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2024年1月11日 星期四

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 1.3

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 1.3

  1. Solve u2x+yuyu=0 with initial condition u(x,1)=x24+1.
  2. SolutionLet F:R5R be defined by

    F(x,y,z,p,q)=p2+yqzfor (x,y,z,p,q)R5.

    Consider the characteristic curve

    x=x(t),y=y(t),z=z(t)=u(x(t),y(t)),p=p(t)=ux(x(t),y(t)),q=q(t)=uy(x(t),y(t)).

    Then we have the following ordinary differential equations

    {dxdt=Fp(x,y,z,p,q)=2p,dydt=Fq(x,y,z,p,q)=y,dzdt=pFp(x,y,z,p,q)+qFq(x,y,z,p,q)=2p2+yq,dpdt=Fx(x,y,z,p,q)pFz(x,y,z,p,q)=p,dqdt=Fy(x,y,z,p,q)qFz(x,y,z,p,q)=0

    with the initial condition

    x(0)=s,y(0)=1,z(0)=u(x(0),y(0))=s24+1.

    From F(x(0),y(0),z(0),p(0),q(0))=0, we have

    p2(0)+q(0)s241=0.

    By the strip condition, we have

    s2=dz(0)ds=p(0)dx(0)ds+q(0)dy(0)ds=p(0)1+q(0)0.

    Thus, p(0)=s/2 and q(0)=1. Now from dpdt=p and dqdt=0, we obtain

    p(t)=p(0)et=s2et,q(t)=q(0)=1.

    Hence, the first three equations become

    dxdt=set,dydt=y,dzdt=s22e2t+y,

    which implies x(t)=x(0)+s(et1)=set and y(t)=y(0)et=et, and

    z(t)=z(0)+s24(e2t1)+(et1)=s24e2t+et=(set2)2+et.

    It is easy to get t=ln(y) and s=x/y, and we obtain

    u(x,y)=x24+yfor xR and y>0.

    Here we note that y=et>0. It is remarkable that this solution can be extended to R2. Function u(x,y)=x24+|y| for (x,y)R2 is a solution in function space C1(R×R)C0(R2), where R=R{0}.

  3. Solve u=xux+yuy+(u2x+u2y)/2 with initial condition u(x,0)=(1x2)/2. [This is an example of Clairaut's equation u=xp+yq+f(p,q).]
  4. SolutionLet F:R5R be defined by

    F(x,y,z,p,q)=zxpyq(p2+q2)/2for (x,y,z,p,q)R5.

    Consider the characteristic curve

    x=x(t),y=y(t),z=z(t)=u(x(t),y(t)),p=p(t)=ux(x(t),y(t)),q=q(t)=uy(x(t),y(t)).

    Then we have the following ordinary differential equations

    {dxdt=Fp(x,y,z,p,q)=xp,dydt=Fq(x,y,z,p,q)=yqdzdt=pFp(x,y,z,p,q)+qFq(x,y,z,p,q)=p(xp)+q(yq)=xpyqp2q2,dpdt=Fx(x,y,z,p,q)pFz(x,y,z,p,q)=(p)p1=0,dqdt=Fy(x,y,z,p,q)qFz(x,y,z,p,q)=(q)q1=0

    with initial condition

    x(0)=s,y(0)=0,z(0)=u(x(0),y(0))=(1s2)/2.

    From F(x(0),y(0),z(0),p(0),q(0))=0, we have

    1s22sp(0)p2(0)+q2(0)2=0.

    By the strip condition, we get

    s=dz(0)ds=p(0)dx(0)ds+q(0)dy(0)ds=p(0)1+q(0)0=p(0).

    Thus, p(0)=s and q(0)=±1. Now from dpdt=dqdt=0, we obtain

    p(t)=p(0)=s and q(t)=q(0)=±1.

    Hence, the first three equations become

    dxdt=x+s,dydt=y1,dzdt=sxys21,

    which implies x(t)=s+(x(0)s)et=s, y(t)=1+(y(0)±1)et=±(et1), and

    dzdt=s2(et1)s21=et.

    Thus, we can solve

    z=z(t)=z(0)+(et1)=et1+s22.

    It is easy to get t=ln(1±y) and s=x, and we obtain

    u(x,y)=1±y1+x22=±y1x22for (x,y)R2.


  5. Consider u=u2x+u2y with the initial condition u(x,0)=ax2. For what positive constants a is there a solution? Is it unique? Find all solutions.
  6. Solution

    When a(0,1/4], there is a solution. More precisely, the partial differential equation admits a unique solution (a=1/4) and two solutions (a(0,1/4)), respectively.

