数理解析研究所講究録
第 1993 巻 2016 年 100-106

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Symmetry-breaking bifurcation of positive solutions to
a one-dimensional Liouville type equation

Satoshi Tanaka
Faculty of Science,
Okayama University of Science

We consider the two-point boundary value problem for the one-dimensional Liouvilletype equation(1)$\{\begin{array}{l}u"+\lambda|x|^{\mathfrak{l}}e^{u}=0, x\in(-1,1) ,u(-1)=u(1)=0,\end{array}$where
and $l>0.$
Jacobsen and Schmitt [2] studied the exact multiplicity of radial solutions of the
problem for the multi-dimensional Liouville type equation$\lambda>0$(2)$\{\begin{array}{ll}\Delta u+\lambda|x|^{l}e^{u}=0 in B,u=0 on\partial B,\end{array}$where $l\geq 0$ and $:=\{x\in R^{n}:|x|<1\}$ . They proved the following $(i)-(iii)$ :(i) if $1\leq N\leq 2$ , then there exists
such that (2) has exactly two radialand no radialsolutions for $0<\lambda<\lambda_{*}$ , a unique radial solution for;solution for$3\leqN<$(ii) if
10 41, then (2) has infinitely many radial solutions for $\lambda=(l+$$2)(N-2)$ and a finite but large number of radial solutions when $|\lambda-(l+2)(N-2)|$
is suﬃciently(iii) if $N\geq 10+4l$ , then (2) has a unique radial solution for $0<\lambda<(l+2)(N-2)$
and no radial solution for $\lambda\geq(l+2)(N-2)$ .
We note here that every solution of (2) is positive in , by the strong maximum principle.
Result $(i)-(iii)$ were established by Joseph and Lundgren [3] for the case $l=0$ , that is,for the Liouville equation$B$$\lambda_{*}>0$$\lambda=\lambda_{*}$$\lambda>\lambda_{*}$$+$$smal1_{\rangle}$

$B$

(3)

$\{\begin{array}{ll}\Delta u+\lambda e^{u}=0 in \Omega,u=0 on \partial\Omega,\end{array}$when $\Omega=B$ . Gidas, Ni and Nirenberg’s theorem ([1]) shows that every positive solution$:=\{x\in$of (3) is radially symmetric when $\Omega=B$ . However, when is an annulus: $a<|x|<b\},$ $a>0$ , problem (3) may has non-radial solutions. Indeed, Lin [4]proved that (3) has infinitely many symmetry-breaking bifurcation points when $N=2$
and $\Omega=A$ . Nagasaki and Suzuki [6] found that large non-radial solutions of (3) when
$N=2$ and $\Omega=A$ . More precisely, for each suﬃciently large $\mu>0$ , there exist
such that $\lambda>0,$ is a non-radial solution of (3) and $\int_{A}e^{u}dx=\mu$ when $N=2$ and$\Omega$$A$$R^{N}$$(\lambda, u)$$u$$\Omega=A.$

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Recently, Miyamoto [5] considered the problem for the Liouville type equation (2)and proved the following result.

Theorem $A$ ([5]). Let
be the largest integer that is smaller thanand let. All the radial solutions of (2) with $N=2$ can be written explicitly as$1+ \frac{l}{2}$$n_{(\}}$$\alpha_{n}:=2\log\frac{2t+4}{l+2-2n}$$\lambda(\alpha)=2(l+2)^{2}(e^{-\alpha/2}-e^{-\alpha}) , U(r_{\}}\alpha)=\log\frac{e^{\alpha}}{(1+(e^{\alpha/2}-1)r^{l+2})^{2}}.$

