SOLUTIONS TO SELECTED EXERCISES. Applied Partial Differential Equations, 3rd Ed. Springer-Verlag, NY (2015)

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1 SOLUTIONS TO SELECTED EXECISES Applied Partial Differential Equations, 3rd Ed. Springer-Verlag, NY (25) J. David Logan University of Nebraska Lincoln

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3 CHAPTE 2 Partial Differential Equations on Unbounded Domains. Cauchy Problem for the Heat Equation Exercise a. Making the transformation r (x y)/ we have u(x, t) 2 4πkt e (x y)2 / dy (x )/ (x+)/ ( erf π e r2 dr ( (x+)/ ) erf ( (x )/ )) Exercise b. We have u(x,t) 4πkt e (x y)2 / e y dy Now complete the square in the exponent of e and write it as (x y)2 Then make the substitution in the integral y x2 2xy +y 2 +y y+2kt x)2 r y +2kt x +kt x Then u(x,t) e kt x π 2 ekt x( erf (2kt x)/ e r2 dr ( (2kt x)/ )) Exercise 2. We have u(x,t) G(x y,t) φ(y) dy M G(x y,t)dy M 3

4 4 2. PATIAL DIFFEENTIAL EQUATIONS ON UNBOUNDED DOMAINS Exercise 3. Use This gives erf(z) 2 z e r2 dr π 2 z ( r 2 + )dr π 2 π (z z3 3 + ) w(x,t) 2 + x π t + Exercise 4. The verification is straightforward. We guess the Green s function in two dimensions to be g(x,y,t) G(x,t)G(y,t) e x2 / e y2 / 4πkt 4πkt 2 4πkt e (x +y 2 )/ where G is the Green s function in one dimension. Thus g is the temperature distribution caused by a point source at (x,y) (,) at t. This guess gives the correct expression. Then, by superposition, we have the solution u(x,y,t) 2 4πkt e ((x ξ)2 +(y η) 2 )/ ψ(ξ,η)dξdη Exercise 6. Using the substitution r x/ we get G(x,t)dx e r2 dr π Exercise 7. Verification is straightforward. The result does not contradict the theorem because the initial condition is not bounded. 2. Cauchy Problem for the Wave Equation Exercise. Applying the initial conditions to the general solution gives the two equations F(x)+G(x) f(x), cf (x)+cg (x) g(x) We must solve these to determine the arbitrary functions F and G. Integrate the second equation to get cf(x)+cg(x) x g(s)ds+c Now we have two linear equations for F and G that we can solve simultaneously.

5 2. CAUCHY POBLEM FO THE WAVE EQUATION 5 Exercise 2. Using d Alembert s formula we obtain u(x,t) 2c x+ct x ct ds +.25s 2 2arctan(s/2) x+ct x ct 2c c (arctan((x+ct)/2) arctan((x ct)/2)) Exercise 4. Let u F(x ct). Then u x (,t) F ( ct) s(t). Then Then F(t) t u(x,t) c Exercise 5. Letting u U/ρ we have s( r/c)dr + K t x/c s(y)dy +K u tt U tt /ρ, u ρ U ρ /ρ U/ρ 2 and u ρρ U ρρ /ρ 2U ρ /ρ 2 +2U/ρ 3 Substituting these quantities into the wave equation gives U tt c 2 U ρρ which is the ordinary wave equation with general solution Then U(ρ,t) F(ρ ct)+g(ρ+ct) u(ρ,t) r (F(ρ ct)+g(ρ+ct)) As a spherical wave propagates outward in space its energy is spread out over a larger volume, and therefore it seems reasonable that its amplitude decreases. Exercise 6. The exact solution is, by d Alembert s formula, u(x,t) 2 (e x ct +e x+ct )+ 2c (sin(x+ct) sin(x ct)) Exercise 7. Use the fact that u is has the same value along a characteristic. Exercise 8. Write where h(s,t) v H(s, t)u(x, s)ds 4πt(k/c2 ) e s2 /(4t(k/c 2 )) which is the heat kernel with k replaced by k/c 2. Thus H satisfies H t k c 2H xx

