MATH 220: Problem Set 3 Solutions

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1 MATH 220: Problem Set 3 Solutions Problem 1. Let ψ C() be given by: 0, x < 1, 1 + x, 1 < x < 0, ψ(x) = 1 x, 0 < x < 1, 0, x > 1, so that it verifies ψ 0, ψ(x) = 0 if x 1 and ψ(x)dx = 1. Consider (ψ j ) j 1 constructed as ψ j (x) = jψ(jx), so that ψ j (x) = 0 if x 1/j, and ψ j(x)dx = 1, for all j 1. (1) Let us show that ι ψj δ 0 in D (). By definition, it means that we need to prove that, for all φ Cc (), ψ j (x)φ(x)dx δ 0 (φ) = φ(0). So let s prove that for every ɛ > 0, there exists j 0 such that for all j j 0, ψ j (x)φ(x)dx φ(0) < ɛ. (2) Let ɛ > 0 and φ Cc (). Since φ is continuous, there exists η > 0 such that for all x < η, φ(x) φ(0) < ɛ. Then consider j 0 such that 1/j 0 < η. Since we know that ψ j (x) = 0 for all x 1/j, ψ j(x)dx = 1, and ψ j (x) 0 for all x, for all j j 0 we have: ψ j (x)φ(x)dx φ(0) = ɛ 1/j ψ j (x)φ(x)dx 1/j 1/j 1/j ψ j (x) φ(x) φ(0) dx ψ j (x)dx = ɛ. ψ j (x)φ(0)dx (2) Let φ C c () be such that φ(x) = 1 for all x < 1. Then, for j 1, ψ j (x) 2 φ(x)dx = 0 1/j (1) (3) 1/j j 2 (1 + jx) 2 dx + j 2 (1 jx) 2 dx = 2j 0 3. (4) 1

2 Therefore, as j +, converge. } + Consequently, {ι ψ 2j j=1 ψ j (x) 2 φ(x)dx +, and does not converge to any distribution since does not converge for the very φ we exhibited. { } + ι ψ 2 j (φ) does not j=1 { } + ι ψ 2 j (φ) { } + (3) We have just shown that ι ψj δ 0, but ι ψ 2 j (φ) does not converge j=1 to any distribution. Therefore there is no continuous extension of the map Q : f f 2 on C() to D (). j=1 Problem 2. We consider the conservation law: u t + (f(u)) x = 0, u(x, 0) = φ(x), (5) with f C 2 (). Since u is continuous and f is C 2, v = f (u) is also continuous. Since u is C 1 apart from jump discontinuities in its first derivatives, away from the jumps, u t (resp. u x ) is perfectly defined and continuous. Therefore, since f is C 1, and away from the discontinuities, v t = f (u)u t (resp. v x = f (u)u x ) is also continuous, v is C 1 apart from jump discontinuities in its first derivatives (the same ones as u). Therefore v has the same properties as u. Moreover, we have, away from discontinuities: v t +vv x = f (u)u t +f (u)f (u)u x = f (u)(u t +f (u)u x ) = f (u)(u t +(f(u)) x ) = 0. (6) So v verifies the Burger s equation (the ankine-hugoniot condition is vacuous: there are no shock since v is continuous). If f > 0, f is strictly convex and f is strictly increasing and therefore the inverse function (f ) 1 exists. We can therefore first solve for v from the Burger s equation: v t + vv x = 0, v(x, 0) = f (φ(x)), (7) and then u = (f ) 1 (v) is solution of the original PDE. Suppose now that u has a jump discontinuity. ankine-hugoniot condition: Then, according to the ξ (t) = f(u +) f(u ) u + u. (8) If v could have been defined as previously, v would have the same discontinuity (v = f (u)) and again by ankine-hugoniot: ξ (t) = v v2 2. = 1 v + v 2 (v + + v ) = 1 2 (f (u + ) + f (u )). (9) 2

