<p>Answer to Q1:<br />
If we re-write FONC as a function of <span class="math-container">$a$</span>: <span class="math-container">$u'(c(a))-V_{a} =0$</span><br />
Differentiate wrt <span class="math-container">$a$</span> (as in Walde 2010): <span class="math-container">$u''(c(a)) c'(a) -V_{aa} =0$</span><br />
We know from SOSC that  <span class="math-container">$u''(c(a))&lt;0$</span>.<br />
If we assume consumption is increasing in wealth <span class="math-container">$c'(a)&gt;0$</span>, then <span class="math-container">$V_{aa}&lt;0$</span></p>
<p><span class="math-container">$
\left[\begin{array}{l}
\rho V(a_{t}) = 
\frac{\gamma (V_a)^{1-\frac{1}{\gamma}}}{1-\gamma} 
- \frac{1}{1-\gamma}
 + V_a \times r a_{t}
\\
V_{a}(a_{t}) &gt;0  \text{ FONC: for max to be well defined}
\\
V_{aa}(a_{t}) &lt;0  \text{ differentiate FONC &amp; SOSC}
\end{array} \right]
$</span></p>
<p>Some comments:</p>
<ul>
<li>I think it's enough for <span class="math-container">$V_a&gt;0, V_{aa}&lt;0$</span> to hold at a single point</li>
<li>I believe if we add these two conditions (increasing &amp; concave value) to <strong>DE</strong> we can pin down the unique optimal sol in this non-generic example.<br />
I cannot prove there is no other sol to <strong>DE</strong> w/ <span class="math-container">$V_a&gt;0, V_{aa}&lt;0$</span>.<br />
I really wanna know how these conditions generalize to more generic econ problems.</li>
<li><span class="math-container">$V_a&gt;0$</span> rules out Sol 1</li>
<li><span class="math-container">$V_{aa}&lt;0$</span> rules out Sols 2 &amp; 3</li>
<li>These conditions don't feel as essential/generic as <span class="math-container">$a(0)=a_0$</span> &amp; TVC.</li>
<li>Since <strong>DE</strong> is a first-order non-linear ode, I believe we either need one  equation, such as <span class="math-container">$V(a_{1})=V_1$</span>, or two inequalities (<span class="math-container">$V_a&gt;0, V_{aa}&lt;0$</span>) to pin down a unique sol.<br />
(I'm not sure about this. It may only be true for &quot;nice&quot; odes. Can someone who knows help?)<br />
Since <strong>DE</strong> is first-order, I don't like relying on a boundary condition w/ a 2nd derivative <span class="math-container">$V_{aa}&lt;0$</span></li>
<li><a href="https://benjaminmoll.com/wp-content/uploads/2019/07/HACT.pdf" rel="nofollow noreferrer">Achdou Hans Lasry Lions Moll</a> add a boundary condition that doesn't bind <span class="math-container">$a_t \geq \underline{a}$</span> where <span class="math-container">$\underline{a}&lt;0$</span> and use a &quot;state-boundary condition&quot; <span class="math-container">$V_a(\underline{a}) \geq u'(r\underline{a})$</span>.<br />
Issue: if <span class="math-container">$\gamma \in (0,1)$</span> then <span class="math-container">$u'(r\underline{a})$</span> may not be a real number.<br />
Is one inequality enough theoretically, even if numerically their method converges to the unique sol?<br />
Does this rule out the affine solutions above?<br />
And how does this generalize to generic problems?</li>
<li>Most importantly, why do the constant &amp; affine solutions above not satisfy the conditions for a viscosity solution? Why does the optimal sol satisfy them?</li>
</ul>

