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HW 4 Mathematics 503, Mathematical Modeling, CSUF
, June 9, 2007
Summer 2007 Compiled on October 14, 2025 at 5:10pm [public]
1 problem: Find extrermals for the following functional:
(b) \(\int _{a}^{b}y^{2}+\left ( y^{\prime }\right ) ^{2}+2ye^{x}\ \ dx\)
Solution:
Assume first that \(y\left ( x\right ) \) has normal conditions on the boundaries. I.e. \(y\left ( a\right ) =y_{a},y\left ( b\right ) =y_{b}\)
\[ L\left ( y,y^{\prime },x\right ) =y^{2}+\left ( y^{\prime }\right ) ^{2}+2ye^{x}\]
We have the functional
\[ J\left ( y\right ) =\int _{a}^{b}L\left ( y,y^{\prime },x\right ) \ dx \]
and we seek to find a function \(y\left ( x\right ) \) which minimizes this functional.
Let the vector space from which we can pick \(y\left ( x\right ) \) from be
\[ V=C^{2}\left [ a,b\right ] \]
And let the set of admissible functions (within \(V\) ) (Is this set a subspace?) be defined
as
\[ A=\left \{ y\left ( x\right ) \ \text {s.t.}\ y\left ( x\right ) \in V\text { and }y\left ( a\right ) =y_{0},y\left ( b\right ) =y_{1}\right \} \]
And let the set of admissible directions \(v\left ( x\right ) \) be
\[ A_{d}=\left \{ v\left ( x\right ) \in V\text { s.t. }v\left ( a\right ) =0,v\left ( b\right ) =0\right \} \]
Use the variational method since the Lagrangian contains a quadratic terms.
\begin{align*} J\left ( y+v\right ) & =\int _{a}^{b}L\left ( \left ( y+v\right ) ,\left ( y+v\right ) ^{\prime },x\right ) \ \ dx\\ & =\int _{a}^{b}\left ( y+v\right ) ^{2}+\left ( \left ( y+v\right ) ^{\prime }\right ) ^{2}+2\left ( y+v\right ) e^{x}\ \ dx\\ & =\int _{a}^{b}\left ( y^{2}+v^{2}+2yv\right ) +\left ( y^{\prime }+v^{\prime }\right ) ^{2}+2\left ( ye^{x}+ve^{x}\right ) \ \ dx\\ & =\int _{a}^{b}y^{2}+v^{2}+2yv+\left ( y^{\prime }\right ) ^{2}+\left ( v^{\prime }\right ) ^{2}+2y^{\prime }v^{\prime }+2ye^{x}+2ve^{x}\ \ dx \end{align*}
rearrange terms
\begin{align*} J\left ( y+v\right ) & =\overset {J\left ( y\right ) }{\overbrace {\int _{a}^{b}y^{2}+\left ( y^{\prime }\right ) ^{2}+2ye^{x}}}+\overset {+ve}{\overbrace {v^{2}\ +\left ( v^{\prime }\right ) ^{2}}}+2yv+2y^{\prime }v^{\prime }+2ve^{x}\ \ dx\\ J\left ( y+v\right ) & =\ J\left ( y\right ) +\overset {+ve}{\overbrace {\int _{a}^{b}v^{2}\ +\left ( v^{\prime }\right ) ^{2}\ dx}}+2\overset {make\ this\ zero}{\overbrace {\int _{a}^{b}yv+y^{\prime }v^{\prime }\ +ve^{x}\ dx}}\end{align*}
Hence if we can find \(\tilde {y}\left ( x\right ) \) which will make the last term above zero, then \(J\left ( y\right ) \) will have been minimized by
this\(\ \tilde {y}\left ( x\right ) \)
Therefor the problem now becomes of solving for \(y\left ( x\right ) \) the following integral equation
\begin{equation} \int _{a}^{b}yv+y^{\prime }v^{\prime }+ve^{x}\ \ dx=0 \tag {1}\end{equation}
We need to try to convert the above into something like \(\int _{a}^{b}f\left ( y,y^{\prime },e^{x}\right ) v\left ( x\right ) \ \ dx=0\) so that we can say that \(f\left ( y,y^{\prime },e^{x}\right ) =0\) , so this
means in (1) we need to do integration by parts on the term \(y^{\prime }v^{\prime }\) . Hence (1) can be written
as
\[ \int _{a}^{b}yvdx+\int _{a}^{b}y^{\prime }v^{\prime }dx+\int _{a}^{b}ve^{x}\ \ dx=0 \]
Now since \(\int u\ dz=\left [ uz\right ] _{a}^{b}-\int _{a}^{b}z\ du\) , now let \(u=y^{\prime }\rightarrow du=y^{\prime \prime },\) and let \(dz=v^{\prime }dx\rightarrow z=v\) , hence we have