6 6 2. PATIAL DIFFEENTIAL EQUATIONS ON UNBOUNDED DOMAINS Then, we have v t kv xx (H t (s,t)u(x,s) kh(s,t)u xx (x,s))ds (H t (s,t)u(x,s) (k/c 2 )H(s,t)u ss (x,s))ds where, in the last step, we used the fact that u satisfies the wave equation. Now integrate the second term in the last expression by parts twice. The generated boundary terms will vanish since H and H s go to zero as s. Then we get v t kv xx (H t (s,t)u(x,s) (k/c 2 )H ss (s,t)u(x,s))ds 3. Well-Posed Problems Exercise. Consider the two problems u t +u xx, x, t > u(x,) f(x), x If f(x) the solution is u(x,t). If f(x) +n sinnx, which is a small change in initial data, then the solution is u(x,t) + n en2t sinnx which is a large change in the solution. So the solution does not depend continuously on the initial data. Exercise 2. Integrating twice, the general solution to u xy is u(x,y) F(x)+G(y) where F and G are arbitrary functions. Note that the equation is hyperbolic and therefore we expect the problem to be an evolution problem where data is carried forward from one boundary to another; so a boundary value problem should not be well-posed since the boundary data may be incompatible. To observe this, note that u(x,) F(x)+G() f(x). u(x,) F(x)+G() g(x) where f and g are data imposed along y and y, respectively. But these last equations imply that f and g differ by a constant, which may not be true.

7 4. SEMI-INFINITE DOMAINS 7 Exercise 3. We subtract the two solutions given by d Alembert s formula, take the absolute value, and use the triangle inequality to get u u 2 2 f (x ct) f 2 (x ct) + 2 f (x+ct) f 2 (x+ct) + 2c x+ct x ct 2 δ + 2 δ + 2c δ + 2c δ 2(2ct) δ +Tδ 2 g (s) g 2 (s) ds x+ct x ct δ 2 ds 4. Semi-Infinite Domains Exercise 2. We have u(x,t) (G(x y,t) G(x+y,t))dy erf (x/ ) Exercise 3. For x > ct we use d Alembert s formula to get u(x,t) 2 ((x ct)e (x ct) +(x+ct)e (x+ct) ) For < x < ct we have from (2.29) in the text u(x,t) 2 ((x+ct)e (x+ct) (ct x)e (ct x) ) Exercise 4. Letting w(x,t) u(x,t) we get the problem w t kw xx, w(,t), t >, ;w(x,), x > Now we can apply the result of the text to get w(x,t) Then (G(x y,t) G(x+y,t))( )dy erf (x/ ) u(x,t) erf (x/ ) Exercise 5. The problem is u t ku xx, x >, t > u(x,) 7, x > u(,t), t > From Exercise 2 we know the temperature is u(x,t) 7 erf (x/ ) 7 2 π x/ e r2 dr

8 8 2. PATIAL DIFFEENTIAL EQUATIONS ON UNBOUNDED DOMAINS The geothermal gradient at the current time t c is Solving for t gives u x (,t c ) 7 πktc t c sec yrs This gives a very low estimate; the age of the earth is thought to be about 5 billion years. There are many ways to estimate the amount of heat lost. One method is as follows. At t the total amount of heat was ρcu dv 7ρc 4 3 π3 2932ρc 3 S where S is the sphere of radius 4 miles and density ρ and specific heat c. The amount of heat leaked out can be calculated by integrating the geothermal gradient up to the present day t c. Thus, the amount leaked out is approximately (4π 2 ) tc Ku x (,t)dt 4π 2 ρck(7) So the ratio of the heat lost to the total heat is tc ρc 2 (.6 2 ) ρc 2 (.6 2 ) 2932ρc % πkt dt Exercise 6. Follow the suggested steps. Exercise 7. The left side of the equation is the flux through the surface. The first term on the right is Newton s law of cooling and the second term is the radiation heating. 5. Sources and Duhamel s Principle Exercise. The solution is u(x,t) 2c Exercise 2. The solution is t x+c(t τ) x c(t τ) sin sds c 2 sinx 2c 2(sin(x ct)+sin(x+ct)) u(x,t) where G is the heat kernel. Exercise 3. The problem t G(x y,t τ)siny dydτ w t (x,t,τ)+cw x (x,t,τ), w(x,,τ) f(x,τ)

