MATH 819 FALL We considered solutions of this equation on the domain Ū, where

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1 MATH 89 FALL. The D linear wave equation weak solutions We have considered the initial value problem for the wave equation in one space dimension: (a) (b) (c) u tt u xx = f(x, t) u(x, ) = g(x), u t (x, ) = h(x) We considered solutions of this equation on the domain Ū, where = (, ), Ū = [, )... Definition. A function u is called a classical solution to () if u C (Ū), and if u satisfies ()... Definition. A function u C (Ū) is a weak solution to () if for every ϕ Cc (Ū) one has () ϕfdxdt ϕ(x, )h(x) ϕt (x, )g(x) ( ) dx = ϕtt ϕ xx u dxdt..3. Theorem (consistency of classical and weak). Every classical solution is a weak solution. (The proof: start with ϕfdxdt = (u tt u xx )ϕdxdt and integrate by parts.).4. Theorem (explicit formula for all weak solutions). If u is a weak solution of () then it is given by (3) u(x, t) = g(x t) g(x t) h(x )dx T (x,t) f(x, t ) dx dt, where T (x, t) is the triangle with vertices (x, t), (x t, ) and (x t, ). About the proof: If u is a classical solution then you can derive (3) by integrating u tt u xx over the triangle T (x, t) and applying Green s theorem. To prove the general case choose a point ( x, t) and an appropriate test function ϕ ε ϕε(x, t) = ψ ( x t x t ) ( x t x t) ψ ε ε where ψ : is a smooth (C ) function which is nondecreasing, and which satisfies ψ(s) = for s, ψ(s) = for s. Let ε and carefully compute the limits of all the integrals you have in (). The result is (3).

2 MATH 89 FALL.5. Theorem (about convergence of weak solutions). Let u n be a sequence of weak solutions with data (f n, g n, h n ). Assume that (f n, g n, h n, u n ) converge uniformly on compact sets to (f, g, h, u). Then u is also a weak solution with data (f, g, h). This theorem is one reason to introduce weak solutions: note that one doesn t assume that the derivatives of the solutions converge, just the solutions u n themselves..6. Theorem (about existence of weak solutions). If f, g, h are continuous functions then the function u defined by (3) is a weak solution of (). About the proof: When f C, h C and g C then you can check that u is in fact a classical solution to (). To prove the general case where (f, g, h) are merely continuous, we used an approximation argument.. Problems se the definitions and theorems above to solve these problems:.. Homogeneous wave equation. Assume f =. (a) Show that every weak solution u can be written as u(x, t) = α(x t) β(x t). for certain functions α, β :. (b) elate the functions α, β to the initial values g, h. (c) Suppose α, β C () are arbitrary continuous functions; is u(x, t) = α(x t) β(x t) then necessarily a weak solution according to our definition.?.. An example of weak solutions. The function u(x, t) = x satisfies the wave equation u tt u xx = for all (x, t) except x =, t >. It also satisfies u(x, ) = x and u t (x, ) = for all x. (a) Find the weak solution to () with f =, g(x) = x and h(x) =. (b) Is u a weak solution to the wave equation?.3. Another example. The function x t x, u(x, t) = x < satisfies the homogeneous wave equation for all (x, t) except x = again. Find the weak solution of u tt u xx = with u(x, ) = x for x >, u(x, ) = for x < and u t (x, ) = for all x..4. An example involving convergence. Consider u n (x, t) = n cos(nx) cos(nt). (a) Show that each u n is a classical solution of () with f =, h = and g n = n cos nx. (b) Show that the u n converge uniformly, and verify that the limit is again a solution. (c) Do the partial derivatives of first and second order converge as n?