\[ \int _{a}^{b}y^{\prime }v^{\prime }dx=\overset {0}{\overbrace {\left [ y^{\prime }v\right ] _{a}^{b}}}-\int _{a}^{b}vy^{\prime \prime }\ dx \]
Hence (1) can be written as
\begin{align*} 0 & =\int _{a}^{b}yv+y^{\prime }v^{\prime }+ve^{x}\ \ dx\\ & =\int _{a}^{b}yv-vy^{\prime \prime }+ve^{x}\ \ dx\\ & =\int _{a}^{b}\left ( y-y^{\prime \prime }+e^{x}\right ) v\ \ dx \end{align*}
Now we apply the standard argument and say that since \(v\left ( x\right ) \) is arbitrary function, and the integral
above is always zero, then it must be that
\[ \left ( y-y^{\prime \prime }+e^{x}\right ) =0 \]
or
\[ \fbox {$y^{\prime \prime }-y=e^{x}$}\]
This is a linear second order ODE with constant coefficients with a forcing function. The
homogeneous ODE will have 2 independent solutions, say \(y_{1}\left ( t\right ) \) and \(y_{2}\left ( t\right ) \) , so the total solution
is
\begin{align*} y & =y_{h}\left ( x\right ) +y_{p}\left ( x\right ) \\ & =c_{1}y_{1}\left ( x\right ) +c_{1}y_{2}\left ( x\right ) +y_{p}\left ( x\right ) \end{align*}
To solve the homogeneous ODE
\[ y^{\prime \prime }-y=0 \]
Assume the solution is \(y=Ae^{mx}\) , hence the characteristic equation is \(m^{2}-1=0\rightarrow m=\pm 1,\) hence the solution is \(y_{1}\left ( x\right ) =e^{x},y_{2}\left ( x\right ) =e^{-x}\) ,
so
\begin{align*} y_{h}\left ( t\right ) & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{x}+c_{2}e^{-x}\end{align*}
Or the solution can be written in hyperbolic sin and cosine
\[ y_{h}\left ( t\right ) =c_{1}\cosh \left ( x\right ) +c_{2}\sinh \left ( x\right ) \]
Now to find particular solution, use variation of parameters. Assume
\[ y_{p}=-y_{1}u_{1}+y_{2}u_{2}\]
where
\begin{align*} u_{1} & =\int \frac {y_{2}e^{x}}{W}dx\\ u_{2} & =\int \frac {y_{1}e^{x}}{W}dx \end{align*}
Where
\begin{align*} W & =y_{1}y_{2}^{\prime }-y_{2}y_{1}^{\prime }=-e^{x}e^{-x}-e^{-x}e^{x}\\ & =-1-1\\ & =-2 \end{align*}
Hence
\begin{align*} u_{1} & =-\frac {1}{2}\int e^{-x}e^{x}dx\\ & =-\frac {1}{2}\int dx\\ & =-\frac {1}{2}x \end{align*}
and
\begin{align*} u_{2} & =-\frac {1}{2}\int e^{x}e^{x}dx\\ & =-\frac {1}{2}\int e^{2x}dx\\ & =-\frac {1}{4}e^{2x}\end{align*}
Hence the solution is
\begin{align*} y & =c_{1}e^{x}+c_{2}e^{-x}+y_{p}\\ & =c_{1}e^{x}+c_{2}e^{-x}+\left ( -u_{1}y_{1}+u_{2}y_{2}\right ) \\ & =c_{1}e^{x}+c_{2}e^{-x}+\left ( \frac {1}{2}xe^{x}+-\frac {1}{4}e^{2x}e^{-x}\right ) \end{align*}
Hence
\[ \fbox {$y\left ( x\right ) =c_{1}e^{x}+c_{2}e^{-x}+\frac {1}{2}xe^{x}-\frac {1}{4}e^{x}$}\]
2 problem: Show that the minimal area of a surface of revolution in a catenoid, that is, the surface
found by revolving a catenary
\[ y=c_{1}\cosh \left ( \frac {x+c_{2}}{c_{1}}\right ) \]
about the \(x\) axis
solution:
First we assume that \(y^{\prime }\left ( x\right ) >0\) over the integration range. And that the lower end of the integration \(x=a\) is
smaller than the upper limit \(x=b\)
If we view a small \(ds\) at \(y\left ( x\right ) \) we see that it has a length of
\[ ds^{2}=dy^{2}+dx^{2}\]
Hence
\begin{align} ds & =\sqrt {dy^{2}+dx^{2}}\nonumber \\ & \fbox {$ds=dx\sqrt {\left ( \frac {dy}{dx}\right ) ^{2}+1}$}\tag {1}\end{align}
or we can also write
\begin{align} ds & =\sqrt {dy^{2}+dx^{2}}\nonumber \\ & \fbox {$ds=dy\sqrt {\left ( \frac {dx}{dy}\right ) ^{2}+1}$}\tag {2}\end{align}
So there are 2 ways to solve this depending if we use (1) or (2). Let us leave the choice open for a
little longer.
