Toronto Math Forum

Solution for a(i) is attached. Please correct me if I am wrong.

Maybe the transform for $e^{-\frac{ax^2}{2}}$ should be $\sqrt{\frac{2\pi}{a}}e^{-\frac{k^2}{2a}}$ instead of $\sqrt{\frac{2\pi}{a}}e^{-\frac{ak^2}{2}}$?

Just wondering, the boundary condition in this question is $X(0)=0$ and $X'(l)+\beta X(l)=0$?

It says we will have quiz 4 on the official website, which covers HW6.

For part a, when we increase $T$, we expand our domain, so $M(T)$ should be at least non-decreasing.
Similarly, for part b, when we expand the domain, $m(T)$ is non-increasing.

Suppose $u=\phi_1(x+c_1t)+\psi_1(x-c_1t)$ for $x>0$ and $u=\phi_2(x+c_2t)+\psi_2(x-c_2t)$ for $x<0$.
Also we can solve $\phi_1(x)=\phi(x)-\frac{1}{2}\phi(0)$, $\psi_1(x)=\frac{1}{2}\phi(0)$ for $x>0$.
And $\phi_2=\psi_2=0$ for $x<0$.
It follows that $u=\phi(x+c_1t)$ for $x>c_1t$ and $u=0$ for $x<-c_2t$.
Now we are interested in $\psi_1(x)$ when $x<0$ and $\phi_2(x)$ when $x>0$.
Notice from boundary condition, for $t>0$
$$\phi_1(c_1t)-\psi_1(-c_1t)=\frac{c_1\beta}{c_2}\phi_2(c_2t)$$
$$\phi_1(c_1t)+\psi_1(-c_1t)=a\phi_2(c_2t)$$
Solve for $\psi_1$ and $\phi_2$, we have
$$\psi_1(x)=\frac{c_1\beta-ac_2}{ac_2+c_1\beta}\phi_1(-x) \enspace \text{for   } x<0$$
$$\phi_2(x)=\frac{2c_2\phi_1(\frac{c_1x}{c_2})}{ac_2+c_1\beta} \enspace \text{for   } x>0$$
Thus, for $-c_2t<x<0$,
$$u=\phi_2(x+c_2t)=\frac{2c_2}{ac_2+c_1\beta}[\phi(\frac{c_1}{c_2}(x+c_2t))-\frac{1}{2}\phi(0)]$$
For $0<x<c_1t$,
$$u=\phi_1(x+c_1t)+\psi_1(x-c_1t)=\phi(x+c_1t)-\frac{1}{2}\phi(0)+\frac{c_1\beta-ac_2}{ac_2+c_1\beta}[\phi(c_1t-x)-\frac{1}{2}\phi(0)]$$

By the hint, we do a change of variable. Let $$x=\frac{1}{2}(\zeta+\eta)$$ $$t=\sinh(\frac{1}{2}(\zeta-\eta))$$
Then we compute $u_{\zeta}$ and $u_{\zeta\eta}$.
By chain rule,
$$u_\zeta=\frac{1}{2}u_x+u_t\cdot t_\zeta=\frac{1}{2}(u_x+u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)))$$
Similarly, $$u_{\zeta\eta}=\frac{1}{2}([u_x]_\eta+[u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)]_\eta)$$
$$[u_x]_\eta=\frac{1}{2}u_{xx}-\frac{1}{2}u_{xt}\cosh(\frac{1}{2}(\zeta-\eta))$$
$$[u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)]_\eta=[u_t]_\eta\cosh(\frac{1}{2}(\zeta-\eta))+u_t[\cosh(\frac{1}{2}(\zeta-\eta))]_\eta=\frac{1}{2}u_{tx}\cosh(\frac{1}{2}(\zeta-\eta))+u_{tt}[\cosh(\frac{1}{2}(\zeta-\eta))]^2-\frac{1}{2}u_t\sinh(\frac{1}{2}(\zeta-\eta))$$
Notice that $$[\cosh(\frac{1}{2}(\zeta-\eta))]^2=1+[\sinh(\frac{1}{2}(\zeta-\eta))]^2=t^2+1$$
We have $$u_{\zeta\eta}=\frac{1}{2}[\frac{1}{2}u_{xx}-\frac{1}{2}(t^2+1)u_{tt}-\frac{1}{2}tu_t]=\frac{1}{4}(u_{xx}-(t^2+1)u_{tt}-tu_t)$$
Combine with the original PDE, we have $u_{\zeta\eta}=0$. So $u=g(\eta)+f(\zeta)$.

Part a)
First we make the coefficient of $y''$ one, which results in
\begin{equation*}
y''(x) - \frac{(2\ln x+3)}{x(\ln x+1)} y'(x) + \frac{(2\ln x+3)}{x^2(\ln x+1)}y(x)=0
\end{equation*}
By Abel' Thm,
\begin{equation}
\ln W=\int{\frac{(2\ln x+3)}{x(\ln x+1)}}dx\label{A}
\end{equation}
Using change of variables to evaluate the integral,let
\begin{equation*}
\mu=\ln x + 1
\end{equation*}
The the integral becomes
\begin{equation*}
\int{\frac{2\mu+1}{\mu}}du =\int{2}du+\int{\frac{1}{\mu}}du \label{B}
\end{equation*}
We have \begin{equation*} \ln W=(2\mu+\ln \mu)+C_1   \end{equation*}
Plug in x back, which is \begin{equation} W= C_2x^2(\ln x + 1) \end{equation}



Part b)
To verify y = x is a solution. Just plug in, we have
\begin{equation*} x^3(\ln x+1)\cdot  0 -(2\ln x+3)x^2 \cdot 1 + (2\ln x+3) x^2  = 0 \end{equation*}
Then to find another independent solution, notice that $W > 0$ for $x > 1$
So we can just let $C_2 = 1$. And we know $W = \left| \begin{matrix} x & y \\1 &  y’ \end{matrix}\right|$
Now we have a new ode
\begin{equation}  y' - \frac{1}{x} y = x(\ln x + 1), \qquad x>1. \end{equation}
Solve the homogeneous ode first,
\begin{equation*} z' - \frac{1}{x}z = 0 \end{equation*}
We get $z=x$. Then we know $y=\mu(x)x$.

And $y'=\mu'x+\mu$, we plug in into (3). We have \begin{equation*} u'=(\ln x + 1) \end{equation*}
Then \begin{equation*} u=\int{\ln x +1}dx \end{equation*}
We use integral by parts to evaluate $\int{\ln x}dx$ first.
\begin{equation*} \int{\ln x} dx = x\ln x - \int{1}dx = x\ln x - 1 \end{equation*}
So we have \begin{equation*} \mu x=\ln x + x + C \end{equation*}
Finally, we have $y_2=(x\ln x+x+C)x$.