    Let F:R5R be defined by

    F(x,y,z,p,q)=p2+q2zfor (x,y,z,p,q)R5.

    Consider the characteristic curve

    x=x(t),y=y(t),z=z(t)=u(x(t),y(t)),p=p(t)=ux(x(t),y(t)),q=q(t)=uy(x(t),y(t)).

    Then we have the following ordinary differential equations

    {dxdt=Fp(x,y,z,p,q)=2p,dxdt=Fq(x,y,z,p,q)=2q,dzdt=pFp(x,y,z,p,q)+qFq(x,y,z,p,q)=p2p+q2q=2(p2+q2)=2z,dpdt=Fx(x,y,z,p,q)pFz(x,y,z,p,q)=p,dqdt=Fy(x,y,z,p,q)qFz(x,y,z,p,q)=q

    with the initial condition

    x(0)=s,y(0)=0,z(0)=u(x(0),y(0))=as2.

    From F(x(0),y(0),z(0),p(0),q(0))=0, we have p2(0)+q2(0)=z(0)=as2. By the strip condition, we get

    2as=dz(0)ds=p(0)dx(0)ds+q(0)dy(0)ds=p(0)1+q(0)0=p(0).

    Thus, p(0)=2as and q(0)=±|s|a4a2R, provided a(0,1/4]. Now from dpdt=p and dqdt=q, we obtain

    p(t)=p(0)et=2aset,q(t)=q(0)et=±|s|eta4a2.

    Hence, the first three equations become

    dxdt=4aset,dydt=±2|s|eta4a2,dzdt=2z,

    which implies

    x(t)=x(0)+4as(et1)=s(1+4aet4a),y(t)=y(0)±2|s|(et1)a4a2=±2|s|(et1)a4a2,z(t)=z(0)e2t=as2e2t.

    • When a=1/4, it is clear that x=set and y=0, and we obtain

      u(x,y)=x24for (x,y)R2.

    • Suppose a(0,1/4). From x=s+4as(et1), we have xs4a=s(et1). On the other hand, y2=4s2(et1)2(a4a2). Combining them, we get 4a2y2=(xs)2(a4a2), and hence s=x±2ya14a. From x=s(14a+4aet), we get

      t=ln(1+xs4as)=ln(2xa4a2±y(4a1)2xa4a2±4ya).

      Therefore, we obtain

      u(x,y)=a(x±(4a1)y2a4a2)2=(xay14a2)2for (x,y)R2.

      Here we have used the fact that

      s2e2t=(set)2=[s(1+x/s14a)]2=(s+xs4a)2=(x±2ya14ay2a4a2)2


  7. To generalize (26) to n variables, let u(x)=u(x1,,xn) and pj=u/xj for j=1,,n. Then consider

    F(x,u,p)=0.

    1. Find the characteristic equations [i.e., generlizations of (29), (30)].
    2. Find the conditions on the initial values of p [i.e., generalization of (31), (32)].
  8. Solution
    1. Let x=x(t)=(x1(t),,xn(t)) be the characteristic curve. Define z(s)=u(x(t)), p(t)=(p1(t),,pn(t))=u(x(t)). Differentiating pi with respect to t yieldsdpidt=ddt[uxi(x(t))]=nj=1uxixj(x(t))dxjdt.On the other hand, we also differentiate F(x,u,u)=0 with respect to xi:

      Fxi(x,u,u)+Fu(x,u,u)uxi+ni=jFpi(x,u,u)uxjxi=0,

      which impliesFxi(x(t),z(t),p(t))+pi(t)Fu(x(t),z(t),p(t))+nj=1Fpi(x(t),z(t),p(t))uxjxi(x(t))=0.Suppose dxidt=Fpi(x(t),z(t),p(t)) for i=1,,n. Then by (1) and (2), we obtain

      dpidt=nj=1uxixj(x(t))Fpj(x(t),z(t),p(t))=Fxi(x(t),z(t),p(t))pi(t)Fu(x(t),z(t),p(t)).

      Differenitaing z=u(x(t)) with respec to t gives

      dzdt=ni=1uxi(x(t))dxidt=ni=1pi(x(t))Fpi(x(t),z(t),p(t)).

    2. Suppose Γ is the noncharacteric curve parameterized by (x,z)=(f(s),g(s))=(f1(s),,fn(s),g(s)). Then h(s)=(h1(s),,hn(s))=u(f(s)) satisfies

      F(f(s),g(s),h(s))=0.

      In addition, from g(s)=u(f(s)), we have the strip condition:

      g(s)=u(f(s))f(s)=h(s)f(s)=ni=1hi(f(s))fi(s).


