# Finding a solution to a PDE

EDIT: This is for a production scheduling problem with quadratic production and linear inventory costs.

The goal is to \begin{equation*} \max_{u} \int_{0}^{T} -(c_{1}u^{2} + c_{2}y) dt \end{equation*} subject to \begin{align*} y' &= u,\\ y(0) &= 0,\\ y(T) &=B.\\ \end{align*}

So the Bellman equation is: \begin{equation*} \max_{u} {-c_{1}u^{2} + c_{2}y + \frac{\partial V}{\partial y}u + \frac{\partial V}{\partial y}} =0. \end{equation*}

By first-order condition, we obtain $u^{*} = \frac{1}{2c_{1}}\left ( \frac{\partial V}{\partial y} \right )$.

We insert this $u^{*}$ into the Bellman equation to obtain:

$$0 = -c_{1}\left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y} \right )^{2} + \frac{\partial V}{\partial y} \left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y} \right ) + \frac{\partial V}{\partial t}$$

And then I obtain the following PDE: $$0 = \frac{1}{4c_{1}}\left (\frac{\partial V}{\partial y} \right )^{2} + \frac{\partial V}{\partial t}$$

I have to solve for $V(y,t)$.

My work: $V = e^{ay + bt}$.

Then I find $V_{y} = ae^{ay+bt}$, and $V_{t} = be^{ay+bt}$.

Then I put it values of $V_{y}$ and $V_{t}$ in the PDE to solve for $b$:

$$\frac{1}{4c_{1}}(ae^{ay+bt})^{2} + be^{ay+bt} = 0$$

$$\Rightarrow a^{2}e^{2(ay+bt)} + be^{ay+bt} = 0$$

$$\Rightarrow a^{2}e^{2} e^{(ay+bt)} = -be^{ay+bt}$$

Assuming that $e^{ay+bt} \neq 0$, we obtain $b = -a^{2}e^{2}$.

Now, substitute the value of $b$ back in the equation for $V$, to obtain $$V(y,t) = e^{ay - a^{2} e^{2}t} = e^{a(y-aAt)},$$ where $A = e^{2}$.

Does this look correct?

Also, the boundary conditions are given with $y(0)=0$, and $y(T)=B$.

• You made a mistake when you thought $$\mathrm{e}^{2x} = \mathrm{e}^{2}\mathrm{e}^x$$ which it doesn't. Commented Apr 13, 2015 at 6:29

The solution which is just a little better than trivial is $$V = \alpha y + \beta t$$ This leads to $$\frac{\alpha^2}{4c_1} + \beta = 0$$ The boundary conditions do not make sense to me. As I was expecting $V(\text{some t},\text{some y})$?

• Hello, I have made an edit so that you can understand the context.
– OGC
Commented Apr 13, 2015 at 23:50

$\dfrac{\partial V}{\partial t}+\dfrac{1}{4c_1}\left(\dfrac{\partial V}{\partial y}\right)^2=0$

$4c_1\dfrac{\partial V}{\partial t}+\left(\dfrac{\partial V}{\partial y}\right)^2=0$

$4c_1\dfrac{\partial^2V}{\partial y\partial t}+2\dfrac{\partial V}{\partial y}\dfrac{\partial^2V}{\partial y^2}=0$

$2c_1\dfrac{\partial^2V}{\partial y\partial t}+\dfrac{\partial V}{\partial y}\dfrac{\partial^2V}{\partial y^2}=0$

Let $U=\dfrac{\partial V}{\partial y}$ ,

Then $2c_1\dfrac{\partial U}{\partial t}+U\dfrac{\partial U}{\partial y}=0$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=2c_1$ , letting $t(0)=0$ , we have $t=2c_1s$

$\dfrac{dU}{ds}=0$ , letting $U(0)=U_0$ , we have $U=U_0$

$\dfrac{dy}{ds}=U=U_0$ , letting $y(0)=f(U_0)$ , we have $y=f(U_0)+U_0s=f(U)+\dfrac{Ut}{2c_1}$ , i.e. $U=F(2c_1y-Ut)$

• Did you get to line 3 by differentiating wrt y?
– OGC
Commented Apr 14, 2015 at 0:44
• Also, I am not familiar with this method. Is it called the method of advection?
– OGC
Commented Apr 14, 2015 at 1:13