CC BY-SA 4.0
2021-07-27T13:34:44.447Z
Hey Albert. Cannot really provide the necessary and/or sufficient conditions to have a coincidence of solutions but may add something to your own answer. $V_a>0$ should be a given by non-satiation of the utility function and the FOC. The additional condition on $V_{aa}(a)$, pointed out in Walde (2010), should precisely rule out solutions of the HJB that are not solutions to the maximization problem. If for all state variable levels, $V_{aa}(a) < 0$, then $c'(a)<0$ and we can argue by contradiction. (continued)
RAGMS
CC BY-SA 4.0
2021-07-27T13:36:04.667Z
For any initial condition $a_0$, take $a_0' > a_0$. $c_t' = c(a_0') < c_t = c(a_0)$ and so we will have that for $t = \epsilon$, $a_{\epsilon}' > a_{\epsilon}$. The same argument applies for any time period, $t \geq 0$. This means that consumption will be lower at all points in time, $c_t' \leq c_t$. We reached our contradiction. Just take in turn the previous consumption plan - also achievable with the new initial condition. By the monotonicity of the integral, the value function would then be higher. Hence $c_t' \quad \forall t \geq 0$, determined by the proposed solution, cannot be optimal.
RAGMS
CC BY-SA 4.0
2021-07-27T14:36:30.903Z
For now I cannot truly rule out solution 2 since $V_{aa}(a)=0$. If $r=\rho$ and $B_0$ respects the equality imposed, we can then solve for $B_1$ using the boundary conditions. For $B_1$ to respect the terminal condition we must have that $B_1 = (ra_0)^{-\gamma}$. This means that $B_1$ cannot be a constant independent of the boundary conditions. (continued)
RAGMS
CC BY-SA 4.0
2021-07-27T14:40:17.453Z
This conclusion is intuitive if we just look at the FOC which sets $c(a)$ to a constant. Take a starting $a_0 \leq 0$ for example. The $\lim_{t \rightarrow \infty} a_t$ diverges as $\dot{a_t}<0 \forall t \geq 0$ and the tvm is not respected. In fact, as long as $a_0 \neq a*$, for a certain $a*$, the limit will always diverge and the tvm will not be respected.
RAGMS
CC BY-SA 4.0
2021-07-26T17:46:08.477Z
2021-07-26T19:40:13.610Z
2021-07-26T19:40:13.610Z
Albert Zevelev
<p>Maybe I am completely wrong here (given that I don't see the need to talk about viscosity solutions at all) but in the standard representation theorems you have a terminal/limit condition that the solution to the HJB has to satisfy for it to be the value function. This involves checking, for any admissible control,
<span class="math-container">$$ \lim_{T \to \infty} E[ e^{-\rho T} V(a_T)] = 0 $$</span>
Your affine solution to DE clearly violates this when <span class="math-container">$r = \rho$</span>.</p>

CC BY-SA 4.0
2021-07-23T21:11:29.090Z
I'm not sure your limit condition is enough to pin down the optimal solution.       

Sol 1: $\lim_{T\to \infty} e^{-\rho T} \frac{1}{\rho (\gamma -1)} =0$, but it is not optimal.

