Math 126: Homework 13 solutions

Transcription

1 Math 6: Homework 3 lutions. (a) The initial data can be parameterized according to = f (s) = s, y = g(s) =, u( f (s), g(s)) = h(s) = sin s. The characteristics are determined by the ODE system Equation gives y = t, = y + y, () =, () = zy 4. (3) which gives = t + t = t + t log = C + log( + t ) for me constant C. Hence, the characteristic passing through (s, ) at t = is = s( + t ). Equation 3 then gives which can be integrated to give = zt 4 z = t 4 log z = t 8 + D for me constant D. By using the initial data, it can be seen that z = Z(s, t) = e t /8 sin s The family of characteristics for different values of s is shown is Fig.

2 3 y Figure : Characteristics of the partial differential equation considered in question, projected into the y plane. (b) From part (a), it can be seen that s = (c) Taking partial derivatives of Eq. 4 gives + y, t = y, ( ) u(, y) = Z(s, t) = e y /8 sin + y. (4) u (, y) = e y /8 ( ) + y cos + y ( u y (, y) = e y /8 y ( ) 4 sin + y y ( )) ( + y ) cos + y Hence and ( ) y + y u = ye y/8 ( + y ) cos + y ( ) y + y u + u y = ye y/8 sin 4 + y = yu 4

3 u y Figure : Three-dimensional plot of the lution u(, y) of question. since the terms involving cosine will cancel each other. Thus u(, y) satisfies the PDE. In addition ( ) u(, ) = e sin + y = sin the lution satisfies the initial data. (d) Figure shows a plot of the lution. It can be seen how the sinuidal initial data is transported along the parabolic characteristics. A small amount of decay is al visible in the y direction, due to the presence of the ep( y /8) term in the lution.. The characteristics will be given by which is equivalent to = y, = = ( y) = + y. 3

4 This has a first integral ψ(, y) = + y and thus the characteristics are given by the curves where ψ(, y) is constant. Hence a typical characteristic is + y = R for me constant R, which corresponds to a circle of radius R. The characteristics are a family of concentric circles centered on the origin. Since the lution is constant along each characteristic a general lution can then be written as u(, y) = G( + y ) for me function G(λ). The initial condition states that and similarly u(, ) = h() = G( ) u(, ) = h( ) = G(( ) ) = G( ). For a lution to eist it is therefore necessary that h() = h( ), h is even. This is reanable, since any characteristic circle of radius R will intersect the initial data at ±R, the values must be consistent. For any even choice of h, a general lution can be written as G(λ) = h( λ) defined for λ, that u(, y) = h( + y ). Hence h being even is both a necessary and sufficient condition for a general lution to eist. 3. (a) The characteristics are given by the ODE system =, = 3, = 3z and the initial data can be can be parameterized according to Hence and therefore = f (s) = s, y = g(s) =, u( f (s), g(s)) = h(s). = X(s, t) = s + t (5) = 3(s + t) y = Y(s, t) = (s + t) 3 s 3 (6) where the constants of integration have been chosen that the characteristics pass through the initial data when t =. 4

5 3 y Figure 3: Characteristics for the partial differential equation considered in question 3. (b) The characteristics are shown in Fig. 3. The Jacobian is given by J(s, t) = X s(s, t) Y s (s, t) X t (s, t) Y t (s, t) = 3(s + t) 3s 3(s + t) = 3(s + t) 3(s + t) + 3s = 3s and thus J(s, ) = 3s. At s =, the Jacobian vanishes, implying that the characteristics and the initial data are tangent, which is in agreement with shapes of characteristics shown in Fig. 3. As described in the tetbook, for a C lution to eist, ( ) a(,, z ) b(,, z rank ) c(,, z ) f () g () h =. () This can be written as h () =. ( rank h () ) = (c) To find the general lution, first consider the differential equation for z given above: = 3z. 5

6 Thus z = 3(s + t) log z = (s + t) 3 + C for me integration constant C. Using the initial condition shows that Equations 5 and 6 show that and the general lution is z = Z(s, t) = h(s)e (s+t)3 s 3. s = S(, y) = 3 3 y t = T(, y) = 3 3 y u(, y) = Z(S(, y), T(, y)) = h( 3 3 y)e y If h() =, then h () = h() =, the condition from part (b) is satisfied. However, u(, y) = ( 3 y) /3 e y ( ) u y (, y) = y + (3 y) /3 u y (, y) = ( ) 3 3 y + ( y)/3 e y which is unbounded as y and thus u is not C in a neighborhood of the ais. This implies that while the condition from part (b) is necessary for a C lution to eist, it is not sufficient. Note that other choices of h(s) would lead to a C lution: if h(s) = s 3, then which is C everywhere. 4. This problem can be epressed as u(, y) = ( 3 y)e y F(, y, u, u, u y ) = u + u y 4u =. e y 6

7 Following the notation of the tetbook, the last two variables of F can be denoted as p = u and q = u y, F(, y, u, p, q) = p + q 4u. (7) The characteristic system is then given by The initial data is dp dq = F p = p (8) = F q = q (9) = pf p + qf q = p + q () = F pf u = 4p () = F y qf u = 4q. () f (s) = cos s, g(s) = sin s, h(s) =. Let ϕ(s) and ψ(s) be the initial conditions on p and q respectively. Equation 7 requires that ϕ + ψ = 4 and the condition h (s) = ϕ(s) f (s) + ψ(s)g (s) requires that This gives one lution as ϕ sin s + ψ cos s =. ϕ = cos s, ψ = sin s and a second lution as ϕ = cos s, ψ = sin s. Consider the first lution for the initial data. Equation shows that q = Ce 4t for me constant C, and to be consistent with the initial data, this leads to q = e 4t sin s. Equation 9 then becomes = 4e4t sin s y = e 4t sin s + C 7

8 for me constant C. Using the initial data g(s) = sin s gives y = e 4t sin s. A similar consideration of Eqs. and 8 gives Thus p = e 4t cos s, = e 4t cos s. = 8e8t (cos s + sin s) = 8e 8t z = e 8t. To find a general lution u(, y), note that and thus + y = e 8t (cos s + sin s) = e 8t u(, y) = z = + y. Now consider the second case for the initial data. Then q = e 4t sin s Eq. 9 becomes = 4e4t sin s, and by considering the initial data g(s) = sin s, Similarly y = ( e 4t ) sin s. p = e 4t cos s, = ( e 4t ) cos s. The equation for / is the same with the same initial data z = e 8t still holds. To find a general lution u(, y), note that and thus the general lution is + y = ( e 4t ) u(, y) = e 4t = + y ( ) + y. Note that etending this lution outside of the disk + y = 4 would require further analysis, since t as + y 4. 8