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

By Pinchover Y., Rubinstein J.

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Extra resources for An introduction to partial differential equations. Extended solutions for instructors

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2(−1)n h . nπ(n2 + h) The constants An will be determined later on. Therefore, ∞ An e−(n w(x, t) = n=1 2 +h)t + 2(−1)n h nπ(n2 + h) sin nx . We proceed to find An from the initial condition ∞ w(x, 0) = An + n=1 2(−1)n h x = sin nx = − nπ(n2 + h) π ∞ n=1 2(−1)n sin nx . nπ 33 Therefore, An = 2(−1)n nπ 1− n2 h +h n = 1, 2, 3, . . It follows that ∞ w(x, t) = n=1 2(−1)n nπ 1− n2 h +h e−(n 2 +h)t + n2 h sin nx , +h and u(x, t) = w(x, t) + v(x, t) = w(x, t) + πx . This solution is not classical at t = 0, since the sine series does not converge to −x/π in the closed interval [0, 1].

This shows that yM is a minimizer, and it is indeed unique. 5 (a) The action is t2 J= t1 D 1 2 1 u − |∇u|2 − V (u) dx. 2 t 2 (b) Taking the first variation and equating it to zero we obtain the nonlinear KleinGordon equation utt − ∆u + V (u) = 0. 7 (a) Introducing a Lagrange multiplier λ, we solve the minimization problem |∇u|2 dxdy + λ 1 − min D u2 dxdy , D for all u that vanish on ∂D. Equating the first variation to zero we obtain the EulerLagrange equation ∆u = −λu x ∈ D, u = 0 x ∈ ∂D. 54) by u and integrate by parts.

Let M be the maximum of u on ∂D. For all (x1 , y1 ) ∈ D u(x1 , y1 ) ≤ vε (x1 , y1 ) ≤ max{vε (x, y) | (x, y) ∈ ∂D} ≤ M + επ 2 . Letting ε → 0, we obtain u(x1 , y1 ) ≤ M . (c) Write w(x, y) := u1 (x, y) − u2 (x, y), where u1 (x, y), u2 (x, y) are two solutions of the problem. We should show that w(x, y) = 0 in D. Notice that the functions ±w(x, y) solve the equation with homogeneous boundary conditions. Therefore, part (b) implies ±w(x, y) ≤ 0 in D, namely w(x, y) = 0 in D. 17 (a) The general solution is of the form ∞ 2 Bn e−2n t sin nx.

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