9 6. LAPLACE TANSFOMS 9 has solution (see Chapter ) w(x,t,τ) f(x ct,τ) Therefore, by Duhamel s principle, the solution to the original problem is u(x,t) t f(x c(t τ),τ)dτ Applying this formula when f(x,t) xe t and c 2 gives u(x,t) t (x 2(t τ))e τ dτ This integral can be calculated using integration by parts or a computer algebra program. We get u(x,t) (x 2t)(e t ) 2te t +2( e t ) 6. Laplace Transforms Exercise 4. Using integration by parts, we have ( t ) ( t ) L f(τ)dτ f(τ)dτ e st dt s s t s F(s) ( t ) d f(τ)dτ ds e st dt f(τ)dτ e st + s f(t)e st dt Exercise 5. Since H for x < a we have L(H(t a)f(t a)) a f(t a)e st dt where we used the substitution τ t a, dτ dt. Exercise 8. The model is u t u xx, x >, t > u(x,) u, x > u x (,t) u(,t) Taking the Laplace transform of the PDE we get f(τ)e s(τ+a) dτ e as F(s) The bounded solution is U xx su u U(x,s) a(s)e xs + u s

10 2. PATIAL DIFFEENTIAL EQUATIONS ON UNBOUNDED DOMAINS The radiation boundary condition gives a(s) s a(s)+ u s or u a(s) s(+ s) Therefore, in the transform domain u U(x,s) s(+ s s) e x + u s Using a table of Laplace transforms we find ( ( ) ] x t+ x u(x,t) u u [erfc 4t ) erfc 4t e x+t where erfc(z) erf(z). Exercise. Taking the Laplace transform of the PDE gives, using the initial conditions, U xx s2 c 2U g sc 2 The general solution is U(x,s) A(s)e sx/c +B(s)e sx/c + g s 3 To maintain boundedness, set B(s). Now U(,s) gives A(s) g/s 3. Thus U(x,s) g s 3e sx/c + g s 3 is the solution in the transform domain. Now, from a table or computer algebra program, L ( s 3 ) t2 2, L (F(s)e as ) H(t a)f(t a) Therefore Hence L ( s 3e xs/c ) H(t x/c) (t x/c)2 2 u(x,t) gt2 2 gh(t x/c)(t x/c)2 2 Exercise. Taking the Laplace transform of the PDE while using the initial condition gives, for U U(x,y,s), The bounded solution of this equation is U yy pu U a(x,s)e y s The boundary condition at y gives su(x,o,s) U x (x,,s) or a a x, or a(x,s) f(s)e xs The boundary condition at x u forces f(s) /s. Therefore U(x,y,s) s e xs e y s

11 7. FOUIE TANSFOMS From the table of transforms u(x,y,t) erf((y x)/ 4t) Exercise 3. Taking the Laplace transform of the PDE gives, using the initial conditions, The general solution is U xx s2 c 2U U(x,s) A(s)e sx/c +B(s)e sx/c To maintain boundedness, set B(s). Now The boundary condition at x gives U(,s) G(s) which forces A(s) G(s). Thus Therefore, using Exercise 4, we get U(x,s) G(s)e sx/c u(x,t) H(t x/c)g(t x/c) 7. Fourier Transforms Exercise. The convolution is calculated from x e x2 (y x)e y2 dy Exercise 2. From the definition we have F (e a ξ )) e a ξ e ixξ dξ e aξ e ixξ dξ + e aξ ixξ dξ + e aξ e ixξ dξ e aξ ixξ dξ a ix e(a ix)ξ + a ix e( a ix)ξ a π a 2 +x 2 Exercise 3a. Using the definition of the Fourier transform F ( ξ) u(x)e i( ξ)x dx F(u)(ξ)

12 2 2. PATIAL DIFFEENTIAL EQUATIONS ON UNBOUNDED DOMAINS Exercise 3b. From the definition, û(ξ +a) F(e iax u)(ξ) u(x)e i(ξ+a)x dx u(x)e iax e iξx dx Exercise 3c. Use 3(a) or, from the definition, F(u(x+a)) u(x+a)e iξx dx u(y)e iξ(y a) dy e iaξ û(ξ) Exercise 6. From the definition û(ξ) e ax e iξx dx e (iξ a)x dx iξ a e(iξ a)x a iξ Exercise 7. Observe that Then xe ax2 d 2a dx e ax2 F(xe ax2 ) 2a ( iξ)f(e ax2 ). Exercise 9. Take transforms of the PDE to get û t ( iξ) 2 û+ ˆf(ξ,t) Solving this as a linear, first order ODE in t with ξ as a parameter, we get û(ξ,t) t e x2 (t τ) ˆf(ξ,τ)dτ Taking the inverse Fourier transform, interchanging the order of integration, and applying the convolution theorem gives u(x, t) t t F [ e x2 (t τ) ˆf(ξ,τ) ]dτ F [ e x2 (t τ) ] f(x,τ)dτ t 4π(t τ) e (x y)2 /4(t τ) f(y,τ)dydτ

