Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 1.3
- Solve u2x+yuy−u=0 with initial condition u(x,1)=x24+1.
- Solve u=xux+yuy+(u2x+u2y)/2 with initial condition u(x,0)=(1−x2)/2. [This is an example of Clairaut's equation u=xp+yq+f(p,q).]
- 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.
- 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(et−1), we have x−s4a=s(et−1). On the other hand, y2=4s2(et−1)2(a−4a2). Combining them, we get 4a2y2=(x−s)2(a−4a2), and hence s=x±2y√a√1−4a. From x=s(1−4a+4aet), we get
t=ln(1+x−s4as)=ln(2x√a−4a2±y(4a−1)2x√a−4a2±4ya).
Therefore, we obtainu(x,y)=a(x±(4a−1)y2√a−4a2)2=(x√a∓y√1−4a2)2for (x,y)∈R2.
Here we have used the fact thats2e2t=(set)2=[s⋅(1+x/s−14a)]2=(s+x−s4a)2=(x±2y√a√1−4a∓y2√a−4a2)2
- 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.
- Find the characteristic equations [i.e., generlizations of (29), (30)].
- Find the conditions on the initial values of p [i.e., generalization of (31), (32)].
- 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))]=n∑j=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+n∑i=jFpi(x,u,∇u)uxjxi=0,
which impliesFxi(x(t),z(t),p(t))+pi(t)Fu(x(t),z(t),p(t))+n∑j=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 obtaindpidt=n∑j=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 givesdzdt=n∑i=1uxi(x(t))dxidt=n∑i=1pi(x(t))Fpi(x(t),z(t),p(t)).
- 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)=n∑i=1hi(f(s))f′i(s).
- Show that the family of spheres Sa given by (x−a)2+y2+z2=1 has as its envelope E the unit cylinder y2+z2=1.
- Consider Clairaut's equation u=xux+yuy+f(ux,uy).
- Verify that u(x,y;a,b)=ax+by+f(a,b) is a complete integral.
- Use envelopes to generate a solution of u=xux+yuy+(u2x+u2y)/2 that is not linear in x and y.
- 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))=x⋅a+b⋅y+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. - 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 w∈C1, 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 a∈R. Then the envelope E can be determined by{u(x,y;a)=0,∂au(x,y;a)=0.
The second equation yields x+1⋅y+(a+a⋅1)=0, which gives a=−x+y2. Plugging into the first equation, we obtainu(x,y)=−x+y2⋅x−x+y2⋅y+12[(x+y2)2+(x+y2)2]=−(x+y)24for (x,y)∈R2.
Note that this solution is not linear in x and y. - 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.
- Verify that condition (34) fails for u(x,t;a,b)=ax−tH(a)+b, yet this is a complete integral.
- Use envelopes to generate a solution of ut+u2x=0 that is not linear in x and t.
- Since u(x,t;a,b)=ax−tH(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)]=[1−H′(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)]=[x−tH′(a)1−H′(a)100] has rank 2. This shows u=u(x,t;a,b) is a complete integral. - From a., we consider the two-parameter solution u(x,t;a,b)=ax−ta2+b. To find the envelope, we set b=w(a) for some w∈C1, which gives a one-parameter solution u(x,t;a):=u(x,t;a,w(a))=ax−ta2+w(a). For the sake of simplicity, we choose w(a)=a for a∈R. Then the envelope E can be determined by
{u(x,y;a)=0,∂au(x,y;a)=0.
The second equation x−2ta+1=0, which gives a=(x+1)/(2t). Plugging into the first equation, we obtainu(x,y)=x+12t⋅x−t⋅(x+1)24t2+x+12t=(x+1)24tfor (x,t)∈R2.
Note that this solution is not linear in x and t. - To describe the wave front produced by an initial disturbance at a point, consider (36) with Γ begin given by f≡g≡h≡0. Describe the solution(s).
- 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)=−log√x2+y2+yxfor x>0.
- Consider the eikonal equation in three dimensions
u2x+u2y+u2z=1.
- Solve the initial value problem with u=k=constant on the plane αx+βy+z=0.
- Find a complete integral.
- Let F:R7→R be defined by
F(x,y,z,u,p,q,r)=p2+q2+r2−1for (x,y,z,u,p,q,r)∈R7.
Consider the characteristic curvex=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 conditionx(0)=s1,y(0)=s2,z(0)=−αs1−βs2,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 have0=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 obtainp(0)=±α√α2+β2+1,q(0)=±β√α2+β2+1,r(0)=±1√α2+β2+1.
From dpdt=dqdt=drdt=0, we getp(t)=±α√α2+β2+1,q(t)=±β√α2+β2+1,r(t)=±1√α2+β2+1.
Hence, the first four equations become{dxdt=±2α√α2+β2+1,dydt=±2β√α2+β2+1,dzdt=±2√α2+β2+1,dudt=2.
It yields thatx(t)=s1±2tα√α2+β2+1,y(t)=s2±2tβ√α2+β2+1,z(t)=−αs1−βs2±2t√α2+β2+1,u(t)=2t+k.
Here (t,s1,s2) can be expressed in terms of (x,y,z) byt=±αx+βy+z2√α2+β2+1,s1=β2x+x−αβy−αzα2+β2+1,s2=α2y+y−αβx−βzα2+β2+1.
Therefore, we obtainu(x,y,z)=k±αx+βy+z√α2+β2+1for (x,y,z)∈R3.
- 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+ysinasinb−zsinbcosacosbsinacosb−sinb1000]
is of rank 3. Therefore, u=u(x,y,z;a,b,c) is a complete integral.
