A R K I V F O R M A T E M A T I K B a n d 7 n r 28
1.67 06 2 Communicated 12 April 1967 by T. NAGELL and L. CARLESON
On the partial differential equation u~uxx+ 2 UxUyUxy+u~uyy=O
B y GUI~NAR ARONSSON
1. Introduction
This pap6r treats various aspects of the partial differential equation
(~u~ 2B2u _~uBu B2u /Bu\2B2u
Bx]--I
~ + 2 - - + t ~ y ) ~ = 0 . (1) Bx By BxByThis equation was derived in [1] where an extension problem was studied, and it turned out that (1) is closely connected to this extension problem (Theorems 6, 7 and 8 in [1]). The equation is quasi-linear and parabolic
(AC-B2=O),
and is not of a n y classical type. The results from [1] will be used very little in this paper. As far as the author knows, the equation (1) has not been treated before, apart from the paper [1].Let
u(x, y)
be a solution of (1) and let C be a trajectory of the vector field grad u.Then it is proved in Section 2 that C is either a convex curve or a straight line, and this result, together with a formula for the curvature of C, is fundamental for the later sections.
I n Section 3 we consider two particular classes of solutions to (1).
Section 4 is devoted to a discussion of the regularity of solutions to (1). I t turns out that a solution for which the trajectories of grad u are convex curves, is infi- nitely differentiahle.
I n Section 5 we consider some differential-geometric aspects of (1).
Section 6 contains an estimate for ]grad u]. A consequence of this estimate is t h a t a nonconstant solution of (1) has no stationary points.
I n Section 7 we consider solutions of (1) outside a compact set and solutions in the whole plane. The latter ones turn out to be linear functions only.
I n Section 8 we consider the behaviour of [ grad u ] near the b o u n d a r y of a region.
Section 9, finally, contains a few results on the Dirichlet problem for (1).
I n this paper, we will only consider classical solutions of (1), that is, solutions in C 2. We will not discuss extensions of the results to the case of more than two inde- pendent variables.
2. S o m e preliminary considerations A l e m m a o n the curvature o f a streamline We introduce the notation
I t is easy to see that
A((I)) = 89 grad {(grad d)) 2} .grad O.
(2)
G. ARO~SSO~, On a partial differential equation
The m e a n i n g of A(O) = 0 is therefore t h a t [grad O[ is c o n s t a n t along every t r a j e c t o r y of the vector field g r a d (I). Such trajectories will be called streamlines in the sequel.
Hence (P is a linear function of the arc length along each streamline. I t also follows f r o m (2) t h a t A((])) is an o r t h o g o n a l invariant. Clearly, a n y function (I) for which [grad (I)[ is constant, satisfies A((I))=0. This particular class of solutions will be discussed in Section 3. These functions correspond to unique solutions of t h e exten- sion problem in [1] (Theorems 3-5).
Consider a function (I)(x, y) E C u in a n e i g h b o u r h o o d of a point P where g r a d (I) # 0 . I n t r o d u c e curvilinear coordinates u = (I) a n d v = a function which is c o n s t a n t along each streamline of (I). Assume also t h a t v E C 2 a n d g r a d v # 0 . This gives (locally) a one-to-one m a p p i n g (x, y ) o ( u , v) for which J = d ( u , v)/d(x, y ) # 0 a n d
u~v~ + u~v~ = O. (*)
N o w we have the formulas x u = J i Vy, X v = - - J 1 u~, Yu = - - J 1 Vx a n d Yv = J1 u~, where J1 = 1/J. The relation (*) can t h u s be written
xuxv + y~yv = O.
Clearly, a streamline of u in the xy-plane is given b y v = constant, a n d arc length along such a curve is given b y
/ u r / ~
x~ + y~ du.
p
N o w t h e condition A ( ( I ) ) = 0 means t h a t this is a linear function of U which is equi- valent to ~/~U(Xu + y~) = 0 , or 2 2 x u x ~ + y~yu~ = 0 .
A careful analysis of the preceding reasoning leads to the result:
w h w h xu + y ~ > 0 , Xv + y , > 0 Let there be given t w o / u n c t i o n s x(u, v), y(u, v) i n C 2 /or " 2 2 2 and which satis/y the system
x~x, + y~yv = O, XuX~ § Y~Y~u = O.
T h e n (each/unction element o/) the i n v e r s e / u n c t i o n u = u ( x , y) satisfies A ( u ) = 0 . E x a m p l e . The functions x = v cos u, y = v sin u satisfy t h e s y s t e m a n d u = arctg (y/x) satisfies A ( u ) = 0 .
Consider again a function u(x, y) E C 2 in a n e i g h b o u r h o o d of a point where g r a d u # 0 . A streamline of u is given b y u y d x - u , d y = O , where t h e coefficients are in C 1. A n y streamline is d e t e r m i n e d b y a n initial point a n d it is a curve in C ~ with continuously v a r y i n g curvature. The same holds for the level lines of u, which are governed b y uxdx + u~dy=O.
The c u r v a t u r e of a streamline is given b y
I g r a d u l u , + y~y a r c t g u ,
1
i gra d ul a (u~u~ - uxu~uzx + u x u ~ u ~ - u~ uxy) 1
21gra d u[3grad {(grad u)2} 9 ( - u~,u~).
Here, the last two expressions hold also at points where uz = 0.
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ARKIV FOR MATEMATIK, B d 7 n r 2 8
F o r the level lines, we h a v e
1 1
, grad u , ( - uy ~ + ux ~ ) arctg ( - U~) - , grad u la (u~ uxx - 2 ux uy uxy + u~ uyy) 1
- I g r a d u I a ((grad u)2 Au - A (u)).
I f u is a solution of
A(u)=0,
then, clearly, the c u r v a t u r e of a streamline can be w r i t t e n_+ I g r a d {(grad u)2}l = -I- I g r a d (I g r a d u I)l 2 Igrad u I s I g r a d u I
L e m m a 1.
Let u(x, y) satis/y A ( u ) = 0 in a domain D and let
g r a du:~O in D. I / C is a streamline o~ u in D, then there are two alternatives:
I. The curvature o / C is 4 0 at all points o/C.
I I .
C is a straight line.
Consequently, the streamlines of u are
convex curves and straight lines.
Proo/.
I t is sufficient t o p r o v e t h e following assertion: I f t h e c u r v a t u r e is zero a t A E C, t h e n there is a (1-dimensional) n e i g h b o u r h o o d of A along C where t h e curva- t u r e is zero. I n t r o d u c e a coordinate s y s t e m such t h a t XA=(U~)A=0 a n d consider a n e i g h b o u r h o o d U of A such t h a t ux 4 0 in U. W e consider the streamlines as solu- tions of a n initial-value p r o b l e mdy/dx=Cg(x, y), y(O)=Y0,
a n d we writey(xl)=Yl =
yl(Xl, Y0)"Since
A(u)
= 0 , we have, for fixed xi,~(Y0) = ]grad u 1(0. y.) = ]grad U](x,. v,) = ~(Yl).
I f x 1 is fixed a n d [xl] is small enough, t h e n
~--~oYl(xDYo)
= exp [S~'Ov(x,y(x))dx] > 0
([3], pp. 25-27; [9], pp. 73-74).Clearly, we also h a v e
Yo=Yo(Yl)
a n ddyo/dy 1
is finite. Therefored ~ / d y l = d ~ / d y o " d y o / d Y l .
Here,
dg/dyo=O
forYo=YA
according t o our f o r m u l a s for the c u r v a t u r e . H e n c edlt?/dyl
= 0 a t YI = Y l ( X l ,YA)
which m e a n s t h a t ]grad ([ g r a d u ])[ = 0 a t (xi,y~(xi, YA)).
This completes t h e proof.
A t the s a m e t i m e we h a v e p r o v e d t h a t if g r a d ( [ g r a d u ] ) = 0 a t a p o i n t of a s t r e a m - line C, t h e n this relation holds a t all points of C, which is t h e n a s t r a i g h t line. I f we consider t h e h o d o g r a p h m a p p i n g
p=ux, q=uy,
it follows t h a tD(u)=-d(p, q)/
d ( x ,
y)=uxxuyy-u~y
- 2 = 0 on C. N o w let C be a c u r v e d streamline a n d let (r, 0) b e p o l a r coordinates in t h e h o d o g r a p h plane. T h e n g r a d r is orthogonal to C a n d n o t zero, a n d g r a d 0 has a nonzero c o m p o n e n t along C. H e n c eD ( u ) 4 0
on C.Along a curved streamline C,
g r a d ( [ g r a d u ] )and D(u) are both nonzero, and
g r a d (] g r a d u] )points to the concave side of C.