13 7. FOUIE TANSFOMS 3 Exercise. Proceeding exactly in the same way as in the derivation of (2.65) in the text, but with k replaced by I, we obtain the solution where u(x,) f(x). Thus u(x,t) 4πit u(x,t) 4πit e (x y)2 /4it f(y)dy e i(x y)2 /4t y 2 dy Here, in the denominator, i denotes the root with the positive real part, that is i (+i)/ 2. Exercise. Letting v u y we have Hence Then u(x, y) v xx +v yy, x, y > ; v(x,) g(x) π y y v(x,y) π v(x, ξ)dξ π y yg(τ) (x τ) 2 +y 2dτ yg(τ) (x τ) 2 +y 2dτdξ y (x τ) 2 +y 2dξdτ g(τ) ( ln ( (x τ) 2 +y 2) ln ( (x τ) 2)) dτ g(x ξ)ln ( ξ 2 +y 2) dτ +C Exercise 2. We have u(x,y) y π π l l dτ (x τ) 2 +y 2 ( l x ( arctan y ) +arctan ( )) l+x y Exercise 3. From the definition of the Fourier transform F(u x ) u x (x,t)e iξx dx u(x,t)e iξx iξ u(ξ,t)e iξx dx iξû(ξ, t) For the second derivative, integrate by parts twice and assume u and u x tend to zero as x ± to get rid of the boundary terms.

14 4 2. PATIAL DIFFEENTIAL EQUATIONS ON UNBOUNDED DOMAINS Exercise 4. In this case where f is a square wave signal, F(f(x)) f(x)e iξx dx a a e iξx dx 2sinξx ξ Exercise 5. Taking the Fourier transform of the PDE gives which has general solution u t Du xx cu x û t (Dξ 2 +iξc)û û(ξ,t) C(ξ)e Dξ2 t iξct The initial condition forces C(ξ) ˆφ(ξ) which gives Using û(ξ,t) ˆφ(ξ)e Dξ2 t iξct F ( e Dξ2 t ) 4πDt e x2 /4Dt and we have F ( û(ξ,t)e iaξ) u(x+a) F ( e iξct e Dξ2 t ) 4πDt e (x+vt)2 /4Dt Then, by convolution, u(x,t) φ 4πDt e (x+vt)2 /4Dt Exercise 6. (a) Substituting u exp(i(kx ωt)) into the PDE u t + u xxx gives iω+(ik) 3 or ω k 3. Thus we have solutions of the form u(x,t) e i(kx+k3 t) e ik(x+k2 t) The real part of a complex-valued solution is a real solution, so we have solutions of the form u(x,t) cos[k(x+k 2 t)] These are left traveling waves moving with speed k 2. So the temporal frequency ω as well as the wave speed c k 2 depends on the spatial frequency, or wave number, k. Note that the wave length is proportional to /k. Thus, higher frequency waves are propagated faster. (b) Taking the Fourier transform of the PDE gives This has solution û t ( iξ) 3 û û(ξ,t) ˆφ(ξ)e iξ3 t where ˆφ is the transform of the initial data. By the convolution theorem, u(x,t) φ(x) F (e iξ3t )

15 7. FOUIE TANSFOMS 5 To invert this transform we go to the definition of the inverse. We have F (e iξ3t ) (3t) /3Ai e iξ3t e iξx dξ cos(ξ 3 t+ξx)dξ ( z 3 cos 3 + zx ) (3t) /3 ) ( x (3t) /3 (3t) /3dz where we made the substitution ξ z/(3t) /3 to put the integrand in the form of that in the Airy function. Consequently we have ( ) y u(x,t) φ(x y)ai dy. (3t) /3 (3t) /3 Exercise 8. The problem is u tt c 2 u xx, x, t > u(x,) f(x), u t (x,), x Taking Fourier transforms of the PDE yields whose general solution is û tt +c 2 ξ 2 û û A(ξ)e iξct +B(ξ)e iξct From the initial conditions, û(ξ,) ˆf(ξ) and û t (ξ,). Thus A(ξ) B(ξ).5ˆf(ξ). Therefore û(ξ,t).5ˆf(ξ)(e iξct +e iξct ) Now we use the fact that to invert each term. Whence F (ˆf(ξ)e iaξ ) f(x a) u(x,t).5(f(x ct)+f(x+ct)) Exercise 2. Notice the left side is a convolution. Take the transform of both sides, use the convolution theorem, and solve for ˆf. Then invert to get f.