Math 5440 Math 5440 Problem Set 4 Solutions Aaron Fogelson Fall, 03 : (Logan,.8 # 4) Find all radial solutions of the two-dimensional Laplace s equation. That is, find all solutions of the form u(r) where r = x + y. Find the steady-state temperature distribution in the annular domain r if the inner circle r = is held at 0 degrees and the outer circle r = is held at 0 degrees. We seek u(r) that satisfies 0 = u r + u r r = { r u }. r r r Integrating, we see that ru r = C for some constant C and integrating again we obtain u(r) = C ln(r) + B where B and C are constants. To satisify the given conditions, we need 0 = u() = B, so B = 0, and 0 = u() = C ln(), so C = 0/ ln(). Hence u(r) = 0 ln r ln.
: (Logan,.9 # ) Classify the PDE u xx + ku xt + k u tt = 0, k = 0. Find a transformation ξ = x+bt, τ = x+dt of the independent variables that transforms the equation into a simpler equation of the form U ξξ = 0. Find the solution to the given equation in terms of two arbitrary functions. The equation u xx + ku xt + k u tt = 0 k = 0, has A =, B = k, and C = k, so D = 4k 4k = 0 and the equation is therefore parabolic. Letting ξ = x + bt and τ = x + dt converts the equation to (+kb+k b )U ξξ +(+k(d+b)+k bd)u ξτ +(+kd+k d )U ττ = 0 and choosing d = /k and b = 0 this becomes Integrating twice, we find that U ξξ = 0. U(ξ, τ) = F(τ)ξ + G(τ) for arbitrary functions F and G. In terms of x and t, the solution is u(x, t) = F(x t/k)x+g(x t/k).
3: (Logan,.9 # 4) Show that the equation can be transformed into the form u tt c u xx + au t + bu x + du = f(x, t) w ξτ + kw = g(ξ, τ), w = w(ξ, τ), by first making the transformation ξ = x ct, τ = x + ct, and then letting u = w exp(αξ + βτ) for some choice of α, β. Let ξ = x ct and τ = x+ct, u(x, t) = U(ξ, τ) = U(x ct, x+ct), and f(x, t) = F(ξ, τ). Then, we see that u xx = U ξξ + U ξτ + U ττ, and u tt = U ξξ c c U ξτ + c U ττ. Hence the PDE f = u tt c u xx + au t + bu x + du, becomes F = 4c U ξτ +(b ac)u ξ +(b+ac)u τ + du. Letting U(ξ, τ) = w(ξ, τ) exp(αξ + βτ), note that U ξ = w ξ exp(αξ + βτ)+αw exp(αξ + βτ), U τ = w τ exp(αξ+ βτ)+ βw exp(αξ+ βτ), and U ξτ = ( w ξτ + βw ξ + αw τ + αβw ) exp(αξ+ βτ). Substituting these into the PDE we get [ F = 4c { w ξτ + βw ξ + αw τ + αβw } +(b ac)(w ξ + αw) + (b+ac)(w τ + βw)+dw] exp(αξ + βτ), or F = [ ] 4c w ξτ +(b ac 4βc )w ξ +(b+ac 4αc )w τ +( 4c αβ+α(b ac)+ β(b+ac)+d)w exp(αξ + βτ) Setting β = (b ac)/4c and α = (b+ac)/4c kills off the first derivative terms and gives us an equation of the desired form. 3
4: (Logan,. # ) Solve the Cauchy problem u t = ku xx, < x < and t > 0, u(x, 0) = φ(x), < x <, for each of the following initial conditions a) φ(x) = if x < and φ(x) = 0 if x >. b) φ(x) = exp( x) if x > 0 and φ(x) = 0 if x < 0. For each part, set k = 0.00 and use Maple to plot the solution as a function of x for t = 0, 0.5,.0,.5. a) For φ(x) = { x <, 0, x >, u(x, t) = = G(x y, t)φ(y)dy e (x y) /4βt dy. 4πβt Letting r = (y x)/ 4βt, this formula becomes u(x, t) = π ( x)/ 4βt (+x)/ 4βt e r dr = ( x)/ 4βt e r dr (+x)/ 4βt e r dr π 0 π 0 ( ) ( ) = erf x (+ x) 4βt erf. 4βt b) For φ(x) = { e x x > 0, 0, x < 0, u(x, t) = = = 0 0 G(x y, t)φ(y)dy e (x y) /4βt e y dy 4πβt e (x y) /4βt y dy 4πβt 4
