Problem source

In a cosmological model, the radius $\rm R$ of the universe is a function of the age $t$ of the universe. The function $\rm R$ satisfies
the three conditions:
$$
\mbox{${\rm R}(0)=0$}, \ \ \ \ \ \ \ \ \
\mbox{${\rm R'}(t)>0$ for $t>0$}, \ \ \ \ \ \ \ \ \ \  
\mbox{${\rm R''}(t)<0$ for $t>0$},
\tag{*}
$$
where ${\rm R''}$ denotes the second derivative of $\rm R$.
 The function ${\rm H}$ is defined by
\[
{\rm H} (t)=  \frac{{\rm R}'(t)}{{\rm R}( t)}\;.
\]
\begin{questionparts}
\item Sketch a graph of ${\rm R} (t)$.  By considering a tangent
to the graph, show that $t<1/{\rm H}(t)$.
\item Observations reveal that ${\rm H}(t) = a/t$, where $a$ is constant. Derive an expression for ${\rm R}(t)$. What range of values of $a$ is consistent with the three conditions $(*)$?
\item Suppose, instead, that observations reveal that ${\rm H}(t)= b t^{-2}$, where $b$ is constant.  Show that this is not consistent with conditions $(*)$  for any value of $b$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ 

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){ln(1+(#1))};
    \def\xl{-0.5};
    \def\xu{10};
    \def\yl{-0.5};
    \def\yu{3};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);

        % \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
        \filldraw (5, {\functionf(5)}) circle (1.5pt) node[below right] {$(t, R(t))$};
        % \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};

        \draw[red, thick, dashed] ({0}, {1/6*(0-5)+\functionf(5)}) -- ({9}, {1/6*(9-5)+\functionf(5)}) node[above, sloped, pos=0.3] {\tiny $y = R'(t)(x-t) + R(t)$};

        \node[below, rotate=15, blue] at (8, {\functionf(8)}) {\tiny $y = R(x)$};
        
        \draw[thick, blue, smooth, domain=0:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

        % \node[blue, above, rotate=70] at (-1, {\functionf(-1)}) {\tiny $y = (x-p)(x-q)(x-r)$}; 
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

Notice the tangent must hit the $y$-axis above the origin, ie 
\begin{align*}
&& 0 &< R'(t)(0-t) + R(t) \\
\Rightarrow && R'(t) t &< R(t) \\
\Rightarrow && t &<  \frac{R(t)}{R'(t)} = \frac{1}{H(t)}
\end{align*}

\item Suppose $H(t) = a/t$ then

\begin{align*}
&& \frac{R'}{R} &= \frac{a}{t} \\
\Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\
\Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\
\Rightarrow && R &= Kt^a
\end{align*}

Since we need $R(t) > 0$, $K > 0$, since $R'(t) > 0$ we need $a > 0$, since $R''(t) < 0$ we need $a(a-1) < 0$ ie $0 < a < 1$

\item Suppose instead $H(t) = bt^{-2}$ then

\begin{align*}
&& \frac{R'}{R} &= \frac{b}{t^2} \\
\Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\
\Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\
\Rightarrow && R &= Ke^{-b/t}
\end{align*}

Since $R > 0$ we must have $K > 0$. 
\begin{align*}
R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\
\Rightarrow && b &> 0 \\
R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\
&&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\
\Rightarrow && b &< 2t\\
\Rightarrow && b &< 2t
\end{align*}

which cannot be true for all $t$, ie there is no $b$ which satisfies this.

\end{questionparts}