The radial solutions can be parameterized by the, it has one turning point
at $\lambda=\lambda(\alpha_{0})=(l+2)/2$ , and \’it blows up as. For each $n\in\{1, 2, \cdots, n_{0}\},$
is a symmetry breaking bifurcation point from which an unboundedbranch consisting of non-radial solutions of (2) with $N=2$ emanates, and $U(r;\alpha)$ isnondegenerate if or$n=0$ , 1,
. Eachradial branch is in)$L^{\infty}-no7^{\neg}m$$\lambda\downarrow 0$$(\lambda(\alpha_{r\iota}), U(r;\alpha_{n}))$$\neq\alpha_{n},$$\{u>0\}\subset R\cross H_{f}^{2}(B)$$(0, \lambda(\alpha_{0}))$$nonarrow$$\cdots,$ $n_{0}$$\cross$.

When $N=2$ , 1’adial solutions of problems (2) and (3) can be written explicitly, andhence, Lin [4] and Miyamoto [5] succeeded to show the existence of bifurcation points.
That is diﬃcult even if we know exact solutions, much more diﬃcult if we do not know
them usually. When $N\neq 2$ , we do not find exact radial solutions of (2). However, thestructure of eigenvalues and eigenfunctions of the linearized problem in the dimension 1
is well-known, and then, by the comparison function introduced in [7]} we can find theMorse indices of even solutions of (1). Then we obtain the existence of a symmetrybreaking bifurcation point of (1).
Let $m(U)$ be the Morse index of a solution to (1), that is, the number of negative
eigenvalues of
$U$

$\mu$

(4)

$\{\begin{array}{ll}\phi"+\lambda|x|^{l}e^{U(x)}\phi+\mu\phi=0, x\in(-1,1) ,\phi(-1)=\phi(1)=0, \end{array}$

A solutionof (1) is said to be degenerate if $\mu=0$ is an eigenvalue of (4). Otherwise,
it is said to be nondegenerate.
The main result is as follows.
$U$

Theorem 1. For each $\alpha>0$ , there exists a uniquesuch that (1) withhas a unique positive even solution $U=U(x;\alpha)$ such that
$Moreover_{Z}$ there exist
andthat
and the followinghold:(i) if $0<\alpha<\alpha_{*f}$ then $m(U)=0$ and $U(x;\alpha)$ is nondegenerate;(ii) if, then $m(U)=0$ and $U(x;\alpha)$ is degenerate;(iii) if
, then $m(U)=1$ and $U(x;a)$ is nondegenerate;(iv) if, then $m(U)=1$ and $U(x, \alpha)$ is degenerate;(v) if, then $m(U\rangle=1, U(x;\alpha)$ is degenerate andis a non-evenpoint,
is,
that
each
existsthere
suchthat is abifurcationfor
non-even positive sogution of (1) and;(vi) if, then $m(U\rangle=2$ and $U(x;\alpha)$ is degenerate;(vii) ifthen $m(U)=2$ and $U(x;\alpha)$ is nondegenerate.
$(\lambda(\alpha), U(x;\alpha))$$\lambda=\lambda(\alpha)$$||U||_{\infty}=\alpha.$$\alpha_{*},$$\alpha_{1_{f}}\alpha_{2}$$\alpha_{3}\mathcal{S}uch$$\alpha_{*}<\alpha_{1}\leq\alpha_{2}\leq\alpha_{3}$$(i)-(vii\rangle$$\alpha=\alpha_{*}$$\alpha_{*}<\alpha<\alpha_{1}$$a=\alpha_{1}$$(U, \lambda)$$\alpha=\alpha_{2}$$\hat{c}>0_{2}$$(\lambda, u)$$|\lambda-\lambda(\alpha_{2})|+\Vert u-U(\cdot, \alpha_{2})\Vert_{\infty}<\epsilon$$\alpha=\alpha_{3}$$\alpha>\alpha_{3z}$

$u$

102
.
Here and Hereafter, we use the notation
For the proof of Theorem 1, see [8]. Here, we give a suﬃcient condition for the second$\Vert U\Vert_{\infty}=\sup_{x\in[-1,]]}U(x)$eigenvalue of the linearized problem to be negative for the following problem(5)$\{\begin{array}{ll}u"+\lambda h(x)f(u)=0, x\in(-1,1) ,u(-1)=u(1)=0.

