9.5 Verify that the solution of the boundary value problem in Example 9.2 at points P(X₁, T₁) and P(X3, 73) shown in Fig
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- 9 5 Verify That The Solution Of The Boundary Value Problem In Example 9 2 At Points P X T And P X3 73 Shown In Fig 1 (14.89 KiB) Viewed 38 times
- 9 5 Verify That The Solution Of The Boundary Value Problem In Example 9 2 At Points P X T And P X3 73 Shown In Fig 2 (70.57 KiB) Viewed 38 times
- 9 5 Verify That The Solution Of The Boundary Value Problem In Example 9 2 At Points P X T And P X3 73 Shown In Fig 3 (17.22 KiB) Viewed 38 times
in the question
9.5 Verify that the solution of the boundary value problem in Example 9.2 at points P(X₁, T₁) and P(X3, 73) shown in Fig. 9.2 can also be written in the form (9.13), where the components of f contain linear combinations of values of u(x, 0).
Example 9.2 Suppose that the 3 x 3 matrix A has eigenvalues A₁ = -1, A2 = 1 and X3 = 2 and that the corresponding eigenvectors of AT are v₁ = [1,0, 1], v₂ = [0, 1, 1] and v3 = [1, 1,0]. Consider the boundary value problem in which the PDE Aux + u, = 0 is to be solved in the first quadrant of the x-1 plane with initial data that specifies the three components of u = [u, v, w] on the line t = 0, and with two boundary conditions (0, 1) = v(0, 1) and w(0, 1) = 0 specified on the positive 1-axis. Use the method of characteristics to determine the solution at the point P(X₂, T₂) with T₂ < X₂ < 27₂. There are three families of characteristics: ₁:₁ = -1, x+1= constant, vu=u+w=constant, x-t=constant, vu=v+w=constant, 1₂:4₂ = 1, 13:3= 2, x-2r=constant, vu=u+v=constant. Note that, as discussed in Sect. 4.1, the fact that two of the three characteristic families (2 and 3) are directed into the domain along the t-axis is the reason why two boundary conditions need to be specified there. The first quadrant is divided into three regions by the two incoming characteristics (x-1=0 and x-2r = 0) that pass through the origin. Typical points P₁ (X₁ > 2T₁), P₂ (T₂ < X₂ <27₂) and P3 (X3 < T3) in each of these regions are shown in Fig. 9.2. P3(X3, T3) P2(X2, T2) KKK P₁(X₁, T₁) x BC Dz Fig. 9.2 The characteristics through points P₁, P2, and P3 for Example 9.2 drawn backwards in time, with reflections when they intersect the t-axis. The dashed lines show the characteristics x-1=0 and x-2r=0 that pass through the origin
VTu (P₂) = f. (9.13) where f = [u(D) + w(D), v(C) + w(C), 2 (u(B) + w(B))]T. Since the coefficient matrix in (9.13) is nonsingular, we deduce that the solution is uniquely specified by the given initial data on the line t = 0.