G. ARONSSOr~, On a partial differential equation
J ) 7~
J
Fig. 1
I] the streamline C is a straight line, then g r a d ( I g r a d u I) and D(u) are both zero along C.
I t should be noticed t h a t D(u) =-zt,x uyy -u2zy <~ 0 always. This means t h a t t h e surface z = u ( x , y) has nonpositive Gaussian curvature. I t follows easily b y considering 2 Uxx + 2u, u~u,y +u~ uyy as a quadratic form in u, a n d uu. Finally, we observe t h a t
Ux
a streamline C c a n n o t t e r m i n a t e inside D, since ]grad u I is c o n s t a n t on C.
Example. We shall illustrate these things b y the C a u c h y - K o w a l e w s k i t h e o r e m ([4], [6], [8]). W e write
1 + u ~ u~)
u x z - 2 (2uzZlyUxy 2 ux
a n d prescribe u a n d u~ on the y-axis, in a n e i g h b o u r h o o d of y = 0 . Choose, for example, u(0, y ) = 0 a n d ux(O, y ) = ~ v ( y ) > 0 , where ~v(y) is analytic for [y] <5. T h e n there is an analytic solution u(x, y) in a n e i g h b o u r h o o d of x = y - 0 . L e t (v'(0) = 0 a n d ~v'(y) 4 0 if y 4 0 . Consider three cases:
1. qv(y) has a m a x i m u m at y = 0 . 2. ~(y) has a m i n i m u m at y = 0.
3. q~'(y) does n o t change sign at y = 0 .
I n each case, the x-axis is a streamline a n d all other streamlines are curved. T h e streamlines are sketched in Figs. 1-3. I n Fig. 3, it is a s s u m e d t h a t ~ ' ( y ) > 0 if y # 0 .
J
J J
- >
Fig. 2 398
ARKIV FOR MATEMATIK. Bd 7 nr 28
Fig. 3
2 2
3. T w o particular c l a s s e s o f s o l u t i o n s to u x u x x - { - 2 ux Uy Uxy-t- Uy Uyy = 0 I n this section we will make a few remarks on t h e class of functions u for which ]grad u[ is constant. We will also determine all harmonic functions u which satisfy
A(u)
= 0 .I t follows from the identity A (u)-: 89 grad ((grad u)2) 9 grad u t h a t those functions u 6 C a, for which I grad u I is constant, constitute a subclass of all solutions of A ( u ) = O.
This is natural from another aspect: among all functions u(x, y) 6 C 2, those which are absolutely minimizing, are characterized b y the differential equation A ( u ) = 0 ([1], Theorem 8). A subclass of all absolute minimals are those functions u which are unique solutions of some extension problem of the t y p e considered in [1]. And the condition for this is t h a t u 6 C 1 and t h a t ]grad u] is constant ([1], Theorems 4 a n d 5).
The differential equation ]grad u] = constant is treated in [4], pp. 88-91 and [6], pp. 4 0 4 3 . I t is well known t h a t if the surface z = u ( x , y) is the tangential developable of a helix with its axis parallel to the z-axis, then ]grad u[ =const. Other basic types of solutions are linear functions and functions of the form
A V(x - xo) 2 + (y - yo) 2 + B.
There also exist (even infinitely differentiable) solutions whose restrictions to different subdomains belong to different basic types (namely the first and the second).
I t should be noticed t h a t a Cl-function u, for which l grad u I is constant, need not be in C 2. This is shown b y the function
x for x > O , y < O, u(x,y) V x i - + y 2 for x>O,y>~O.
Here, a2u/~y ~ is discontinuous across the x-axis. However, we have the following result:
Theorem 1. Let u(x, y) 6 Cl(gl) and let ]grad u [ = M = constant in ~1.
Then ~u/~x and ~u/~y satis/y Lipschitz conditions on each compact subset o/ gl.
G. AnONSSON, On a partial differential equation
Fig. 4
Proo/. W e k n o w f r o m L e m m a 1 in [1] t h a t the streamlines of u are straight lines w i t h o u t c o m m o n points in ~ . Consider a c o m p a c t set K c ~ a n d p u t d = t h e distance f r o m K t o ~ . T a k e t w o points P, Q E K such t h a t d i = ~ < d (Fig. 4). L e t P b e fixed a n d Q variable in the circle ~QQ < d. Clearly, grad u(P). grad u(Q):V O, a n d f r o m t h e c o n t i n u i t y we h a v e g r a d u(P).grad u(Q)> 0. Hence the angle between grad u(P) a n d g r a d u(Q) is equal to t h e smallest angle g between t h e corresponding stream- lines, a n d we get
dl ~ dl
H e n c e lux(P)-ux(Q)l<~M.o~<~(Myr/2d).dl. If PQQ>d, t h e n I%(P) - u~(Q) I <. 2 M <~ ( 2 M i d ) " d 1.
I t follows t h a t u~ satisfies a Lipschitz condition with the c o n s t a n t 2M/d, a n d similarly for %.
N o t e t h a t % a n d % need n o t satisfy a Lipschitz condition, or even be u n i f o r m l y continuous in ~ . This can be seen f r o m the example u = V ~ + y 2 in ~ : (x -- 1) 2 +y~ < 1.
T h e differential equation
A(u)=--u~%x + 2%%Uxy + uy%y 2 = 0 (1)
has a formal resemblance to t h e differential equation of minimal surfaces,
2 )uxx _ 2u~%%u + (1 + ux )%~ - 0 2 _ (2) (I + %
a n d addition of these equations gives [1 + (grad u)2J(uxz + % y ) = 0 , or
Au = 0. (3)
I t is clear t h a t a function u(x, y), which satisfies a n y t w o of t h e differential e q u a t i o n s (1), (2), a n d (3), also satisfies t h e remaining one. We will n o w determine these com- m o n solutions.
Theorem 2. 1 / u ( x , y) is harmonic in a domain D and i/ A ( u ) = 0 in D, then u is either a linear/unction, u = A x + B y + C, or it can be written
u = D arctg y - Yo + E
X - - X 0
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ARKIV FOR MATEMATIK. B d 7 n r 28 /or some point (Xo, Yo), such that a r e t g [ ( y - y o ) / ( x - x o ) ] is one-valued and continuous in D.
(In p a r t i c u l a r , i t follows t h a t (xo, Yo)r D.)
Proo/3 W r i t e z = x + i y a n d let v b e t h e c o n j u g a t e h a r m o n i c f u n c t i o n of u in a s i m p l y c o n n e c t e d s u b d o m a i n D 1 of D. T h i s gives a n a n a l y t i c f u n c t i o n w ( z ) = u + i v in n 1. W e h a v e w ' ( z ) = ~ u / ~ x - i ( ~ u / ~ y ) , w h i c h m e a n s t h a t w'(z) = g r a d u ] .
P u t w (z)=T. On a s t r e a m l i n e of u we h a v e v = c o n s t a n t a n d T = c o n s t a n t . The f u n c t i o n s w(z) a n d T =w'(z) m a p D 1 o n t o D w a n d D~, r e s p e c t i v e l y . W e a s s u m e t h a t D 1 is chosen such t h a t w'(z)4=0 in D 1 a n d such t h a t t h e m a p p i n g D I - ~ D w is one-to-one. F u r t h e r , we m a y a s s u m e t h a t 0 = a r t T is o n e - v a l u e d a n d c o n t i n u o u s in D~.
N o w t h e f u n c t i o n log Iv I = l o g ] w ' ( z ) [ is h a r m o n i c in D , . B u t t h e i n v e r s e of w(z), z = z ( w ) , is a n a l y t i c in Dw, a n d hence h ( w ) = l o g [T[z(w)][ is h a r m o n i c in Dw:
~2h/~u2 + ~ h / ~ v 2 = 0. Since ~h/~u = 0, we g e t ~2h/~u2 = ~2h/(~u ~v) = ~2h/~v2 = 0 in D w.
