# Download An introduction to partial differential equations. Extended by Pinchover Y., Rubinstein J. PDF

By Pinchover Y., Rubinstein J.

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Example text

The time step is always ∆t = ∆x/4. The results are presented in the Table below. Notice that adding terms into the partial sums of the Fourier representation adds very little to the accuracy. 74) is given by Ui,n+1 − Ui,n Ui+1,n − 2Ui,n + Ui−1,n Ui+1,n+1 − 2Ui,n+1 + Ui−1,n+1 = + , ∆t 2(∆x)2 2(∆x)2 where Ui,n = u(xi , tn ), condition leads to 1 ≤ i ≤ N − 1, n ≥ 0, and N = π/∆x + 1. The initial Ui,0 = f (xi ) 0 ≤ i ≤ N − 2, xi = i∆x. Observe that the solution at the boundary point x = 0 is determined by the boundary condition U0,n = 0 n ≥ 1.

18, u solves the heat equation in D. 19 (a) The mean value theorem for harmonic functions implies u(0, 0) = 1 2π π u(r, θ) dθ −π for all 0 < r ≤ R. Substitute r = R into the equation above to obtain u(0, 0) = 1 2π π u(R, θ) dθ = −π 1 2π π/2 sin2 (2θ) dθ = −π/2 1 . 4 (b) This is an immediate consequence of the strong maximum principle. This principle implies u(r, θ) ≤ max u(R, ψ) = 1 ψ∈[−π/2,π/2) for all r < R, and the equality holds if and only if u is constant. Clearly our solution is not a constant function, and therefore u < 1 in D.

24 Substituting the expansion ∞ u(x, t) = Tn (t) cos nx n=0 36 into the PDE we obtain ∞ (Tn ) (t) cos nx + n2 Tn (t) cos nx = cos 2t cos 3x , n=0 leading to (Tn ) (t) + n2 Tn (t) = 0 (T3 ) (t) + 9 T3 (t) = cos 2t n=3, n=3. 25) n = 0, 3. Therefore, ∞ 1 u(x, t) = cos 2t cos 3x + (A0 t + B0 ) + (An cos nt + Bn sin nt) cos nx . 5 n=1 The first initial condition ∞ 1 1 An cos nx = cos2 x = (cos 2x + 1) u(x, 0) = cos 3x + B0 + 5 2 n=1 implies 1 1 A 3 = − , A2 = , 5 2 The second initial condition B0 = 1 , 2 An = 0 ∀n = 0, 2, 3.