3 MATH 89 FALL 3 3. Semilinear equations Let f : be a function, and consider the semilinear wave equation (4) u tt u xx = f(u), u(x, ) = g(x) and u t (x, ) = h(x). Examples are the Klein-Gordon equation u tt u xx = m u, with m > a constant, and the Sine-Gordon equation u tt u xx = sin u. 3.. Definition. A function u C (Ū) is a weak solution to (4) if u is a weak solution to the equation u tt u xx = F (x, t), where F (x, t) = f(u(x, t)). u(x, ) = g(x) and u t (x, ) = h(x). 3.. Theorem. (Existence) If f : is Lipschitz continuous then (4) has a unique weak solution u C (Ū) for any initial data g, h C (). The proof proceeds by applying Picard-iteration to the integral equation u(x, t) = u (x, t) f(u(x, t ))dx dt. where T (x,t) u def = g(x t) g(x t) h(x )dx 3.3. Theorem. (Local Existence) Let g, h C () be given. If f : is continuously differentiable, then there exist a T >, and a u C ( [, T )) such that u is a weak solution to (4) on [, T ). This theorem follows by applying the previous theorem to a modified f. Let f(u) for u M, f M (u) = f(m) for u > M and f( M) for u < M The f M is Lipschitz continuous and the initial value problem u tt u xx = f M (u), u(x, ) = g(x) and u t (x, ) = h(x). has a unique solution u M C (Ū). If M > sup g(x) then there exists a T > such that u M (x, t) < M for all t [, T ). Hence u M is a weak solution of the original equation (4) for (x, t) [, T ) Theorem. (niqueness)let u, u be weak solutions on [, T ) of (4) with initial data (g, h ) and (g, h ) respectively. If g (x) = g (x) and h (x) = h (x) on an interval ( x t, x t), then u = u on the triangle (x, t) : t t, x x < t t.

4 4 MATH 89 FALL Let u be the weak solution of u(x, t) = f(x, t), 4. egularity u(x, ) = g(x), u t (x, ) = h(x). so that u is given by (4). If u were a smooth solution, and if f, g, h were differentiable functions, then you could differentiate the PDE w.r.t. x or t. You would find that v = u x and w = u t satisfy (5) v = f x, with initial conditions v = g x, v t = h x and (6) w = f t, with initial conditions w = h, w t = g xx f(x, ) If the data (f, g, h) are merely continuous then the solution u need not be differentiable. In general it can happen that the solution is differentiable (u C ), but that the derivatives v = u x and w = u t are not weak solutions of the above equations (6) and (5). The explicit solution (3) implies that for arbitrary continuous f, g, h one has a weak solution which is continuous. By differentiating the integrals in (5) one can prove that to get a solution u C (i.e. one for which u x and u t are continuous) you only have to assume that g C (): 4.. Theorem. If u C (Ū) is the weak solution to (), then (7) f, h C, g C = u C Proof: you have u(x, t) = g(x t) g(x t) h(x )dx t t t f(x, t ) dx dt. If f, h C, g C then you can differentiate under the integral and conclude that u x, u t are continuous on Ū. This Theorem does not imply that (5) and (6) hold. To prove that it turns out that you also need f x, f t, h x, g x and g xx to exist and to be continuous. (8) h C, f C, g C = u C and u t and u x are weak solutions of (5) and (6). 4.. Proof that u x satisfies (5). There are at least two proofs: you can use the integral form (3) of the solution, or you can directly use the definition. of weak solution. To prove that u x satisfies (5) we must show that for every test function ϕ Cc (Ū) one has (9) ϕ u x dxdt ϕ t g x ϕh x dx = ϕf x dxdt. Integrate by parts twice in the first double integral: ϕ u x dxdt = ϕ x udxdt.

5 MATH 89 FALL 5 If ϕ is a test function then so is ϕ x, so, since u is a weak solution we get ϕ x udxdt = ϕ xt g ϕ x hdx ϕ x fdxdt. Integrate by parts in the x direction again, and you find ϕ x udxdt = ϕ t g x ϕh x dx ϕf x dxdt. This implies (9). 5. Problems 5.. An example. Give an example of a solution of u =, u(x, ) = g(x), u t (x, ) = h(x) for which u C (Ū) v = u x is not a weak solution of v =, u(x, ) = g(x), u t (x, ) = h(x) for any g, h C (). Can you find such a solution u if h(x) =? 5.. Proof that u t satisfies (6). Above in 4. we proved half of (8). Prove the other half, i.e. prove directly from the definition of weak solution that u t satisfies (6) under the hypotheses in (8) Higher regularity. Show by induction on n that if u is a weak solution of u = f(x, t), u(x, ) = g(x), u t (x, ) = h(x) then f C n (Ū), g Cn (), h C n () = u C n (Ū) egularity for semilinear equations. Let n. Suppose that f : is a C n function, and let u C ( [, T )) be a maximal weak solution of u = f(u), u(x, ) = g(x), u t (x, ) = h(x) with g C n () and h C n (). Show that u C n ( [, T )). In other words, show that the solution is as smooth as the initial data. (Hint: use the previous problem and proceed with induction on n.)