H e n c e h=co+ClV , w i t h r e a l c o n s t a n t s Co, c 1. F r o m t h e C a u c h y - R i e m a n n e q u a t i o n s we g e t ~O / ~u = - cl, ~O / ~v = 0, w h i c h gives 0 - c 2 - c1 u, a n d log T = h + iO = (c o + i%) - ic l(u + iv) = C O - ic 1 w in D~.
H e n c e T = C 1 e x p ( - i c l w ) , w h e r e C 1 4 0 . I f c 1 = 0 , we g e t W = C l z + C 2, w h i c h gives u = A x + B y + C.
I f c 1 ~ 0 , t h e n d z / d w = l / C l e x p ( i c l w ) a n d Z - Z o = C 2 e x p (iclw), C 2 4 0 , z o ~ D 1.
T a k i n g t h e a r g u m e n t s of b o t h m e m b e r s , we g e t a r g ( z - % ) = C l U + c a.
This p r o v e s t h e r e s u l t for D1, a n d t h e g e n e r a l r e s u l t follows b y a n a l y t i c conti- n u a t i o n .
T h e r e is a c o n s e q u e n c e of t h i s t h e o r e m t h a t m a y b e of s o m e i n t e r e s t .
W e m a y i n t e r p r e t u(x, y) as a h y d r o d y n a m i c p o t e n t i a l : Consider a t w o - d i m e n s i o n a l , s t e a d y p o t e n t i a l flow of a n ideal liquid. I f each p a r t i c l e h a s c o n s t a n t speed, t h e n t h e flow is e i t h e r a u n i f o r m t r a n s l a t i o n or o t h e r w i s e t h e p a r t i c l e s m o v e in c o n c e n t r i c c i r c u l a r o r b i t s w i t h t h e s p e e d C/r, w h e r e r is t h e d i s t a n c e f r o m t h e c e n t e r of t h e circles, a n d t h e c o n s t a n t C > 0 is c o m m o n t o all p a r t i c l e s . (The s p e e d of a p a r t i c l e is t h e m o d u l u s of its v e l o c i t y vector.)
2 2
4. T h e differentiability p r o p e r t i e s o f s o l u t i o n s to ux uxx + 2 ux Uy uxy + uy Uyy = 0.
This section c o n t a i n s a r e s u l t on t h e r e g u l a r i t y of a class of solutions t o A ( u ) = 0 . I t is p r o v e d b y a p p l i c a t i o n of t h e h o d o g r a p h m a p p i n g w h i c h w o r k s o n l y if uxxuy~- u~y 4 0 . I t is s h o w n b y a n e x a m p l e t h a t t h e r e s u l t is false w i t h o u t this r e s t r i c t i o n . L e m m a 2. Let u(x, y ) E C 2 in a region D and let A ( u ) = 0 in D. A s s u m e / u r t h e r that g r a d u 4=0 in D and that Uxx%y-U2xy :~0 in D.
T h e n u E C~176
Proo/. A p p l y t h e h o d o g r a p h m a p p i n g p - u ~ , q =u~. (See [6], p. 521, or [2], p. 12.) W e h a v e J - d ( p , q)/d(x, y ) = U x x U y y - u ~ 4 0 . H e n c e t h e m a p p i n g (x, y)->(p, q) is o n e - t o - o n e a n d b i c o n t i n u o u s in a n e i g h b o u r h o o d of a n a r b i t r a r y p o i n t in D. W e r e s t r i c t our a t t e n t i o n t o such a n e i g h b o u r h o o d . T h e f u n c t i o n s x = x(p, q) a n d y = y(p, q) a r e in C 1 a n d we h a v e t h e w e l l - k n o w n r e l a t i o n s
qy : gxp; py : -- Jxq; qx = -- JY~ a n d px = Jyq.
1 This proof was suggested by Professor Bengt J. Andersson, Stockholm. The author's original proof is more complicated.
G. AaoNssor~, O n a p a r t i a l d i f f e r e n t i a l e q u a t i o n
N o w we introduce the function (Legendre transform) q a ' = x p + y q - u . Clearly, q~
is a Cl-function of p a n d q, a n d
7 - = x + p x p + qy~, - u x x v - % ~,~, = x , o p
~tF
77_ = p x q + qyq + y - u~ xq - % yq = y .
Hence Re is a C2-function of p a n d q, a n d
?x 1 1
~ ' P - ~ p - ~1 q~ = [1 u ~ ,
~ ~ x - 1 1
qeqq ~ Y 1 1
2 9
The equation u:: Ux,: + 2u~ uy u~y + u~ %y = 0 is thus t r a n s f o r m e d into
q2~-Fpp -- 2 p q W p q + p~Wq,~ = 0
(1)
a n d we get a linear differential equation in the h o d o g r a p h plane. Next, we introduce polar coordinates (r, 0) in the h o d o g r a p h plane. This can be done, since we h a v e assumed grad u # 0 . The equation (1) is t h e n t r a n s f o r m e d into
~F00 + rq e, = 0.
A n o t h e r substitution, ~)= - l o g r, gives tFoo = q ~ ,
which is the well-known heat equation. I t is k n o w n t h a t every solution of the h e a t equation is infinitely differentiable ([7], p. 314,[5], p. 74). Hence, ~F E C ~176 as a func- tion of (~, 0), a n d consequently also as a function of (p, q). Thus, x =tFp a n d y =VFq are infinitely differentiable functions of p a n d q. I t follows t h a t p = u x a n d q = % are infinitely differentiable with respect to x a n d y. This completes t h e proof.
R e m a r k . I t will be p r o v e d later t h a t the condition g r a d u 4 0 is a l w a y s satisfied, unless u is constant. This condition can therefore be omitted. 1 However, t h e condi- tion U x x % y - U 2 x ~ ~ - 0 c a n n o t be omitted. One w a y to show this is to c o n s t r u c t a func- tion f i x , y ) in C ~, b u t n o t in C a, for which I g r a d / I is constant. This can be done b y using the geometric properties of such functions which were discussed in Section 3, b u t it involves some calculations a n d we prefer to describe a quite different example.
Compare Theorems 6 and 7.
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ARKIV FOR MATEMATIK. B d 7 n r 2 8
Example. Consider a Cauchy problem for A ( u ) = 0 . We write the equation as uz~ = - 1/u2z (2Ux% u~y + u~ uyy) and prescribe u and ux on the y-axis, in a neighbourhood of y=O, b y u(0, y)=O, uz(O, y ) = l § a. From the Cauchy-Kowalewski theorem it follows that there is an analytic solution u(x, y) in a neighbourhood of the origin.
This corresponds to case 3 of the example in Section 2. I t is clear that the x-axis is a streamline and that the other streamlines are curved. Along the x-axis, we have u x = l , % = 0 ; further, A ( u ) = 0 gives u ~ = 0 and from U~xU~y-u~ =0 2 we get u ~ = 0 . Along the y-axis we have ux > 0, uy = 0, and we know that every curved streamline turns the convex side downwards. Therefore, we must have uy(x, y ) > 0 if x >0, y 4 0 and uy(x, y ) < 0 if x <0, y 4 0 . Since uy = 0 on the x-axis we must have uyy = 0 there.
Consequently, u - x , ux = 1 and % = U x x - u , ~ - u y y = 0 along the x-axis.
Form the function u l ( x , y ) = l u(x'y) for y>~O,
(x
for y~<O.I t follows from above t h a t u 1E C 2 and A(Ul) =0. From our initial conditions it follows that u~yy=0 and u~y~y=6 at the origin. Therefore, %yy(x, 0 ) 4 0 if 0 < Ix[ < 8 and this shows t h a t (ul)yyy does not exist at these points. Hence, u I does not belong to C a in a n y neighbourhood of the origin. I t also follows t h a t a continuation o / a solution o/ A(u) = 0 need not be unique.
5. Application of differential geometry
This section contains a discussion of the "intrinsic" geometric properties of a surface S: z=~P(x, y) where A((:I))=0.
Theorem 3. Let (I)(x,y) satis/y A((I))=0, and g r a d ( I ) 4 0 . Consider the sur/ace S : z=O(x, y) and the projections on this sur/ace o/the streamlines o/OP(x, y).
These image curt,es are both asymptotic curves on S, and
helices with a common axis. namely the z-axis.
(For definition of a helix, see [10], p. 33).
Proo/. The asymptotic directions on a n y surface z = (I)(x, y) are determined by
~Pxx(dx) ~ + 2@xydx dy § r 2 = O.
