Non-linear Schrodinger Equation

One form of the nonlinear Schrodinger equation is

$i\partial_tu(x,t)+\Delta u(x,t)+\epsilon u(x,t)|u(x,t)|^{p-1}=0$

where $u:\mathbb{R}^d\times\mathbb{R}\rightarrow\mathbb{C}$ is a continuous function whose regularity is to be further clarified, and $\epsilon=1, -1$. And $p$ is a positive real number greater than $1$. This equation describes a Hamiltonian system, and if $\epsilon=1$, the force in the system is attractive, while if $\epsilon=-1$, the force is repulsive. The linear terms in this equation describes the free propagation of a wave(without any external forces), while the nonlinear term is in fact the potential of the system. This post is to show the existence of some particular solutions to this equation and their stability. In this post, we consider only the case $\epsilon=1$. The other case is not so similar, and we don’t consider it here.

The first simplification is to just consider the soliton waves. Here a soliton wave is a stationary wave, which means we can write it as $u(x,t)=Q(x)e^{ict}$. And then we hope that, just like in the situation of linear PDE, a general solution can be approximated by these solitons. One great characteristic of the soliton solution is that it conserves energy during the propagation. After this simplification, the original equation can be written as $-Q+\Delta Q+Q|Q|^{p-1}$.

The simplest case for this equation is perhaps when $d=1$, the one-dimensional case. In this case, we get that $Q''-Q+Q|Q|^{p-1}=0$. One routine way to do this is multiply both sides by $Q'$, but in this way it is not easy to handle the term $|Q|^{p-1}$. So, we first assume that $Q$ is a real-valued function.We get that $(Q'^2/2-Q^2/2+|Q|^{p+1}/(p+1))'=0$. In physics, we call the term in the bracket the first integral, which is a constant, giving a conservative quantity of the system. Thus we get that $Q'^2/2-Q^2/2+|Q|^{p+1}/(p+1)=const=E$, where $E$ denote, in some sense, the energy of the system(this is possible since this is a Hamiltonian system). Then the next step is perhaps to look at the phase portrait of this equation(the evolution of the system in the phase space, here its phase space is of dimension $2$, just $Q',Q$). There are some obvious solution to this equation, the first ones are $Q(x)=0, Q(x)=1, Q(x)=-1$. We can see that there is an orbit that tend to the origin as $|x|\rightarrow\infty$. In fact, this is the only possible solution which belongs to $L^2(\mathbb{R})$. Indeed, using the first integral, we can find the solution

$Q(x)=(\frac{p+1}{2cosh^2(x(p-1)/2)})^{1/(p-1)}=0$

Of course, after a translation transformation, $Q(x)$ is still a solution. But all solutions lying in $L^2$ is a translation of this $Q(x)$. Note that this solution is always positive, and decays exponentially to $0$ as $|x|\rightarrow\infty$.

So, for the case of dimension $1$, things are not so hard. We have used the phase space to give us some intuition, and then use the first integral to get the solution we are after.

Now turn to high dimensions. Can we use the same method as in dimension $1$ in this situation? What are the functions on $\mathbb{R}^d$ that a very simple? Considering the Laplacian in the Schrodinger equation(which is spherical-invariant), perhaps we should first give a try on the spherical symmetric functions, that is $f:\mathbb{R}^d\rightarrow\mathbb{R}, (x_1,x_2,...,x_d)\mapsto f(r)$ where $r^2=x_1^2+...+x_d^2$. It is easy to check that $\Delta Q(x_1,...,x_d)=\frac{d-1}{r}Q'+Q''$. Plugging this into the original equation, we get that

$Q''+\frac{d-1}{r}Q'-Q+Q|Q|^{p-1}=0$.

So things return to one dimension, but this time with an extra term, $\frac{d-1}{r}Q'$. Note that since $Q$ is spherical symmetric, we have that $Q(r)'|_{r=0}=0$. This time the equation is more difficult, because it is no longer Hamiltonian, the added term shows that there is some non-conservative force in system, thus it is not so easy to find a first integral. But we still want to know if there is a positive solution which decays exponentially.

One method to solve the problem is to use the variational method. It is really not obvious which minimization problem to consider, nor which function space to consider. Here we give directly the theorem

Suppose that $d>1, M>0,1. We set $H^1_r=H^1_r(\mathbb{R}^d)=\{f\in H^1(\mathbb{R}^d)|f(x_1,...,x_d)=f(r)\}$, $A_M=\{f\in H^1_r|\int_{\mathbb{R}^d}|f|^{p+1}=M\}$, and $I_M=\inf_{f\in A_M}||f||_{H^1}$. Then, there exists $latex f_M\in A_M$ such that $||f_M||_{H^1}=I_M$; what is more, every such $f_M$ for fixed $M$, it can be written as $f_M=e^{ia}(\frac{M}{I_M})^{1/(p-1)}Q(r)$ where $Q(r)$ a spherical symmetric function on $\mathbb{R}^d$ satisfies $\Delta Q-Q+Q^p=0$, $Q(r)>0$ and $Q\in H^1_r$.

