Solve PDE using method of characteristics with non-local boundary conditions. Given the population model by the following linear first order PDE in $u(a,t)$ with constants $b$ and $\mu$ :
$$u_a + u_t = -\mu t u\,\,\,\,\,a,t>0$$
$$u(a,0)=u_0(a)\,\,\,a≥0$$
$$u(0,t)=F(t)=b\int_0^\infty u(a,t)\,\mathrm{d}a$$
We can split the integral in two with our non-local boundary data:
$$F(t)=b\int_0^t u(a,t)\,\mathrm{d}a+b\int_t^\infty u(a,t)\,\mathrm{d}a$$
Choosing the characteristic coordinates $(\xi, \tau)$ and re-arranging the expression to form the normal to the solution surface we have the following equation with initial conditions:
$$\bigl(u_a, u_t, -1\bigr) \bullet  \bigl(1, 1, -\mu t u \bigr)=0$$
$$x(0)=\xi, \,\,\,t(0)= 0,\,\,\, u(0)=u_0(\xi)$$
Characteristic equations:
$$\frac{\mathrm{d}a}{\mathrm{d}\tau}=1, \,\,\,\frac{\mathrm{d}t}{\mathrm{d}\tau}=1, \,\,\,\frac{\mathrm{d}u}{\mathrm{d}\tau}=-\mu tu$$
Solving each of these ODE's in $\tau$ gives the following:
$$(1)\int \mathrm{d}a=\int \mathrm{d}\tau \,\,\,\,\,\,\,\,\,\,(2)\int \mathrm{d}t=\int \mathrm{d}\tau\,\,\,\,\,\,\,\,\,\,(3)\int \mathrm{d}u=-\int \mu tu\,\mathrm{d}\tau$$
$$a = \tau + F(\xi)\,\,\,\,\,\,\,\,\,\,t=\tau + F(\xi)$$
$$\therefore a=\tau + \xi \,\,\,\,\,\,\,\,\,\,\therefore t=\tau$$
$$\int \mathrm{d}u=-\int \mu \tau u\,\mathrm{d}\tau$$
$$\int \frac{1}{u}\,\mathrm{d}u=-\int \mu \tau \,\mathrm{d}\tau$$
$$\ln u = - \frac{1}{2}\mu \tau ^2+ F(\xi)$$
$$u = G(\xi)\,e^{-\frac{1}{2}\mu \tau^2}$$
$$\therefore u = u_0(\xi)\,e^{-\frac{1}{2}\mu \tau^2}$$
Substituting back the original coordinates we can re-write this expression with a coordinate change:
$$\xi = a-t \,\,\,\,\,\,\,\,\,\,\tau = t$$
$$\therefore u(a,t)=u_0(a-t)\,e^{-\frac{1}{2} t^2}$$
Now this is where I get stuck, how do I use the boundary data to come up with a well-posed solution?
$$u(0,t)=u_0(-t)\,e^{-\frac{1}{2}\mu t^2}=b\int_0^t u(a,t)\,\mathrm{d}a+b\int_t^\infty u(a,t)\,\mathrm{d}a$$
Very unsure how I can evaluate this integral...
 A: $$u_a + u_t = -\mu t u$$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{da}{1}=\frac{dt}{1}=\frac{du}{-\mu t u}$$
First characteristic equation, from $\frac{da}{1}=\frac{dt}{1}$ :
$$a-t=c_1$$
Second characteristic equation, from $\frac{dt}{1}=\frac{du}{-\mu t u}$ :
$$ue^{\mu t^2/2}=c_2$$
General solution of the PDE on the form of implicit equation $c_2=\Phi(c_1)$ :
$$ue^{\mu t^2/2}=\Phi(a-t)$$
where $\Phi$ is an arbitrary function, to be determined according to a boundary condition.
$$u(a,t)=e^{-\mu t^2/2}\Phi(a-t)$$
$$ $$
CONDITION : $u(a,0)=u_0(a)=e^{-\mu 0^2/2}\Phi(a-0)=\Phi(a)$
Now the function $\Phi(x)=u_0(x)$ is determined. We put it into the above general solution where $x=a-t$ :
$$\boxed{u(a,t)=e^{-\mu t^2/2}u_0(a-t)}$$
The solution of the PDE fitting to the condition $u(a,0)=u_0(a)$ is fully determined if the function $u_0(a)$ is known. In this case the initial condition $u(0,t)=F(t)$ is superfluous. Moreover, this condition might introduce a contradiction with the condition $u(a,0)=u_0(a)$.
So, if the  two conditions are specified without relationship between them, the problem has no solution in general.
The problem is likely to have a solution if the next relationship was satisfied :
$$u(0,t)=b\int_0^\infty u(a,t)\,\mathrm{d}a= e^{-\mu t^2/2}u_0(0-t)=e^{-\mu t^2/2}u_0(-t)$$
$$b\int_0^\infty e^{-\mu t^2/2}u_0(a-t)\,\mathrm{d}a=e^{-\mu t^2/2}u_0(-t)$$
$$b\int_0^\infty u_0(a-t)\,\mathrm{d}a=u_0(-t)$$
Conclusion :
If $u_0(a)$ is not a function of general form, but is a particular function which satisfies the equation $b\int_0^\infty u_0(a-t)\,\mathrm{d}a=u_0(-t)$ the problem has a solution which is : $u(a,t)=e^{-\mu t^2/2}u_0(a-t)$ .
A: substitute $u(a,t)$ by $u_0(a-t)\exp(...)$ in the integral, and then you may find your way out.