Sol 2: $\lim_{T\to \infty} e^{-\rho T} (B_0 + B_1 a_{T}) =a_{0} B_{1}  \lim_{T\to \infty} e^{-(r-\omega)T} = 0$, but it's not optimal
Albert Zevelev
CC BY-SA 4.0
2021-07-23T21:18:51.300Z
I agree that this does not rule out Sol 1 - but for Sol 2, it must hold for ANY admissible control, not just the optimal one. So i can take $c_t = 0$ for all $t$
user36504
CC BY-SA 4.0
2021-07-23T20:22:13.453Z
2021-07-23T20:22:13.453Z
user36504
<p>Consider the deterministic consumption-savings problem:<br />
<span class="math-container">$
V(a_t) 
= 
\underset{c}{\max}
\int_{\tau =t}^{\tau = \infty} e^{-\rho (\tau - t) } u(c_{\tau}) d\tau
$</span>  w/ <span class="math-container">$u(c)=\frac{c^{1-\gamma}-1}{1-\gamma}$</span> and <span class="math-container">$\gamma, \rho &gt;0$</span><br />
s.t.<br />
<span class="math-container">$\frac{da}{d\tau} = \left( r a_{\tau}  - c_{\tau} \right)  $</span><br />
Initial condition: <span class="math-container">$a(0)=a_0$</span> given<br />
Terminal condition:
<span class="math-container">$ \underset{t\to\infty}{\lim} e^{-\rho t} \lambda(t) a_{t} 
=\underset{t\to\infty}{\lim} e^{-\rho t} u'(c_{t}) a_{t}
=\underset{t\to\infty}{\lim} e^{-\rho t} V_{a}(a_{t}) a_{t}
=0$</span></p>
<p>Let's call the optimization problem above &quot;Sequential Problem&quot; <strong>SP</strong> &amp; denote the set of solutions
<span class="math-container">$\text{Sol}(\textbf{SP}):=
\{V(a_t), c(a_t): \text{solve }\textbf{SP} \}$</span>.<br />
Under routine assumptions <span class="math-container">$\text{Sol}(\textbf{SP})$</span> is a set w/ one element.</p>
<p>Using the Hamiltonian we can solve a system of two BV-ODEs in closed form.<br />
Define: <span class="math-container">$\omega \equiv \left(\frac{r-\rho}{\gamma}\right)$</span><br />
<span class="math-container">$c(a_{t}) = \left(r - \omega \right) \times a_{t}$</span><br />
<span class="math-container">$a_{t}=a_{0}e^{\omega t}$</span><br />
<span class="math-container">$\text{TVC: } \underset{t\to\infty}{\lim} e^{-\rho t} u'(c_{t}) a_{t} 
=
\underset{t\to\infty}{\lim} 
e^{-\rho t} \left( (r - \omega)a_{t} \right)^{-\gamma} a_{t}
=
\underset{t\to\infty}{\lim} 
e^{-\rho t} \left( a_{t} \right)^{1-\gamma} 
=
\underset{t\to\infty}{\lim} 
e^{-\rho t} \left( a_{0}e^{\omega t} \right)^{1-\gamma} 
=
\underset{t\to\infty}{\lim} 
e^{-\rho t} e^{(1-\gamma) \omega t}
=
\underset{t\to\infty}{\lim} 
e^{-(\rho - (1-\gamma) \omega ) t}
=0
\Leftrightarrow 
\rho - (1-\gamma) \omega &gt;0
$</span><br />
Note: <span class="math-container">$\rho - (1-\gamma) \omega = r - \omega$</span>
TVC holds <span class="math-container">$\Leftrightarrow \rho - (1-\gamma) r &gt;0$</span><br />
<span class="math-container">$V(a_{t})  = \frac{-1}{(1-\gamma)\rho} + 
\frac{1}{\rho - (1-\gamma)\omega} 
\frac{\left( \left(r - \omega \right) \times a_{t} \right)^{1-\gamma} }{1-\gamma}$</span><br />
Note:
<span class="math-container">$V_{a}=
\frac{\left(r - \omega \right)}{\rho - (1-\gamma)\omega}
 \left( \left(r - \omega \right) \times a_{t} \right)^{-\gamma}
=
\left( \left(r - \omega \right) \times a_{t} \right)^{-\gamma}
\Rightarrow 
c(a_t) =u'^{-1}(V_a)
$</span></p>
<p><strong>HJB</strong>:
<span class="math-container">$\rho V(a_t) = \underset{c}{\max}
\{ u(c_{t}) + V_a \times (r a_{t}  - c_{t}) \}$</span><br />
Denote the set of solutions <span class="math-container">$\text{Sol}(\textbf{HJB})$</span>.<br />
FONC: <span class="math-container">$
u'(c_{t}) - V_a =0
\Rightarrow
c(a_t) = (V_a)^{-\frac{1}{\gamma}}
\Rightarrow V_a &gt; 0
$</span><br />
Plugin:
