PARTIAL DIFFERENTIAL EQUATIONS MIDTERM

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1 PARTIAL DIFFERENTIAL EQUATIONS MIDTERM ERIN PEARSE. For b =,,..., ), find the explicit fundamental solution to the heat equation u + b u u t = 0 in R n 0, ). ) Letting G be what you find, show u 0 x) = lim t 0 + Gx, t; y, 0)u 0 y) dy, R n 2) when u 0 is bounded and continuous on R n. Taking the Fourier transform with respect to x) of the equation gives ξ 2 û + ib ξû û t = 0 ξ 2 + ib ξ ) û = û t = û = ce ξ 2 +ib ξ)t, so we need to take the inverse Fourier transform of this solution. [ ] F ce ξ 2 +ib ξ)t = e ix ξ ce ξ 2+ib ξ)t dξ 2π) n R n n = c e ξ2 j t+it+x j )ξ j dξ j 3) 2π = c j= n j= R t+x j ) 2 2π e 4 e t ξ j it+x j) 2t R 2 dξ j 4) where 3) follows by independence the integrand is symmetric under permutation of the coordinates) and 4) follows by completing the square: ) ξj 2 t + it + x j)ξ j = t ξj 2 it+x j)ξ j t ) ) 2 = t ξj 2 it+x j)ξ j it+xj ) + t+x j) 2. t 2t 4t We evaluate the integral in 4) by the change of variables y j = ξ j t it+x j ) 2 t t dy j = dξ j

2 so that e 2π R ξ it+x j ) j t 2 t Substituting this back in to 4), we obtain 2 dξ j = 2π e y2 j dyj = t R 2π 2π = t 2πt). /2 [ ] F ce ξ 2 +ib ξ)t = c n j= 2πt) /2 e t+x j )2 /4t c = e n 2πt) n/2 c = e x+b t 2 /4t. 2πt) n/2 j= t+x j) 2 /4t Thus, the fundamental solution to the original problem is ux, t) = c 2πt) n/2 e x+b t 2 /4t. Other solutions may be computed by convolving against it, i.e., using the kernel Gx, t; y, s) = c 2πt s)) n/2 e x y+b t s) 2 /4t s). Now we show u 0 x) = lim Gx, t; y, 0)u 0 y) dy t 0 + R n by showing that the difference of the two goes to 0 as t 0 +. First, note that I = Gx, t; y, 0)u 0 y) dy u 0 x) = Gx, t; y, 0) u 0 y) u 0 x)) dy R n R n because R n Gx, t; y, 0) dy =. For any fixed t > 0, we can split the integral as I = J + J 2 J = J 2 = x+bt y δ x+bt y >δ for 2πt) n/2 0 y) u 0 x) dy 2πt) n/2 0 y) u 0 x) dy. 2

3 Split J again: J = x+bt y δ 2πt) n/2 0 y) u 0 x + bt) + u 0 x + bt) u 0 x) dy x+bt y δ 2πt) n/2 0 y) u 0 x + bt) dy }{{} + x+bt y δ 2πt) n/2 0 x + bt) u 0 x) dy }{{} 2 To control, we make δ small enough to ensure that This gives x + bt y < δ = 0 y) u 0 x + bt) < a x + bt y. = a = a = aδ x+bt y δ x+bt y δ δ 0 x+bt y =δ δ 0 e x+tb y 2 /4t 2πt) n/2 0 y) u 0 x + bt) dy x+bt y =δ e x+tb y 2 /4t x + bt y dy 2πt) n/2 e x+tb y 2 /4t x + bt y dy 2πt) n/2 e δ2 /4t dy 2πt) n/2 = ăδ e δ2 /4t 2πt) n/2 vol nbx, δ) t Note that a ă because of some Jacobian that I m too lazy to work out. However, it doesn t contain t or y, so we don t have to worry about it. On to 2 : 2 = 2πt) n/2 0 x + bt) u 0 x) dy x+bt y δ = 0 x + bt) u 0 x) = 0 x + bt) u 0 x) c t 0 + 0, x+bt y δ 2πt) n/2 by the continuity of u 0. The constant c is just the integral of the Gaussian, which does not depend on t. dy 3

4 Back to J 2. Since u 0 is continuous on this bounded domain, we know 0 M, and accordingly 0 y) u 0 x) 2M. Hence, J 2 = x+bt y >δ 2πt) n/2 0 y) u 0 x) dy 2M dy x+bt y >δ 2πt) n/2 e x y+tb 2 /8t = 2M dy x+bt y >δ 2πt) n/2 e x y+tb 2 /8t 2M e δ2 /4t dy 2πt) n/2 x+bt y >δ = 2Mde δ2 /4t t

