AMATH 353 Lecture 9. Weston Barger. How to classify PDEs as linear/nonlinear, order, homogeneous or non-homogeneous.

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1 AMATH 353 ecture 9 Weston Barger 1 Exam What you need to know: How to classify PDEs as linear/nonlinear, order, homogeneous or non-homogeneous. The definitions for traveling wave, standing wave, wave train How to find the dispersion relation of a linear PDEs, group velocity, phase velocity How to tell if an equation is dispersive. How to tell from the dispersion relation if solutions experience growth or decay. Characteristics of the advection and wave equation. How to draw the (x, t wave profile for piecewise initial data. I did an example in class. d Alembert s solution for the full line. d Alembert s solution for the half-line boundary value problem. Separation of variables. How to compute Fourier coefficients. How to define orthogonality on [, ]. Change of variables/chain rule. How to do the homework. continuing lecture 8... Doing a similar calculation (which you should do, we get u yy = cos θ r u θθ sin θ cos θ r u θ + cos θ u r + sin θu rr + r cos θ sin θ u r θr 1

2 Adding these two together, we get u xx + u yy = u rr + 1 r u r + 1 r u θθ. We can now reformulate our problem in polar coordinates: laplacepolar} u xx + u yy = u rr + 1 r u r + 1 r u θθ, r (, R, θ [, π u(r, θ = f (θ. (.1 form As we have done on other finite domains, we apply separation of variables. We look for a solution of the Inserting the above equation into (.1 gives u(r, θ = S(rT (θ. S T + 1 r S T + 1 r T S = S + r S S + T T = S + r S S = T T = λ. Since our equation for T is the most simple, we solve it first. Again, we break into cases the sign of λ: λ =. Then T = T = a + bθ. We want u to be a single valued function in r and θ. Thus, u must be periodic in θ. Therefore, b = and we have T = 1, ignoring the multiplicative constant a. In S, we have + rs = rs + S = (rs = rs = c 1 S (r = c 1 ln(r + c. We require that S does not blow up as r, as this would not satisfy our PDE (or our physical situation. Therefore, c 1 =. Our full solution is thus u (r, θ = c

3 λ >. We let λ = α, with α >. Then we get T αt = T (θ = a cos(αθ + b sin(αθ. We need T to be π periodic, so this implies α = n for n = 1,, 3,. Then λ = n. This gives Next we examine S. Well, we have T n (θ = a n cos(nθ + b n sin (nθ. n + rs n n S n =. This is a well known ODE called the Euler ODE. We solve this by looking for solutions of the form S n (r = r p. Inserting this expression into the ODE above yields This gives that p(p 1 + p n = p = ±n. S n (r = c n r n + d n r n. Since we require that S n stays bounded as r, we get that d n =. Therefore, u n (r, θ = r n (a n cos(nθ + b n sin (nθ. Suppose that λ = α, for α >. Then This solution is not periodic, so a n = b n =. The general solution is thus u(r, θ = a + T (θ = a n e αθ + b n e αθ. r n (a n cos (nθ + b n sin (nθ. We can now apply the boundary condition u(r, θ = f (θ. This results in f (θ = a + R n (a n cos (nθ + b n sin (nθ. So, using the Fourier series formula, we get that a = 1 π f (θ dθ π a n = 1 π πr n f (θ cos (nθ dθ b n = 1 π πr n f (θ sin (nθ dθ. This completely determines aplace s equation on the circle. 3

4 eq:polar} Proposition.1. et R > be a positive number and define the change of variable ( r = x + y y, θ = arctan. (. x Using (., the problem u xx + u yy =, (x, y (ξ, η ξ + η < R } u(x, y = f (x, y, for x + y = R becomes u rr + 1 r u r + 1 r u θθ =, r (, R, θ [, π u(r, θ = f (θ whose solution is a = 1 π b n = 1 πr n u(r, θ = a + r n (a n cos (nθ + b n sin (nθ π f (θ dθ, a n = 1 π πr n f (θ cos (nθ dθ, π f (θ sin (nθ dθ 3 Complex Form of the Fourier Series eq:fourier} :fouriercoef} eq:euler} Recall the Fourier series for a function f (x on [, ] f (x = a + a n = 1 ( nπx f (x cos dx ( a n cos ( nπx + b n sin ( nπx (3.1 b n = 1 ( nπx f (x sin dx, (3. Here, I have rewritten the coefficients so that a and a n are defined by the same integral and b =. Recall that cos θ = eiθ + e iθ Inserting the definitions (3.3 into (3.1, we see that, sin θ = eiθ e iθ. (3.3 i From (3., we see that f (x = a + 1 (a n ib n e inπx / + 1 (a n + ib n e inπx /. a ( n = a n, b ( n = b n. 4

5 mplexfourier} xfouriercoef} So, we can rewrite etting we have that f (x = a + 1 = a + 1 c n = a n + ib n (a n ib n e inπx / + 1 inπx / (a n + ib n e n= 1 (a n + ib n e inπx / + 1 (a n + ib n e inπx /. n= 1 c n = a n + ib n, f (x = c n e inπx / (3.4 n= = 1 = 1 ( ( nπx ( nπx f (x cos + i sin dx f (x e inπx / dx. (3.5 The equation (3.4 is known as the complex Fourier series. 3.1 Complex Orthogonality When we derived Fourier series, we determined the coefficients a n and b n in (3. by the orthogonality relationships of sin(nπx / and cos(nπx /. The same can be done to derive the complex Fourier series coefficients c n (3.5. Recall that the definition of the complex conjugate for a complex number z = a + ib is Therefore, we derive the relationships z = (a + ib = a ib. e iθ = cos(θ i sin(θ = cos( θ + i sin( θ = e iθ. We define the inner product of two complex valued functions f and g. f (z g(z dz. We note that for integers n and m, from the orthogonality relationships for sin and cos, we get ( ( e inπx / e imπx / nπx ( nπx ( ( mπx ( mπx dx = cos i sin cos + i sin dx = δ n,m + δ n,m = δ n,m. 5