Hence, from A((I))=0, it follows that d y : d x - % : u x corresponds t o asymptotic curves. We also know that u is a linear function of arc length on each streamline in the xy-plane, du/ds = C. From this it follows that we have, on the corresponding space curve, dz/ds 1 : C/~l + ~ = c o n s t . This completes t h e proof.
Remark. Suppose that O,zqbyy-(I)~x~<0 in a domain D. Then the surface S has only hyperbolic points, and S has two distinct families of asymptotic curves. Ac- cording to the theorem, one of these families consists of helices with a common axis. Further, the geodesic curvature of such a curve is not zero.
The geometric features of the case (Px~(I)yy-(I)2xy = 0 (where [grad u[ is constant) were treated in Section 3 and the references cited there.
The condition grad (I) ~=0 is in fact no restriction, in view of Theorem 6.
I n the converse direction we have
a. ARO~SSOr~, On a partial differential equation
(
Y
Fig. 5
T h e o r e m 4. Let the sur/ace S be in C ~, and let S be s i m p l y covered by a /amily ~4 o / a s y m p t o t i c curves with nonvanishing geodesic curvature. A s s u m e now that all curves in A are helices with a common axis 1. Introduce an orthonormal coordinate system such that the z-axis is parallel to l.
The sat]ace S will then de/ine a ]unction z = O(x, y), at least locally, and this ]unction satis/ies A (69) = O.
Proo]. Consider a curve C E A, a point P E C, a n d let _t, n be respectively t a n g e n t a n d principal n o r m a l to C at P. (Note t h a t _n is well-defined, since the geodesic c u r v a t u r e is n o t zero.) T h e n _t._n=0, a n d since C is a helix with the axis parallel to the z-axis, we have n . e ~ = 0 . F u r t h e r , since C is an a s y m p t o t i c curve we h a v e N ' n = 0, where N is t h e surface normal. Hence, _t, N a n d % lie in a plane (the recti- fying plane of C).
W e m u s t h a v e ]_t-_e z ] < 1 since, otherwise, C would h a v e to be a straight line.
Because of the relation _t.N = 0, this means t h a t N.e~ =4=0. This shows t h a t S defines, locally, a function z =q)(x, y ) E C 2 a n d further, since grad (I) m u s t lie in the inter- section of the rectifying plane of C a n d t h e xy-plane, it follows t h a t C is t h e image of a streamline of (P. Finally, C is an a s y m p t o t i c curve, which means t h a t O~ Ox~+2OxO~Ox~ + O ~ O ~ y = O . (Compare t h e proof of the previous theorem.) 2
I t m a y perhaps be of some interest to h a v e a n o t h e r derivation of the p a r a m e t r i c r e p r e s e n t a t i o n t h a t was derived in Section 2.
- O x y < O . Consider a surface S : z =d)(x, y), where A(O) = 0 , g r a d 9 4=0, a n d Ox~Oyy 2 T h e n g r a d ( I g r a d ( I ) ] ) # 0 a n d we have locally a one-to-one m a p p i n g (x, y ) o ( ( I ) , grad (I)). N o w we introduce p a r a m e t e r s u, v on S b y writing u = ( I ) , v = grad (I) . T h e n F = 0 , E = l / v 2, where the n o t a t i o n is t h a t of [10], p. 58.
I t also follows t h a t F ~ I = 0 ([10], p. 107). F r o m Theorem 3 we k n o w t h a t v = c o n - s t a n t corresponds to a s y m p t o t i c curve3.
H e n c e e = 0 a n d the Gauss equation for _xu~ takes the f o r m _xu~=F~lx, ([10], pp.
74-75, 107). If we write F ~ I = B ( u , v) a n d x=q)l(u, v), y = ~ 2 ( u , v), z = ~ a ( u , v ) - u , we obtain
(~,)~ - B ( u , v) (q~z)~ = O, i = 1, 2, 3.
F u r t h e r , the p a r a m e t r i c lines are orthogonal on S, as well as their projections on t h e xy-plane. Hence (~1)~(~1)~ + (~2)~(~2), = 0 .
A n a r g u m e n t in the converse direction leads to the following result:
4 0 4
ARKIV FOR MATEMATIK. Bd 7 nr 28
Suppose t h a t there are given two functions
qpl(u,
v), ~ ( u , v) in C 2 which satisfy(q)~)uu-B(u,
v)(~)~ = 0, i = l, 2, whereB(u, v)
is an arbitrary continuous function.Suppose also t h a t
a n d
in the region in question.
B y the relations
( ~ ) u ( ~ l ) , + ( ~ ) u ( ~ ) v = 0 ( ~ ) ~ ( ~ ) , - ( ~ ) ~ ( ~ ) u =~ 0
x = ~ l ( u , v ) , y = ~ 2 ( u , v ) , z = u
z is defined, at least locally, as a function of x and y. This function
z =@(x, y)
satisfies A(O) = 0 .I t is easy to see t h a t this is identical with the result on parametric representation t h a t was derived in Section 2. First, if the above conditions are satisfied, then
(~1)~(~1)~u + ( ~ ) u ( 9 ~ ) ~ = B(~, v)[(~l)u(~)~ + ( ~ ) u ( ~ ) ~ ] = 0.
And if the conditions in Section 2 are satisfied, then the vector
(Xu, y~) (40)
is ortho- gonal to (xv, yv) and to(X~u, y~)
which means t h a t the latter ones are parallel:Xuu= B(u, v)x~, y ~ = B(u, v)y~.
6. An inequality for I grad u I" Nonexistence of stationary points
An i m p o r t a n t feature of the differential equation
A ( u ) = 0
is t h a t a nonconstant solution has no stationary points. This is obtained below as a consequence of an estimate for I grad u ]. We begin with a lemma.L e m m a 3.
Consider a /unction /(x)ECI[A, B/ with
/ ( A ) = 0and
/ ( B ) = I .Then there is a sequence o/ open intervals
(Iv}on
[A, B],]inite or denumerably infinite, such
that:(1)
The intervals 1~ are pairwise disjoint,
(2)/'(x)
> 0on all lv,
(3)
i / I v = ( a ~ , by), then X,(/(b,)-/(av) )
= 1 ,(4)
i / y e l v / o r some v, and x
< y ,then/(x) </(y).
(In particular, x e Iv, y e Iz, v 4 # implies t h a t / ( x )4/(y).)
(5) 0 < ] ( x ) < l
on all Iv.
Proo/.
F o r m the function~(X)=SUpA<t<x ](t).
Since/(x)
satisfies a Lipschitz condition, so does ~F(x). Hence ~F(x) is absolutely continuous. Further, ~F(x) is non-decreasing. WriteI = [ A , B].
We m a y assume t h a t m a x 1 / ( x ) = l since, other- wise, we can consider the interval [A, X], where X = m i n {xI](x)=
1). The result for this subinterval also gives the result for I . Hence W ( A ) = 0 , ~F(B)= l, ~F'(x)/>0 a.e. a n d~BI'F'(x)dx=I.
Now consider
E = { x ] A < x < B , ~ F ' ( x ) > O ) .
LetxoEE.
Clearly,~F(Xo)>~/(xo),
a n d if we h a d ~F(x0)>/(x0), then ~F(x) would necessarily be constant in a neigh- bourhood of %. Hence~F(Xo)=/(Xo),
and ifx>~x o,
then~F(x)=SUpxo~<x/(t).
F r o m this, and from ~F'(x0)>0 it follows t h a t ]'(xo)>0. Choose a ~ > 0 such t h a t
G. AROr~SSON, On a partial differential equation
~F(Xo-(5)<~F(Xo)<~F(Xo+(~) and such t h a t / ' ( x ) > 0 for Xo-(~x<~Xo+(~. Choose also a ~1, 0<~1 ~(~, such t h a t / ( x o - ~ l ) >~F(x0-(~ ). On the interval x o-51 <~X~Xo+(~
we thus have
/(x)
= s u p/(t) >1/(Xo-
~1) > ~ ( x 0 - 8) = s u p / ( t )and hence ~F(x) =/(x).
Hence, E is an open set and ~F'(x)=/'(x) on E. Write E = [.J,(a,, b,). We get
II v
d a v
Hence, the assertion (3) is true, and it is very easy to see t h a t (1), (2), (4), and (5) also hold. This completes the proof.
