By Pinchover Y., Rubinstein J.
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Additional resources for An introduction to partial differential equations. Extended solutions for instructors
13, Legendre polynomials with different indices are orthogonal to each other on E(−1, 1). Furthermore, since Pn is an n-degree polynomial, we infer that Pn (t) satisfies 1 −1 tl Pn (t) dt = 0 ∀l = 0, 1, 2, . . , n − 1. 48), together with the normalization Pn (1) = 1 determines the Legendre polynomials uniquely. Set 1 dn n Qn (t) := n t2 − 1 . n 2 n! dt Clearly, Qn is an n-degree polynomial. Repeatedly integrating by parts it follows that 1 Q (s)Qm (s) ds = 0 for n = m. Moreover, Qn (1) = 1. Therefore, Pn = Qn .
2 An e−n t cos nx . w(x, t) = A0 + n=1 34 Evaluating the sum at t = 0 ∞ w(x, 0) = A0 + An cos nx = π cos 2x + n=1 1 cos 2001x , 20014 and comparing coefficients we get A2 = π, A2001 = 1 , 20014 An = 0 n = 2, 2001. Finally we write u(x, t) = πe−4t cos 2x+ 1 −20012 t 1 t e cos 2001x+ − 4 2 2001 2001 20014 cos 2001x. 22 Write v(x, t) = a(t)x2 + b(t)x + c(t) to obtain from the boundary conditions the function v(x, t) = x2 /2 + c(t). If we demand v to solve the homogeneous PDE too, we further find vt − 13vxx = c (t) − 13 = 0, =⇒ c(t) = 13t.
R a Substituting λn into the equation for T we derive Tn (t) = exp(−n2 π 2 t/a2 ), and the solution takes the form ∞ An e− w(r, t) = n2 π 2 t a2 n=1 1 nπr sin . r a The initial conditions then imply ∞ w(r, 0) = An sin n=1 nπr = r (r − a). e. 2 An = a a 0 nπr 4 a2 r (r − a) sin dr = − 3 3 [1 − (−1)n ]. 1 Select ψ = v ∇u in Gauss’ theorem: ∇ · ψ(x, y) dxdy = D ψ(x(s), y(s)) · n ˆ ds. 3 We solve by the separation of variables method: u(x, y) = X(x)Y (y). We obtain X Y + Y X = kXY ⇒ −Y Y = X − k = λ.