A\cos(\omega _x x)+B\sin(\omega _x x)$$
Applying BCs we see that
$$X(0)=A=0 \Rightarrow X(x)=B\sin(\omega _x x) \Rightarrow X(a)=B\sin(\omega _x a)=0 \Rightarrow \omega _x =\frac{n_x\pi}{a} $$
Similarly for $Y(y)$ we can get that
$$Y(y)=C\cos(\omega _y y)+D\sin(\omega _y y)$$
$$Y(0)=C=0 \Rightarrow Y(y)=D\sin(\omega _y y) \Rightarrow y(b)=B\sin(\omega _y b)=0 \Rightarrow \omega _y =\frac{n_y\pi}{b} $$
Note that $\omega _x=\frac{n_x\pi}{a}=\frac{n_y\pi}{b}=\omega _y$. Hence, $\frac{a}{b}=\frac{n_x}{n_y}$ so
$$  \frac{a}{b} \in \mathrm{Q}$$

HA9 / HA9 Problem 1
« on: March 24, 2015, 10:56:40 PM »
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
u= z^4 + az^2(1-x^2-y^2-z^2)+b(1-x^2-y^2-z^2)^2
with unknown coefficients $a,b$.

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?

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