# It is desirable to make sense out of the equation by

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Chapter 10 / Exercise 45
Applied Calculus
Berresford/Rockett
Expert Verified
), it is desirable to make sense out of the equation by reinterpreting it when discontinuities develop. To this end, let φ = φ ( x, t ) be a C 1 function. Let D be a rectangle in the xt plane determined by a x a and 0 t T , such that φ ( x, t ) = 0 for x = ± a, x = T , and for all ( x, t ) in the upper half plane outside D . Let u be a “genuine” solution of equation ( 8.5.15 ).
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Chapter 10 / Exercise 45
Applied Calculus
Berresford/Rockett
Expert Verified
8.5 Green’s Functions 103 (a) Show that t 0 [ t + f ( u ) φ x ] dxdt + t =0 u 0 ( x ) φ ( x, 0) dx = 0 . (8.5.16) (HINT: Start with D [ u t + f ( u ) x ] φ dxdx = 0.) Thus, if u is a smooth solution, then equation ( 8.5.16 ) holds for all φ as above. We call the function u a weak solution of equation ( 8.5.15 ) if equation ( 8.5.16 ) holds for all such φ . (b) Show that if u is a weak solution that is C 1 in an open set Ω in the upper half of the xt plane, then u is a genuine solution of equation ( 8.5.15 ) in Ω. 6. ( The jump condition , which is also known in gas dynamics as the Rankine-Hugoniot condition .) The definition of a weak solution given in Exercise 5 clearly allows discontinuous solutions. However, the reader shall now determine that not every type of discontinuity is admissible, for there is a connection between the discontinuity curve and the values of the solution on both sides of the discontinuity. Let u be a (weak) solution of equation ( 8.5.15 ) and suppose Γ is a smooth curve in the xt plane such that u “jumps” across a curve Γ; that is, u is of a class C 1 except for jump discontinuity across Γ. We call that Γ a shock wave . Choose a point P Γ and construct, near P , a “rectangle” D = D 1 D 2 , as shown in Figure 8.5.3 . Choose φ to vanish on D and outside D . R t x P Q D 1 D 2 Γ Figure 8.5.3. The solution u jumps in value from u 1 to u 2 across Γ. (a) Show that D [ t + f ( u ) φ x ] dxdt = 0
104 8 Integral Theorems of Vector Analysis and D 1 [ t + f ( u ) φ x ] dxdt = D 1 [( ) t + ( f ( u ) φ ) x ] dxdt. (b) Suppose that u jumps in value from u 1 to u 2 across Γ so that when ( x, t ) approaches a point ( x 0 , t 0 ) on Γ from ∂D i , u ( x, t ) approaches the value u i ( x 0 , t 0 ) . Show that 0 = ∂D 1 φ [ u dx + f ( u ) dt ] + ∂D 2 φ [ u dx + f ( u ) dt ] and deduce that 0 = Γ φ ([ u ] dx + [ f ( u )] dt ) where [ α ( u )] = α ( u 2 ) α ( u 1 ) denotes the jump in the quantity α ( u ) across Γ. (c) If the curve Γ defines x implicitly as a function of t and ∂D intersects Γ at Q = ( x ( t 1 ) , t 1 ) and R = ( x ( t 2 ) , t 2 ), show that 0 = R Q φ ([ u ] dx + [ f ( u )] dt ) = t 2 t 1 φ [ u ] dx dt + [ f ( u )] dt. (d) Show that at the point P on Γ, [ u ] · s = [ f ( u )] , (8.5.17) where s = dx/dt at P . The number s is called the speed of the discontinuity. Equation ( 8.5.17 ) is called the jump condition ; it is the relationship that any discontinuous solution will satisfy. 7. (Loss of uniqueness) One drawback of accepting weak solutions is loss of uniqueness. (In gas dynamics, some mathematical solutions are extraneous and rejected on physical grounds. For example, dis- continuous solutions of rarefaction shock waves are rejected because they indicate that entropy decreases across the discontinuity.) Consider the equation u t + u 2 2 x = 0 , with initial data u ( x, 0) = 1 , x 0 1 , x < 0 .
8.5 Green’s Functions 105 Show that for every α 1 , u α is a weak solution, where