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 $:

$${u}_{a}+{u}_{t}=-\mu tu\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}a,t>0$$

$$u(a,0)={u}_{0}(a)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}a\ge 0$$

$$u(0,t)=F(t)=b{\int}_{0}^{\mathrm{\infty}}u(a,t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}a$$

We can split the integral in two with our non-local boundary data:

$$F(t)=b{\int}_{0}^{t}u(a,t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}a+b{\int}_{t}^{\mathrm{\infty}}u(a,t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}a$$

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:

$$(}{u}_{a},{u}_{t},-1{\textstyle )}\bullet {\textstyle (}1,1,-\mu tu{\textstyle )}=0$$

$$x(0)=\xi ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t(0)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}u(0)={u}_{0}(\xi )$$

Characteristic equations:

$$\frac{\mathrm{d}a}{\mathrm{d}\tau}=1,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}t}{\mathrm{d}\tau}=1,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}u}{\mathrm{d}\tau}=-\mu tu$$

Solving each of these ODE's in $\tau $ gives the following:

$$(1)\int \mathrm{d}a=\int \mathrm{d}\tau \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(2)\int \mathrm{d}t=\int \mathrm{d}\tau \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(3)\int \mathrm{d}u=-\int \mu tu\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau $$

$$a=\tau +F(\xi )\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t=\tau +F(\xi )$$

$$\therefore a=\tau +\xi \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\therefore t=\tau $$

$$\int \mathrm{d}u=-\int \mu \tau u\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau $$

$$\int \frac{1}{u}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}u=-\int \mu \tau \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau $$

$$\mathrm{ln}u=-\frac{1}{2}\mu {\tau}^{2}+F(\xi )$$

$$u=G(\xi )\phantom{\rule{thinmathspace}{0ex}}{e}^{-\frac{1}{2}\mu {\tau}^{2}}$$

$$\therefore u={u}_{0}(\xi )\phantom{\rule{thinmathspace}{0ex}}{e}^{-\frac{1}{2}\mu {\tau}^{2}}$$

Substituting back the original coordinates we can re-write this expression with a coordinate change:

$$\xi =a-t\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\tau =t$$

$$\therefore u(a,t)={u}_{0}(a-t)\phantom{\rule{thinmathspace}{0ex}}{e}^{-\frac{1}{2}{t}^{2}}$$

Now this is where I get stuck, how do I use the boundary data to come up with a well-posed solution?

$$u(0,t)={u}_{0}(-t)\phantom{\rule{thinmathspace}{0ex}}{e}^{-\frac{1}{2}\mu {t}^{2}}=b{\int}_{0}^{t}u(a,t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}a+b{\int}_{t}^{\mathrm{\infty}}u(a,t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}a$$