0.8 0.8 0.8 0.8 0.6 0.6 0.6 0.6 0.4 0.4 0.4 0.4 0. 0. 0. 0. 4 0 4 4 0 4 4 0 4 4 0 4 x x x x Figure 0.: Plots of solution at times 0, 0.5,, and.5. We complete the square in the exponent y xy+x 4βt y { y = (4βt x)y+x } 4βt { (y+βt x) = + 4βtx 4β t } 4βt = (y+βt x) 4βt Substituting this into the formula above for u(x, t) we get u(x, t) = e βt x 4πβt Letting s = (y+βt x)/ 4βt, we get u(x, t) = e βt x e s ds π βt x 4βt = eβt x { erf 0 x+βt. e (y+βt x) /4βt dy. ( )} βt x. 4βt 5
0.8 0.6 0.4 0. 0.4 0. 0. 0.4 x Figure 0.: Plots of solution at times 0, 0.5,,.5, and.0. 6
5: (Logan,. # ) If φ(x) M for all x, where M is a positive constant, show that the solution u to the Cauchy problem u t = ku xx, < x < and t > 0, u(x, 0) = φ(x), < x <, satisfies u(x, t) M for all x and t > 0. Hint: Use the calculus fact that the absolute value of an integral is less than or equal to the integral of the absolute value. The solution to the IVP u t = βu xx for < x < and t > 0, with u(x, 0) = φ(x) for < x < is u(x, t) = G(x y, t)φ(y)dy. If φ(x) M for all x, then u(x, t) = = = M. G(x y, t)φ(y)dy G(x y, t)φ(y) dy G(x y, t) φ(y) dy G(x y, t)mdy The last step uses the fact that G(x y, t)dy = for all x and any t > 0. 7
6: (Logan,. # 4) Show that if u(x, t)and v(x, t) are any two solutions to the onedimensional heat equation u t = βu xx, then w(x, y, t) = u(x, t)v(y, t) solves the twodimensional heat equation, w t = β(w xx + w yy ). Guess the solution to the twodimensional Cauchy problem w t = β(w xx + w yy ), < x <, < y <, t > 0, w(x, y, 0) = ψ(x, y), < x <, < y <. Suppose that u(x, t) and v(x, t) solve the one-dimensional diffusion problem u t = βu xx. Let w(x, y, t) = u(x, t)v(y, t). Then w t = uv t + u t v = uβv yy + βu xx v = β((uv) xx + (uv) yy ) = β(w xx + w yy ). We have therefore shown that w(x, y, t) satisfies the two-dimensional diffusion equation. If we let u(x, t) = G(x, t) and v(y, t) = G(y, t), then w(x, y, t) = G(x, t)g(y, t) solves the two-dimensional diffusion equation and w(x, y, 0) = δ(x)δ(y), so this function is the fundamental solution of the two-dimensional diffusion equation. The solution to the two-dimensional initial value problem w t = β(w xx + w yy ) for < x <, < y <, and t > 0, with w(x, y, 0) = ψ(x, y) is w(x, y, t) = G(x x, t)g(y y, t)ψ(x, y )dx dy. 8
7: Suppose u tt = c u xx for < x < and t > 0, and u t (x, 0) = g(x) = 0 and { x / for x < u(x, 0) = f(x) = 0 for x. Let c = and find the solution to this problem. t=0,,,3,4,5,6. Use Maple to plot the solution at Since g(x) = 0 for all x, the solution to this problem is u(x, t) = { f(x ct)+ f(x+ct)}. There are various cases depending on whether x ct < and whether x + ct <. } { x+ct + x ct, if x+ct < and x ct < { } x+ct, if x+ct < and x ct u(x, t) = { } x ct, if x+ct and x ct < 0, if x+ct and x ct. 5 4 3 0 x+ct = Problem Set 4 #7 Regions x+ct = D x ct = x ct = B C E F A 0 8 6 4 0 4 6 8 0 Figure 0.3: Plot in xt-plane. In region A, x ct < and x + ct < ; in region B, x ct > and x+ct < ; in region C, x ct < and x+ct > ; in regions D,E, and F, x ct > and x+ct >. Figure 0.4: Plots of solution at times 0,,, 3, 4, 5, and 6. 9
8: Suppose u tt = c u xx for < x < and t > 0, and u(x, 0) = f(x) = 0 and { for x < u t (x, 0) = g(x) = 0 for x. Let c = and find the solution to this problem. t=0,,,3,4,5,6. Use Maple to plot the solution at Since f(x) = 0, the D Alembert solution to the wave equation is u(x, t) = c For the given initial velocity, we obtain x+ct x ct g(s)ds. 0 if x ct < and x+ct <, c (+ x+ct) if x ct < and x+ct, u(x, t) = c if x ct < and < x+ct, t if x ct and x+ct, c ( x+ct) if x ct and < x+ct, 0 if < x ct and < x+ct. Solution at t = 0 is identically 0. Figure 0.5: Plots of solution at times,, 3, 4, 5, and 6. 0