<span class="math-container">$
\Rightarrow
u(c) - cV_a
=
\frac{\gamma (V_a)^{1-\frac{1}{\gamma}}}{1-\gamma} - \frac{1}{1-\gamma}
$</span><br />
SOSC: <span class="math-container">$u''(c_{t})&lt;0
\Rightarrow u''\left( (V_a)^{-\frac{1}{\gamma}} \right)&lt;0
$</span><br />
We can combine these eqns (maximized HJB &amp; FONC)<br />
<span class="math-container">$
\left[\begin{array}{l}
\rho V(a_{t}) = 
u(c_{t}) + V_{a}(a_{t})\times \left(ra_{t} - c_{t} \right)  
\\
c(a_{t}) = (V_{a}(a_{t}))^{-\frac{1}{\gamma}}
\end{array} \right]
\Leftrightarrow
\left[\begin{array}{l}
\rho V(a_{t}) = 
\frac{\gamma (V_a)^{1-\frac{1}{\gamma}}}{1-\gamma} 
- \frac{1}{1-\gamma}
 + V_a \times r a_{t}
\end{array} \right]
$</span><br />
<strong>DE</strong>:
<span class="math-container">$\rho V(a_t) 
= 
\frac{\gamma (V_a)^{1-\frac{1}{\gamma}}}{1-\gamma} - \frac{1}{1-\gamma}
 + V_a \times r a_{t}   $</span><br />
Denote the set of solutions <span class="math-container">$\text{Sol}(\textbf{DE})$</span>.</p>
<p><strong>DE</strong> is a non-linear equation (first-order ODE) w/ multiple solutions  <span class="math-container">$V(a_t)$</span>.</p>
<ul>
<li>Sol 1: <span class="math-container">$V(a)=\frac{1}{\rho (\gamma-1)}$</span>, solves <strong>DE</strong>, if <span class="math-container">$\gamma&gt;1$</span><br />
Note: while Sol 1 solves <strong>DE</strong> it does not solve the <span class="math-container">$\max$</span> part of <strong>HJB</strong>, e.g. adding <span class="math-container">$V_a &gt;0$</span> to <strong>DE</strong> will rule out Sol 1, but this condition isn't enough to pin down the sol to <strong>SP</strong>.</li>
<li>Sol 2: <span class="math-container">$V(a)=B_0 + B_1 a$</span>, solves <strong>DE</strong>, if <span class="math-container">$r=\rho$</span> and <span class="math-container">$\rho B_0 = \frac{\gamma (B_1)^{1-\frac{1}{\gamma}}}{1-\gamma} - \frac{1}{1-\gamma}$</span>.<br />
Note: condition <span class="math-container">$V_a&gt;0$</span> is satisfied if <span class="math-container">$B_1 &gt;0$</span>.</li>
<li>Sol 3: <a href="https://www.macro.economics.uni-mainz.de/files/2018/11/Waelde2010DeterministicTwinProblem.pdf" rel="nofollow noreferrer">Walde 2010</a> claims this problem also has a strictly convex solution (I think) in a note I don't fully understand</li>
<li>Sol 4: <span class="math-container">$V(a_{t}) = \frac{-1}{(1-\gamma)\rho} + 
\frac{1}{\rho - (1-\gamma)\omega} 
\frac{\left( \left(r - \omega \right) \times a_{t} \right)^{1-\gamma} }{1-\gamma}$</span>, solves <strong>DE</strong></li>
</ul>
<p>We have:
<span class="math-container">$\text{Sol}(\textbf{SP})
\subseteq \text{Sol}(\textbf{HJB})
\subseteq \text{Sol}(\textbf{DE})
$</span></p>
<p><strong>Q1</strong>: what conditions does <span class="math-container">$V(a)$</span> need to satisfy s.t. the solution to <strong>DE</strong> is also the solution to the optimization problem <strong>SP</strong>?</p>
<ul>
<li>My understanding of <a href="https://benjaminmoll.com/wp-content/uploads/2020/02/viscosity_for_dummies.pdf" rel="nofollow noreferrer">viscosity for dummies</a> is that while <strong>DE</strong> has many solutions, the optimal solution is the unique &quot;viscosity solution&quot;</li>
</ul>
<p><strong>Q2</strong> why do the constant &amp; affine solutions above not satisfy the conditions for a viscosity solution?<br />
-the authors of &quot;viscosity for dummies&quot; don't provide a simple example of an HJB w/ multiple closed form solutions &amp; show only the optimal solution satisfies the properties</p>