(The range of $p$ is not very strange once we note that $2, the range that appear many times in Sobolev embeddings.)To show the existence of a minimal element in $A_M$, we need a lemma concerning the compact injection of $H^1_r$ into $L^p$.

The injection $H^1_r(\mathbb{R}^r)\rightarrow L^p(\mathbb{R}^d)$ is compact with $2.

In fact, for any smooth $f\in C^{\infty}_c(\mathbb{R}^d)\bigcap H^1_r$, we have that $f(r)^2=-\int_r^{\infty}df(s)^2$. Note that for a spherical symmetric function $f$, we have that $\int_{\mathbb{R}^d}f(x)dx=c_d\int_0^{\infty}f(r)r^{d-1}dr$, thus $||f||_{L^2}=c_d\int_0^{\infty}|f(s)|^2r^{d-1}dr$. So, we have that $f(r)^2=-2\int_r^{\infty}\frac{f(s)s^{(d-1)/2}f'(s)s^{(d-1)/2}}{s^{d-1}}ds\leq \frac{1}{r^{d-1}}\sqrt{\int s^{d-1}|f(s)|^2ds\int s^{d-1} |f(s)'|^2ds}$ $\leq 2\frac{1}{r^{d-1}}||f||_{L^2}||f'||_{L^2}\leq \frac{1}{r^{d-1}}||f||_{H^1}$. So, we have that, for a regular spherical symmetric function, the value at the point $|x|=r$ decreases polynomially. So, if a sequence $f_n$ converges weakly to $f$ in $H^1_r$, we see that the map $H^1_r\rightarrow L^p(\{x||x| is compact(using the norm $||f'||_{L^p}$ to control that of $f$ and then Arzela-Ascoli theorem). So, for fixed $R>0$, there is a subsequence $f_{\phi(n)}$ which converges to $f$ under the norm $||g||_{L^p(|x|. Then for the remaining region, $|x|>R$, we have that $\int_{|x|>R} |g|^p\leq ||g||_{L^{\infty}(|x|>R)}^{p-2}\int |g|^2\leq (||g||_{H^1}/R^{d-1})^{(p-2)/2}||g||_{H^1}^2$. So this show that for the $H^1$-bounded sequence $f_{\phi(n)}$, we have that $||f_{\phi(n)}-f||_{L^p(|x>R|)}$ goes to zero as $R\rightarrow \infty$. Then use the diagonal method, we can find a subsequence $f_{h(n)}$ which converges to $f$ under the norm $L^p$.

Now return to the existence of a minimal element in $A_M$. Suppose that $f_n\in A_M$ such that $||f_n||_{H^1}\rightarrow I_M\geq0$. Thus, this sequence is $H^1$-bounded, so according to the weak compacity of the unit ball in a Hilbert space, we can find a subsequence $f_{\phi(n)}$ which converge weakly to $f\in H^1_r$. Using the above lemma, we can again find a subsequence $f_{\alpha(n)}$ which converges to $f$ under the norm $L^p$. Note that, this means $||f||_{L^{p+1}}=M^{1/(p+1)}>0$, thus $||f||_{H^1}>0$. But we have that $||f||_{H^1}\leq \inf\lim ||f_{\alpha(n)}||=I_M$, thus $||f||_{H^1}=I_M$(the last inequality can be seen by observing that, if in a Hilbert space $H$ with an scalar product $(,)$, a sequence $a_n$ converges weakly to $a$, then $(a_n-a,a_n-a)=(a_n,a_n)+(a,a)-2(a_n,a)$, so $2(a_n,a)-(a,a)=(a_n,a_n)-(a_n-a,a_n-a)\leq (a_n,a_n)$. Taking the $\inf\lim$, we have that $(a,a)=\inf\lim 2(a_n,a)-(a,a)\leq \inf\lim (a_n,a_n)$). A very important step in the above argument is to show that the minimal element is not trivial, or it belongs $A_M$ which is equivalent to its existence.

