Math 9 - Partial Differential Equations Homework 5 and Answers. The one-dimensional shallow water equations are h t + (hv) x, v t + ( v + h) x, or equivalently for classical solutions, h t + (hv) x, (hv) t + (hv + h ) x, where v is the horizontal velocity and h is the height. how that the above two systems are strictly hyperbolic for h >. olution: The both systems have eigenvalues: v ± h which are real and distinct for h >.. Use characteristics to find the solutions of (a) u t + u x u, u t +x, x R, t >. (b) u t + txu x, u t +x, x R, t >. (c) u t + uu x, u t x, x R, t >. olution: (a) The characteristic passing through (x, ) is x x + t. du(x(t), t) dt u, so u(x, t) u(x(), )e t u (x )e t u (x t)e t (b) The characteristic x x(t) passing through (x, ) is: dx dt tx, x() x, which gives x x e t, so u(x, t) u (x ) u (xe t ). (c) The characteristic x x(t) passing through (x, ) is: and du(x(t),t) x +t. 3. Prove that dx dt u, x() x, e t +(x t). dt, then u(x(t), t) u(x(), ) u (x ) x, thus x(t) x + x t, and u(x, t) x u(x, t) is a weak solution to the Riemann problem u t + uu x, u t u (x), if x < t,, if x > t,, if x <,, if x >.
olution: For a test function φ C(R R + ), assume suup φ [, a] [, T ] with a >, T >, ) (uφ t + u φ x dxdt + u (x)φ(x, )dx T a ) a (uφ t + u φ x dxdt + u (x)φ(x, )dx T t ( φ t + ) φ x dxdt + φ(x, )dx T T T T φ t dtdx + φ t dtdx + x (φ(x, T ) φ(x, ))dx + φ(x, x)dx + T T T T t φ x dxdt + (φ(x, T ) φ(x, x))dx + φ( t, t)dt, T or use the divergence formula, ) (uφ t + u φ x dxdt + u (x)φ(x, )dx T a ) a (uφ t + u φ x dxdt + u (x)φ(x, )dx T t ( φ t + ) φ x. x t t dxdt + φ(x, )dx (x,t) ( φ, φ)dxdt + φ(x, )dx ( φ, φ) 5 (, )dl + 4. Consider the nonhomogeneous conservation law t,<x< u t + f(u) x g(u, x, t), φ(x, )dx (φ( t, t) φ(, t))dt + φ(x, )dx ( φ, φ) (, )dl + φ(x, )dx where g is a source term. Define u(x, t) L (R R + ) to be a weak solution if (uϕ t + f(u)ϕ x + gϕ) dx dt + u(x, )ϕ(x, ) dx, for all test functions ϕ C (R R + ). how that the Rankine-Hugoniot jump condition is the same as the homogeneous case when g. olution: uppose that a curve : x x(t) divides the upper half plane into two parts V ± and u(x, t) is a piecewise smooth solution with jump discontinuity along. That is, u is a weak solution in the sense of distributions and also a C solution in both V + and V, and at each point (x, t ) on the curve, the limits of u(x, t) as (x, t) (x, t ) in both V + and V exist but are not necessarily equal. Choose a test function ϕ R R + with compact support in some open region V R R +, and
ϕ(x, ) but not vanishing along the curve. Then u t + f(u) x g in V ± V and (uϕ t + f(u)ϕ x + gϕ) dx dt (uϕ t + f(u)ϕ x + gϕ) dx dt + (uϕ t + f(u)ϕ x + gϕ) dx dt V V + ((uϕ) t + (f(u)ϕ) x (u t + f(u) x g)ϕ) dx dt V + ((uϕ) t + (f(u)ϕ) x (u t + f(u) x g)ϕ) dx dt V + (x,t) (f(u)ϕ, uϕ) dx dt + (x,t) (f(u)ϕ, uϕ) dx dt V V + (f(u )ϕ, u ϕ) ν dl + (f(u + )ϕ, u + ϕ) ( ν) l (f(u )ν + u ν )ϕ dl (f(u + )ν + u + ν +)ϕ dl ((f(u ) f(u + ))ν + (u u + )ν ) ϕ dl, for all test functions ϕ, thus the jump condition is the same as the homogeneous case. 5. Construct the entropy solution of the following IVP: u t + ( u)u x, x R, t >,, if x <, u t u (x), if x. olution: The characteristics passing through (x, ) have slope if x < and have slope if x >. The flux is f(u) u u f(u+) f(u ). The shock speed is σ u + u (u ++u ), o the solution is u(x, t), x < t,, x > t. 6. Find the traveling wave solutions u(x, t) to the following equation u t + ( u)u x εu xx, ε > : constant, such that lim u(x, t), lim x u x(x, t) ; x lim u(x, t), lim x and also find the limit lim u(x, t). ε Compare the limit with the solution in the previous exercise. olution: If u(x, t) f(x σt) f(ξ) with ξ x σt, then u x(x, t), x σf + ( u)f εf, f( ), f(), f ( ) f (). Integrating it from to ξ, we get f f σf + σ εf, 3
taking ξ, we obtain σ. o and thus for a constant C. Then and εf (f )(f ), df (f )(f ) ε dξ, f(ξ) + u(x, t) + lim u(x, t) ε which is the weak solution in the previous problem. + Ce ε ξ, + Ce ε (x+ t),, if x + t <,, if x + t >, 7. Construct the entropy solution of the following IVP: u t + uu x, x R, t >,, if x <,, if x <, u t u (x), if x,, if x >. olution: The solution is sketched in the figure. t 5 3 R 4 x The line is a shock passing through point (, ), with left state u, right state u +, speed σ, equation: x + t, crossing the t-axis at point (, ). The line is a shock passing through point (, ), with left state u, right state u +, speed σ, equation: x + t, ending at point (, ). The region R is the rarefaction wave with value u x t. 4
The shock wave and the rarefaction R starts interaction at point (, ) and produce a shock wave curve 3 : x x(t) with left state u and right state u + x t, so x (t) ( + x t ), x(), which gives the equation of the curve 3 : x t t. The shock wave and the rarefaction R starts interaction at point (, ) and produce a shock wave curve 4 : x x(t) with left state u x t and right state u +, so x (t) ( x t + ), x(), which gives the equation of the curve 4 : x t. The shock waves 3 and 4 meets at point (8, 6), which is obtained by solving x t t t. Their interaction yields another shock wave 5 passing through point (8, 6), with left state u and right state u +. The equation of 5 is x t. 5