<optimal-control><continuous-time>
Multiple solutions to an HJB, how to pin down the optimal "viscosity" solution?
Albert Zevelev
# Multiple solutions to an HJB, how to pin down the optimal "viscosity" solution?

(Asked by: username_0 on 2021-07-23)

Consider the deterministic consumption-savings problem:  
$
V(a\_t)
=
\underset{c}{\max}
\int\_{\tau =t}^{\tau = \infty} e^{-\rho (\tau - t) } u(c\_{\tau}) d\tau
$ w/ $u(c)=\frac{c^{1-\gamma}-1}{1-\gamma}$ and $\gamma, \rho >0$  
s.t.  
$\frac{da}{d\tau} = \left( r a\_{\tau} - c\_{\tau} \right) $  
Initial condition: $a(0)=a\_0$ given  
Terminal condition:
$ \underset{t\to\infty}{\lim} e^{-\rho t} \lambda(t) a\_{t}
=\underset{t\to\infty}{\lim} e^{-\rho t} u'(c\_{t}) a\_{t}
=\underset{t\to\infty}{\lim} e^{-\rho t} V\_{a}(a\_{t}) a\_{t}
=0$

Let's call the optimization problem above "Sequential Problem" **SP** & denote the set of solutions
$\text{Sol}(\textbf{SP}):=
\{V(a\_t), c(a\_t): \text{solve }\textbf{SP} \}$.  
Under routine assumptions $\text{Sol}(\textbf{SP})$ is a set w/ one element.

Using the Hamiltonian we can solve a system of two BV-ODEs in closed form.  
Define: $\omega \equiv \left(\frac{r-\rho}{\gamma}\right)$  
$c(a\_{t}) = \left(r - \omega \right) \times a\_{t}$  
$a\_{t}=a\_{0}e^{\omega t}$  
$\text{TVC: } \underset{t\to\infty}{\lim} e^{-\rho t} u'(c\_{t}) a\_{t}
=
\underset{t\to\infty}{\lim}
e^{-\rho t} \left( (r - \omega)a\_{t} \right)^{-\gamma} a\_{t}
=
\underset{t\to\infty}{\lim}
e^{-\rho t} \left( a\_{t} \right)^{1-\gamma}
=
\underset{t\to\infty}{\lim}
e^{-\rho t} \left( a\_{0}e^{\omega t} \right)^{1-\gamma}
=
\underset{t\to\infty}{\lim}
e^{-\rho t} e^{(1-\gamma) \omega t}
=
\underset{t\to\infty}{\lim}
e^{-(\rho - (1-\gamma) \omega ) t}
=0
\Leftrightarrow
\rho - (1-\gamma) \omega >0
$  
Note: $\rho - (1-\gamma) \omega = r - \omega$
TVC holds $\Leftrightarrow \rho - (1-\gamma) r >0$  
$V(a\_{t}) = \frac{-1}{(1-\gamma)\rho} +
\frac{1}{\rho - (1-\gamma)\omega}
\frac{\left( \left(r - \omega \right) \times a\_{t} \right)^{1-\gamma} }{1-\gamma}$  
Note:
$V\_{a}=
\frac{\left(r - \omega \right)}{\rho - (1-\gamma)\omega}
\left( \left(r - \omega \right) \times a\_{t} \right)^{-\gamma}
=
\left( \left(r - \omega \right) \times a\_{t} \right)^{-\gamma}
\Rightarrow
c(a\_t) =u'^{-1}(V\_a)
$

**HJB**:
$\rho V(a\_t) = \underset{c}{\max}
\{ u(c\_{t}) + V\_a \times (r a\_{t} - c\_{t}) \}$  
Denote the set of solutions $\text{Sol}(\textbf{HJB})$.  
FONC: $
u'(c\_{t}) - V\_a =0
\Rightarrow
c(a\_t) = (V\_a)^{-\frac{1}{\gamma}}
\Rightarrow V\_a > 0
$  
Plugin:
$
\Rightarrow
u(c) - cV\_a
=
\frac{\gamma (V\_a)^{1-\frac{1}{\gamma}}}{1-\gamma} - \frac{1}{1-\gamma}
$  
SOSC: $u''(c\_{t})<0
\Rightarrow u''\left( (V\_a)^{-\frac{1}{\gamma}} \right)<0
$  
We can combine these eqns (maximized HJB & FONC)  
$
\left[\begin{array}{l}
\rho V(a\_{t}) =
u(c\_{t}) + V\_{a}(a\_{t})\times \left(ra\_{t} - c\_{t} \right)
\\
c(a\_{t}) = (V\_{a}(a\_{t}))^{-\frac{1}{\gamma}}
\end{array} \right]
\Leftrightarrow
\left[\begin{array}{l}
\rho V(a\_{t}) =
\frac{\gamma (V\_a)^{1-\frac{1}{\gamma}}}{1-\gamma}
- \frac{1}{1-\gamma}
+ V\_a \times r a\_{t}
\end{array} \right]
$  
**DE**:
$\rho V(a\_t)
=
\frac{\gamma (V\_a)^{1-\frac{1}{\gamma}}}{1-\gamma} - \frac{1}{1-\gamma}
+ V\_a \times r a\_{t} $  
Denote the set of solutions $\text{Sol}(\textbf{DE})$.