Now a size of a differential area \(dA\) of a strip of width \(ds\) and a length given by the circumference of the
circle generated by rotation is
\[ dA=2\pi y\left ( x\right ) ds \]
Hence the total surface area is the integral of the above over the range which \(y\left ( x\right ) \) is defined at. Let
this be from \(x=a,\) to \(x=b\) is given by
\begin{align*} A & =\int _{x=a}^{x=b}dA\\ & =2\pi \int _{x=a}^{x=b}y\left ( x\right ) ds \end{align*}
Since we have \(y\left ( x\right ) \) already in the Lagrangian, let then pick expression (2) from the above.
\[ A=2\pi \int _{x=a}^{x=b}y\left ( x\right ) \sqrt {\left ( \frac {dx}{dy}\right ) ^{2}+1}dy \]
Now I need to change the limits. Let \(y\left ( a\right ) =y_{1}\) , and let \(y\left ( b\right ) =y_{2}\) hence
\[ A=2\pi \int _{y=y\left ( a\right ) }^{y=y\left ( b\right ) }y\left ( x\right ) \sqrt {\left ( \frac {dx}{dy}\right ) ^{2}+1}dy \]
If we write the above in the more standard format, we have
\begin{align*} A & =2\pi \int _{y=y\left ( a\right ) }^{y=y\left ( b\right ) }y\left ( x\right ) \sqrt {1+\left ( x^{\prime }\right ) ^{2}}dy\\ & =2\pi \int _{y=y\left ( a\right ) }^{y=y\left ( b\right ) }L\left ( y,x\left ( y\right ) ,x^{\prime }\left ( y\right ) \right ) \end{align*}
Remember now that \(y\) is the independent variable, and \(x\) is the dependent variable. This is different
from the normal way.
Hence the Lagrangian \(L\) is
\begin{equation} L\left ( y,x,x^{\prime }\right ) =y\sqrt {1+\left ( x^{\prime }\right ) ^{2}} \tag {3}\end{equation}
If I had picked expression (1) instead, I would have obtained the Lagrangian as
\begin{equation} L\left ( x,y,y^{\prime }\right ) =y\sqrt {1+\left ( y^{\prime }\right ) ^{2}} \tag {4}\end{equation}
Both will give the same answer but with (3) we have \(\frac {\partial L}{\partial x}-\frac {d}{dy}\left ( \frac {\partial L}{\partial x^{\prime }}\right ) =0\) and the first term \(\frac {\partial L}{\partial x}=0\) since \(L\) does not
depend on \(x\) , and now we can just say that \(\frac {\partial L}{\partial x^{\prime }}=\) constant,. While with (4) we have \(\frac {\partial L}{\partial y}-\frac {d}{dx}\left ( \frac {\partial L}{\partial y^{\prime }}\right ) =0\) now \(\frac {\partial L}{\partial y}\) is not
zero.
Now we continue, and we will use (3) as our lagrangian.
We start the solution of the problem. We seek a function \(\tilde {x}\left ( y\right ) \) which minimizes \(J\left ( x\right ) =\int _{y=y\left ( a\right ) }^{y=y\left ( b\right ) }L\left ( y,x,x^{\prime }\right ) dy\) .