2 2
Theorem 5. Let u(x, y)EC~(D) and let ux uxx +2u~UyUxy +u~uyy = 0 in D. Let A, B be two points in D such that the closed segment A B lies in D, and such that grad u # 0 on this segment. Then
- - : r l 2
]grad u(A)] ~ < : ~ + ~ ,
log L gra d u(B)]
where A--B=the distance/rom A to B, and d=the distance ]rom the segment A B to ~D.
Proo/. The idea of the proof is the following: If we p u t v = ]grad u], then the curvature of a streamline can be written (apart from sign)
dO ]grad v ] _ d v / d n
ds v v
where dv/dn denotes differentiation along a level line. Let us consider the function v along some fixed level line l. Suppose t h a t the curvature of the streamlines inter- secting l is bounded: (dv/dn)/v <~K, or dv/v ~ K d n . Integration gives S(dv/v)~K~dn, or log (V2/V1)~K.L, where L is the length of l, and V 1, V 2 are the corresponding values of v. This is an inequality of the desired type. Now the curvature of the streamlines need not be uniformly bounded, and the vague reasoning above does not give anything. However, since the streamlines are convex curves (or straight lines), it is possible to estimate the integrated curvature along (part of) a streamline. Accord- ingly, the proof is based upon an integration of the formula dO/ds = I grad v I/v.
P u t M = ]grad u(A)[, M 1 = [grad u(B)[, and assume t h a t M > M 1. Let l denote arc length along the segment A B , increasing towards B. Consider a sub-interval 1 of the segment, with endpoints K, L, such t h a t (d/dl) ]grad u [ < 0 on I . Let I be an open interval. Let {~} be the family of streamlines which intersect I . Note t h a t none of these streamlines can intersect I tangentially, since we would have
(d/dl) [ grad u [ = 0 at such a point.
Suppose each streamline in (~} to be continued in both directions from I inde- finitely or until it approaches ~D. If E is the point set thus covered b y these curves, 406
ARKIV FOR MATEMATIK. B d 7 n r 2 8
Fig. 6
t h e n E is obviously open. L e t s be arc length along the curves in {~}, for each of these curves, let 8 = 0 in t h e intersection with I , a n d let 8 a n d u increase in t h e same direct ion. Compare Figure 6.
:Put Vi = g r a d u L, V~ = [grad u [ K. L e t G be the subset of E which is defined b y S 1 < s < S 2, V 1 < g r a d u < V s. Here, S 1 < 0, S 2 ) 0, a n d S 1 [, [ S 2 are b o t h less t h a n the distance from I to ~D.
Clearly, G is open a n d connected. W r i t e v = [ g r a d u [ a n d consider first the m a p p i n g (x, y)-> (u, v) f r o m G onto a set G 1 in t h e (u, v)-plane. Clearly, it is one-to-one, a n d w e h a v e
This follows f r o m
A(u)=0.
Sinced(u, v)/d(x, y)
is continuous a n d n o t zero, the same sign m u s t hold in all of G. F u r t h e r , it follows t h a t G 1 is also open, a n d t h a t t h e inverse m a p p i n g is also in C 1. We getd(x,y) _ +_ 1 d(u,v)
[grad u i l g r a d v]"Clearly, t h e sets G a n d G1 are also in one-to-one correspondence with a set G~ in the (8, v)-plane. The m a p p i n g
(s, v)-~ (u, v)
follows the formulau = ~ ( v ) + 8 . v
(v = v).
Here, ~(v) is the value of u in the corresponding point on I. Clearly, ~(v)E C 1. T h u s
d(x,y) d(s,v)
:Now we have
d(u,v)_ 8+~l'(V ) =
0 v > O .
Clearly, the m a p p i n g s between (s,v) a n d
(u,v)
are inC 1,
a n d we get_ _ = d ( x , y ) d(u,v)_ • ! l
d(u,v) d(8,v) [gad uiigrad
vi v =Igrad vl"
f f G I f s~<s<s~ dsdv mG = I dxdy
= i gra d v l.J J V , < v < V ~
G. ARor~ssor~,
On a partial differential equation
Fig. 7
Before e s t i m a t i n g t h e last integral, we choose $ 1 = - S ~ a n d such t h a t , for a n y p o i n t P on the s e g m e n t
AB,
t h e closed circle f~e with center a t P a n d radius S~lies in D. T h e n the t o t a l (integrated) c u r v a t u r e of a n y curve in {y), calculated for
SI<S<S2,
is n o t greater t h a n 27~. This is e a s y to see, since y(P) is c o n v e x a n d t h e continuation of y~ b e y o n d s = S~, s = $2, m u s t m e e t ~f~p w i t h o u t intersecting itself.W e leave the details.
N o w we can e s t i m a t e
raG:
f f dsdv ~v, ~s, ds mG
= ]gradv I iv, dv
i s , [grad v I"- -
F r o m t h e Schwarz inequality, we getJs, ]grad vl"
1Wow consider ySs~ g r a d rids. T h e f o r m u l a for c u r v a t u r e of streamlines, Section 2, reads
dO/ds
= + [gradvl/v ,
a n d since v is c o n s t a n t in this integral, we getf[gradv]d =f d =vf] d <2=v.
W i t h t h e preceding inequality, we get
fs " ds (82-81) 2 28~
1 ]grad vl >~ 2~e.v ~ ' v a n d finally
mG >~
(2S~/ze)
log (V2/Vl).Now, if
dv/dl<O
on the whole s e g m e n tAB,
or on a finite n u m b e r of p o r t i o n s ofAB,
t h e n we can a p p l y the previous reasoning to the whole s e g m e n tA B
or to e a c h of these portions, respectively. H o w e v e r , we do n o t k n o w w h e t h e r(d/dl)]grad
u I changes sign a finite or infinite n u m b e r of times onAB,
a n d therefore, we m u s t a p p l y L e m m a 3 to the function Igrad u I on the s e g m e n tA B
(with obvious modi- fications). This gives a sequence of intervals I v a n d we also get corresponding sets G~. N o t e t h a t S1, S 2 are defined i n d e p e n d e n t l y of I . F r o m the condition 4 in the l e m m a it follows t h a t {Gv} are pairwise disjoint. W r i t e p ~ = i n f x ~ l g r a d u [ , 408ARKIV FOR MATEMATIK. Bd 7 nr 28 qv= supi~ [grad u I . Then all Pv, qv are positive and the preceding estimate gives mG, >1 (2S~/~) log (q~/p~) and m ( U G~) >~(2S~/z) Z , log (qv/P~). Now the intervals (p~, q~) are disjoint, they are sub-intervals of (/1/1, M) and Z ~ ( q ~ - p , ) = M - M 1.
With the aid of Levi's theorem we infer from this t h a t Zv log (q~/pp) = l o g (M/M1).
Hence we arrive at
M
log ~11~< 2 ~ m ( U ~ G ~ ) .
Now the distance from a point in
U,,G,,
to the segment A B is at most $2, which means t h a t m( U v Gv) ~< 2S2" A B § HenceM .< ~ A B ~s log ~ $2 2 "
Here, S 2 can be a n y positive number less t h a n d, and if we let S 2 tend to d, t h e n we obtain the desired estimate.
I t would not be difficult to formulate and prove a corresponding estimate for a n y two points A, B in D t h a t are connected b y a smooth curve in D instead of a straight segment. We omit these details.
2 2
Theorem 6. Let u(x,y)EC2(D) and let uxux~+2ux%Ux~+U~Uy~=O in D. Then grad u 4 0 in all o/ D, unless u is constant in all o / D .
Proo/. This is an immediate consequence of the previous theorem.
Theorem 7. Let u(x, y ) E C 2 in a domain D and let A ( u ) = 0 in D. Assume also that
. , 2 9
uxx%y-ux~ :4-0 ~n D. Then uEC~176
Proo]. This follows immediately from L e m m a 2 and the previous theorem.
7. S o l u t i o n s i n a n e i g h b o u r h o o d o f
infinity.
Global s o l u t i o n sI t is possible to find all solutions of A ( u ) = 0 in a neighbourhood of infinity, t h a t is, all functions satisfying A ( u ) = 0 outside some compact set.
Theorem 8. Let F be a compact set in the plane ( R ~) and let G be the convex hull o]
F. Further, let A(u) =0 in R 2 - F.
Then [ grad u ] = constant in R 2 - G, and only these cases are possible:
1. u = A x + B y + C in R ~ - G .