  9. Show that the family of spheres Sa given by (xa)2+y2+z2=1 has as its envelope E the unit cylinder y2+z2=1.
  10. SolutionLet w(x,y,z;a)=(xa)2+y2+z21. Then the evelope E is the intersection

    {w(x,y,z;a)=(xa)2+y2+z21=0,a(x,y,z;a)=2(xa)=0

    From the second equation, we get x=a, and hence y2+z2=1.

  11. Consider Clairaut's equation u=xux+yuy+f(ux,uy).
    1. Verify that u(x,y;a,b)=ax+by+f(a,b) is a complete integral.
    2. Use envelopes to generate a solution of u=xux+yuy+(u2x+u2y)/2 that is not linear in x and y.
  12. Solution
    1. Since u(x,y;a,b)=ax+by+f(a,b), it is easy to get ux(x,y;a,b)=a and uy(x,y;a,b)=b. Thus, u=u(x,y;a,b) satisfies the partial differential equation:

      xux(x,y;a,b)+yuy(x,y;a,b)+f(ux(x,y;a,b),uy(x,y;a,b))=xa+by+f(a,b)=u(x,y;a,b).

      Moreover, the independence of the parameters follows from the invertiblity of the matrix

      [uxa(x,y;a,b)uya(x,y;a,b)uxb(x,y;a,b)uyb(x,y;a,b)]=[1001].

      This shows u(x,y;a,b)=ax+by+f(a,b) is a complete integral.
    2. From a., we consider the two-parameter solution u(x,y;a,b)=ax+by+(a2+b2)/2. To find the envelope, we set b=w(a) for some wC1, which gives a one-parameter solution

      u(x,y;a):=u(x,y;a,w(a))=ax+w(a)y+(a2+(w(a))2)/2.

      For the sake of simplicity, we choose w(a)=a for aR. Then the envelope E can be determined by

      {u(x,y;a)=0,au(x,y;a)=0.

      The second equation yields x+1y+(a+a1)=0, which gives a=x+y2. Plugging into the first equation, we obtain

      u(x,y)=x+y2xx+y2y+12[(x+y2)2+(x+y2)2]=(x+y)24for (x,y)R2.

      Note that this solution is not linear in x and y.

  13. The Hamilton-Jacobi equation ut+H(xu)=0, where H depends only on xu=(ux1,ux2,,uxn), arises frequently in physics, such as geometric optics (cf. the eikonal equation). Here we assume n=1.
    1. Verify that condition (34) fails for u(x,t;a,b)=axtH(a)+b, yet this is a complete integral.
    2. Use envelopes to generate a solution of ut+u2x=0 that is not linear in x and t.
  14. Solution
    1. Since u(x,t;a,b)=axtH(a)+b, it is easy to get ux(x,t;a,b)=a and ut(x,t;a,b)=H(a). Thus, u=u(x,t;a,b) satisfies the partial differential equation:

      ut(x,t;a,b)+H(ux(x,y;a,b))=H(a)+H(a)=0.

      Despite the matrix [uxa(x,t;a,b)uta(x,y;a,b)uxb(x,t;a,b)utb(x,t;a,b)]=[1H(a)00] is non invertible, a and b are independent paramters because [ua(x,t;a,b)uxa(x,t;a,b)uta(x,t;a,b)ub(x,t;a,b)uxb(x,t;a,b)utb(x,t;a,b)]=[xtH(a)1H(a)100] has rank 2. This shows u=u(x,t;a,b) is a complete integral.
    2. From a., we consider the two-parameter solution u(x,t;a,b)=axta2+b. To find the envelope, we set b=w(a) for some wC1, which gives a one-parameter solution u(x,t;a):=u(x,t;a,w(a))=axta2+w(a). For the sake of simplicity, we choose w(a)=a for aR. Then the envelope E can be determined by

      {u(x,y;a)=0,au(x,y;a)=0.

      The second equation x2ta+1=0, which gives a=(x+1)/(2t). Plugging into the first equation, we obtain

      u(x,y)=x+12txt(x+1)24t2+x+12t=(x+1)24tfor (x,t)R2.

      Note that this solution is not linear in x and t.

  15. To describe the wave front produced by an initial disturbance at a point, consider (36) with Γ begin given by fgh0. Describe the solution(s).
  16. SolutionLet F:R5R be defined by

    F(x,y,z,p,q)=c2(p2+q2)1for (x,y,z,p,q)R5,

    where c>0 is a constant. Consider the characteristic curve

    x=x(t),y=y(t),z=u(x(t),y(t)),p=p(t)=ux(x(t),y(t)),q=q(t)=uy(x(t),y(t)).

