System of PDE with integration conditions interior of domain

Question

I am trying to solve this system of pde numerically. I am unable to find any method in Mathematica that can handle this problem. The issue is having this integration condition in the domain. I tried the method of lines but failed because NDSolve does not work with extra equations except initial and boundary conditions. And, I need to solve this for positive $u(t,x)$ and $v(t,x)$.

$\frac{\partial u(t,x)}{\partial t}=\frac{\partial^2 u(t,x)}{\partial x^2}-(\frac{2x+1}{1+x})u(t,x)(1-u(t,x)-v(t,x)); (0,1)$

$\frac{\partial v(t,x)}{\partial t}=\frac{\partial^2 v(t,x)}{\partial x^2}-v(t,x)(1-v(t,x)-u(t,x)); (0,1)$

boundary conditions $\frac{\partial u(t,0)}{\partial x}=0$, $\frac{\partial u(t,1)}{\partial x}=0$, $\frac{\partial v(t,0)}{\partial x}=0$, $\frac{\partial v(t,1)}{\partial x}=0$

initial conditions $u(0,x)=a$ (where a is a parameter), $v(0,x)=0.5$

interior conditions $\int_0^1{u(t^*,x)+v(t^*,x)dx=1}$ for all $t^*\in[0,1]$

This is what I tried:

p[x_]:=(2x+1)/(1+x);

u0 = 0.5 % this should be the parameter a%;

v0 = 0.5;

X = 1;

T = 1;

Tplot = 1;

sol = NDSolve[{D[u[t,x],t]==D[u[t,x],x,x]-p[x]*u[t,x]*(1-u[t,x]-v[t,x]), D[v[t,x],t]==D[v[t,x],x,x]-v[t,x]*(1-u[t,x]-v[t,x]), (D[u[t,x],x]/.x->0) == 0,(D[u[t,x],x]/.x->X) == 0,(D[v[t,x],x]/.x->0) == 0,(D[v[t,x],x]/.x->X) == 0, u[0,x]==u0,v[0,x]==v0},{u,v},{t,0,T},{x,0,X}]

Plot3D[{Evaluate[u[t,x]]/.sol[[1,1]],Evaluate[v[t,x]]/.sol[[1,2]]},{t,0,Tplot},{x,0,X},PlotRange->All,AxesLabel->{"t","x","Sol"}, PlotLegends->{"U","V"}]

Obviously, this is not what I should do. I have not included the integration condition (interior condition) in the code and not implemented a technique to find positive solutions.

Answer 1

If you provide enough initial and boundary conditions for a PDE, then there are various theorems proving that the solution is uniquely determined. For your equations, a solution is uniquely determined if you provide two initial conditions (since there are two functions and the highest derivatives in $t$ are first-order) and four boundary conditions (since there are two functions and the highest derivatives in $x$ are second order.)

You have provided a sufficient number of initial and boundary conditions, so there will be one unique solution without imposing the "interior condition." If this unique solution satisfies your "interior condition", great! But NDSolve will generally find (a numerical approximation to) this unique solution without your needing to impose this extra condition by hand. If, on the other hand, the unique solution does not satisfy the "interior condition", then there are simply no functions that satisfy all of the equations you have imposed, and you must weep and wail and gnash your teeth.

So in principle, you could just run NDSolve without imposing these conditions and then check whether the interior condition holds. In this case, you would run the code:

intcond[t_] := Integrate[(u[x, t] + v[x, t]) /. First[sol], {x, 0, 1}] - 1
Plot[intcond[t], {t, 0, 1}]

If the interior condition is satisfied at all $t$, then this plot should be close to 0. And it is!

In fact, the solution to your equations without imposing the interior condition is $u(x,t) = v(x,t) = \frac12$ for all $x$ and $t$; and NDSolve returns (a numerical approximation to) this solution when you run your code. It is fairly obvious to see that it satisfies the boundary conditions.

However, for a general set of initial conditions $u(x,0) = u_0(x)$ & $v(x,0) = v_0(x)$, it is unlikely that this interior condition will be satisfied. In particular, if the interior condition holds for all $t$, then we have \begin{align} \frac{d}{dt} \left[ \int_0^1 (u+v) dx \right] &= 0 \\ \int_0^1 (\partial_t u+ \partial_t v) dx &= 0\\ \int_0^1 \left[ \partial_{xx} u + \partial_{xx} v - \frac{2x+1}{1+x} u(1 - u - v) - v (1 - u - v) \right] dx &= 0\\ - \int_0^1 \left[ \left( \frac{2x+1}{1+x} u + v \right)(1 - u - v) \right] dx &= 0 \end{align} (In the third step I have applied the ODEs; in the fourth step I have integrated the first two terms, which then vanish because $\partial_x u$ and $\partial_x v$ vanish on the boundaries.) It is unlikely that an arbitrary pair of functions $u_0$ and $v_0$ will satisfy this condition. However, I think that if they do satisfy this condition at $t = 0$ then the solution will automatically satisfy the interior condition for all $t \in [0,1]$.