Solution
Let F:R5→R be defined byF(x,y,z,p,q)=p2+yq−zfor (x,y,z,p,q)∈R5.
Consider the characteristic curvex=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 conditionx(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 havep2(0)+q(0)−s24−1=0.
By the strip condition, we haves2=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 obtainp(t)=p(0)et=s2et,q(t)=q(0)=1.
Hence, the first three equations becomedxdt=set,dydt=y,dzdt=s22e2t+y,
which implies x(t)=x(0)+s(et−1)=set and y(t)=y(0)et=et, andz(t)=z(0)+s24(e2t−1)+(et−1)=s24e2t+et=(set2)2+et.
It is easy to get t=ln(y) and s=x/y, and we obtainu(x,y)=x24+yfor x∈R 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}.Solution
Let F:R5→R be defined byF(x,y,z,p,q)=z−xp−yq−(p2+q2)/2for (x,y,z,p,q)∈R5.
Consider the characteristic curvex=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)=−x−p,dydt=Fq(x,y,z,p,q)=−y−qdzdt=pFp(x,y,z,p,q)+qFq(x,y,z,p,q)=p(−x−p)+q(−y−q)=−xp−yq−p2−q2,dpdt=−Fx(x,y,z,p,q)−pFz(x,y,z,p,q)=−(−p)−p⋅1=0,dqdt=−Fy(x,y,z,p,q)−qFz(x,y,z,p,q)=−(−q)−q⋅1=0
with initial conditionx(0)=s,y(0)=0,z(0)=u(x(0),y(0))=(1−s2)/2.
From F(x(0),y(0),z(0),p(0),q(0))=0, we have1−s22−sp(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 obtainp(t)=p(0)=−s and q(t)=q(0)=±1.
Hence, the first three equations becomedxdt=−x+s,dydt=−y∓1,dzdt=sx∓y−s2−1,
which implies x(t)=s+(x(0)−s)e−t=s, y(t)=∓1+(y(0)±1)e−t=±(e−t−1), anddzdt=s2−(e−t−1)−s2−1=−e−t.
Thus, we can solvez=z(t)=z(0)+(e−t−1)=e−t−1+s22.
It is easy to get t=−ln(1±y) and s=x, and we obtainu(x,y)=1±y−1+x22=±y−1−x22for (x,y)∈R2.
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:R5→R be defined byF(x,y,z,p,q)=p2+q2−zfor (x,y,z,p,q)∈R5.
Consider the characteristic curvex=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)=p⋅2p+q⋅2q=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 conditionx(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 get2as=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|√a−4a2∈R, provided a∈(0,1/4]. Now from dpdt=p and dqdt=q, we obtainp(t)=p(0)et=2aset,q(t)=q(0)et=±|s|et√a−4a2.
Hence, the first three equations becomedxdt=4aset,dydt=±2|s|et√a−4a2,dzdt=2z,
which impliesx(t)=x(0)+4as(et−1)=s(1+4aet−4a),y(t)=y(0)±2|s|(et−1)√a−4a2=±2|s|(et−1)√a−4a2,z(t)=z(0)e2t=as2e2t.
Solution
Solution
Let w(x,y,z;a)=(x−a)2+y2+z2−1. Then the evelope E is the intersection{w(x,y,z;a)=(x−a)2+y2+z2−1=0,∂a(x,y,z;a)=−2(x−a)=0
From the second equation, we get x=a, and hence y2+z2=1.Solution
Solution
Solution
Let F:R5→R be defined byF(x,y,z,p,q)=c2(p2+q2)−1for (x,y,z,p,q)∈R5,
where c>0 is a constant. Consider the characteristic curvex=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)=p⋅2c2p+q⋅2c2q=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 conditionx(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)=c−1coss and q(0)=c−1sins for s∈[0,2π). Now from dpdt=dqdt=0, we obtainp(t)=p(0)=c−1coss,q(t)=q(0)=c−1sins.
Hence, the first three equations becomedxdt=2ccoss,dydt=2csins,dzdt=2,
which impliesx(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=tan−1(y/x), and we obtainu(x,y)=±√x2+y2cfor (x,y)∈R2.
Solution
Let F:R5→R be defined byF(x,y,z,p,q)=c2(x,y)(p2+q2)−1for (x,y,z,p,q)∈R5.
Consider the characteristic curvex=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)=p⋅2c2(x,y)p+q⋅2c2(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 byx(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 get0=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 havedx2x2p=−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=s2−x2(t)s2x2(t), which implies p(t)=±√s2−x2(t)sx(t). Hence, the first equation becomesdxdt=±2x√s2−x2s,
which can be integrated by separation of variables:−tanh−1(√s2−x2(t)s)=∫t0sx(t)√s2−x2(t)dxdtdt=±∫t02dt=±2t.
This gives x(t)=√s2−s2tanh2(∓2t)=|s|sech(∓2t). On the other hand, the second equation becomesdydt=±2s⋅sech2(∓2t),
which can be solved byy(t)=y(0)−stanh(∓2t)=−stanh(∓2t).
Finally, we have z(t)=z(0)+2t=2t. By tanh2(θ)+sech2(θ)=1 for all θ∈R, we have x2+y2=s2, which givess=√x2+y2andt=±12tanh−1(yx2+y2)=±14ln(√x2+y2+y√x2+y2−y).
Therefore, we obtainu(x,y)=±12ln(√x2+y2+y√x2+y2−y)=±12ln((√x2+y2+y)2(x2+y2)−y2)=±ln(√x2+y2+y|x|)for x≠0,y∈R.
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