    Then we have the following ordinary differential equations

    {dxdt=Fp(x,y,z,p,q)=2c2p,dydt=Fq(x,y,z,p,q)=2c2q,dzdt=pFp(x,y,z,p,q)+qFq(x,y,z,p,q)=p2c2p+q2c2q=2,dpdt=Fx(x,y,z,p,q)pFz(x,y,z,p,q)=0,dqdt=Fy(x,y,z,p,q)qFz(x,y,z,p,q)=0

    with the initial condition

    x(0)=f(s)0,y(0)=g(s)0,z(0)=u(x(0),y(0))=h(s)0.

    From F(x(0),y(0),z(0),p(0),q(0))=0, we have c2(p2(0)+q2(0))=1. On the other hand, the strip condition gives nothing but 0=0. Thus, p(0)=c1coss and q(0)=c1sins for s[0,2π). Now from dpdt=dqdt=0, we obtain

    p(t)=p(0)=c1coss,q(t)=q(0)=c1sins.

    Hence, the first three equations become

    dxdt=2ccoss,dydt=2csins,dzdt=2,

    which implies

    x(t)=x(0)+2ctcoss=2ctcoss,y(t)=y(0)+2ctsins=2ctsins,z(t)=z(0)+2t=2t.

    It is easy to get t=±x2+y2/(2c) and s=tan1(y/x), and we obtain

    u(x,y)=±x2+y2cfor (x,y)R2.


  17. Instead of assuming the propagation speed is constant, consider c=c(x,y) in (36) and derive the characteristic equations. In the special case c=|x| with initial condition u(x,0)=0, find the solution to be

    u(x,y)=logx2+y2+yxfor x>0.

  18. SolutionLet F:R5R be defined by

    F(x,y,z,p,q)=c2(x,y)(p2+q2)1for (x,y,z,p,q)R5.

    Consider the characteristic curve

    x=x(t),y=y(t),z=z(t)=u(x(t),y(t)),p=p(t)=ux(x(t),y(t)),q=q(t)=uy(x(t),y(t)).

    Then we have the following ordinary differential equations

    {dxdt=Fp(x,y,z,p,q)=2c2(x,y)p,dydt=Fq(x,y,z,p,q)=2c2(x,y)q,dzdt=pFp(x,y,z,p,q)+qFq(x,y,z,p,q)=p2c2(x,y)p+q2c2(x,y)q=2,dpdt=Fx(x,y,z,p,q)pFz(x,y,z,p,q)=2c(x,y)cx(x,y)(p2+q2)=2cx(x,y)c(x,y),dqdt=Fy(x,y,z,p,q)qFz(x,y,z,p,q)=2c(x,y)cy(x,y)(p2+q2)=2cy(x,y)c(x,y).

    Now we suppose that c(x,y)=|x|. Then the five ordinary differential equations become

    {dxdt=2x2p,dydt=2x2q,dzdt=2,dpdt=2sgn(x)|x|=2x,dqdt=0.

    In addition, the noncharacteristic curve can be parameterized by

    x(0)=s,y(0)=0,z(0)=u(x(0),y(0))=0.

    From F(x(0),y(0),z(0),p(0),q(0))=0, we have s2(p2(0)+q2(0))=1. By the strip condition, we get

    0=dz(0)ds=p(0)dx(0)ds+q(0)dy(0)dt=p(0)1+q(0)0.

    Thus, p(0)=0 and q(0)=±1/s. From dqdt=0, we get q(t)=q(0)=±1/s. Using the first and fourth equations, we have

    dx2x2p=xdp2ordxx3=pdp.

    Thus, we obtain 12x2(0)12x2(t)=p2(0)p2(t)2. By the initial condition, we get p2(t)=1x2(t)1s2=s2x2(t)s2x2(t), which implies p(t)=±s2x2(t)sx(t). Hence, the first equation becomes

    dxdt=±2xs2x2s,

    which can be integrated by separation of variables:

    tanh1(s2x2(t)s)=t0sx(t)s2x2(t)dxdtdt=±t02dt=±2t.

    This gives x(t)=s2s2tanh2(2t)=|s|sech(2t). On the other hand, the second equation becomes

    dydt=±2ssech2(2t),

    which can be solved by

    y(t)=y(0)stanh(2t)=stanh(2t).

    Finally, we have z(t)=z(0)+2t=2t. By tanh2(θ)+sech2(θ)=1 for all θR, we have x2+y2=s2, which gives

    s=x2+y2andt=±12tanh1(yx2+y2)=±14ln(x2+y2+yx2+y2y).

