PDEs - Solutions (1)

8/3/2019 PDEs - Solutions (1)

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MATH20401(PDEs) Tony Shardlow Answer Sheet Part 1

1.

Function Comment Conclusion

u(x, y) = A(y) uy = A(y) False

u(x, y) = A(y) uxy = 0 True

u(t, x) = A(x)B(t) uxt = A(x)B(t) = 0 False

u(t, x) = A(x)B(t) uuxt = ABAB = uxut True

u(t,x,y) = A(x, y) ut = tA(x, y) = 0 True

u(x, t) = A(x+ct) + B(xct) utt + c2uxx = 2c

2(A + B) False

u(x, y) = ekx sin(ky) uxx = k2ekx sin(ky), uyy = k

2ekx sin(ky) True

2. These are not the only possible cases; you might find other PDEs:

(a) u(t, x) = A(x+ct) + B(xct): ut = cA(x+ct) cB(xct), ux = A

(x+ct) +B(xct), utt = c

2A(x+ct) + c2B(xct), uxx = A(x+ct) + B(xct) so utt

c2uxx = 0 (wave equation).

(b) u(t, x) = A(x) + B(t): ut = B(t) so utx = 0.

(c) u(t, x) = A(x)/B(t) : l n u = ln A(x) ln B(t) so (ln u)tx = 0. Now (ln u)t = ut/uand (ln u)tx = utx/u utux/u

2 . Finally, uutx utux = 0.

(d) u(t, x) = A(xt): ut = xA

(xt), ux = tA

(xt), so tut xux = 0.(e) u(t, x) = A(x2t): ut = x

2A(x2t), ux = 2xtA(x2t) so 2tut xux = 0.

(f) u(t, x) = A(x2/t): ut = x2

t2 A(x2/t), ux =

2xt A

(x2/t) so 2tut + xux = 0.

3. (a) u(0, x) = A(x) + B(x) = f0(x) so A(x) + B(x) = f0(x). There is not enough infor-mation to determine both A( ) and B( ).

(b) u(0, x) = A(x) + B(0) = f0(x) so A(x) = f0(x) B(0). We would only need to knowone value of B, namely B(0) to determine A. However, the initial conditions givesno information about B.

(c) u(0, x) = A(x)/B(0) = f0(x) so A(x) = B(0)f0(x). We would only need to know onevalue of B, namely B(0) to determine A, but we have no information about B.

(d) u(0, x) = A(0) = f0(x) so A(0) = f0(x). This tries to set a constant A(0) tosomething that is not constant f0(x), which is not possible!

(e) (exactly the same)

(f) u(0, x) = A() = f0(x) so A() = f0(x). This tries to set a constant A() tosomething that is not constant f0(x), which is again not possible!

4. (a) u(1, x) = A(x + c) + B(x c) = f1(x) so A(x + 1) + B(x 1) = f1(x). Again moreinformation is needed to find both A( ) and B( ).

(b) u(1, x) = A(x) + B(1) = f1(x) so A(x) = f1(x) B(1). We only need to know B(1)to determine A. We obtain no information about B.

MATH20401(PDEs): Answer Sheet I: Page 1


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(c) u(1, x) = A(x)/B(1) = f1(x) so A(x) = B(1)f1(x). We only need to know B(1) todetermine A. There is no information about B.

(d) u(1, x) = A(x) = f1(x) so A(x) = f1(x).

(e) u(1, x) = A(x2) = f1(x) so A(x2) = f1(x). This determines A() for 0 provided

f1() = f1().

(f) u(1, x) = A(x2) = f1(x) so A(x2) = f1(x). Again, this determines A() for 0provided f1() = f1().