The next step, which is very natural, is to study these minimal elements in $A_M$. An observation is that,

If $f\in A_M$ such that $||f||_{H^1}=I_M$, then so is $|f|$

We know that $||f||_{L^2}=|||f|||_{L^2}$. Moreover, for a regular function $f$, we have that $|\partial_{x_i}f|\geq |\partial_{x_i}|f||$(for the case $f(P)\neq0(P\in\mathbb{R}^d)$, just write the real part and the imaginary part of $f$; for the case $f(P)=0$, just use the definition of $\partial_{x_i}f(0)=\lim_{t\rightarrow 0}\frac{f(te_i)}{t}$).

So things are simplified again, this time we can consider only those non-negative minimal elements in $A_M$. Note that this minimization problem is under constraint, which reminds us to use the Lagrange multiplicator method.

First, if $f\in A_M$ is a non-negative minimal element, then for any $h\in H^1_r$, we want to construct a family of elements $f_t\in A_M$. Note that any function on $f,h,t$ is again spherical symmetric. First consider $f+th$, but its $L^{p+1}$-norm is not $M^{1/(p+1)}$, We can divide it by its norm, $\frac{f+th}{||f+th||_{L^{p+1}}}$. This time, the $L^{p+1}$-norm is $1$, so we can set $f_t=\frac{M^{1/(p+1)}}{||u+th||_{L^{p+1}}}(u+th)$. Now we set $t<<1$, then we have that $f_t=(f+th)(1-\frac{t}{M}\int |f|^ph)=f+t(h-c(f,h))$(where $c(f,h)=\int |f|^ph$).Now consider the $H^1$-norm, $||f_t||^2_{H^1}=||f||_{H^1}^2+2t\int f(h-c(f,h))+\nabla f\nabla (h-c(f,h))$. Here there is a little problem: the constant function $c(f,h)$ doesn’t belong to any $A_M,H^1,L^p(p<\infty)$, so perhaps we can first choose those $h\in H^1_r$ such that $c(f,h)=0$. If we have done so, then since $||f_t||_{H^1}\geq I_M=||f||_{H^1}$, so we must have that $0=\int fh+\nabla f\nabla h=\int (f-\Delta f)h$. So, for two linear functionals $L_1:H^1_r\rightarrow \mathbb{C}, h\mapsto _{L^2}$, $L_2:H^1_r\rightarrow\mathbb{C}, h\mapsto <|f|^p,h>_{L^2}$, we have that if $L_2(h)=0$, then $L_1(h)=0$, thus $ker(L_2)\subset ker(L_1)$. We can show that these two functionals are continuous(the continuity of $L_1$ is easy, but the second is not so obvious), thus we have that $(ker(L_1))^{\perp}\subset(ker(L_2))^{\perp}$, since they are not trivial, so their co-dimensions are both equal to $1$, thus coincide, and as a result they differ by a scalar, that is $L_1=\lambda L_2$. This show that $f-\Delta f=\lambda |f|^p$. Note that, if we integrate both side with respect to $f$, we get that $\int (f-\Delta f)f=\lambda\int f^{p+1}$(since $f$ is non-negative), that is to say $||f||_{H^1}=\lambda ||f||_{L^{p+1}}^{p+1}$, hence $\lambda=I_M/M$. So after a scalar transformation, we proved that there is a non-negative spherical symmetric solution to the non-linear Schrodinger equation

$-f+\Delta f+f^p=0$.

Then the problem is, is this solution unique? Not exactly, but if we pose some condition, we can ensure that it is the case, as the following result(Kwong,1987) shows

There is an unique solution to the equation, $-u+\Delta u+Q^p=0$ such that $u\in H^1_r$ with $u(r)>0(\forall r\geq0)$.

So, the effective condition is that $u$ never takes the zero value.

Now look at another minimal element $v$ in $A_M$. Suppose that $v$ never takes the value $0$. Then, since $||v||_{H^1}=|||v|||_{H^1}=I_M$, we must have that $v=e^{i\alpha}|v|$ due to the fact that $v$ is never $0$. Now that $|v|$ is positive, a minimal element in $A_M$, thus according to the above result, after a scalar transformation, $|v|$ must be equal to that unique solution. So, in this way, we proved the whole theorem, with the exception that we accepted the theorem of Kwong, which is too difficult.

The condition that $u$ is never zero sounds familiar. Right, in quantum mechanics, in particular the one dimensional infinite potential problem, the ground state wave function is never zero, either. In some sense, this $u$ behaves like the ground state of the system, any other reasonable state must have some ‘energy’ higher than this one. The difference between these two concepts is that the state $u$ is not stable, while the ground state in quantum mechanics is usually stable. This shows that, $u$ is not really a state of the system, but only a phase transition, a critical state.