This post is about the existence and uniqueness of the solution to the linear Schrodinger equation. That is to say, $latex i\partial_{t}u+\Delta u=0, (t,x)\in\mathbb{R}\times\mathbb{R}^{d}, u(t,x)\in\mathbb{C}$ with the initial condition The main tool is due to Strichartz, in particular his norms. This post, serves to be an introduction to this method in using the linear Schrodinger equation.

First of all, since this equation is linear, the first idea comes to mind is to use Fourier transforms with respect to . In fact, there is a group structure in the evolution process. That is what we are going to do right now.

Take the Fourier transform of the above Schrodinger equation, then we obtain that

where we denote the Fourier transform of . So, we get that . We solve this ordinary equation very easily, that is . Then we take the inverse Fourier transform, we obtain that where the Dirac distribution at the origin. The group structure of this evolution can be seen very easily using the Fourier transform. The real problem is the regularity.

This is not very hard. Note that, we have $latex |\hat{u} (t,\xi )|=|\overline{u_{0}}(\xi)|$. Have we seen something like this before which concerns the norm of the Fourier transform of a function? Yes, in the Sobolev spaces. In deed, if we suppose that , then we see easily that due to the above comment for fixed time . Note that here we do not require that the time be non-negative. This also guarantees that the evolution process hasa group structure. Now we can have a closer look at the operator . We have seen that it is in fact an evolution operator. This said, we have that if there are two initial conditions, , then using the norm for the Hilbert space , we have that . Using the group structure, we see that the conjugate of is just . All the way around, we have not seen any explicite formulation of . In fact, we have seen that . In general, we can not calculate the inverse Fourier transform of directly, since it doesn’t lie in . So, we must solve this problem in the sense of distribution. We can construct a sequenceof Gaussian-like functions .This has an inverse Fourier transform. And, this sequence converges to in the topology of (that is to say, for any , we have due to the dominant convergence theorem).After some calculations, we get that . What is more, we verify easily that as , we have that in the topology of . This, combined with the comments above about the isometricity of , shows that the application is continuous. This result is very natural, and in some sense it has to be so, since we aresolving a differential equation, so the solution must be continuous with respect to the temporal variable. Now we have that . What does this expression mean? Note that, for the initial data, , its Fourier transform has an amplitude for the frequency . But for the time , the absolute value of this amplitude doesn’t change, yet it oscillates much, in fact it oscillates in a fashion proportional to . We can interpret this oscillation as the propagation of the component at the frequency of the original wave. In other words, the higher the frequency, the faster its propagation.

Yet the problem is to control this ‘dispersion’. But how? We should use which norms? It is a very good yet hard question. Perhaps we can use the norms. Which ? For example, for the , we know already that the norm doesn’t change with time, that is . There is another norms, which we can use, it is the or . Note that is in fact a convolution operator, so this reminds us of the inequality of Young, that is . Note that, . So, we have some estimations on the and , so now we can use the interpolation theorems on exposants, that is, if , then we have that . Note that , so we have that .

This estimation is for every instant . Can we say something global? That is to say, can we have some norm to measure the whole time global behavior of ?

This is what Strichartz did. He defines a norm on the space . That is, for a smooth function , we define for and similarly for . Is this norm surprising? Not really. First, we want to know the information of for each instant . So, we get a positive valued function on . Then we want to control this function using some norm . So, this is a very natural approach. Now perhaps some dimensional analysis is helpful. Suppose that , then we can verify easily that . So, if we want that there be some inequality like , we must also have it valid for . Note that, for the second term, we have that . While for the first term, we have that (this is really due to the fact that in the expression of , in the exponent, there is , while in the rational term, there is ). Thus, according to the above remark, we have that . So, for the inequality to be held for any reasonable , we must have that , or . The easiest case for is of course . And thus, we have the inequality of Strichartz,

**If , then we have that for some constant independent of with and .**

Note that this result says that the linear operator, is continuous, where it is easy to verify that is a Hilbert space and is a Banach space with the norms defined above. This important observation reminds us to use something operator theory. To show that is bounded, we can in some way consider its adjoint, . That is to say, . There is a general result from operator theory,

**Suppose that is a linear operator from a Hilbert space(its scalar product is , and for the scalar product induced on , we write it ) to a Banach space(the action of a linear functional on the elements is ), and its adjoint defined by . Then we have that .**

This is not hard to prove. We just have to utilize the expression like and things like that. We will not prove this result here. One remark is that the composition doesn’t make sense in general.

Now we want to see what are in our context. It is not hard to see that . One one hand, . One the other hand, .But we have seen above that . So, we get that , thus we see that . Thus, , just a convolution. So to prove the continuity of , we have to show the continuity of or . But for , perhaps the difficulty is the same as . So, we first consider , the most complex one. For , we have that . So, we first evaluate . Note that for almost all , , so using the estimation on the dispersion above, we get that . Now we get another convolution, . Note that , so all in all, we have that . Now comes the last thing, .

Now we have a convolution which involves a special function . This function doesn’t lie in any . Perhaps we can use the inequality of Hardy-Littlewood-Sobolev,

**If such that then for all , we have that for some constant independent of .**

Now apply this theorem to our context, here (indeed lies in ). Besides, we have that . What a relief, it satisfies this inequality, and we conclude that . That is to say, the operator is continuous, and so is . And we proved the result. Note that in the above argument, we use implicitly that (a condition also in the Hardy-Littlewood-Sobolev inequality). For the case, , we must have that , thus where we have used the isometricity of for the norm . So, this case is also proved. And thus the whole result is finished. What is more, reexamine the condition in the inequality of Hard-Littlewood-Sobolev, we have to ensure that . This means that . So, the above result is really valid for the case , just like what the condition says in the theorem.

Now we want to generalize the above analysis for the case where the equation is not homogeneous, that is:

** with initial condition **

First we can write the explicit formula for this solution . A little digression here. We see that serves as an integral kernel, which means that it takes into account the effects of stimulation. For example, if . At time , the effect of the stimulation is taken to be . If at first, we add some stimulation, , then its effect at time becomes , and if at time we add another stimulation , then this stimulation will have an effect at time . So, the whole effect of all the stimulation will be . Then why don’t we take a similar integral for the stimulations ? The point is that , so the effect of these are uniform, that is why we consider only the starting point, that is enough. Now return to our work, our result for the inhomogeneous case is

**Suppose that satisfy and . Then the solution to the equation with initial condition satisfies that for any satisfying the same condition as as above. What is more, for each such pair, there is a constant independent of such that .**

The main idea of the proof for this result is that, we first show that for some constant using the same argument as above, and then for another constant using the continuity of the operator . So, we can apply the interpolation theorems to obtain that for any , for some constant depending on . As for the case , we will prove it later.

So, all this way, we have shown the global property of the solution to the linear Schrodinger equation.