# PDE with method of characteristics

Let $F(z,t)=\sum_{n=r}^k z^n p_{r,n}(t)$ with $|z|<1$ a generating function of probability. $F$ satisfies the following PDE: $$\frac{\partial }{\partial t}F(z,t) + (z-1)(az+b)\frac{\partial }{\partial z}F(z,t) = (z-1)(ak+\frac{br}{z})F(z,t).$$ Solved with the initial condition $F(z,0)=z^n$, the solution of the PDE is $$F(z,t)=z^r \left \{ \frac{az(1-e^{-(a+b)t)})+ae^{-(a+b)t}+b}{a+b} \right \}^{k-r}$$ I tried to solve using the method of characteristics, obtaining $$\frac{dt}{1} = \frac{dz}{(z-1)(az+b)}=\frac{dF(z,t)}{(z-1)(ak+\frac{br}{z})F(z,t)}$$ but I don't really know how the solution above comes up. Any idea?

Here is a solution that directly uses the method of characteristics.

Before giving the solution, it will help to recall the general strategy.

The PDE describes a surface in $(t,z,F)$-space. The method of characteristics gives differential equations which describe a family of curves, such that this surface is the union of those curves.

Concretely, solving the characteristic equations lets you write $z = z(t)$ and $F = F(t)$, but with unknown constants of integration in each formula. These are curves lying on the solution surface, and you want to try to rework them (and remove the constants of integration!) so that you get a formula for the surface that they lie on. The key to doing this is to use the initial conditions.

Namely, if you plug in $t = 0$, you get a formula for $z(0)$ in terms of one constant of integration, and a formula for $F(0)$ in terms of the other constant of integration. Now, if you have the initial condition, you know the relationship between $F(0)$ and $z(0)$, and so you can solve for the second constant of integration in terms of the first. Also, using the formula for $z(t)$, you can express the first constant of integration in terms of $z$ and $t$. Putting this together, you can express $F$ as a function of $z$ and $t$.

In practice, you can simplify some of these steps. E.g. in your particular example, the first characteristic equation lets us solve for $z$ in terms of $t$, while the second characteristic equation involves just $F$ and $z$, so we can solve for $F$ in terms of $z$. There is then no need to explicitly plug in the formula for $z$ in terms of $t$ to write $F$ in terms of $t$; we could do that, but it would just complicate the algebra. In fact, we also won't explicitly write $z$ in terms of $t$ (because the formula is a bit complicated); instead we will just directly proceed to writing the constant of integration in terms of $t$ and $z$.

But, despite these short cuts, if you look at what I do below, you will see that I am following the general strategy described above.

Now here is the solution:

Solving your first characteristic equation gives

$$A e^{(a+b)t} = 1 - \dfrac{a+b}{az+b},$$ which we rewrite as $$A = \bigl( 1 - \dfrac{a+b}{az+b}\bigr) e^{-(a+b)t}.$$

Solving your second characteristic equation gives

$$F = B z^r (az +b)^{k-r}.$$

(Both these equations are straightforward, if somewhat painful, calculus.)

Here $A$ and $B$ are the constants of integration. According to the general strategy, we have to use the initial condition to eliminate $A$ and $B$ (and hence express $F$ as a function of $t$ and $z$).

The initial condition tells us that at the point on this curve whose $t$-coord. is zero, if we let $z_0$ be the $z$-coord., then the $F$-coord. is $z_0^r$. In other words, the curve contains a point $(0,z_0,z_0^r)$, for some (unknown) value of $z_0$. We use this to eliminate $A$ and $B$. (You'll see that it doesn't matter that we don't know what $z_0$ is; just having its existence is enough to eliminate $A$ and $B$.)

