1535-Damped solution.sagews

var('beta, rho, xi, mu, nu, sigma_b, sigma_m, sigma_n, K, L, Q');
M(mu, sigma, t) = exp(mu*t+sigma^2*t^2/2);
ssc=-M(beta,sigma_b,-xi)+M(beta,sigma_b,rho-1)*((K+L)*(rho+xi)-L*(1-M(nu,sigma_n,-(rho+xi)))/(1-M(nu,sigma_n,-1))+L*M(mu,sigma_m,1-(rho+xi)))+Q^2/2*(rho+xi)*(rho+xi-1)*M(beta,sigma_b,rho-2);
ssc

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{2} \, {\left(\rho + \xi - 1\right)} {\left(\rho + \xi\right)} Q^{2} e^{\left(\frac{1}{2} \, {\left(\rho - 2\right)}^{2} \sigma_{b}^{2} + {\left(\rho - 2\right)} \beta\right)} + {\left({\left(\rho + \xi\right)} {\left(K + L\right)} + L e^{\left(\frac{1}{2} \, {\left(\rho + \xi - 1\right)}^{2} \sigma_{m}^{2} - {\left(\rho + \xi - 1\right)} \mu\right)} - \frac{{\left(e^{\left(\frac{1}{2} \, {\left(\rho + \xi\right)}^{2} \sigma_{n}^{2} - {\left(\rho + \xi\right)} \nu\right)} - 1\right)} L}{{\left(e^{\left(\frac{1}{2} \, \sigma_{n}^{2} - \nu\right)} - 1\right)}}\right)} e^{\left(\frac{1}{2} \, {\left(\rho - 1\right)}^{2} \sigma_{b}^{2} + {\left(\rho - 1\right)} \beta\right)} - e^{\left(\frac{1}{2} \, \sigma_{b}^{2} \xi^{2} - \beta \xi\right)}

@interact
def rp(beta=5, sigma_b=1, mu=15, nu=0.3, K=0.1, L=0.1, Q=0.9):
    p = plot(ssc(xi=0.0, beta=beta, sigma_b=sigma_b, nu=nu, sigma_n=nu/5, mu=mu, sigma_m=mu/5, K=K, L=L, Q=Q), (rho,0,2));
    p.show(ymin=-1, ymax=1)

p = plot([ssc(xi=0.25, beta=bn, sigma_b=1, K=0.1, L=0.1, nu=0.3, sigma_n=0.06, mu=15, sigma_m=3, Q=0.9) for bn in [1,3,5,8,12]], (rho,0,2));
p.show(ymin=-1, ymax=1)

p = plot([ssc(xi=0.25, beta=5, sigma_b=1, K=0.1, L=0.1, nu=0.3, sigma_n=0.06, mu=mn, sigma_m=3, Q=0.9) for mn in [5,10,15,20]], (rho,0,2));
p.show(ymin=-1, ymax=1)

p = plot([ssc(xi=0.25, beta=5, sigma_b=1, K=0.1, L=0.1, nu=nn, sigma_n=0.06, mu=15, sigma_m=3, Q=0.9) for nn in [0.1,0.3,0.6,0.9,1.2,2]], (rho,0,2));
p.show(ymin=-1, ymax=1)

var('n, k');
R(n) = beta*(rho - n) + (rho - n)^2*sigma_b^2/2;
F(n) = k*(beta + (rho - n) * sigma_b^2);
ls = exp(-sigma_b^2*k^2/2)*(-cos(k *beta) + Q*exp(R(1))*(rho*cos(F(1)) - k*sin(F(1))) + Q^2/2*exp(R(2))* ((rho*(rho - 1) - k^2)*cos(F(2)) - k*(2*rho - 1)*sin(F(2)))) - k^2*Q^2/2*exp(R(2));
lt = L*exp(R(1))*((1 - exp(-nu + sigma_n^2/2))^(-1)*exp(-rho*nu + rho^2*sigma_n^2/2)*(cos(k*(nu - rho*sigma_n^2))* exp(-k^2*sigma_n^2/2) - 1) + exp((1 - rho)*mu + (1 -rho)^2*sigma_m^2/2)*(cos(k*(mu + (1 - rho)*sigma_m^2))*exp(-k^2 *sigma_m^2/2) - 1));
ll = ls+lt;

