MATH 124A Solution Key HW 05

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1 3. DIFFUSION ON THE HALF-LINE Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A MATH 24A Solution Key HW 5 3. DIFFUSION ON THE HALF-LINE. Solve u t ku x x ; u(x, ) e x ; u(, t) on the half-line < x <. SOLUTION. By the method of odd extensions: consider the diffusion equation on the whole line v t kv x x on < x <, < t <, e x, if x >, v(x, ) φ odd (x), if x, e x if x <. Our construction makes it plaint that v(x, t) is an odd function, hence; v(, t), and u(x, t) v(x, t) when x >. An appeal to section 2.4 on page 49 in our text reveals that the solution for v(x, t) is given by v(x, t) S(x y, t)φ odd (y) d y S(x y, t)e y d y + S(x y, t)e y d y S(x y, t)e y d y+ S(x y, t)e y d y + S(x y, t)( e y ) d y S(x y, t)e y d y S(x y, t)e y d y S(x+y)e y d y S(x y, t) S(x + y, t) e y d y e (x y) 2 /4kt e (x+y)2 /4kt e y d y. In fact, we may recast our solution in terms of r f (x) and obtain u(x, t) 2 ekt x r f 2kt x 2kt + x 4kt 2 ekt+x r f. 4kt 3. Derive the solution formula for the half-line Neumann problem w t κw x x for < x <, < t < ; w x (, t) ; w(x, ) φ(x).

2 3. DIFFUSION ON THE HALF-LINE Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A SOLUTION. By the method of even extensions: consider the diffusion equation on the whole line v t kv x x on < x <, < t <, φ(x) if x v(x, ) φ even (x) φ( x) if x, By our construction v(x, t) defines an even function, hence; v x (, t), and w(x, t) v(x, t) when x >. Once again, the solution for v(x, t) was established in section 2.4, that is, v(x, t) S(x y, t)φ even (y) d y S(x y, t)φ(y) d y + S(x y, t)( φ( y)) d y S(x y, t)φ(y) d y + S(x+y, t)φ(y) d y S(x y, t) + S(x + y, t) φ(y) d y e (x y) 2 /4kt + e (x+y)2 /4kt φ(y) d y. 4. Consider the following problem with a Robin boundary condition: DE: u t ku x x on the half-line < x < and < t <, IC: u(x, ) x for t and < x <, (*) BC: u x (, t) 2u(, t) for x. The purpose of this exercise is to verify the solution formula for (*). Let f (x) x for x >, let f (x) x + e 2x for x <, and let v(x, t) e (x y)2 /4kt f (y) d y. (a) What PDE and initial condition does v(x, t) satisfy for < x <? (b) Let w v x 2v. What PDE and initial condition does w(x, t) satisfy for < x <? (c) Show that f (x) 2f (x) is an odd function for x. (d) Use Exercise 2.4. to show that w is an odd function of x. 2

3 3. DIFFUSION ON THE HALF-LINE Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A (e) Deduce that v(x, t) satisfies (*) for x >. Assuming uniqueness deduce that the solution of (*) is given by u(x, t) e (x y)2 /4kt f (y) d y. SOLUTION. Solving the diffusion equation subject to the Robin boundary condition on the half-line involves the same technique we saw in the Dirichlet and Neumann conditions, respectively. That is, we introduce a function that s defined on the whole line with an extension of the initial conditions that result in the boundary conditions being satisfying on the region of interest. More specifically, we aim to extend u(x, t) to a function v(x, t) that satisfies the diffusion equation for all x, with v(x, ) for x > subject to v x (, t) 2v(, t). In particular, this shall indeed by satisfied once v x (x, t) 2v(x, t) is an odd function in the variable x. With that objective in mind let x for x >, f (x) x + e 2x for x < and v(x, t) e (x y)2 /4kt f (y) d y. (a) A routine calculation involving / t and 2 / 2 makes it plain that v indeed satisfies x the diffusion equation. It remains to establish that the initial condition is satisfied. To see this, write v(x, t) e p2 f (x 4kt p) dp π by making a change of variables y(p) x 4kt p. Ultimately, we obtain v t kv x x x, t >, v(x, ) f (x). (b) Since w is a linear combination of v and v x, it follows immediately that w must satisfy the diffusion equation. w t kw x x, x, t > w(x, ) f (x) 2 f (x) (c) A direct calculation makes it plain that f (x) 2 f (x) 2x, x > 2x, x < Please note that Theorem 2 on page 42 A.3 is absolutely crucial here. 3

4 3.2 REFLECTION OF WAVES Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A if we really want to be pedantic we could use the formal definition of the derivative and proceed with an ɛ-proof as in Math 8B by letting ɛ + and ɛ. Similarly, f + 2x, x < ( x) 2 f ( x) + 2x, x > which in turn proves that f (x) 2 f (x) is indeed an odd function in x. (d) This follows from the fact that any solution that satisfies the diffusion equation with an initial data that is odd in x must be odd in x as well. (We highly recommend that you prove this for yourself if you have some spare time.) (e) Since w(x, t) is an odd function of x, with w(, t) for t >, then v x (, t) 2v(, t). In addition, v(x, ) f (x) x for x >, hence; for x >, v(x, t) satisfies our original PDE of interest with the given Robin boundary conditions. Since we may assume that our solution is unique, it must be given by u(x, t) e (x y)2 /4kt f (y) d y. 3.2 REFLECTION OF WAVES. Solve the Neumann problem for the wave equation on the half-line < x <. SOLUTION. By method of even extensions: consider the reformulated PDE v t t c 2 v x x < x, t < v(x, ) φ even (x) v t (x, ) ψ even (x) φ(x), x φ( x), x ψ(x), x ψ( x), x. By our construction, v defines an even function, hence; v x (, t), and u(x, t) v(x, t) for all x >. By d Alembert s formula, v(x, t) 2 φeven (x + ct) + φ even (x ct) + 2c Accordingly, we have the following two cases. 4 x+ct x ct ψ even (s) ds.