Where \(\tilde {x}\left ( y\right ) \in V\) s.t. \(C^{2}\left [ a,b\right ] \) , and let the set of admissible functions
\(A\left ( x\left ( y\right ) \right ) =\left \{ x\left ( y\right ) |x\in V\text { and }x\left ( a\right ) =x_{1},x\left ( b\right ) =x_{2}\right \} \) ,
and let the set of admissible directions \(v\left ( y\right ) =\left \{ v\left ( y\right ) |v\in V\text { and }v\left ( a\right ) =0\text { and }v\left ( b\right ) =0\right \} \)
Now that we have written down all the formal definitions, we can just solve this by applying
Euler-Lagrange equation since the Lagrangian above meets the conditions of using Euler-Lagrange
equations (\(L\) is a function of \(x,x^{\prime },y\) and \(x\) is defined at the boundary conditions with a dirichlet type
boundary conditions).
The Euler Lagrangian equation is
\[ \frac {\partial L}{\partial x}-\frac {d}{dy}\left ( \frac {\partial L}{\partial x^{\prime }}\right ) =0 \]
Since \(L\) does NOT depend on \(x\) then \(\frac {\partial L}{\partial x}=0\) , and the above reduces to
\[ \frac {d}{dy}\left ( \frac {\partial L}{\partial x^{\prime }}\right ) =0 \]
Since the derivative is zero, then we can write that
\[ \frac {\partial L}{\partial x^{\prime }}=c_{1}\]
Where \(c_{1}\) is some constant. So the above becomes
\begin{align*} y\frac {2x^{\prime }}{2\sqrt {1+\left ( x^{\prime }\right ) ^{2}}} & =c_{1}\\ \frac {yx^{\prime }}{\sqrt {1+\left ( x^{\prime }\right ) ^{2}}} & =c_{1}\\ \frac {\left ( yx^{\prime }\right ) ^{2}}{1+\left ( x^{\prime }\right ) ^{2}} & =c_{1}^{2}\end{align*}
Hence we have
\begin{align*} \left ( yx^{\prime }\right ) ^{2} & =c_{1}^{2}\left ( 1+\left ( x^{\prime }\right ) ^{2}\right ) \\ & =c_{1}^{2}+c_{1}^{2}\left ( x^{\prime }\right ) ^{2}\\ \left ( x^{\prime }\right ) ^{2}\left ( y^{2}-c_{1}^{2}\right ) & =c_{1}^{2}\end{align*}
Hence the final ODE is
\[ \fbox {$x^{\prime }\left ( y\right ) =\frac {c_{1}}{\sqrt {\left ( y^{2}-c_{1}^{2}\right ) }}$}\]
This is a linear ODE Its solution is found by integration both side as follows
\begin{align*} \int x^{\prime }\left ( y\right ) dy & =\int \frac {c_{1}}{\sqrt {\left ( y^{2}-c_{1}^{2}\right ) }}dy\\ x\left ( y\right ) & =c_{1}\int \frac {dy}{c_{1}\sqrt {\left ( \frac {y}{c_{1}}\right ) ^{2}-1}}\\ x\left ( y\right ) & =\int \frac {dy}{\sqrt {\left ( \frac {y}{c_{1}}\right ) ^{2}-1}}\end{align*}
Let \(\frac {y}{c_{1}}=u\,\ \) hence \(dy=c_{1}du\) and the above becomes (do not need to worry about limits of integrations, as I will flip
back the earlier variable in a minute)
\[ x\left ( y\right ) =c_{1}\int \frac {\ du}{\sqrt {u^{2}-1}}\]
Which from table is given by
\[ x\left ( y\right ) =c_{1}\ln \left ( u+\sqrt {u^{2}-1}\right ) +c_{2}\]
Where \(c_{2}\) is constant of integration. Hence going back to our variables, we have
\begin{equation} x\left ( y\right ) =c_{1}\ln \left ( \frac {y}{k}+\sqrt {\left ( \frac {y}{k}\right ) ^{2}-1}\right ) +c_{2} \tag {3}\end{equation}
From tables I found that \(\cosh ^{-1}\left ( z\right ) =\ln \left ( z+\sqrt {z^{2}-1}\right ) \)
Hence (3) can now be written as
\begin{align*} x\left ( y\right ) & =c_{1}\cosh ^{-1}\left ( \frac {y}{k}\right ) +c_{2}\\ \frac {x\left ( y\right ) -c_{2}}{c_{1}} & =\cosh ^{-1}\left ( \frac {y}{c_{1}}\right ) \end{align*}
Or
\[ \frac {y}{c_{1}}=\cosh \left ( \frac {x-c_{2}}{c_{1}}\right ) \]
Then
\[ \fbox {$y\left ( x\right ) =c_{1}\cosh \left ( \frac {x-c_{2}}{c_{2}}\right ) $}\]
So the above curve will minimize the surface area.