\end{array}$where $\lambda>0$ and $h\in C^{1}([-1,0$ ) $\cup(0,1])$ $\cap C[-1, 1],$ $h(-x)=h(x)$ , $h(x)>0$ and$h’(x)\geq 0$ for $x>0,$ $f\in C^{1}[0, \infty$ ), $f(s)>0$ and $f’(s)\geq 0$ for $s>0$ . Namely we willshow the following result, which plays a crucial role in the proof of Theorem 1.

Proposition 1. Assume that, for each suficiently large $\alpha>0$ , there exist
and $U(x;\alpha)$ such that $U(x;\alpha)$ is a positive even solution of (5) atandsuch that. Assume moreover that there exist$\lambda(\alpha)>$$\lambda=\lambda(\alpha)$$0$$\Vert U(.

(6)

and

$\delta>0$$\mathcal{S}_{0}>0$; \alpha)\Vert_{\infty}=\alpha$$\frac{l(x)(g(s)-1)-4}{g(s)+i(x)+3}\geq\delta, x\in(0,1], s\geq s_{0},$where $l(x)=xh’(x)/h(x)$ and $g(s)=sf’(s)/f(s)$ . Let(7)$\mu_{2}(\alpha)$be the second eigenvalue of$\{\begin{array}{ll}\phi"+\lambda(\alpha)h(x)f’(U(x;\alpha))\phi+\mu\phi=0, x\in(-1,1\phi(-1)=\phi(1)=0.

Then$\mu_{2}(\alpha)<0$

for all suﬃciently large\end{array}$$\alpha>0.$

In the case where $h(x)=|x|^{l},$ $l>0$ and $f(s)=e^{8}$ , it follows that $l(x)=xh’(x)/h(x)=$for $x\in(O, 1] and g(s)=sf’(s)/f(s)=s$ , and hence (6) is satisfied.
, then $m(U)=0.$
We conclude that if is a positive even solution of (1) andbe an eigenfunction corresponding
Indeed, let
be the first eigenvalue of (4) and let$\phi_{1}(x)>0$$(-1,1)$.
Integrating the equalityonto . We may assume that
$l$

$U$

$\Vert U\Vert_{\infty}\leq 1$$\phi_{1}$$\mu_{1}$$\mu_{1}$$(\phi_{1}(x)U’(x)-\phi_{1}’(x)U(x))’=\mu_{1}\phi_{1}(x)U(x)+\lambda|x|^{l}e^{U(x)}\phi_{1}(x)(U(x)-1)$

on [-1, 1], we have$\mu_{1}\int_{-1}^{1}\phi_{1}(x)U(x)dx=\lambda\int_{-1}^{1}|x|^{l}e^{U(x)}\phi_{1}(x)(1-U(x))dx>0.$

Consequently, we have $\mu_{1}>0$ , which means $m(U)=0$ . By applying Proposition 1, we
can conclude that $m(U(\cdot\cdot\alpha))=0$ for $0<\alpha\leq 1$ and $m(U(. ; \alpha))\geq 2$ for all suﬃciently
large $\alpha>1$ . Then, using the Leray-Schauder degree, we can find a bifurcation point.
To prove Proposition 1, we need the following two lemmas.
Lemma 1. Let
be an eigenfunction corresponding to the second eigenvalue(7). Thenis odd, $\phi_{2}(0)=\phi_{2}(1)=0$ andfor $x\in(O, 1)$ .
$\phi_{2}$$\phi_{2}(x)\neq 0$$\phi_{2}$Proof Let

