Math 5440 Problem Set 4 Solutions

Transcript

1 Math 544 Math 544 Problem Set 4 Solutions Aaron Fogelson Fall, 5 : (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 = p + y. Find the steady-state temperature distribution in the annular domain r if the inner circle r = is held at degrees and the outer circle r = is held at degrees. We seek u(r) that satisfies = u u r + 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 = u() = B, so B =, and = u() = C ln(), so C = / ln(). Hence u(r) = ln r ln.

2 : (Logan,.9 # ) Classify the PDE u + ku t + k u tt =, k 6=. Find a transformation ξ = + bt, τ = + dt of the independent variables that transforms the equation into a simpler equation of the form U ξξ =. Find the solution to the given equation in terms of two arbitrary functions. The equation u + ku t + k u tt = k 6=, has A =, B = k, and C = k, so D = 4k 4k = and the equation is therefore parabolic. Letting ξ = + bt and τ = + dt converts the equation to ( + kb + k b )U ξξ + ( + k(d + b) + k bd)u ξτ + ( + kd + k d )U ττ = and choosing d = /k and b = this becomes Integrating twice, we find that U ξξ =. U(ξ, τ) = F(τ)ξ + G(τ) for arbitrary functions F and G. In terms of and t, the solution is u(, t) = F( t/k) + G( t/k).

3 3: (Logan,.9 # 4) Show that the equation can be transformed into the form u tt c u + au t + bu + du = f (, t) w ξτ + kw = g(ξ, τ), w = w(ξ, τ), by first making the transformation ξ = ct, τ = + ct, and then letting u = w ep(αξ + βτ) for some choice of α, β. Letting ξ = ct and τ = + ct, and u(, t) = U(ξ, τ) = U( ct, + ct), we see that u = 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 + au t + bu + du, becomes f = 4c U ξτ + (b ac)u ξ + (b + ac)u τ + du. Letting U(ξ, τ) = w(ξ, τ) ep(αξ + βτ), note that U ξ = w ξ ep(αξ + βτ) + αw ep(αξ + βτ), U τ = w τ ep(αξ + βτ) + βw ep(αξ + βτ), and U ξτ = w ξτ + βw ξ + αw τ + αβw ep(αξ + βτ). Substituting these into the PDE we get h f = 4c w ξτ + βw ξ + αw τ + αβw + (b ac)(w ξ + αw) + (b + ac)(w τ + βw) + dw] ep(αξ + βτ), or f = h i 4c w ξτ + (b ac 4βc )w ξ + (b + ac 4αc )w τ + ( 4c αβ + α(b ac) + β(b + ac))w ep(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 4: (Logan,. # ) Solve the Cauchy problem u t = ku, < < and t >, u(, ) = φ(), < <, for each of the following initial conditions a) φ() = if jj < and φ() = if jj >. b) φ() = ep( ) if > and φ() = if <. For each part, set k =. and use Maple to plot the solution as a function of for t =,.5,.,.5. a) For φ() = ( jj <,, jj >, u(, t) = = Z G( y, t)φ(y)dy e ( y) /4βt p dy. 4πβt Letting r = (y )/ p 4βt, this formula becomes u(, t) = p π Z ( )/ p 4βt (+)/ p 4βt e r dr Z p ( )/ 4βt Z p = p e r dr p (+)/ 4βt e r dr π π!! = erf p 4βt erf ( + ) p. 4βt b) For φ() = ( e >,, <, u(, t) = = = G( y, t)φ(y)dy e ( y) /4βt p e y dy 4πβt e ( y) /4βt y p dy 4πβt 4

5 Figure.: Plots of solution at times,.5,, and.5. We complete the square in the eponent y y + 4βt y y = (4βt )y + 4βt (y + βt ) + 4βt 4β t = 4βt (y + βt ) = + βt. 4βt Substituting this into the formula above for u(, t) we get u(, t) = e βt p 4πβt Letting s = (y + βt )/ p 4βt, we get Z u(, t) = e βt p e s ds π βt p 4βt = eβt ( erf e (y+βt ) /4βt dy.!) βt p. 4βt 5

6 Figure.: Plots of solution at times,.5,,.5, and.. 6

7 5: (Logan,. # ) If jφ()j M for all, where M is a positive constant, show that the solution u to the Cauchy problem u t = ku, < < and t >, u(, ) = φ(), < <, satisfies ju(, t)j M for all and t >. 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 for < < and t >, with u(, ) = φ() for < < is If φ() M for all, then u(, t) = ju(, t)j = j = = M. G( y, t)φ(y)dy. G( y, t)φ(y)dyj jg( y, t)φ(y)jdy G( y, t)jφ(y)jdy G( y, t)mdy The last step uses the fact that R G( y, t)dy = for all and any t >. 7