5 2. Find three solutions to the boundary value problem { u = 0, in R n +; u =, on R n +. Why does uniqueness fail? Are there more than 3 solutions? 5) It is clear that u = + cx n satisfies the problem, for any c R. The essential reason is that a multivariable polynomial which contains no more than a single power of any of its n variables will be killed off by the Laplacian, and that + x n fx) = on the boundary, so long as fx) doesn t blow up on the boundary. So we leverage this idea as follows. a) u = + cx n, for any c R. b) u = cx n j n x j. c) u = cx n j <j 2 n x j x j2. d) More generally, for any 0 k n, we have k u = + cx n x jm. j <j 2 < <j k n m= e) Most generally, Let M k denote the set of monomials of degree k which contain no more than one power of any given coordinate x j. M k = {x i x i2... x ik. i < < i k n }. Now consider the polynomials { n } P = c mk m k x). m k M k k= Then + x n px) will be a solution for any p P. The number of solutions resulting from this construction is n ) n = 2 n. k k= It is clear that all of these satisfy the boundary conditions, because x n = 0 on the boundary, and polynomials are bounded away from. To see again why these satisfy the Laplacian, apply the linearity of the Laplacian and it suffices to consider a general term: ) Dj 2 k cx j x j2... x jm ) = j k j k cx j x j2... x jm ) ) = cxj j k x j2... x jk x jk+... x jm = 0, since cx j x j2... x jk x jk+... x jm is constant with respect to x jk. 5

6 Similarly, we take another solution of the Laplacian in R n and multiply it by x n ; let x = x, x 2,..., x n, 0) and consider Then we have u = x n ux) = + c x. n 3 n Dj 2 u j= n = Dj 2 u + D2 n u j= = cx n n ) x + c n 3 x n 3 D2 nx n = cx n 0 + c x 0 n 3 = 0, since x 3 n is a solution of the Laplacian in R n as shown previously, and D 2 n x n = 0. Altogether, this makes 2 n solutions. Uniqueness fails because the domain is not bounded. Hence, solutions can behave in wildly different ways as x n. 6

7 3. Let u be a smooth solution to the wave equation u tt u + m 2 u = 0 in R n. What is the energy? Prove the principle of causality: the domain of dependence is the backward light cone in space time. Hint: mimic the case m = 0 in 9., IBP over cone.) We find the energy. First, t 2 u2 t + m2 2 u2 + 2 u 2 ) u t u) = u tt u t + m 2 uu t + u t u u t u u t u = u tt + m 2 u u)u t = 0 since u is a solution. 6) Integrating over R n gives [ [ ] ] t 2 u2 t + m2 2 u2 + 2 u 2 u t u) = 0 R n [ ] t 2 u2 t + m2 2 u2 + 2 u 2 u t u) = 0 7) R n R [ n ] t 2 u2 t + m2 2 u2 + 2 u 2 = 0 R n [ ] 2 u2 t + m2 2 u2 + 2 u 2 = E, R n where E is the energy. The second integral in 7) vanishes because of the divergence theorem: u t u) = u t u) n = u t u) n = 0. R n R n Now that we have the energy, we rewrite the energy identity 6) as where D u t u x ) + + D n u t u xn ) + D t 2 u2 t + m2 2 u2 + 2 u 2 ) = w w = u t u x. u t u xn 2 u2 t + m2 2 u2 + 2 u 2 Rn+. Denote the outward unit normal vector by ν = ν,..., ν n, ν t ) R n+. Now take F to be a solid cone frustum in n + )-dimensional spacetime with top T this is the 7

8 level curve where t = t ), bottom B the level curve where t = 0), and side/mantle where x x 0 = t t 0 ). 0 = F w = F w ν = T w ν + B w ν + w ν. 8) On T, ν = 0,..., 0, ), so that w ν = 2 u2 t + m2 u 2 + u 2 ) and w ν = 2 T T u 2 t + m2 u 2 + u 2 ). 9) On B, ν = 0,..., 0, ), so we similarly have since t = 0 on B. B w ν = 2 = 2 B B u 2 t + m2 u 2 + u 2 ) ψ 2 + m 2 ϕ 2 + ϕ 2 ), 0) On, the outward unit normal vector following p. 27) at x 0, t 0 ) is ν = ϕ ϕ = x x 0,..., x n x 0n, t t ) 0, 2 r r t t 0 where r = x x 0 = x x 0 ) x n x 0n ) 2. Thus we get the integral w ν = 2 [ n j= u t u j ) x j x 0j r 2 u 2 t + m 2 u 2 + u 2)]. ) If we let ρ = n j= x j x 0j r = x x 0 r = x x 0 x x 0, then the radial derivative is u r = n j= u j x j r = n j= u j x j x 0j r = ρ u, 8