$M_{1}$

be the first eigenvalue of$\{\begin{array}{ll}\Phi"+\lambda(\alpha)h(x)f’(U(x;\alpha))\Phi+M\Phi=0, x\in(O, 1) ,\Phi(0)=\Phi(1)=0, \end{array}$$\mu_{2}(\alpha)$

of

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and let$\Phi_{1}$$\Phi_{1}(x)\neq 0$

be an eigenfunction corresponding to$M_{1}$. Then $\Phi_{1}(O)=\Phi_{1}(1)=0$ andon $(0,1)$ . Set$\Phi(x)=\{\begin{array}{ll}\Phi_{1}(x) , x\in[0, 1 ],-\Phi_{1}(-x) , x\in[-1, 0) .\end{array}$

Noting the fact that $\lim_{xarrow-0}\Phi"(x)=\lim_{xarrow-く\}}(-\Phi_{1}"(-x))=-\Phi_{1}"\langle 0$ ) $=0$ , we easily
check that is a solution of$\Phi$$\{\begin{array}{ll}\Phi^{l/}+\lambda(\alpha)h(x)f’(U(x;\alpha))\Phi+M_{1}\Phi=0, x\epsilon(-1,1) ,\Phi(-1)=\Phi(1)=0, \end{array}$andand$M_{1}$

is odd, $\Phi(x)\neq(j on ((\delta, 1)$ and $\Phi(0)=0$ . Therefore,
is an eigenvalue of (7)is an eigenfunction corresponding to. Since has exactly one zero in $(-1,1)$ ,$c\Phi(x)$nust beand hencemust befor some $c\neq 0.$$\Phi$$M_{1}$$\Phi$$M_{1}$$\Phi$$\square$$\phi_{2}(x)$$\mu_{2}$Lemma 2. Assume that $w\in C[a, b]$ is positive and concave on$\langle a,$$b$). Let $p\in(O, 1/2)$ .

Then $w(x) \geq p\max_{\zeta\in[a,b]}w(\xi)$ for $x\in[(1-\rho)a+\rho b, \rho a+(1-\rho)b].$

Proof. We take $c\in[a, b]$ for which $w(c)= \max_{\xi\in[a,b]}w(\xi)$ . Thenpositive and concave on$w\langle c$) $>0$ . Since

$w$

is

we have$(a,$ $b$$w(x) \geq\frac{w(c)(x-a)}{c-a}\geq\frac{w(c)(x-a)}{b-a}=:l_{1}(x) , x\in[a, c],$and$w(x) \geq\frac{w(c)(b-x)}{b-\{^{n}}\geq\frac{w(c)(b-x)}{b-(x}=:i_{2}(x) , x\in[c, b].$

Hence
then

$u\rangle(x)\geq\iota Jzin\{l_{1}(x\rangle, l_{2}(x)$} on$[a, b]$. We conclude that if $x\in[(1-p)a+\rho b, (a+b)/2],$$\min\{l_{1}(x), l_{2}(x)\}=l_{1}(x)\geq l_{1}((1-p)a+\rho b)=\rho w(c)$,and if $x\in[(a+b)/2, \rho a+(1-p)b]$ , then$\min\{l_{1}(x), l_{2}(x)\}=l_{2}(x)\geq l_{2}(pa+(1-\rho)b)=\rho w(c)$.

The proof is complete.

$\square$Now we are ready to show Proposition 1.

Proof of Proposition 1. Let$\alpha>0$

be suﬃciently large. We use the following comparison

function $y(x)$ introduced in [7]:$y(x)=xU(x;\alpha)-(x-1)^{2}U’(x;\alpha)$.

