HOMEWORK 5. Proof. This is the diffusion equation (1) with the function φ(x) = e x. By the solution formula (6), 1. e (x y)2.

Transcription

1 HOMEWORK 5 SHUANGLIN SHAO. Section 3.. #. Proof. This is the diffusion equation with the function φx e x. By the solution formula 6, vx, t e x y e x+y φydy e x y e x+y e x y y dy e y dy e x+y y dy To compute the first integral, e x y y dy e kt x e kt x π ekt x e x +y xy+y dy e x +y +kt xy+kt x kt x dy e 4k t x kt x kt x e y+kt x dy e y+kt x e y dy e y dy π ekt x π π ekt x kt x π Erfkt x Erf kt x. dy e y dy

2 Similarly for the second integral, So e x+y y dy ex+kt Erf kt + x. vx, t ekt x Erf kt x ekt+x Erf kt + x.. Section 3.. # Proof. Let vx, t ux, t. Then the system of equation is By 6, Hence vx, t v t kv xx, vx,, v, t. π + π. e x y e x y e x+y φydy e x+y e x y dy + ux, t. dy e x+y dy 3. Section 3.. #3. Proof. We make an even extension of φ: { φx, for x, φ even x φ x, for x. We solve the diffusion equation with the Neumann condition u t ku xx, ux, φ even x, u x, t. By the solution formula for the diffusion equation, ux, t Sx y, tφ even ydy.

3 Since φ even is an even function in y, so u x, t ux, t. Let wx, t ux, t for x. If we differentiate both sides and set x, we obtain u x x,. So w x x,. Next we replace φ even by φ in the formula above to derive a solution formula for the original problem. For x, wx, t Sx y, tφydy + Sx y, tφydy + Sx y, t + Sx + y, t φydy. Sx y, tφ ydy, Sx + y, tφ ydy, 4. Section 3.. #. Proof. We make an even extension of φ and ψ: { φx, for x, φ even x φ x, for x. and ψ even x { ψx, for x, ψ x, for x. Thus by the solution for the wave equation on the real line vx, t [φ evenx + ct + φ even x ct] + c Since φ even and ψ even are even in y, so This implies that v x, t vx, t. v x, t. Let wx, t vx, t for x.therefore w x, t, ψ even ydy. which satisfies the boundary condition. For x, we divide the discussion into 3 cases because x, c and < x <. 3

4 Case. x + ct and x ct. vx, t [φ evenx + ct + φ even x ct] + c [φx + ct + φx ct] + c Case. x + ct and x ct <. vx, t [φ evenx + ct + φ even x ct] + c [φx + ct + φct x] + c [φx + ct + φct x] + c ψydy. ψ even ydy ψ even ydy ψ ydy + c ct x ψydy + c ψydy ψydy Case 3. x ct x + ct <. vx, t [φ evenx + ct + φ even x ct] + c [φ x ct + φct x] + c [φ x ct + φct x] + c ct x ψ ydy ψydy. ψ even ydy Combining these three cases, we obtain the solution formula for v. 5. Section 3.. # 5. Proof. This is the Dirichlet problem for the wave equation over the half line with φx, ψx and v, t. By the solution formula on page 6, for x, vx, t φ odd x + ct + φ odd x ct + c φ odd x + ct + φ odd x ct. When x + ct and x ct, Thus φ odd x + ct, φ odd x ct. vx, t. 4 ψ odd ydy

5 When x + ct and x ct <, vx, t +. When x + ct x ct, vx, t Section 3.3. #. Proof. We make an odd extension of φ: φx, for x, φ odd x φ x, for x,, for x. and we make an odd extension of f: fx, for x, f odd x f x, for x,, for x. Now we solve the system of equations: u t ku xx fx, t, ux, φ odd x. By the solution formula, we obtain ux, t Sx y, tφ odd ydy + t Sx y, t sf odd y, sdyds. The first integral is an odd function of x since φ odd S is an even function. Indeed, S x y, tφ odd ydy S x + y, tφ odd ydy This implies that when x, Sx y, tφ odd ydy. S y, tφ odd ydy. 5 is an odd function, and

6 When x in the second integral, because f odd proves that t t. S y, t sf odd y, sdyds 4πks e y 4kt s f odd y, sdyds is an odd function and Sx, t is an even function in x. This u, t. The second integral is independent of φ; For the first integral, Thus ux, t Sx y, tφ odd ydy Sx y, tφydy + Sx y, tφydy Sx y, t φ ydy Sx + y, tφydy Sx y, t Sx + y, t φydy. Sx y, t Sx + y, t φydy+ t Sx y, t sfy, sdyds. 7. Section 3.3. #. Proof. Let ux, t vx, t ht. Then vx, t ux, t + ht. Then from the equations satisfied by v, we have u t ku xx fx, t h t, u, t, ux, φx h. Then by Exercise in this section, ux, t + Sx y, t Sx + y, t φy h dy t Sx y, t s fy, s h t dyds. 6

7 Then we have vx, t ux, t + ht + t +ht. 8. Section 3.4, #. Proof. By using the formula 3, since φ ψ, we have ux, t c c c x t x+ct s t t t s t x ct s x+ct s x ct s ysdyds ydy sxct sdyds st sds x t sds t 3 x t3 3 xt3 6. t s ds 7

8 9. Section 3.4. #. Proof. By using the formula 3, since φ ψ, we have ux, t c a a t x+ct s t t eax+act a eax+act a eax a c x ct s e ay x+ct s x ct s ds e ay dyds e ax+ct s e ax ct s ds t t e acs ds eax act e acs ds a ac e acs t eax act a ac eacs t e act eax e act a c eax a c + eax+act a c + eax act a. c. Section 3.4. # 3. Proof. We know that ux, sin x and u t x, + x. solution formula, we have Then by the ux, t [φx + ct + φx ct]+ c For the first term, ψydy+ c t x+ct s x ct s [φx + ct + φx ct] [sinx + ct + sinx ct]. For the second term, c For the third term, c + ydy 4c t x+ct s x ct s + x + ct + x ct t + x. cos ydyds cosx + ct cosx ct. c 8 cos ydyds.

9 Hence ux, t [sinx + ct + sinx ct] + t + x + cosx + ct cosx ct. c Department of Mathematics, KU, Lawrence, KS address: slshao@math.ku.edu 9