2
$\begingroup$

An age structured population with distribution $u(a,t)$ over age $a$ has a death rate increasing linearly with time and constant birth rate $b$, $u(\alpha,0) = u_{0}(\alpha)$.

Model is $u_{\alpha} + u_{t} = -\mu t u, t>0$ $u(a,0) = u_{0}(a), a \geq 0$ $u(0,t) = F(t) = b\int_{0}^{\infty} u(a,t) da$(which is the non-local boundary condition), $\mu,b$ are constants.

We need to find the solution $u(\alpha,t)$ for $\alpha > t > 0$ and for $0<a<t$.

Seeing the non-local boundary condition I thought of $F(t) = b\int_{0}^{t} u(a,t)da +b\int_{t}^{\infty} u(a,t) da$ (in case it may be useful)

I am clueless about how to obtain $u(a,t)$.

Thinking of method of characteristics, we have $\frac{d\alpha}{1} = \frac{dt}{1} = \frac{du}{-\mu t}$ and taking last equality we have $u = -\mu t^2 +c$

$\endgroup$
3
$\begingroup$

Since the equation is linear, let's define $$u(\alpha,t)=f(\alpha)g(t)$$then by substituting in the equation we obtain$$f'(\alpha)g(t)+f(\alpha)g'(t)=-\mu t f(\alpha)g(t)$$after dividing by $u(\alpha ,t)$ and solving the outcome ODEs we finally conclude that:$$u(\alpha,t)=Ae^{k(\alpha-t)}e^{-{1\over 2}\mu t^2}$$where $k$ and $A$ are complex constants. Since the equation is linear, then any linear combination of the above answers is also itself an answer, therefore$$u(\alpha,t)=\int_{0}^{\infty}A(\omega)e^{k(\omega)\cdot (\alpha-t)}e^{-{1\over 2}\mu t^2}d\omega$$Now if we decide to have a real answer with the assumption that $u(a,0)$ has a Fourier Transform, then only the values of $k(\omega)$ that are pure imaginary are needed. In that case, $$\Re\{A(-\omega)\}=\Re\{A(\omega)\}\\\Im\{A(-\omega)\}=-\Im\{A(\omega)\}$$Also without loss of generality, we assume $k(\omega)=i\omega$. These constraints assure that our final answer has the following real-valued form:$$u(\alpha,t)=e^{-{1\over 2}\mu t^2}\int_0^\infty C(\omega)\cos \omega (\alpha-t)+D(\omega)\sin \omega (\alpha-t)d\omega$$now if we expand $u(a,0)$ evenly around zero we obtain$$D(\omega)=0\\C(\omega)={2\over \pi}\int_0^\infty u(a,0)\cos \omega ada$$and the final answer becomes:$$u(\alpha,t)=e^{-{1\over 2}\mu t^2}\int_0^\infty C(\omega)\cos \omega (\alpha-t)d\omega$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.