5 3.2 REFLECTION OF WAVES Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A (i) If x ct >, then φ even (x f t) φ(x f t) and ψ even (s) ψ(s) for every s in the interval x ct s x + ct, hence; u(x, t) 2 φ(x + ct) + φ(x ct) + 2c x+ct x ct (ii) Conversely, if x ct <, then φ even (x ct) φ(ct x), so u(x, t) 2 φ(x + ct) + φ(ct x) + 2c x+ct ψ(s) ds + 2c ψ(s) ds. ct x ψ(s) ds. 2. The longitudinal vibrations of a semi-infinite flexible rod satisfying the wave equation u t t c 2 u x x for x >. Assume that the end x is free u x ; it is initially at rest but has a constant initial velocity V for a < x < 2a and has zero initial velocity elsewhere. Plot u versus x at the times t, a/c, 3a/2c, 2a/c, and 3a/c. SOLUTION. 3. A wave f (x + ct) travels along a semi-infinite string, < x <, for t <. Find the vibrations u(x, t) of the string for t > if the end x is fixed. SOLUTION. Given that our string is fixed at the end x, if follows that we are in a situation of the homogeneous Dirichlet condition u(, t) as expressed in (3.2.) in our text. Therefore, the vibrations u(x, t) of the string for t > is given by the odd reflection formula with initial data u(x, ) φ(x) f (x), u t (x, ) ψ(x) c f (x) that is, u(x, t) φodd (x + ct) + φ 2 odd (x ct) + 2c x+ct x ct ψ odd (y) d y. 4. Repeat exercise if the end is free. SOLUTION. 5

6 3.3 DIFFUSION WITH A SOURCE Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A 3.3 DIFFUSION WITH A SOURCE. Solve the inhomogeneous diffusion equation on the half-line with Dirichlet boundary condition: u t ku x x f (x, t) < x <, < t <, u(, t) u(x, ) φ(x) using the method of reflection. SOLUTION. By the method of reflection: consider the inhomogeneous diffusion equation on the whole line f (x, t), x >, v t kv x x f odd(x, t), x, f ( x, t), x <. with data φ(x), x >, v(x, ) φ odd (x), x, φ( x), x <. By our construction v(x, t) indeed defines an odd function, hence; v(, t), and u(x, t) v(x, t) for x >. Furthermore, the solution for v(x, t) was presented on page 67 in section 3.3 of our textbook. That is, v(x, t) + S(x y, t)φ odd (y) d y + t S(x y, t)φ(y) d y + t S(x y, t s)f (y, s) d yds + By changing variables from y to y we obtain v(x, t) S(x y, t s)f odd (y, s) d yds S(x y, t) φ( y) d y t S(x y, t) S(x + y, t) φ(y) d y+ t S(x y, t s) f ( y, s) d yds. 2. Solve the completely inhomogeneous diffusion problem on the half-line v t kv x x f (x, t) for < x <, < t <, v(, t) h(t), v(x, ) φ(x), 6 S(x y, t s) S(x + y, t s) f (y, s) d yds.

7 3.3 DIFFUSION WITH A SOURCE Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A by carrying out the subtraction method begun in the textbook. SOLUTION. The calculations are routine. There is virtually no meat to this problem. 3. Solve the inhomogeneous Neumann diffusion problem on the half-line w t kw x x for < x <, < t <, w x (, t) h(t) w(x, ) φ(x), by the subtraction method indicated in the text. SOLUTION. First, put v(x, t) w(x, t) xh(t) and notice that our function v satisfies v t kv x x xh (x) on < x <, t >, v(x, ) φ(x) xh(), v x (, t). Second, we proceed by the method of even extension by letting xh (t), x >, u t ku x x f even (x), x xh (t), x <, with initial data φ(x) xh(), x >, u(x, ) φ even (x), x φ( x) + xh(), x <. It follows that the solution u(x, t) shall be an even function in the variable x which in turn guarantees that u satisfies the boundary condition u x (, t). In addition, v(x, t) u(x, t) for x >. Now, the solution to the inhomogeneous diffusion equation on the whole line is presented on page 67 of section 3.3 in our textbook. That is, for x >, we obtain v(x, t) + S(x y, t)φ even (y) d y + t t S(x y, t s)f even (y, s) d yds S(x y, t) + S(x + y, t) φ(y) yh() d y S(x y, t s) + S(x + y, t s) yh (s) d yds. 7

8 3.3 DIFFUSION WITH A SOURCE Solutions prepared by Jon Tjun Seng Lo Kim Lin, TA Math 24A More specifically, w(x, t) v(x, t) + xh(t) t S(x y, t) + S(x + y, t) φ(y) yh() d y S(x y, t s) + S(x + y, t s) yh (s) d yds + xh(t). 8