**DE** is a non-linear equation (first-order ODE) w/ multiple solutions $V(a\_t)$.

* Sol 1: $V(a)=\frac{1}{\rho (\gamma-1)}$, solves **DE**, if $\gamma>1$  
  Note: while Sol 1 solves **DE** it does not solve the $\max$ part of **HJB**, e.g. adding $V\_a >0$ to **DE** will rule out Sol 1, but this condition isn't enough to pin down the sol to **SP**.
* Sol 2: $V(a)=B\_0 + B\_1 a$, solves **DE**, if $r=\rho$ and $\rho B\_0 = \frac{\gamma (B\_1)^{1-\frac{1}{\gamma}}}{1-\gamma} - \frac{1}{1-\gamma}$.  
  Note: condition $V\_a>0$ is satisfied if $B\_1 >0$.
* Sol 3: [Walde 2010](https://www.macro.economics.uni-mainz.de/files/2018/11/Waelde2010DeterministicTwinProblem.pdf) claims this problem also has a strictly convex solution (I think) in a note I don't fully understand
* Sol 4: $V(a\_{t}) = \frac{-1}{(1-\gamma)\rho} +
  \frac{1}{\rho - (1-\gamma)\omega}
  \frac{\left( \left(r - \omega \right) \times a\_{t} \right)^{1-\gamma} }{1-\gamma}$, solves **DE**

We have:
$\text{Sol}(\textbf{SP})
\subseteq \text{Sol}(\textbf{HJB})
\subseteq \text{Sol}(\textbf{DE})
$

**Q1**: what conditions does $V(a)$ need to satisfy s.t. the solution to **DE** is also the solution to the optimization problem **SP**?

* My understanding of [viscosity for dummies](https://benjaminmoll.com/wp-content/uploads/2020/02/viscosity_for_dummies.pdf) is that while **DE** has many solutions, the optimal solution is the unique "viscosity solution"

**Q2** why do the constant & affine solutions above not satisfy the conditions for a viscosity solution?  
-the authors of "viscosity for dummies" don't provide a simple example of an HJB w/ multiple closed form solutions & show only the optimal solution satisfies the properties

**Tags:** optimal-control, continuous-time


---

## 2 answers

### Answer by username_0 on 2021-07-26. Score: 2

Answer to Q1:  
If we re-write FONC as a function of $a$: $u'(c(a))-V\_{a} =0$  
Differentiate wrt $a$ (as in Walde 2010): $u''(c(a)) c'(a) -V\_{aa} =0$  
We know from SOSC that $u''(c(a))<0$.  
If we assume consumption is increasing in wealth $c'(a)>0$, then $V\_{aa}<0$

$
\left[\begin{array}{l}
\rho V(a\_{t}) =
\frac{\gamma (V\_a)^{1-\frac{1}{\gamma}}}{1-\gamma}
- \frac{1}{1-\gamma}
+ V\_a \times r a\_{t}
\\
V\_{a}(a\_{t}) >0 \text{ FONC: for max to be well defined}
\\
V\_{aa}(a\_{t}) <0 \text{ differentiate FONC & SOSC}
\end{array} \right]
$

Some comments:

* I think it's enough for $V\_a>0, V\_{aa}<0$ to hold at a single point
* I believe if we add these two conditions (increasing & concave value) to **DE** we can pin down the unique optimal sol in this non-generic example.  
  I cannot prove there is no other sol to **DE** w/ $V\_a>0, V\_{aa}<0$.  
  I really wanna know how these conditions generalize to more generic econ problems.
* $V\_a>0$ rules out Sol 1
* $V\_{aa}<0$ rules out Sols 2 & 3
* These conditions don't feel as essential/generic as $a(0)=a\_0$ & TVC.
* Since **DE** is a first-order non-linear ode, I believe we either need one equation, such as $V(a\_{1})=V\_1$, or two inequalities ($V\_a>0, V\_{aa}<0$) to pin down a unique sol.  
  (I'm not sure about this. It may only be true for "nice" odes. Can someone who knows help?)  
  Since **DE** is first-order, I don't like relying on a boundary condition w/ a 2nd derivative $V\_{aa}<0$
* [Achdou Hans Lasry Lions Moll](https://benjaminmoll.com/wp-content/uploads/2019/07/HACT.pdf) add a boundary condition that doesn't bind $a\_t \geq \underline{a}$ where $\underline{a}<0$ and use a "state-boundary condition" $V\_a(\underline{a}) \geq u'(r\underline{a})$.  
  Issue: if $\gamma \in (0,1)$ then $u'(r\underline{a})$ may not be a real number.  
  Is one inequality enough theoretically, even if numerically their method converges to the unique sol?  
  Does this rule out the affine solutions above?  
  And how does this generalize to generic problems?
* Most importantly, why do the constant & affine solutions above not satisfy the conditions for a viscosity solution? Why does the optimal sol satisfy them?

> **Comments:**
> **username_1**: Hey username_0. Cannot really provide the necessary and/or sufficient conditions to have a coincidence of solutions but may add something to your own answer. $V_a>0$ should be a given by non-satiation of the utility function and the FOC. The additional condition on $V_{aa}(a)$, pointed out in Walde (2010), should precisely rule out solutions of the HJB that are not solutions to the maximization problem. If for all state variable levels, $V_{aa}(a) < 0$, then $c'(a)<0$ and we can argue by contradiction. (continued)  
> **username_1**: For any initial condition $a_0$, take $a_0' > a_0$. $c_t' = c(a_0') < c_t = c(a_0)$ and so we will have that for $t = \epsilon$, $a_{\epsilon}' > a_{\epsilon}$. The same argument applies for any time period, $t \geq 0$. This means that consumption will be lower at all points in time, $c_t' \leq c_t$. We reached our contradiction. Just take in turn the previous consumption plan - also achievable with the new initial condition. By the monotonicity of the integral, the value function would then be higher. Hence $c_t' \quad \forall t \geq 0$, determined by the proposed solution, cannot be optimal.  
> **username_1**: For now I cannot truly rule out solution 2 since $V_{aa}(a)=0$. If $r=\rho$ and $B_0$ respects the equality imposed, we can then solve for $B_1$ using the boundary conditions. For $B_1$ to respect the terminal condition we must have that $B_1 = (ra_0)^{-\gamma}$. This means that $B_1$ cannot be a constant independent of the boundary conditions. (continued)  
> **username_1**: This conclusion is intuitive if we just look at the FOC which sets $c(a)$ to a constant. Take a starting $a_0 \leq 0$ for example. The $\lim_{t \rightarrow \infty} a_t$ diverges as $\dot{a_t}<0 \forall t \geq 0$ and the tvm is not respected. In fact, as long as $a_0 \neq a*$, for a certain $a*$, the limit will always diverge and the tvm will not be respected.  

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### Answer by username_2 on 2021-07-23. Score: 0

Maybe I am completely wrong here (given that I don't see the need to talk about viscosity solutions at all) but in the standard representation theorems you have a terminal/limit condition that the solution to the HJB has to satisfy for it to be the value function. This involves checking, for any admissible control,
$$ \lim\_{T \to \infty} E[ e^{-\rho T} V(a\_T)] = 0 $$
Your affine solution to DE clearly violates this when $r = \rho$.

> **Comments:**
> **username_0**: I'm not sure your limit condition is enough to pin down the optimal solution.         Sol 1: $\lim_{T\to \infty} e^{-\rho T} \frac{1}{\rho (\gamma -1)} =0$, but it is not optimal.  Sol 2: $\lim_{T\to \infty} e^{-\rho T} (B_0 + B_1 a_{T}) =a_{0} B_{1}  \lim_{T\to \infty} e^{-(r-\omega)T} = 0$, but it's not optimal  
> **username_2**: I agree that this does not rule out Sol 1 - but for Sol 2, it must hold for ANY admissible control, not just the optimal one. So i can take $c_t = 0$ for all $t$  

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