Problem 1 (section 3.3,#14, page 177)
answer:
\(y\left ( 0\right ) =Y,y\left ( T\right ) =0\)
\[ E=\int _{0}^{T}e^{-\beta t}U\left ( r\left ( t\right ) \right ) dt \]
but
\[ r\left ( t\right ) =\alpha y-y^{\prime }\]
Hence
\[ E=\int _{0}^{T}e^{-\beta t}U\left ( \alpha y\left ( t\right ) -y^{\prime }\left ( t\right ) \right ) dt \]
Hence the Lagrangian is
\[ \fbox {$L\left ( t,y,y^{\prime }\right ) =e^{-\beta t}U\left ( \alpha y\left ( t\right ) -y^{\prime }\left ( t\right ) \right ) $}\]
since \(y\left ( t\right ) \) is defined at boundaries of the interval, we can use Euler-Lagrange
equations
\[ \frac {d}{dy}L\left ( t,y,y^{\prime }\right ) -\frac {d}{dt}\left ( \frac {d}{dy^{\prime }}L\left ( t,y,y^{\prime }\right ) \right ) =0 \]
Now the first term above is
\begin{align*} \frac {d}{dy}L\left ( t,y,y^{\prime }\right ) & =e^{-\beta t}U^{\prime }\frac {d}{dy}\left ( \alpha y\left ( t\right ) -y^{\prime }\left ( t\right ) \right ) \\ & =\alpha e^{-\beta t}U^{\prime }\end{align*}
and the second term is
\begin{align*} \frac {d}{dt}\left ( \frac {d}{dy^{\prime }}L\left ( t,y,y^{\prime }\right ) \right ) & =\frac {d}{dt}\left ( e^{-\beta t}U^{\prime }\frac {d}{dy^{\prime }}\left ( \alpha y\left ( t\right ) -y^{\prime }\left ( t\right ) \right ) \right ) \\ & =\frac {d}{dt}\left ( -e^{-\beta t}U^{\prime }\right ) \end{align*}
Hence our E-L equations now looks like
\begin{equation} \alpha e^{-\beta t}U^{\prime }+\frac {d}{dt}\left ( e^{-\beta t}U^{\prime }\right ) =0\tag {1}\end{equation}
Since \(U\) is a function of \(r\left ( t\right ) \) , then
\[ \frac {d}{dt}\left ( e^{-\beta t}U^{\prime }\right ) =-\beta e^{-\beta t}U^{\prime }+e^{-\beta t}U^{\prime \prime }r^{\prime }\left ( t\right ) \]
And (1) now becomes
\begin{align*} \alpha e^{-\beta t}U^{\prime }-\beta e^{-\beta t}U^{\prime }+e^{-\beta t}U^{\prime \prime }r^{\prime }\left ( t\right ) & =0\\ \left ( \alpha -\beta \right ) U^{\prime }+U^{\prime \prime }r^{\prime }\left ( t\right ) & =0 \end{align*}
This is separable ODE, hence
\begin{align*} \frac {U^{\prime \prime }}{U^{\prime }}\frac {dr}{dt} & =-\left ( \alpha -\beta \right ) \\ \frac {U^{\prime \prime }}{U^{\prime }}dr & =-\left ( \alpha -\beta \right ) dt \end{align*}
Integrate both sides
\begin{align*} \ln \left ( U^{\prime }\left ( r\right ) \right ) & =-\left ( \alpha -\beta \right ) \int \ dt\\ \ln \left ( U^{\prime }\left ( r\right ) \right ) & =-\left ( \alpha -\beta \right ) t+k \end{align*}
where \(k\) is constant on integration
\begin{align*} U^{\prime }\left ( r\right ) & =e^{-\left ( \alpha -\beta \right ) t+k}\\ & =ce^{-\left ( \alpha -\beta \right ) t}\end{align*}
where \(c=e^{k}\) is another constant
But \(U\left ( r\right ) =2\sqrt {r}\) hence \(\frac {dU}{dr}=\frac {1}{\sqrt {r}}\) and the above becomes
\begin{align*} \frac {1}{\sqrt {r}} & =ce^{-\left ( \alpha -\beta \right ) t}\\ \frac {1}{r} & =c^{2}e^{-2\left ( \alpha -\beta \right ) t}\\ & \fbox {$r\left ( t\right ) =c_{2}e^{2\left ( \alpha -\beta \right ) t}$}\end{align*}
Where since \(c^{-2}\) is constant, I call it \(c_{2}\)