2. There is a simple, closed curve F, with continuous curvature, and enclosing a region H D G, such that
u(R) = D . d ( R , F ) + E / o r each R C H . Further, F is convex (not necessarily strictly).
(Here, A .... , E are constants and d(R, F) =the distance/rom R to F.)
Proo/. (1) Let F be contained in an open circular disk D O with b o u n d a r y C. Unless u - constant, there exist M > 0 a n d ~ > 0 such t h a t [u [ ~ M and [grad u [ ~> ~ on C.
Now let ~ be a streamline of u which intersects C in (at least) two points Q, R.
a. ARONSSON,
On a partial differential equation
Then the length of y, measured between Q and
R,
is not greater t h a n2M/5.
This is clear, since L(y).(~ •lu(Q) - u ( R ) ] <2M.
Let ~ be an open annulus which is concentric with C. L e t r 0 be the radius of C and rl, r z the inner and outer radii of s respectively. Suppose t h a t r I > r o
+M/c~.
Consider m a x h I g r a d u ], which is t a k e n at Po C ~ . We claim t h a t g r a d ( I g r a d u ]) = 0 at P0. I f this is not the case, then Po lies on ~s
(2) First, let Po lie on the inner boundary of s I t follows from our choice of r 1 t h a t Y(P0) must extend to infinity at least in one direction. But then Y(Po) has to meet ~ , and at these points grad ( ] grad u [ ) = 0. However, according to our results in Section 2, this contradicts the fact t h a t grad (] grad u I ) # 0 at Po.
(3) Then assume t h a t Po lies on the outer boundary of s and grad (] grad u I) =#0 at Po. Since I grad u] is maximal at Po, it follows t h a t grad (] grad u [) is perpendicular to ~ at Po, a n d it is also clear t h a t this vector points towards the exterior of ~ . L e t l be the tangent of ~s at Po- Then Y(Po) is a convex curve, and if U is a suitable neighbourhood of Po, then Y(Po) fl U and s N U lie on different sides of l. (Compare Section 2). However, since Y(P0) is convex, it also follows t h a t Y(Po) does not meet l, except at P0, and hence Y(Po) extends to infinity in both directions. I t is not diffi- cult to see t h a t the total curvature of Y(P0) is not greater t h a n ~. Further, Y(P0) obviously separates the plane into two parts, D and D ' , one of which (D) is convex.
Clearly, a n y streamline in D belongs to the curved type, extends to infinity in both directions and has total curvature ~ ~r.
(4) Let P be an a r b i t r a r y point in D, and let y+(P) be the p a r t of y(P) where
u>~u(P).
Let C be the level line of u through P and let Q, R be points on C N D such t h a t V I = ]grad u(Q)] < ]grad u(P)] < ]grad u(R)] = V2 and such t h a t the p a r t of C between Q a n d R belongs to D. Then v = ] grad u ] is strictly monotonic along Con and v can be used as a p a r a m e t e r on t h a t curve. Further, let s denote the arc length along a n y streamline y of u which intersects Con. Let s - 0 in the point of intersection, and letduffs
> 0 on y.Consider the set covered b y y+(P') when P ' varies along
CQn.
This set is in one- to-one correspondence with the set in the (s,v)-plane defined b yVI~v<~V2,
0 ~ < s < ~ . Further, the mapping is in C 1 in both directions, andd(x, y)/d(s,
v ) = _+ 1/]grad v] (compare the proof of Theorem 6). We want to estimate the measure of the set G in the xy-plane which corresponds toV l < v < V 2,
0 < s < S , where S > 0 is arbitrary. As in Theorem 6 we haveraG= l d x d y =
v j 0 < s < s Ig t a d v]and, applying Schwarz's inequality and the formula for the curvature of a stream- line, we find
s ds S 2 S 2
[grad v[ ~> v 0 ~ ) >
vO(v)"
Here,
01(v )
is the total curvature of the corresponding streamline (y(v)), evaluated between the limits s = 0 ands - S ,
andO(v)
is the total curvature of y(v) between the limits s - 0 and s - o o . (We haveO<Ol(V )
<0(v)~7~.) Hence, we obtain 1f v, S2 dv 2 i" v~ dv ma>
g~v.O(v)
1 T h e f u n c t i o n O(v) is m e a s u r a b l e , since i t is l o w e r s e m i - c o n t i n u o u s .
410
ARKIV FOR MATEMATIK. B d 7 n r 2 8
H o w e v e r , it is clear t h a t G is contained in a circle with center a t Q a n d with t h e radius (d-[-S), where d is the length of CQR. H e n c e we get
S ~ ( v , dv <~ .< + S ) 2 . J r , v" O(v) m G ..~ ~(d
H e r e , Va, V2, O(v) a n d d are i n d e p e n d e n t of S, a n d S can be choosen a r b i t r a r i l y large. Consequently we m u s t h a v e
dv
~,~ v . O(v)
Choose u 1 > u ( P ) , a n d consider the set in the (s, v)-plane defined b y s ' v = u 1 - u ( P ) , V~<~v<~ V 2. I n the xy-plane, it corresponds to a level line C x : u = u 1 intersecting 7+(P) a t a p o i n t P r N o w t h e a b o v e a r g u m e n t applies to P1 a n d G 1 as well as t o P a n d G. H e n c e we o b t a i n
f
~, v " O(v,u~) v, dv ~ ,where the m e a n i n g of the n o t a t i o n is obvious. Here, u 1 m a y be chosen a r b i t r a r i l y large, a n d it is clear t h a t lim . . . . O(v, Ul) = 0, for e v e r y fixed v. Consequently,
f
v~ dvlim ~ ,
~ - . ~ ~ v . O(v, u O which is a contradiction to t h e a b o v e inequality.
(5) W e m a y therefore conclude t h a t g r a d ( I g r a d ~ 1 ) = 0 a t P0 where Po is a n y p o i n t which realizes m a x ~ I g r a d u ] . I t follows t h a t 7(P0) is a s t r a i g h t line, a n d e x t e n d s to infinity a t least in one direction.
L e t D 1 be a n open circular disk w i t h center on Y(Po) a n d such t h a t D a D ~). Suppose t h a t there exists a point P r such t h a t 7(P) belongs the c u r v e d type. I t follows f r o m our choice of ~ t h a t 7(P) e x t e n d s to, infinity in one direction a t least, for instance 7+(P). W e m a y t h e n assume t h a t u(P) > m a x s u . L e t U be a n e i g h b o u r h o o d of P such t h a t we h a v e u > m a x ~ u in U a n d such t h a t g r a d ( ] g r a d u l ) # 0 in U.
Thus, if Q E U, it follows t h a t 7(Q) belongs to the c u r v e d t y p e a n d t h a t 7+(Q) ex- tends to infinity. :Now we are in a position to a p p l y the s a m e reasoning as in (4).
W e only h a v e to find a n e s t i m a t e for the t o t a l c u r v a t u r e of a streamline V in R 2 - / 9 1 . H o w e v e r , 4 z will do, a n d this follows easily f r o m t h e fact t h a t T c a n n o t m e e t the straight line 7(P0).
W e arrive a t a contradiction, as in 4), a n d m a y conclude t h a t there is no c u r v e d streamline in R 2 - / 9 1 . Consequently, ] g r a d u ] = c o n s t a n t in R 2 /9~.
(6) W e h a v e to p r o v e t h a t I g r a d u I = c o n s t a n t in R z - G. I t is sufficient to p r o v e t h a t g r a d ( [ g r a d u I ) = 0 in R 2 - G. Suppose t h e n t h a t there is a point P (~ G a t which this relation does n o t hold. L e t I be a " h a l f - r a y " f r o m P to infinity, such t h a t l N G = (D, a n d let Q be the p o i n t on l, closest to P, a t which gra d ( ] g r a d u I ) = 0. Since Q ~ G, 7(Q) m u s t e x t e n d to infinity in one direction a t least, for instance v+(Q). L e t Q' be a p o i n t on 7+(Q), such t h a t Q ' ~ / ) I , a n d p u t QQ---'=S. L e t {R~} be a sequence of points on PQ, tending to Q. T h e n {7(R,)} belong to the c u r v e d type. F u r t h e r , t h e points {R; } on 7+(Rv) which correspond to the are length s = S , are defined for
G. AaO~SSON, On a partial differential equation
E2.