3 But in general, f(u + ) f(u ) u + u. 1 2 (f (u + ) + f (u )) (10) (consider for instance f(x) = e x ) and the statement is FALSE. Problem 3. Consider Burger s equation u t + uu x = 0, u(x, 0) = φ(x), (11) with initial condition 0, x < 1, 1 x, 1 < x < 0, φ(x) = 1 + x, 0 < x < 1, 0, x > 1. (12) (1) To build the weak solution, it is very convenient to draw the characteristic curves. In the Burger case, we know that the solution u is constant along the characteristic curves x r (t) = φ(r)t + r. As long as the characteristic curves don t intersect, we have: If r < 1, then φ(r) = 0, which implies that the characteristic curves are and the solution u(x, t) = 0 along those curves, x r (t) = r, r < 1, (13) If 1 < r < 0, then φ(r) = 1 r, which implies that the characteristic curves are x r (t) = ( 1 r)t + r, 1 < r < 0, (14) and the solution u(x, t) = 1 r = 1 x + t 1 t = x + 1 along those t 1 curves. If 0 < r < 1, then φ(r) = 1 + r, which implies that the characteristic curves are x r (t) = ( 1 + r)t + r, 0 < r < 1, (15) and the solution u(x, t) = 1 + r = 1 + x + t 1 + t = x 1 along those t + 1 curves. If r > 1, then φ(r) = 0, which implies that the characteristic curves are and the solution u(x, t) = 0 along those curves. x r (t) = r, r > 1, (16) 3

4 To sum up, we have that, for t small, 0, x < 1, x + 1 u(x, t) = t, x 1 t + 1, 0, x > 1. x+t 1 < x < t, ( 1 < 1 t < 0) x+t 1 < x < t, (0 < 1+t < 1) (17) Now, from the sketch of the characteristic curves and/or the condition 1 < x < t of the solution, we can see that the characteristic curves don t intersect while t < 1. Therefore the above solution is valid for t < 1. The curves intersect at t = 1 Beyond that time, we therefore consider a weak solution satisfying the ankine-hugoniot condition. At the level of the discontinuity ξ(t), we have that u (ξ(t), t) = 0 and u + (ξ(t), t) = ξ(t) 1 t + 1, and ξ(1) = 1. Therefore the ankine-hugoniot condition is 0 1 ξ 2 (t) = ( ) 2 ξ(t) 1 t+1 0 ξ(t) 1 t+1 Hence, ξ verifies the following ODE: ξ (t) = 1 ξ(t) 1 2 t + 1, ξ(1) = 1. = 1 ξ(t) 1 2 t + 1. (18) (19) Solve it (say, by separation of variables) and you get ξ(t) = 1 2(1 + t). Therefore, for t 1, the solution is 0, 1 < x < 1 2(1 + t), x 1 u(x, t) = t + 1, 1 2(1 + t) < x < 1, (20) 0, x > 1. (2) There are two cases for t to consider. For 0 t < 1, And for t > 1, Therefore u(x, t)dx = t 1 u(x, t)dx = x t 1 dx + t 1 1 2(1+t) u(x, t)dx is indeed constant. x 1 dx = 1. (21) t + 1 x 1 dx = 1. (22) t + 1 4

5 (3) Let us call E(t) = For 0 t < 1, we have E(t) = w(x, t)dx = t 1 w(x, t)dx, where w = u 3. ( ) 3 x dx + t 1 t ( ) 3 x 1 dx = 1 t (23) So E(t) is indeed constant before a shock develops, but for t > 1, E(t) = w(x, t)dx = 1 and E(t) is no longer a constant. 1 2(1+t) Let s now explain what is going on here. ( ) 3 x 1 dx = 1 t + 1 t + 1, (24) Following Problem 2, if we define g(x) = 3 4 x4/3, we have that u = g (w). From Problem 2, we know that w t + (g(w)) x = 0. After the shock develops, we have for u: d u(x, t)dx = d 1 u(x, t)dx dt dt ξ(t) = = 1 ξ(t) 1 ξ(t) (u + ) t ξ (t)u + (ξ(t), t) ( ) 1 2 u2 + ξ (t)u + (ξ(t), t) x = 1 2 u +(ξ(t), t) 2 ξ (t)u + (ξ(t), t) = 0, (25) from the ankine-hugoniot jump condition (remember u = 0). Meanwhile for w we have: d w(x, t)dx = d dt dt = = 1 ξ(t) 1 ξ(t) 1 ξ(t) w(x, t)dx (w + ) t ξ (t)w + (ξ(t), t) (g(w + )) x ξ (t)w + (ξ(t), t) = g(w + (ξ(t), t)) ξ (t)w + (ξ(t), t) 0, (26) because w does not satisfy the same ankine-hugoniot condition. Problem 4. 5