    Therefore, we obtain

    u(x,y)=±12ln(x2+y2+yx2+y2y)=±12ln((x2+y2+y)2(x2+y2)y2)=±ln(x2+y2+y|x|)for x0,yR.


  19. Consider the eikonal equation in three dimensions

    u2x+u2y+u2z=1.

    1. Solve the initial value problem with u=k=constant on the plane αx+βy+z=0.
    2. Find a complete integral.
  20. Solution
    1. Let F:R7R be defined by

      F(x,y,z,u,p,q,r)=p2+q2+r21for (x,y,z,u,p,q,r)R7.

      Consider the characteristic curve

      x=x(t),y=y(t),z=z(t),u=u(t)=u(x(t),y(t),z(t)),p=p(t)=ux(x(t),y(t),z(t)),q=q(t)=uy(x(t),y(t),z(t)),r=r(t)=uz(x(t),y(t),z(t)).

      Then we have the following ordinary differential equations

      {dxdt=Fp(x,y,z,u,p,q,r)=2p,dydt=Fq(x,y,z,u,p,q,r)=2q,dzdt=Fr(x,y,z,u,p,q,r)=2r,dudt=pFp(x,y,z,u,p,q,r)+qFq(x,y,z,u,p,q,r)+rFr(x,y,z,u,p,q,r)=2p2+2q2+2r2=2,dpdt=Fx(x,y,z,u,p,q,r)pFu(x,y,z,u,p,q,r)=0,dqdt=Fy(x,y,z,u,p,q,r)qFu(x,y,z,u,p,q,r)=0,drdt=Fz(x,y,z,u,p,q,r)rFu(x,y,z,u,p,q,r)=0

      with the initial condition

      x(0)=s1,y(0)=s2,z(0)=αs1βs2,u(0)=u(x(0),y(0),z(0))=k.

      From F(x(0),y(0),z(0),u(0),p(0),q(0),r(0))=0, we have p2(0)+q2(0)+r2(0)=1. By the strip condition, we have

      0=dz(0)ds1=p(0)dx(0)ds1+q(0)dy(0)ds1+r(0)dz(0)ds1=p(0)1+q(0)0+r(0)α=p(0)αr(0),0=dz(0)ds2=p(0)dx(0)ds2+q(0)dy(0)ds2+r(0)dz(0)ds2=p(0)0+q(0)1+r(0)β=q(0)βr(0).

      Thus, we obtain

      p(0)=±αα2+β2+1,q(0)=±βα2+β2+1,r(0)=±1α2+β2+1.

      From dpdt=dqdt=drdt=0, we get

      p(t)=±αα2+β2+1,q(t)=±βα2+β2+1,r(t)=±1α2+β2+1.

      Hence, the first four equations become

      {dxdt=±2αα2+β2+1,dydt=±2βα2+β2+1,dzdt=±2α2+β2+1,dudt=2.

      It yields that

      x(t)=s1±2tαα2+β2+1,y(t)=s2±2tβα2+β2+1,z(t)=αs1βs2±2tα2+β2+1,u(t)=2t+k.

      Here (t,s1,s2) can be expressed in terms of (x,y,z) by

      t=±αx+βy+z2α2+β2+1,s1=β2x+xαβyαzα2+β2+1,s2=α2y+yαβxβzα2+β2+1.

      Therefore, we obtain

      u(x,y,z)=k±αx+βy+zα2+β2+1for (x,y,z)R3.

    2. Let u(x,y,z;a,b,c)=xcosasinb+ysinasinb+zcosb+c. It is easy to get

      ux(x,y,z;a,b,c)=cosasinb,uy(x,y,z;a,b,c)=sinasinb,uz(x,y,z;a,b,c)=cosb

      and verify that u2x(x,y,z;a,b,c)+u2y(x,y,z;a,b,c)+u2z(x,y,z;a,b,c)=1, which means u(x,y,z;a,b,c) is a solution. Moreover, the matrix

      [ua(x,y,z;a,b,c)uxa(x,y,z;a,b,c)uya(x,y,z;a,b,c)uza(x,y,z;a,b,c)ub(x,y,z;a,b,c)uxb(x,y,z;a,b,c)uyb(x,y,z;a,b,c)uzb(x,y,z;a,b,c)uc(x,y,z;a,b,c)uxc(x,y,z;a,b,c)uyc(x,y,z;a,b,c)uzc(x,y,z;a,b,c)]=[xsinasinb+ycosasinbsinasinbcosasinb0xcosacosb+ysinasinbzsinbcosacosbsinacosbsinb1000]

      is of rank 3. Therefore, u=u(x,y,z;a,b,c) is a complete integral.

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