9 and we can rewrite the integrand in ) as follows w ν = 2 2 u 2 t + m 2 u 2 + u 2) u t u r = 2 2 u2 t u tu r + 2 u2 r 2 u2 r + 2 m2 u u 2 = 2 2 = u t u r ) 2 + m 2 u 2 + u 2 u 2 r u t u r ) 2 + m 2 u 2 + u u r ρ 2 2) Now plugging 9), 0), and 2) into 8), we get w ν + w ν = w ν T B 2 u2 t + m2 u 2 + u 2 ) 2 ψ2 + m 2 ϕ 2 + ϕ 2 ). T B If ϕ ψ 0 on B, then the integral on the right vanishes, forcing the integral on the left to vanish, and thus indicating by the obvious nonnegativity of the integrand) that we must have u 0 on T as well. Recall that T is the level curve for a fixed t 0, t 0 ) which is entirely arbitrary. Hence, u 0 on every slice of F orthogonal to the t axis, i.e., u 0 on the entire solid cone. In particular, ux 0, t 0 ) = 0. Suppose u, v are solutions to u tt u+m 2 u = 0 which agree on B. Then w = u v is a solution which vanishes on B, and hence on all of the cone by the above argument. Thus u = v on all of the cone. 9

10 4. Let D be a bounded C domain, g C D). Prove that the solution to { u = 0 in D, u = g on D, 3) n is unique up to a constant. What is a necessary condition on g so that the above problem has a solution? Suppose that we have two solutions u and v which both satisfy the boundary value problem. Let w := u v. Then w = u v) = u v = 0 0 = 0 in D, and w = u u v) = v = g g = 0 on D, n n n n so w is a solution to the homogeneous Laplacian. By the maximum principle, the maximum of w occurs on D and is hence 0. Similarly, the minimum principle i.e., maximum principle applied to w) implies the minimum of w occurs on D and thus also 0. So w 0 on D, whence u = v. For a necessary condition, consider the Poisson problem { u = f in D, u = g n We can relate f to g via Green s formula: f = u = D D on D. U u = n With f = 0, this shows that for 3) to have a solution, it must be that g = 0. U U g. 0

11 Okay, so that s the midterm. However, I did have a couple of questions about other attempts, that I never had time to ask about in class. So here they are.. For b =,,..., ), find the explicit fundamental solution to the heat equation u + b u u t = 0 in R n 0, ). 4) Letting G be what you find, show u 0 x) = lim t 0 + Gx, t; y, 0)u 0 y) dy, R n 5) when u 0 is bounded and continuous on R n. By the definition of b, this is n j= D2 k + D k)u u t = 0. We make the change of variables u = ve αb x+βt. With D k := x k, v k = D k v, and D t :=, this becomes Then t D k u = v k + αv)e αb x+βt, D 2 k u = v kk + 2αv k + α 2 v)e αb x+βt, and D t u = v t + βv)e αb x+βt. Now 4) becomes [ n v jj + 2αv j + α 2 v) + j= n j= ] n v j + αv) βv v t e αb x+βt = 0 j= ) v jj + 2α + )v j + α + α 2 )v βv v t = 0 By taking α = 2, the v j terms will drop out. Then α + α 2 = 4, so define β = n 4 to cancel out the v terms, and the previous equation becomes simply v v t = 0. We know the fundamental solution to this is vx, t) = e x y 2 4πt s)) n/2 4t s) u0 y) dy, so the solution to the original problem must be e 2b x+nt)/4 ux, y) = e x y 2 4πt s)) n/2 4t s) u0 y) dy. Now we must show 5), i.e., e 2b x+nt)/4 4πt) n/2 R n e x y 2 /4t u 0 y) dy t 0+ u 0 x).

12 We showed in class that e x 2 /4t dx =. 4πt) n/2 Therefore, we can say x, t) u 0 x) e x y 2 /4t u0 y) u 4πt) n/2 0 x)e 2b x+nt)/4 dy. R n No, try ux, y) = e 2b x+nt)/4 e x y 2 4πt s)) n/2 4t s) u0 y) dy R n = e 2b x y +nt)t s) 4πt s)) n/2 4t s) e x y 2 4t s) u0 y) dy R n Stuck!? However, convergence is difficult to prove for this solution, so we relent and turn to mimicking the solution of the Cauchy problem as discussed in class... I talked to you about this one briefly, but I couldn t figure out how to make it work... 2