def spectrumu(beta=5, sigma_b=1, K=0.1, L=0.05, nu=0.5, sigma_n=0.1, mu=15, sigma_m=3, Q=0.7):
    rhou=find_root(ssc(xi=0, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q)==0, 0.9,1.5);
    print "rho = ", rhou ;
    u(k)=ll(xi=xi, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q, rho=rhou);
    return plot(u(k),(k,0,6))

def spectrum(beta=5, sigma_b=1, K=0.1, L=0.05, nu=0.5, sigma_n=0.1, mu=15, sigma_m=3, Q=0.7):
    try:
        rhol=find_root(ssc(xi=0, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q)==0, 0,1)
        rhou=find_root(ssc(xi=0, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q)==0, 0.9,2)
        print "rho = ", rhol, " , ",rhou 
        l(k)=ll(xi=xi, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q, rho=rhol)
        u(k)=ll(xi=xi, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q, rho=rhou)
        return plot([l(k),u(k)],(k,0,6))
    except RuntimeError:
        rhou=find_root(ssc(xi=0, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q)==0, 0,2);
        print "rho = ", rhou ;
        u(k)=ll(xi=xi, beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q, rho=rhou);
        plot(u(k),(k,0,6)).show()
    p = plot(ssc(xi=0, beta=beta, sigma_b=sigma_b, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, K=K, L=L, Q=Q), (rho,0,2));
    p.show(ymin=-1, ymax=1)

spectrum(beta=5, sigma_b=1, K=0.1, L=0.01, nu=0.5, sigma_n=0.1, mu=15, sigma_m=3, Q=0.9)

rho = 0.614660720798 , 1.37804000962

@interact
def _(beta=5, sigma_b=1, K=0.1, L=0.05, nu=0.5, sigma_n=0.1, mu=15, sigma_m=3, Q=0.9):
    show(spectrum(beta=beta, sigma_b=sigma_b, K=K, L=L, nu=nu, sigma_n=sigma_n, mu=mu, sigma_m=sigma_m, Q=Q))

Damped solution

We will now look at solutions that are scale-invariant but not time-independent. They take the form

$$u(x,t)=e^{-\rho x} f(x-\ln(t)/\xi)$$

for arbitrary function $f$. Substituting this Ansatz into the jump-growth equation (predation only to keep equations short) gives

$$\frac{e^{\xi r}}{\xi}\frac{df(r)}{dr}=\int dy\,s(y)\Big(e^{\rho y}f(r)f(r-y)+e^{-\xi y}f(r)f(r+y)-\int dz\,d(y,z)e^{\rho(y+z)+\xi z}f(r-z)f(r-y-z)\Big).$$

This is now an ordinary integro-differential equation for $f(r)$, but still difficult to solve in general. But a particular solution is given by

$$f(r)=f_0 e^{\xi r}.$$

This is a solution provided

$$1/f_0=\int dy\,s(y)\Big(e^{(\rho-\xi)y}+1-\int dz\,d(y,z)e^{(\rho-\xi)(y+z)}\Big).$$

In terms of $u$ this solution reads

$$u(x,t)=f_0e^{-(\rho-\xi)x}/t.$$

Note how the exponents $\rho$ and $\xi$ only appear in the combination $\hat{\rho}=\rho-\xi$.

To get a feel for the prefactor $f_0$ we can use $s(y)=s_0g_{\beta}(y)$ where $g_\beta$ is the Gaussian with mean $\beta$ and variance $\sigma_\beta^2$. We will also use $d(y,z)=\delta(z-\psi(y))$ where $\psi(y)=\log(1+Qe^{-y})$ and then we will expand to second order in $Qe^{-y}$. We can then write the equation for $f_0$ in terms of the moment generating function $M_\beta(a)=\exp(a\beta+a^2\sigma_\beta^2/2)$. This gives

$$1/f_0=s_0\Big(1-\hat{\rho}QM_\beta(\hat{\rho}-1)-\frac{\hat{\rho}(\hat{\rho}-1)}{2}Q^2M_\beta(\hat{\rho}-2)\Big).$$

This shows that, for $\hat{\rho}$ not too far from $1$,  $f_0$ will be positive, as required.