9: A spherical wave is a solution of the three-dimensional wave equation of the form u(r, t) where r is the distance from the origin. The wave equation takes the form u tt = c (u rr + ) r u r. () which is called the spherical wave equation. a) Change variables v = ru to get the equation for v: v tt = c v rr. b) Solve for v using the general solution of the wave equation and thereby solve the spherical wave equation (). c) Use the D Alembert solution of the wave equation to solve Eq() with initial conditions u(r, 0) = f(r) and u t (r, 0) = g(r). Assume that both f(r) and g(r) are even functions of r. a) Let v = ru. Then u tt = (/r)v tt, u r = (/r)v r (v/r ), and u rr = (/r )(rv rr v r ) (/r 4 )(r v r rv). Hence, u rr +(/r)u r = v rr, and v(r, t) therefore satisfies the equation v tt = c v rr. b) Consider v tt = c v rr for all r. The solution has the form v(r, t) = F(r ct)+g(r+ct) for arbitrary functions F and G. Hence, u(r, t) = r F(r ct)+ r G(r+ct). c) Since u(r, 0) = f(r), v(r, 0) = r f(r). Since u t (r, 0) = g(r), v t (r, 0) = rg(r). The problem v tt = c v rr with these initial conditions has the solution v(r, t) = {(r ct) f(r ct)+(r+ct) f(r+ct)}+ c Since u = v/r, the solution u(r, t) is u(r, t) = r r+ct r ct sg(s)ds. r+ct {(r ct) f(r ct)+(r+ct) f(r+ct)}+ sg(s)ds. cr r ct Since u(r, t) = v(r, t)/r, there is a problem as r 0, unless v(r, t) = 0. But the fact that f(r) and g(r) are even functions implies that r f(r) and rg(r) are odd functions. Therefore, v(0, t) = { ct f( ct)+ct f(ct)}+ c +ct ct sg(s)ds = 0, as required.
0: Solve the problem 4u tt + 3u xt u xx = 0 with u(x, 0) = x and u t (x, 0) = e x. (Hint: One way to do this is to factor the PDE. Another is to use a transformation based on the characteristic coordinates.) Write the PDE in factored form as ( 4 t )( x t + ) u = 0. x Let v = u t + u x. We want 4v t v x = 0, or v t /4v x = 0. This implies that v(x, t) = F(x + t/4) for any function F. We then want u t + u x = F(x+t/4). () We look for a particular solution to this in the form u(x, t) = H(x+ 4 t). Since u t = H /4 and u x = H, we get a particular solution by choosing H such that 5/4H = F. Since F is arbitrary, so is H. The general solution of the nonhomogeneous equation () for u is this particular solution plus the general solution of the corresponding homogeneous equation, so u(x, t) = H(x+t/4)+G(x t) for arbitrary functions H and G. To satisfy the initial conditions we want and x = u(x, 0) = H(x)+G(x), e x = u t (x, 0) = /4H (x) G (x). Differentiating the first of these two equations gives us two equations for H and G, namely H + G = x /4H G = e x Hence 5/4H (x) = x+e x and H(x) = 4/5(x + e x )+C for constant C. Then G(x) = x H(x) = /5x 4/5e x C. So the solution u(x, t) is given by ( u(x, t) = H(x+/4t)+G(x t) = 4/5 (x+t/4) + e x+t/4) ( + /5 (x t) 4e x t). Another way to solve the problem is to make a change of variables: Let ξ = x+bt, τ = x+dt, and U(ξ, τ) = u(x, t). We then find that the PDE can be written (4b + 3b )U ξξ +(4d + 3d )U ττ +(8bd+3(b+d) )U ξτ = 0 Choosing b = ( 3+ 5)/8 = /4 and d = ( 3 5)/8 =, the coefficients of the U ξξ and U ττ terms are 0, and the coefficent of the U ξτ term is nonzero. So the equation becomes U ξτ = 0. This has solution U(ξ, τ) = H(ξ)+G(τ) for arbitrary differentiable functions H and G. Hence u(x, t) = H(x+t/4)+G(x t). Then H and G are determined from the initial conditions as above.