~4
Fig. 8
large enough, and tend to Q'. (To see this, consider a streamline as a solution of a differential system with initial conditions x =x o, y =Yo. The solution depends con- tinuously on (x 0, Y0). Compare [9], p. 105, and P a r t 1 of the proof of Theorem 10.) Consequently, R: {~/~1 f o r ~ large enough, and this contradicts the fact t h a t I grad u I = constant in R 2-111. This proves t h a t ]grad u] is constant in R 2 - G.
(7) The next step is to find the form of the solution u.
A. First, let there exist a streamline l which does not meet G. P u t (A, B) = (grad u)t.
We claim t h a t u = A x + By + C in R 2 - G, for some C. Let/1 be a straight line, paral- lel to l, and separating the plane into half-planes El, E 2 such t h a t E~ N G=(I),
l 1 N G::~(I ) (Fig. 8). I t is obvious t h a t u = A x + B y + C in E~. Further, if there is a streamline in El, which does not meet G, then it follows, b y considering the relative positions of the streamlines, t h a t u = A x + B y + C in R 2 - G . Consider then the opposite case, and let ~ be a circle with center at some point Q E G N 11 and such t h a t G c ~ . P u t C = ~ N El. Consider the function ~F(P) = g r a d u(P). QP for a point PEC. I t is continuous, and since y(P) meets G, we m u s t have ~tz(P) # 0 .
Consequently, q~(R) and ~F(S) are of the same sign, which is an obvious contradic- tion (compare Fig. 8). This proves the assertion in the case A.
B. Each streamline meets G. I n this case, the streamlines constitute a family of
"half-rays", emerging from G and covering R 2 - G. The curve 1 ~ will be obtained as a level line of u, enclosing G, and it is fairly obvious from geometric reasons t h a t I ~ m u s t be convex. The details can be filled in as follows: Let Q be a n y point in G and let ~ be a circle, containing G, and with center at Q. P u t ~ F ( P ) = g r a d u(P).QP for P E ~ . Clearly, ~F # 0 and ~t z has fixed sign, for instance LF>0. This means t h a t each streamline is oriented from G to infinity. Let E be the smallest closed circle with center at Q such t h a t E D G. Clearly, u m a y be supposed defined and con- tinuous on ~E. Write M ~ m a x o ~ u . Let M 0 be a n y n u m b e r > M . I f P E ~ , let P1 be the point for which PIP=(1/2~)(u(P)-Mo)grad u(P). When P varies over ~ , P~ clearly describes the set {P' ] u ( P ' ) = M 0, P ' ~G}. I t follows t h a t this level line is a simple, closed curve F, twice continuously differentiable. Further, F encloses a region H and it is obvious t h a t HDG. Next, we claim t h a t u(R)=Mo+2d(R, F) for a n y R CH. Consider a n y R in the exterior of H, and let ~(R) intersect F in P.
Clearly, u(R) = M 0 +;tP--R, and we need only prove t h a t P---R =d(R, F).
I f this is not true, then there is a P x E F such t h a t R-P~=d(R, F)<R--P. B u t this means t h a t RET(P~) and hence ?(P) and ~(P~) intersect at R, which is impossible.
This proves t h a t u(R)=Mo+~d(R, F), and it also proves t h a t d(R, F) is taken on /or only one point on C, and this point (P) is the point where T(R) intersects F.
412
ARKIV FOR MATEMATIK. Bd 7 nr 28
Finally, fix a point P E 1 ~ and let
RE•(P).
The tangent l of F at P divides R 2 into half-planesEl, E2,
and let E 2 contain the unbounded p a r t of 7(P). When R approaches infinity along y(P), it follows from the preceding statement (italicized) t h a t E 2 f3 1 ~ = (I). Consequently, 1 ~ is convex. This completes the proof.Remark.
I f F is a closed, convex curve (not necessarily strictly convex) with continuous curvature, enclosing a region H, then one can verify t h a tv(P)=d(P,
F) is regular (in C 2) in the exterior of H, a n d ]grad v I = 1.We will not discuss the further geometric relations between F and G.
Finally, it should be pointed out t h a t the assumptions in the theorem do
not
imply t h a t I grad u I is constant in R 2 - F. This can be deduced from the example a t the end of Section 4.Theorem
9. I / u ( x , y) EC2(R 2) and A(u)
= 0in the whole plane, then u = A x + B y + C .
Proo/.
I t follows from the previous theorem t h a t ]grad u] is constant. Hence the streamlines are non-intersecting straight lines, which means t h a t t h e y are parallel a n d thus grad u is constant. This completes the proof.A consequence of this theorem is, for instance, t h a t no polynomial of degree > 1 can satisfy
A(u)=0
in a n y domain.The same theorem also holds for the minimal surface equation, compare [2], p. 60.
8. A n estimate for l grad u
I
in L j a p a n o v regionsL e t
u(x, y)
satisfyA(u)=0
in a region D. Theorem 6 states t h a t grad u=~0 in D, unless u =constant. This section treats the question whether I grad u I can be arbi:trarily small near ~D. I t turns out t h a t this cannot occur, if D is " s m o o t h " a n d bounded. I n the opposite case, it m a y happen t h a t i n f , ]grad u I = 0.
L e m m a 4.
Let a(t) be de/ined a.e. on
T 0 < t <T1, and let it be bounded, positive, and measurable. Assume that
f~ a(t) ~ < C(~(T)F
]or almost all TE(To,
T1).Here,
T0>~0 ,T1, C>O and
p > lare constants. Then
T 0 > 0and 1
lOgToT
1 ~< ~PC (ess tliT~m 0 ~(t)) p-1.
Proo/.
Obviously, we m a y assume T O > 0. Define the functionfl(t)>~ 0
b y1 fT
dt
(fl(Tll'=~ JT ~(t) ~" (1)
1 ess limt...Tl_Oae(t ) =limt,~T,_ 0 (ess inft,<t<T,~(t)).
a. AROr~SSON,
On a partial differential equation
~R
A
f
Fig. 9
Then
fl(t)
is increasing andfl(t)
> 0 for t > T 0. Furtherfl(t) <. ~(t)
a.e. Differentiation of (1) givesP dT" (fl(T))~-~ - ~ T ) ~ fl(T) d~ ~ - ~ a.e.
Hence
pCfi'(T)(fl(T)) ~ 29 l / T ,
t h a t isL/pc ) d
d T \ p - 1 (fl(T))"
1 ~ dT (log T) a.e.Consequently,
log T~ 0TI-<~ 'To ~r'ddT
~p[pC-~I ( f i ) ' - I ~ d T = - P C 1 f i ( T i ) ' - ~ ] p -
and since fl(T1)~ esslimt-+Tl-o~(t),
we get the desired result.Theorem 10.
Let u(x, y)be a nonconstant solution o / A ( u ) = 0 in a bounded region D which satisfies the/ollowing conditions (see Fig. 9): There exist constants K 1 > O, K s > 0 and ~, 0 <~ ~ 1, such that i/ B is an arbitrary point on ~D, then there is a coordinate system with the origin at B such that the part o / ~ D which lies in
R = {(x, y)[ Ix I <~K1, - K I < ~ y < 3 K 1 }
414ARKIV F6R MATEMATIK. B d 7 n r 2 8
A
Fig. 10
can be represented as the graph o / a / u n c t i o n y = / ( x ) E C I [ - K 1 , K 1 ] / o r which/'(0) = 0 and such that the angle between the tangents to ~D at the points (x~, /(Xl) ) and (x2,/(x2) ) is not greater than m i n (K 21Xl - x2 ] ~, 88 finally, all points in R with y >/(x) belong to D, and those with y < / ( x ) do not belong to D. ~ Then ] g r a d u I is bounded away/rom zero in D, and
log M1~<217e+ (K1) 2'l, m
where m = i n f D
]grad
u I and M 1 = i n f I g r a d u I , this infimum being taken over those points in D / o r which the distance to the boundary ~D is at least K~.This will be p r o v e d b y the s a m e t y p e of e s t i m a t e s as those used in T h e o r e m 5, b u t here we m u s t s t u d y the b e h a v i o u r of I g r a d u I a t the b o u n d a r y . Therefore, we c a n n o t use a n y u n i f o r m lower b o u n d for t h e length of a streamline, a n d we m u s t find new e s t i m a t e s for the t o t a l c u r v a t u r e a n d for t h e meamlre of a set which is covered b y " s h o r t " streamlines. Such e s t i m a t e s can be o b t a i n e d using the c o n v e x i t y of t h e streamlines a n d the s m o o t h n e s s of ~D. H o w e v e r , this can be done only if t h e streamline t u r n s its c o n v e x side to ~D, as suggested in :Fig. 10, A. (Fig. 10, B, suggests a case t h a t is to be avoided). Consequently, m u c h a t t e n t i o n m u s t be p a i d to the position of t h e s t r e a m l i n e s relative to ~D.