This function $y(x)$ satisfies $y(O)=y(1)=0,$ $y(x)>0$ on $(0,1)$ , and$y”+\lambda(\alpha)h(x)f’(U(x;\alpha))y=\lambda(\alpha)x^{-1}h(x)H\langle x;\alpha)f(U(x;\alpha))$for $x\in(O, 1]$ , where$H(x;\alpha)=(1-x)^{2}l\langle x)+x(3x-4)+x^{2}g(U(x;\alpha$104
Let $h(x, \alpha)$ be an eigenfunction corresponding to. From Lemma 1, it follows thatandfor $x\in(0,1)$ . Without loss of generality, we$x\in(O,1)$
may assume thatfor
. We observeand$\mu_{2}(\alpha)$$\phi_{2}(0;\alpha)=\phi_{2}(1;\alpha)=0$$\phi_{2}(x;\alpha)\neq 0$$\phi_{2}(x;\alpha)>0$$t1_{1}at$$\max_{\zeta\in[0,1]}\phi_{2}(\xi;\alpha)=1$$(y’\phi_{2}-y\phi_{2}’)’=\mu_{2}(\alpha)\phi_{2}y+\lambda(\alpha)x^{-1}h(x)H(x,\alpha)f(U(x;\alpha))\phi_{2}, x\in(O, 1].$

Integrating this equality on $(0,1)$ , we obtain(8)$\mu_{2}(\alpha)\int_{0}^{1}\phi_{2}(x;\alpha)y(x)dx+\lambda(\alpha)\int_{0}^{1}x^{-1}h(x)H(x;\alpha)f(U(x;\alpha))\phi_{2}(x;\alpha)dx=0.$

Since$H(x)=[g(U(x, \alpha))+l(x)+3](x-\frac{l(x)+2}{g(U(x;\alpha))+l(x)+3})^{2}$$+ \frac{l(x)[g(U(x;\alpha))-1]-4}{g(U(x;\alpha))+l(x)+3}$$\geq\frac{l(x)[g(U(x;\alpha))-1]-4}{g(U(x;\alpha))+l(x)+3},$

we have(9)$\int_{0}^{1}x^{-1}h(x)H(x;\alpha)f(U(x, \alpha))\phi_{2}(x;\alpha)dx$$\geq\int_{0}^{1}x^{-1}h(x)\frac{l(x)[g(U(x;\alpha))-1]-4}{g(U(x;\alpha))+l(x)+3}f(U(X|\alpha))\phi_{2}(x, \alpha)dx.$

Since $U”(x;\alpha)=-\lambda(\alpha)h(x)f(U(x;\alpha))<0$ on $(0,1], we$ find that $U’(x;\alpha)$ is decreasing
in $x\in(O, 1].$ From $U’(O;\alpha)=0$ it follows that $U’(X|\alpha)<0$ for $x\in(O, 1$ ], which implies
thatis also decreasing in $x\in(0,1].$ Then there exists $x(\alpha)\in(0,1)$ such that$U(x, \alpha)\geq s_{0}$ for $x\in[0, x(\alpha)]$ andfor $x\in(x(\alpha), 1$ ]. Sinceisconcave on $(0,1)$ , we conclude that$U(x, \alpha)$$U(x;\alpha)<\mathcal{S}_{0}$$U(x, \alpha)$$U(x\cdot\alpha)\geq\alpha(1-x) , x\in[O, 1 ],$which shows that if $x\in[O, (\alpha-s_{0})/\alpha]$ , then $U(x;\alpha)\geq s_{0}$ . Therefore,
which implies(10)

We take(11)$x(\alpha)\geq(\alpha-s_{0})/\alpha,$$\lim_{\alphaarrow\infty}x(\alpha)=1.$$s_{1}\geq s_{0}$for which $x(\alpha)\geq 3/4$ for$\alpha\geq s_{1}$. If$\alpha\geq s_{1}$, then (6) implies$\int_{0}^{x(\alpha)}x^{-1}h(x)\frac{l(x)[g(U(x;\alpha))-1]-4}{g(U(x_{)}\cdot\alpha))+l(x)+3}f(U(x;\alpha))\phi_{2}(x;\alpha)dx$$\geq\delta f(\mathcal{S}_{0})\int_{0}^{x(\alpha)}x^{-1}h(x)\phi_{2}(x;\alpha)dx$$\geq\delta f(s_{0})\int_{1/4}^{3/4}x^{-1}h(x)\phi_{2}(x;\alpha)dx.$

