Lecture 7: The wave equation: higher dimensional case
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1 Lecture 7: The wave equation: higher dimensional case Some auxiliary facts The co area formula. Let be a bounded domain and be a function of class. Denote by : the level set of. Let be any integrable function. Then for almost every the set is a regular hypersurface and, where is the surface measure on. In fact, the exterior integral should be taken only over the interval of those for which is non empty. Example. When is the distance function we have Denote by : the dimensional ball centered at of radius. In this case is non empty only for 0 and the level sets are spheres. Then applying the co area formula gives. This formula is well known (the integral in polar coordinates). The divergence theorem. Let be an open subset with a piecewise smooth boundary and let be a continuously differentiable vector field defined in a neighborhood of. Then div Where is the unit outward normal vector field along the boundary.
2 Spherical means We return to t he wave equation in and consider the following initial value problem 0, 0,, 0,, 0 Define the spherical means, o r averages, of a function, by,,, where is the dimensional area of the dimensional sphere of radius. Here is the dimensional area of the unit sphere : For instance, 2 Γ 2. 2, 2, 4, 2. Similarly we define the averages of t he initial data:,,,, Remark. We recall that the volume of the unit ball and the area of the unit surface are related by This fact is, in principle, well known but we can derive this relation immediately from the co area formu la. Ind eed, from formula in Example we have for : Hence we obtain which yields the required relation for.
3 Lemma (Euler Poisson Darboux equation). Fix and let satisfy the initial value problem above. Then,, 0; and 0, 0, 0, 0,,,, 0,,, 0 Proof. The fact that,, is two times continuously differentiable in follows from standard facts on differentiability of an integral depending on parameters. The situation with parameter requires however a further analysis because the set itself depends on. Step : We show that,, is continuously differentiable in and,,,. We consider an auxiliary function and fo r 0 we have,,,,,, where \ is the spherical slab with the oriented boundary (equipped with outward normals):, see the picture below: In order to apply the divergence theorem to the latter integrals we observe the following identity. From Example above we know that unit vector field : coincides with the gradient of the distant function:. For the further convenience we notice that the divergence of this vector field is
4 div div. On the other hand, the unit normal v ector at is found as. In particular, we have for the scalar products:. Summarizing this and applyin g the divergence theorem we obtain:,, div, Applying the formula from Example again, we see that div, and it follows easily that the following limit exists lim ε div, Similarly the left limit is shown to exist, hence is (continuously) differentiable and div,. In order to transform the latter integral, we apply the above formula for div and recall ag ain that the unit normal coincides with : div,,,, div, Hence applying the divergence theorem we get,, On the other hand, Returning to the definition of,, we obtain,.,,,.
5 It follows also from the above argument that differentiability of is equivalent to that of integral,. But this integral is easily seen to be differentiable (see again formula in Example ) because the Laplacian, is a continuous function in all parameters. Step 2. We return to our formula for,, and apply to it our wave equation:,,,, Differentiation of the latter identity w.r.t. yields by the co area formula (see Example ),,, which implies the desired equation:,., The initial conditions, 0,, and, 0,, follow immediately from the definition. Casen=3 The idea how the above method w orks is the following simple relation, lim,, which makes it possible to connect any solution of the symmetrized equation to the corresponding solution of the initial wave equation. It turns out however that the method works differently for even and odd dimensions. Below we describe briefly the 3 dimensional case. We suppose th at, 0, solves the following initial value problem: We set additionally, 0,, 0,,,. Lemma 2. Function solves,, the following two dimensional initial value problem 0, 0, 0, 0,,,, 0,,,, 0 0, 0.
6 Proof. Indeed, in our case 3, hence 2 Applying the reflection principle (see formula on page 4 in Lectur e 7) we find for 0 :,, 2,, 2,. Since, lim,,, we find tha t,,, lim lim,,. Hence we arrive at the Kirchhoff formula, 2,, 2 4 4, Case n=2 We give only the final result, the so called Poisson formula. Namel y, in the above notation,, 4, where and 0.
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