6 (1) u xx u xy 2u yy = 0. (27) ( ) 1 1/2 We have A =. Therefore det(a) = 2 1/4 = 9/4 < 0, 1/2 2 and T r(a) = = 3 < 0. Therefore the eigenvalues of A are non zero and of opposite signs: Hyperbolic PDE. (2) u xx 2u xy + u yy = 0. (28) ( ) 1 1 We have A =. Therefore det(a) = 0. Therefore at least one 1 1 of the eigenvalues of A is zero: Degenerate PDE. (3) u xx + 2u xy + 2u yy = 0. (29) ( ) 1 1 We have A =. Therefore det(a) = 2 1 = 1 > 0 and T r(a) = > 0. Therefore the eigenvalues of A are non zero and of the same sign: Elliptic PDE. Problem 5. (1) Let s find the general C 2 solution of the PDE u xx u xt 6u tt = 0, (30) by reducing it to a system of first order PDEs (by the way, this is an elliptic PDE). We are looking for a, b, c, d that formally verify: xx xt 6 tt = (a x + b t ) (c x + d t ) = ac xx + (ad + bc) xt + bd tt. So we get the (under-determined) system: ac = 1 ad + bc = 1 bd = 6. (31) (32) From the first equation, let us simply take a = c = 1. Then the system reduces to { d + b = 1 (33) bd = 6, which gives b = 2 and d = 3. 6

7 Therefore we can write u xx u xt 6u tt = 0 as ( x + 2 t ) ( x 3 t ) u = 0. Now let v = ( x 3 t ) u. Then v verifies v x + 2v t = 0 (first order linear PDE!). And we know that the solution writes v(x, t) = h(t 2x) for some h C 1. Now for u we have the system: u x 3u t = h(t 2x). (34) Using the method of characteristics, we get the following equations: x r(s) = 1, x r (0) = 0, t r(s) = 3, t r (0) = r, v r(s) = h(t r (s) 2x r (s)), v r (0) = φ(r), (35) for some function φ C 2. Therefore we have x r (s) = s, t r (s) = 3s + r, and and by integrating from s = 0, we get: v r(s) = h( 3s + r 2s) = h( 5s + r), (36) v r (s) = s 0 = 1 5 h( 5s + r)ds + φ(r) 5s+r r h(y)dy + φ(r), (37) after a change of variables. Now, since we have s = x and r = t + 3x, we finally get: for some f, g C 2. t+3x u(x, t) = 1 h(y)dy + φ(t + 3x) 5 t 2x = f(t + 3x) + g(t 2x), (38) eciprocally, we verify that u of the form u(x, t) = f(t + 3x) + g(t 2x) for f, g C 2 indeed solves the PDE. (2) For an arbitrary φ Cc ( 2 ) we have to show that u(φ xx φ xt 6φ tt ) = v(φ xx φ xt 6φ tt ) + w(φ xx φ xt 6φ tt ) = 0. (39) But from Problem 2 of Pset 2, we have: v(φ xx φ xt 6φ tt ) = v(( x 3 t ) (φ x + 2φ t )) = 0, (40) and similarly, w(φ xx φ xt 6φ tt ) = w(( x + 2 t ) (φ x 3φ t )) = 0. (41) 7

8 Problem 6. Let us solve (in the strong sense): { uxx + 3u x, y, xy 4u yy = xy, u(x, x) = sin x, u x (x, x) = 0. (42) Same strategy here, we reduce it to a system of first order PDEs (by the way, this is an hyperbolic PDE!). We are looking for a, b, c, d that formally verify: xx + 3 xy 4 yy = (a x + b y ) (c x + d y ) = ac xx + (ad + bc) xy + bd yy. So we get the (under-determined) system: ac = 1 ad + bc = 3 bd = 4. (43) (44) From the first equation, let us simply take a = c = 1. Then the system reduces to { d + b = 3 (45) bd = 4, which gives, say, b = 1 and d = 4. Therefore we can write u xx + 3u xy 4u yy = 0 as ( x y ) ( x + 4 y ) u = 0. Now let v = ( x + 4 y ) u. Then v verifies v x v y = xy (first order semilinear PDE!), and (u x + u y ) (x,x) = sin (x) = cos(x), u x (x, x) = 0 imply that u y (x, x) = cos(x) and v(x, x) = 4 cos(x). Therefore v satisfies the following PDE: { vx v y = xy, (46) v(x, x) = 4 cos(x). The ODEs for the characteristics are then: x r(s) = 1, x r (0) = r, y r(s) = 1, y r (0) = r, v r(s) = x r (s)y r (s), v r (0) = 4 cos(r). After solving, we get: x r (s) = s + r, y r (s) = r s, (47) (48) v r (s) = r 2 s s cos(r). Therefore we get the PDE for u: ( ) 2 ( ) x + y x y u x + 4u y = v(x, y) = + 1 ( ) 3 ( ) x y x + y + 4 cos, u(x, x) = sin(x). (49) 8