Proo/ o/ the theorem
(1) L e t B be a n a r b i t r a r y p o i n t on ~D, a n d introduce t h e corresponding coordinate system. L e t A be the p o i n t (0, 2 K 0 a n d R = ( ( x , y)] ]x I <K1, - K I < y < 3 K I } . W r i t e
M = l g r a d u ( A ) t ~ inf Igrad u(P)] = M 1 ,
d(P,OD)~ Kt
a n d m l = inf Igrad u(O,y)l.
0<y<2K1
I f m 1 = M , there is nothing to prove. I f m I < M , t a k e s > 0 such t h a t m 1 + s < M , a n d let C be a p o i n t on the s e g m e n t A B such t h a t I g r a d u(C)] =ml +s. P u t v(x, y) =
]grad u(x, Y) I" W e will consider the function v(y)=v(0, y) for yc<~y<~yA = 2 K x.
According to L e m m a 3 2 there is a finite or d e n u m e r a b l e sequence of open i n t e r v a l s I~ on [Yc, YA] such t h a t
1 A region satisfying these conditions is often called a Ljapunov region.
2 Applied to ~0(t) = - v( - t).
G. ARONSSON, On a partial differential equation (A) The intervals I , are pairwise disjoint.
(B) dv/dy>O a n d m x + e < v ( y ) < M on U = [J~l~.
(C) If 1~ = (p,, q~), t h e n 2 ~ ( v ( q ~ ) - v ( p ~ ) ) = M - m 1 - ~ . (D) I f ylE U a n d yl<y2<~yA, t h e n v(yl) <v(y2).
F o r a n a r b i t r a r y y E [Yc, YA], let ?(y) be the streamline of u passing t h r o u g h (0, y), a n d we consider it only in R/3 D. L e t it be continued u p to the b o u n d a r y of R/3 D in b o t h directions f r o m (0, y), a n d let ~(y) be t h e length of 7(Y)" T h e case :r is n o t y e t excluded. However, we claim t h a t :r is lower semicontinuous:
~(Y0) ~<lim~_~y, ~(y). To see this, consider a streamline as a solution of a differential s y s t e m
dx ux
ds ]grad u I' dy _ uu ds ]grad u I'
with initial conditions X(So)=Xo, y(So)=y o. W e k n o w f r o m o r d i n a r y differential e q u a t i o n s ([9], p. 105) t h a t t h e solution depends continuously on (x0, Y0), which gives the result.
(2) Next, we claim t h a t each streamline of u, considered in R/3 D, has total curva- ture ~< 3zt a n d finite length. Consider a streamline y a n d assume t h a t the total curva- ture is > 3ze. T h e n there are t w o successive vertical (parallel to the y-axis) t a n g e n t s ll, 12, such t h a t t h e curve ~ between the points of t a n g e n c y A, B, is c o n v e x down- wards (y"(s) >0). (See Fig. 11). L e t C, D be t h e points where ll, 12 intersect the line y = 3 K 1. N o w ~ a n d the segments A D , DC a n d C B enclose a convex d o m a i n
~ = R/3 D, a n d ~ c D. The arcs of y in ~ h a v e finite length, since u is b o u n d e d in
~ . Therefore y, continued b e y o n d A a n d B, m u s t h a v e well-defined endpoints on
~ , a n d it follows easily f r o m the c o n v e x i t y of y t h a t these endpoints are situated on CD. I t also follows t h a t the total c u r v a t u r e of the p a r t s of y between these end- points a n d A, B c a n n o t be greater t h a n 89 for each.
Hence, the total c u r v a t u r e of y in R f~ D is n o t greater t h a n 2~r, c o n t r a r y to our a s s u m p t i o n t h a t it is greater t h a n 3~. So the total c u r v a t u r e of a n y streamline y in R N D is n o t greater t h a n 3~r, a n d it follows from this t h a t the length of y is finite.
I n particular, the function ~(y) i n t r o d u c e d a b o v e is always finite. Finally, e v e r y streamline in R/3 D has well-defined endpoints on ~(R/3 D).
(3) Consider t h e set U o = ( y ] y e U , ~ ( y ) < K 1 } . (If this set is e m p t y , t h e n t h e estimates u n d e r (6) below are unnecessary.) W e k n o w t h a t ~(y) has well-defined e n d p o i n t s on ~(R f3 D) = E 1 U E 2 U E 3 U E~, where
E l = t h e g r a p h of y=/(x), for -KI<~x<~K1, E~ = {x = - K~,/( - K1) < y < 3 K 1 } ,
E ~ = { - K I <x<~K 1, y = 3 K 1 } a n d E a = {x = K 1 , / ( K I ) < y < 3K1}.
I t is clear t h a t if yoE Uo, t h e n b o t h endpoints of Y(Y0) are situated on E 1. L e t us s t u d y this case a little further. W e k n o w t h a t x'(s) # 0 at P = (0, yo), 1 where (x(s), y(s)) are coordinates along ?(Y0). W e can assume t h a t Y is oriented in such a w a y t h a t
1 This follows from dv/dy~O on U ~ U 0.
416
7)
z, 4
J
ARKIV FOR MATEMATIK. B d 7 n r 2 8
f
Fig. I1
x'(s) > 0 at (0, Yo)- Now we want to show that the endpoint/or decreasing s is situated at a point (xl, /(Xl) ) with x 1 <0 and the endpoint /or increasing s is situated at (x2,/(x2) ) /or some x 2 > 0. (See Fig. 12.) I t is sufficient to prove the result for ~+. Now v is con- stant along ~(Y0), and thus it follows from property D of U I , , t h a t ~+(Yo) cannot meet the segment AP. Since ~(Y0)<K1, and since the distance from a n y point on A D to E 1 is >K1, it is clear that ~+ cannot meet the segment AD(*). I t remains to show t h a t ~+ cannot meet the (open) segment BP. But this follows from the con- vexity of ~(Yo), if we note t h a t v'(yo)>0 implies that Y(Y0) turns its concave side upwards at P.
This proves our assertion regarding the endpoints. Next, we will prove t h a t the whole curve ~(Yo) can be represented by the graph o/ a/unction y - g ( x ) , where g(x) E CI[Xl, x2] , and g(xi) :/(Xl) , g(x2) =/(X2).
I t is obvious that ~(Yo) can be represented as y=g(x) in a neighbourhood of P = (0, Y0), and that g'(x) is increasing. Let X be the greatest number, such that ~+
can be represented b y y =g(x) on (0, X). Obviously, we need only consider the case limx-.x_0 g'(x) = + ~ , Y =limx_.x_0 g(x) >/(X). The means that ~+ has a vertical tan- gent at Q = ( X , Y). But then the set g(x)<y<3K1, 0 < x < X , is a domain ~, con- taining (x(sQ § y(sQ-~e)) for e small enough, and ~,+ cannot meet the boundary of this domain (compare the reasoning above). But this contradicts the fact that the endpoint of ~,+ lies in the exterior of ~2. Hence limx-.z o g'(x) < ~ , and it is clear t h a t limx~,x o g(x)=/(X). Write x 2 = X . The same reasoning as above leads to a number x 1 < 0 with analogous properties.
Now we claim that the statements regarding the endpoints of ~(Y0) and the func- tion representation of Y(Yo) are valid also for y(Y0), if Yo E U and Y0 <Y0. First of all, r(Yo) must be contained in ~(Yo)= ((x, y) lxl <x<x2,/(x) <y <g(x)}. Therefore, the endpoints of y(Yo) lie on E 1. Now, all the above arguments regarding ~(Yo) apply to ~(Y0) except one, which is labelled b y (*). But that argument is not needed, in view of the inclusion ~,(y0)c ~(Y0).