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Recalling$\max_{\zeta\epsilon[0_{)}1]}\phi_{2}(\xi)=1$(12), we have$\int_{x\langle\alpha)}^{1}x^{-1}h(x)\frac{l(x)[g(U(x;\alpha))-1]-4}{g(U(x;\alpha))+l(x)+3}f(U(x;\alpha))\phi_{2}(x;\alpha)dx$$\geq-\int_{x((x)}^{1}x^{-1}h(x)\frac{(i(x)+4)f(U(x;\alpha))\phi_{2}(x;\alpha)}{g(U(x;\alpha))+l(x)+3}dx$$\geq-f(s_{0})\int_{x(\alpha)}^{1}x^{-1}h(x)\frac{l(x)+4}{l(x)+3}dx.$

Now we will slxow that there exists
to the contrary that there exists
and
Since
and

such thatsuch that$s_{2}\geq s_{1}$$\{\alpha_{n}\}_{n=1}^{\infty}$

for
and

$\mu_{2}(\alpha)<0$$\mu_{2}(\alpha_{n})\geq 0$$a\geq s_{2}$$\alpha_{n}\geq s_{1}$. Assume

for $n\in N$$\lim_{narrow\infty}\alpha_{n}=\infty.$$\phi_{2}(x, \alpha_{n})>0$$\phi_{2}"(x;\alpha_{n})=-h(x)f’(U(x;\alpha_{n}))\phi_{2}(x;\alpha_{n})-\mu_{2}(\alpha_{n})\phi_{2}(x,\alpha_{n})\leq 0,$

we find that$\phi_{2}(x;\alpha_{n})$is concave on $(0,1)$ .

$Fron\backslash$$x\in(O, 1)$,

Lemma 2 with $\rho=1/4,$ $a=0$ and$b=1$ , it follows that$\phi_{2}(x]\alpha_{n})\geq\frac{1}{4}\max_{\epsilon\xi[0,1)}\phi_{2}(\xi;\alpha_{n})=\frac{1}{4}\} x\epsilon\{\begin{array}{l}l3\overline{4}’\overline{4}\end{array}\}$By (11), we have(13)$\int_{0}^{x(\alpha)}x^{-1}h(x)\frac{t(x)[g\langle U(x;\alpha))-1]-4}{9(U(x;\alpha))+l(x)+3}f\langle U(x;\alpha))\phi_{2}(x;\alpha)dx$$\geq\frac{\delta f(s_{0})}{4}l_{/4}^{3/4}x^{-1}h(x)dx.$

Combining (8) with (9), (12) and (13), we have
$0 \geq-\mu_{2}(\alpha_{n})\int_{0}^{1}\phi_{2}(x;\alpha_{n})y(x)dx$$\geq\lambda(\alpha_{n})f(s_{0})[\frac{\delta}{4}l_{/4}^{3/4}x^{-1}h(x)dx-\int_{x(\alpha_{n})}^{1}x^{-1}h(x)\frac{l(x)+4}{l(x)+3}dx],$

which implies$\prime_{x(\alpha_{n})}^{3}x^{-1}h(x)\frac{l(x)+4}{l(x)+3}dx\geq\frac{\delta}{4}\int_{1/4}^{3/4}x^{-\lambda}h(x)dx>0, n\in N.$

This contradicts the fact (10). Consequently, there exists. This mpletes the proof.
for$\alpha\geq s_{2}$$\infty$$Rgl^{i\prime}$$s_{2}\geq s_{1}$such that $\mu_{2}(\alpha)<0$$\square$

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preparation.

ma$[ I\int\Phi\Re\star\not\cong\cdot \mathfrak{B}ae_{p}\Re$$ﬄ^{\{}F$$\Re$