9 Writing once again the characteristic ODEs for the PDE, we get: x r(s) = 1, x r (0) = r, y r(s) = 4, y r (0) = r, v r(s) = v(x r (s), y r (s)), v r (0) = sin(r). (50) After solving, we get: x r (s) = s + r, y r (s) = 4s + r, v r (s) = sin(r) + s 0 ( 1 3 ( 3 2 τ ) 3 = 9 32 s s4 5 2 s3 r 3 4 s2 r sin Therefore (and finally): u(x, t) = 1 16 ( ) 2 ( ) ( y x 5x + y 5y + 13x ( ) 2 ( ) ) 5τ + 2r 3τ 5τ + 2r cos dτ 2 ( ) 5s + 2r sin (r). (51) ) + 8 ( ) x + y 5 sin 3 ( 4x y 2 5 sin 3 (52) ). Problem 7. with and Let s solve the wave equation on the line: u tt c 2 u xx = 0, u(x, 0) = φ(x), u t (x, 0) = ψ(x), (53) 0, x < 1, 1 + x, 1 < x < 0, φ(x) = 1 x, 0 < x < 1, 0, x > 1, 0, x < 1, ψ(x) = 2, 1 < x < 1, 0, x > 1. (54) (55) We know from the course (method of characteristics) that the solution is: u(x, t) = 1 2 (φ(x + ct) + φ(x ct)) + 1 2c x+ct ψ(y)dy. (56) Now, notice first that since c, t 0, then x ct x + ct. As you can see in the figure, there are 10 cases to consider for the solution: 1. If x + ct < 1 and x ct < 1 (domain 1), then u(x, t) = 0. 9

10 2. If 1 x + ct < 0 and x ct < 1 (domain 2), then u(x, t) = 1 2 (1 + x + ct + 0) + 1 x+ct ( 1 2dy = 2c ) (1 + x + ct). (57) c 3. If 0 x + ct < 1 and x ct < 1 (domain 3), then u(x, t) = 1 2 (1 +0)+ 1 2c 1 x+ct 4. If 1 x + ct and x ct < 1 (domain 4), then 1 u(x, t) = c 2dy = 1 2 (1 x ct)+ 1 (1 + x + ct). c (58) 1 5. If 1 x + ct < 0 and 1 < x ct < 0 (domain 5), then 1 u(x, t) = 1 2 (1 + x + ct x ct) + 1 2c 2dy = 2 c. (59) x+ct 6. If 0 x + ct < 1 and 1 < x ct < 0 (domain 6), then u(x, t) = 1 2 (1 x ct x ct) + 1 2c x+ct 7. If 1 x + ct and 1 x ct < 0 (domain 7), then u(x, t) = 1 2 (0+1+)+ 1 2c 1 8. If 0 x + ct < 1 and 0 x ct < 1 (domain 8), then u(x, t) = 1 2 (1 x ct + 1 x + ct) + 1 2c 2dy = 1 + x + 2t. (60) 2dy = 1 ct + 2t. (61) 2dy = 1 2 (1 + x ct)+ 1 (1 x + ct). c (62) x+ct 9. If 1 x + ct and 0 x ct < 1 (domain 9), then u(x, t) = 1 2 (1 x ct) + 1 2c 1 2dy = 1 x + 2t. (63) 2dy = 1 2 (1 + x ct) + 1 (1 x + ct). c (64) 10. If 1 x + ct and 1 x ct (domain 10), then u(x, t) = 0. Finally, as we can now see, u(x, t) vanishes in region (1) and (10). Since φ(x) = 0 and ψ(x) = 0 in x > 1, this result corresponds to Huygens principle. Moreover, u(x, t) is C 1 except on the lines x + ct = 1, x + ct = 0, x + ct = 1, x ct = 1, x ct = 0 and x ct = 1. Since φ(x) and ψ(x) are C 1 everywhere except at x = 1, 0, 1, this result corresponds to the propagation of singularities: u(x, t) is C 1 near (x, t) if φ and ψ are such near x ± ct. 10