Now, Since
\[ y^{\prime }\left ( t\right ) =\alpha y\left ( t\right ) -r\left ( t\right ) \]
Then
\[ \fbox {$y^{\prime }\left ( t\right ) -\alpha y\left ( t\right ) =c_{2}e^{2\left ( \alpha -\beta \right ) t}$}\]
The solution is
\[ y=y_{h}+y_{p}\]
Assume \(y_{h}=Ae^{mt}\) , hence \(Ame^{mt}-\alpha Ae^{mt}=0\rightarrow m=\alpha \)
So the solution is
\[ y_{h}=c_{1}e^{\alpha t}\]
For the particular solution, guess a solution. Since the forcing function is of the form \(ce^{t}\) , guess
\[ y_{p}=Ae^{kt}\]
so \(y_{p}^{\prime }=Ate^{kt}\)
and we substitute this solution in the ODE above, we obtain
\begin{align*} Ake^{kt}-\alpha Ae^{kt} & =c_{2}e^{2\left ( \alpha -\beta \right ) t}\\ A\left ( k-\alpha \right ) e^{kt} & =c_{2}e^{2\left ( \alpha -\beta \right ) t}\end{align*}
so by comparing exponents, we see that \(k=2\left ( \alpha -\beta \right ) \) and \(A\left ( k-\alpha \right ) =c_{2}\) hence \(A=\frac {c_{2}}{k-\alpha }=\frac {c_{2}}{2\left ( \alpha -\beta \right ) -\alpha }=\frac {c_{2}}{\alpha -2\beta }\)
Therefore
\[ y_{p}=\frac {c_{2}}{\alpha -2\beta }e^{2\left ( \alpha -\beta \right ) t}\]
Hence, since
\[ y=y_{h}+y_{p}\]
Then
\[ \fbox {$y\left ( t\right ) =c_{1}e^{\alpha t}+\frac {c_{2}}{\alpha -2\beta }e^{2\left ( \alpha -\beta \right ) t}$}\]
We now find \(c_{1}\) and \(c_{2}\) from I.C. At \(t=0,y=Y\) , hence
\begin{align} Y & =c_{1}+\frac {c_{2}}{\alpha -2\beta }\nonumber \\ c_{1} & =Y-\frac {c_{2}}{\alpha -2\beta }\tag {2}\end{align}
at \(t=T,y=0\) , hence
\begin{align*} 0 & =\left ( Y-\frac {c_{2}}{\alpha -2\beta }\right ) e^{\alpha T}+\frac {c_{2}}{\alpha -2\beta }e^{2\left ( \alpha -\beta \right ) T}\\ c_{2} & =\frac {\left ( \alpha -2\beta \right ) Ye^{\alpha T}}{e^{\alpha T}-e^{2\left ( \alpha -\beta \right ) T}}\end{align*}
so from (2)
\[ c_{1}=Y-\frac {\left ( \alpha -2\beta \right ) Ye^{\alpha T}}{\left ( \alpha -2\beta \right ) \left ( e^{\alpha T}-e^{2\left ( \alpha -\beta \right ) T}\right ) }\]
Hence
\[ \fbox {$y\left ( t\right ) =\left ( Y-\frac {\left ( \alpha -2\beta \right ) Ye^{\alpha T}}{\left ( \alpha -2\beta \right ) \left ( e^{\alpha T}-e^{2\left ( \alpha -\beta \right ) T}\right ) }\right ) e^{\alpha t}+\frac {\left ( 1-\alpha +2\beta Y\right ) e^{\left ( -\alpha +2\beta \right ) T}}{\alpha -2\beta }e^{2\left ( \alpha -\beta \right ) t}$}\]
and
\[ \fbox {$r\left ( t\right ) =\left ( 1-\alpha +2\beta Y\right ) e^{\left ( -\alpha +2\beta \right ) T}\ e^{2\left ( \alpha -\beta \right ) t}$}\]
Analysis on results:
These are 3 plots showing \(y\left ( t\right ) \) and \(r\left ( t\right ) \) . The first is for \(\alpha =0.03\)
This one is for \(\alpha =0.1\)
We notice that the higher the interest rate \(\alpha \) is the more capital will accumulate, which means to
achieve the goal of zero capital at death the rate \(r\left ( t\right ) \) is more steep near the end. If the
money hardly accumulate during life time, i.e. when the interest rate is very low, then
we should expect a straight line for \(y\left ( t\right ) \) , which is verified by this plot below when I set
\(\alpha =0.001\)