5. (a) To confirm that u = AT1/2(Tt)1/2ex2/4(Tt) is a solution of ut = uxx :

ut =12AT

1/2(Tt)3/2ex2/4(Tt) 14AT

1/2x2(Tt)5/2ex2/4(Tt) ,

ux = 12AT

1/2x(Tt)3/2ex2/4(Tt) ,

uxx = 12AT

1/2(Tt)1/2ex2/4(Tt) + 14AT

1/2x2(Tt)5/2ex2/4(Tt) = ut

so that ut = uxx.

(b) For this solution |u(0, x)| = AT1/2

T1/2exp(x2/4(T 0)) A as exp(x2/4T) < 1.

Further as t T, u(t, 0) .

(c) We have shown that u(x, t) is a soln of the backward heat eqn, it tends to infinity ast T, and initially u(x, 0) = A exp(x2/T) with |u(0, x)| A.

Recall that the soln of the backward heat eqn with zero initial condition is the zerosoln u(x, t) = 0 for all x, t . By taking A = 1, change the zero initial conditionto u(x, 0) = exp(x2/T). This is a small change of size to the initial condition.But, as we saw above, the solution now becomes infinite at time t = T and this is farfrom the behaviour of the zero solution. Because of this, the backward heat equationis ill-posed for t > 0 when subject to initial conditions at t = 0 . A small change ininitial condition causes a drastic change in the solution.

(d) The backward heat equation is well posed when subject to final conditions at sometime (say) t = tf for times before the final time, t < tf.

6. (a) To confirm that u = cosh(y/)cos(x/) is a solution: At y = 0 we have u(x, 0) = cosh(0) cos(x/) = cos(x/) and uy(x, 0) =sinh(0) cos(x/) = 0 so the boundary conditions are satisfied. Also, uxx =

1 cosh(y/)cos(x/) and uyy = 1 cosh(y/)cos(x/) so

that the PDE uxx + uyy = 0 is satisfied.Hence u = cosh(y/)cos(x/) is a solution satisfying the conditions.

(b) Now u(x, 0) = cos(x/) and |u(x, 0)| . Hence, take = .

Also u(x, y) = cosh(y/) cos(x/) and cosh(z) = (ez + ez)/2 is unbounded asz . Hence, maxx,y |u(x, y)| as 0 (we can always find x so that

cos(x/) > 1/2)Hence small changes in the boundary conditions lead to large changes in the solution.This means the problem is ill posed. In general, Laplaces equation is ill-posed if itis subjected to conditions on a boundary that does not entirely surround the domainin which the equation is to be satisfied.

7. (a) ut (x2 + u)uxx = x t is second order and quasilinear (because the second order

term includes uuxx , the multiplier depends on u but not second derivatives of u)).

(b) u2utt 12u

2x + (uux)x = e

u is second order and quasilinear (because the second orderterm is u2utt + uuxx , the multipliers of utt and uxx depend on u but not secondderivatives of u).



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(c) ut uxx = u3 is second order and semilinear (because the second order term uxx is

linear and u3 is nonlinear).

(d) (uxy)2 uxx + ut = 0 is second order and fully nonlinear (because the second order

term u2xy is non-linear).

(e) ut + ux uy = 10 is first order, linear and inhomogeneous.

8. We consider the operators to contain the terms autt + butx + cuxx (or different subscriptsfor different independent variables):

(a) ut + utx uxx + u2x = sin u. b

2 4ac = 1 > 0 so: hyperbolic and semilinear.

(b) ux + uxx + uy + uyy = sin(xy). b2 4ac = 4 < 0 so: elliptic, linear and inhomoge-

neous.

(c) ux + uxx uy uyy = cos(xyu). b2 4ac = 4 > 0 so: hyperbolic and semilinear.

(d) utt + xuxx + ut = f(x, t). b2 4ac = 4x so: elliptic for x > 0, hyperbolic for x < 0,

parabolic for x = 0, linear and inhomogeneous.

(e) ut + uuxx + u2utt utx = 0. b

2 4ac = 1 4u3 so: elliptic for u3 > 1

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, hyperbolicfor u3 < 14 , parabolic for u

3 = 14 , and quasilinear.


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