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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...

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2 Answers 2

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$$u_a + u_t = -\mu t u$$ 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 : $u(a,t)=e^{-\mu t^2/2}u_0(a-t)$ .

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  • $\begingroup$ Perfect. Is there any way I can solve that integral though? $\endgroup$ Mar 16, 2019 at 4:19
  • $\begingroup$ This is a very different mathematical problem that the problem raised in your question. This is an integral equation of this kind : Find all functions $y(x)$ such as $$b\int_0^\infty y(z+x)dz=y(x)$$ You should open another question specifically on this subject. Note that $y(x)=c\:e^{-bx}$ is a particular solution. $\endgroup$
    – JJacquelin
    Mar 16, 2019 at 5:55
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substitute $u(a,t)$ by $u_0(a-t)\exp(...)$ in the integral, and then you may find your way out.

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  • $\begingroup$ By a change of variable, $y(x)=b\int_0^\infty y(z+x) dz=b\int_x^\infty y(z)dz$, and then this integral equation is in fact an ode. The solution proposed by JJacquelin is the only kind of solutions. $\endgroup$
    – Bruno
    Apr 20, 2019 at 10:03

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