8 6: (Logan,. # 4) Show that if u(, t)and v(, t) are any two solutions to the onedimensional heat equation u t = βu, then w(, y, t) = u(, t)v(y, t) solves the twodimensional heat equation, w t = β(w + w yy ). Guess the solution to the twodimensional Cauchy problem w t = β(w + w yy ), < <, < y <, t >, w(, y, ) = ψ(, y), < <, < y <. Suppose that u(, t) and v(, t) solve the one-dimensional diffusion problem u t = βu. Let w(, y, t) = u(, t)v(y, t). Then w t = uv t + u t v = uβv yy + βu v = β((uv) + (uv) yy ) = β(w + w yy ). We have therefore shown that w(, y, t) satisfies the two-dimensional diffusion equation. If we let u(, t) = G(, t) and v(y, t) = G(y, t), then w(, y, t) = G(, t)g(y, t) solves the two-dimensional diffusion equation and w(, y, ) = δ()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 + w yy ) for < <, < y <, and t >, with w(, y, ) = ψ(, y) is w(, y, t) = G(, t)g(y y, t)ψ(, y )d dy. 8

9 7: Suppose u tt = c u for < < and t >, and u t (, ) = g() = and u(, ) = f () = ( jj/ for jj < for jj. Let c = and find the solution to this problem. t=,,,3,4,5,6. Use Maple to plot the solution at Since g() = for all, the solution to this problem is u(, t) = f f ( ct) + f ( + ct)g. There are various cases depending on whether j ctj < and whether j + ctj <. u(, t) = 8 n o j+ctj + j ctj, if j + ctj < and j ctj < n o >< j+ctj, if j + ctj < and j ctj n o j ctj, if j + ctj and j ctj < >:, if j + ctj and j ctj ct = Problem Set 4 #7 Regions +ct = D ct = ct = B C E F A Figure.3: Plot in t-plane. In region A, j ctj < and j + ctj < ; in region B, j ctj > and j + ctj < ; in region C, j ctj < and j + ctj > ; in regions D,E, and F, j ctj > and j + ctj > Figure.4: Plots of solution at times,,, 3, 4, 5, and 6. 9

10 8: Suppose u tt = c u for < < and t >, and u(, ) = f () = and u t (, ) = g() = ( for jj < for jj. Let c = and find the solution to this problem. t=,,,3,4,5,6. Use Maple to plot the solution at Since f () =, the D Alembert solution to the wave equation is u(, t) = c Z +ct ct g(s)ds. For the given initial velocity, we obtain 8 if ct >< c ( + ct) if ct and + ct u(, t) = c if ct and + ct, c ( + ct + ) if + ct and ct >: if + ct. Solution at t = is identically Figure.5: Plots of solution at times,, 3, 4, 5, and 6.

11 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, ) = f (r) and u t (r, ) = 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) + G(r + ct). r c) Since u(r, ) = f (r), v(r, ) = r f (r). Since u t (r, ) = g(r), v t (r, ) = rg(r). The problem v tt = c v rr with these initial conditions has the solution v(r, t) = f(r ct) f (r ct) + (r + ct) f (r + ct)g + c Since u = v/r, the solution u(r, t) is u(r, t) = r f(r ct) f (r ct) + (r + ct) f (r + ct)g + cr Z r+ct r ct Z r+ct r ct sg(s)ds. sg(s)ds. Since u(r, t) = v(r, t)/r, there is a problem as r!, unless v(r, t) =. 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(, t) = f ct f ( ct) + ct f (ct)g + c Z +ct ct sg(s)ds =, as required.

12 : Solve the problem 4u tt + 3u t u = with u(, ) = and u t (, ) = e. (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 t + u =. Let v = u t + u. We want 4v t v =, or v t /4v =. This implies that v(, t) = F( + t/4) for any function F. We then want u t + u = F( + t/4). () We look for a particular solution to this in the form u(, t) = H( + 4 t). Since u t = H /4 and u = 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(, t) = H( + t/4) + G( t) for arbitrary functions H and G. To satisfy the initial conditions we want and = u(, ) = H() + G(), e = u t (, ) = /4H () G (). Differentiating the first of these two equations gives us two equations for H and G, namely H + G = /4H G = e Hence 5/4H () = + e and H() = 4/5( + e ) + C for constant C. Then G() = H() = /5 4/5e C. So the solution u(, t) is given by u(, t) = H( + /4t) + G( t) = 4/5 ( + t/4) + e +t/4 + /5 ( t) 4e t.