Namely, the first equation, plugging in $t = 0$ and $z = z_0$, gives $$A = 1 - \dfrac{a+b}{az_0 + b},$$ while in the second equation, plugging in $z = z_0$ and $F = z_0^r$ gives $$B(az_0 + b)^{k-r} = 1,$$ or $$B = (a z_0 +b)^{-(k-r)}.$$ Now using our expression for $A$ in terms of $z_0$, we may rewrite our first equation as $$\dfrac{1}{a z_0 + b} = \dfrac{1}{a+b} \Bigl(1 - \bigl(1 - \dfrac{a+b}{a z + b} \bigr) e^{-(a+b)t}\Bigr).$$ Raising this to the $(k-r)$th power, and using our expression for $B$ in terms of $z_0$, we find that $$B = \Bigl( \dfrac{1}{a+b}\Bigl(1 - \bigl(1 - \dfrac{a+b}{a z + b}\bigr) e^{-(a+b)t}\Bigr) \Bigr)^{k-r}.$$ Finally, we can plug this into our second characeristic equation to find that $$F = z^r \Bigl(\dfrac{1}{a+b}\Bigl(az + b - \bigl(az+b - (a+b)\bigr)e^{-(a+b)t} \Bigr)\Bigr)^{k-r}$$ $$= z^r \Bigl(\dfrac{1}{a+b}\bigl( az(1 - e^{-(a+b)t}) + a e^{-(a+b)t} + b\bigr)\Bigr)^{k-r},$$ as desired.

Let's search a particular solution $G(z)$ depending of $z$ only : $$(z-1)(az+b)G'(z) = (z-1)(ak+\frac{br}{z})G(z)$$

(corresponding to second term $=$ third term of your characteristic system) so that $$(\ln(G(z)))'=\frac{ak+\frac{br}{z}}{az+b}$$

I'll let you prove that an integral of the term at the right is : $\displaystyle r\cdot\ln(z)+\ln(z+\frac ba)\cdot (k-r)$

so that a solution depending of $z$ only is given by :

$$r\cdot\ln(z)+\ln(z+\frac ba)\cdot (k-r)= \ln(G(z))$$ or $$G(z)=z^r(z+\frac ba)^{k-r}$$ (multiplied by a constant factor if needed)

Let's solve the 'truncated' equation (rigth term of your PDE replaced by $0$) :

$$\frac{\partial }{\partial t}F(z,t) + (z-1)(az+b)\frac{\partial }{\partial z}F(z,t) =0$$

The characteristic equation for this is : $$\frac{dt}{1} = \frac{dz}{(z-1)(az+b)}$$ (or first term $=$ second term of your characteristic system)

Let's decompose the right member in simple elements to get $\frac{dz}{(z-1)(az+b)}=\frac{dz}{a+b}\left(\frac{1}{z-1}-\frac{a}{az+b}\right)$. We may then integrate both terms to get : $$t= C+\frac{1}{a+b}\left(\ln(z-1)-\ln(z+\frac ba)\right)$$ that we will rewrite (multiplying by $(a+b)$ and after exponentiation) : $$H(z,t)= D\frac{z-1}{z+\frac ba}e^{-(a+b)t}$$

The general solution of the 'truncated' PDE is obtained as $F_0(z,t)=\Psi(H(z,t))$ with $\Psi$ an arbitrary function.

The general solution of your PDE is the product of the general $F_0(z,t)$ and specific $G(z)$ that is : $F(z,t)=G(z)\cdot \Psi(H(z,t))$

$$F(z,t)=z^r(z+\frac ba)^{k-r}\cdot \Psi \left(\frac{z-1}{z+\frac ba}e^{-(a+b)t}\right)$$

We want $F(z,0)=z^r$ ($t=0$) that is : $$\frac{1}{(z+\frac ba)^{k-r}}=\Psi \left(\frac{z-1}{z+\frac ba}\right)\text{ for any }z$$

Using $1-\displaystyle \left(\frac{z-1}{z+\frac ba}\right)=\frac{1+\frac ba}{z+\frac ba}$ we conclude that following function $\Psi$ should work : $$\Psi(u)=\left(\frac{1-u}{1+\frac ba}\right)^{k-r} \text{and}$$

$$F(z,t)=z^r\cdot \left(\frac{(z+\frac ba)-(z-1)e^{-(a+b)t}}{1+\frac ba}\right)^{k-r}$$