Analytic Majorants and the Cauchy–Kovalevskaya Theorem

Write $f=\sum a_{\alpha }{z}^{\alpha }$, $g=\sum c_{\alpha }{z}^{\alpha }$, $F=\sum A_{\alpha }{z}^{\alpha }$, $G=\sum C_{\alpha }{z}^{\alpha }$. For every $\alpha$, $\mathrm{\mid }{a}_{\alpha }+c_{\alpha }\mathrm{\mid }\le A_{\alpha }+C_{\alpha }$ and $\mathrm{\mid }{\sum }_{\beta +\gamma =\alpha }{a}_{\beta }{c}_{\gamma }\mathrm{\mid }\le {\sum }_{\beta +\gamma =\alpha }{A}_{\beta }{C}_{\gamma }$. Each convolution is finite, including degree zero, proving the sum and product claims.

The coefficient of $z^{\alpha }$ in ${\mathrm{\partial }}_{j}f$ is $({\alpha }_j+1)a_{\alpha +e_j}$; its modulus is at most $({\alpha }_j+1)A_{\alpha +e_j}$, the coefficient in ${\mathrm{\partial }}_{j}F$. Thus differentiation preserves the relation.

In degree at most $q$, a product $h^{\beta }$ can contribute only when $\mathrm{\mid }\beta \mathrm{\mid }\le q$, since every factor has order at least one. There are finitely many such $\beta$ and finitely many decompositions of any fixed multi-index. Repeated application of step 1.1 bounds each coefficient of $a_{\beta }{h}^{\beta }$ by the corresponding coefficient of $A_{\beta }{H}^{\beta }$. Adding these finitely many bounds proves the composition claim.

For convergent inner series choose a positive smaller polyradius $s$ for which the sums $S_i(s)={\sum }_{\alpha }\mathrm{\mid }[z^{\alpha }]h_i\mathrm{\mid }{s}^{\alpha }$ are strictly below the outer convergence radii. Such an $s$ exists because $S_i(s)\to 0$ as all coordinates of $s$ decrease to zero. The absolute sum of the expanded substitution is bounded by ${\sum }_{\beta }\mathrm{\mid }{a}_{\beta }\mathrm{\mid }{\prod }_i{S}_i(s{)}^{{\beta }_i}<\mathrm{\infty }$. Absolute rearrangement therefore identifies the formal substitution with the actual function composition; the same argument applies to products. On still smaller polydiscs F2 identifies the formal derivatives with derivatives of the sums.

Write $f=a+h$ with $a\mathrm{\ne }0$ and $h(0)=0$. Shrink $s$ until ${\sum }_{\alpha }\mathrm{\mid }[z^{\alpha }](h\mathrm{/}a)\mathrm{\mid }{s}^{\alpha }<1$. F3 bounds the absolute sum of $a^{-1}{\sum }_{k\ge 0}(-h\mathrm{/}a{)}^k$ by $\mathrm{\mid }a{\mathrm{\mid }}^{-1}\mathrm{/}(1-S)$ with $S<1$. The finite geometric identity gives $(a+h)a^{-1}{\sum }_{k=0}^K(-h\mathrm{/}a{)}^k=1-(-h\mathrm{/}a{)}^{K+1}$; the last term tends uniformly to zero on this polydisc. Hence the convergent series is exactly $1\mathrm{/}f$. This proves the reciprocal and all asserted analytic-operation claims.

If the family is empty take $M=r=1$. Otherwise F1 gives a common complex polydisc of positive polyradius $R$; take $r=\frac{1}{2}{\min }_i{R}_i$. The finitely many complexifications are continuous on the compact distinguished boundary of the $r$-polydisc. Let $M$ be the larger of $1$ and their finitely many boundary suprema. F2 and $[z^{\alpha }]f_j=D^{\alpha }{f}_j(0)\mathrm{/}\alpha !$ give the asserted bound.

The coefficient of $z^{\alpha }$ in $M{\sum }_{q\ge 0}((z_1+\cdots +z_n)\mathrm{/}r{)}^q$ is $M{r}^{-\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\alpha \mathrm{\mid }!\mathrm{/}\alpha !$. The multinomial factor counts arrangements of a multiset and is at least one, including $\alpha =0$. Hence it dominates the bound in step 1.1. Subtracting the respective constants leaves coefficient zero at degree zero and the same inequality in positive degrees, proving the second comparison.

Complexify $f$ to $F$ on a conjugation-invariant polydisc by F1. Its complex Jacobian at the real point $a$ is the same real matrix as $Df(a)$, so its determinant is nonzero. F2 supplies a holomorphic inverse $G$ on an open neighborhood of $f(a)$. Shrink its domain to a conjugation-invariant polydisc $V$ on which both $G(w)$ and $\overline{G(w)}$ lie in the injectivity neighborhood of $F$. This is possible by continuity at $f(a)$, since both limits are $a$.

Real coefficients give $F(\bar{z})=\overline{F(z)}$. Thus $F(\overline{G(w)})=\bar{w}=F(G(\bar{w}))$ for $w\in V$. Injectivity gives $G(\bar{w})=\overline{G(w)}$, so $G$ maps the real slice into the real slice. Each component is continuous and separately holomorphic (restrict its complex differential to a coordinate line), so F3 expands it at the real centre $f(a)$. Conjugating this series and using the displayed identity and F4 shows every coefficient equals its conjugate. The expansion therefore restricts to a real analytic inverse.

Set $H(x,y)=(x,P(x,y))$. Its derivative at $(a,b)$ has block matrix $\left(\begin{array}{cc}I& 0\\ D_{x}P& D_{y}P\end{array}\right)$, whose determinant is $\det D_{y}P\mathrm{\ne }0$. Steps 1.1–2.1 give a real analytic inverse $K$ near $(a,0)$. The first coordinate identity $H(K(x,z))=(x,z)$ forces $K(x,z)=(x,k(x,z))$. Define $g(x)=k(x,0)$ after shrinking to a product neighborhood.

The identity $H(K(x,0))=(x,0)$ implies $P(x,g(x))=0$ and $g(a)=b$. Conversely, if $(x,y)$ is in the chosen inverse neighborhood and $P(x,y)=0$, then $H(x,y)=(x,0)$, hence $(x,y)=K(x,0)$ and $y=g(x)$. This proves both the graph description and local uniqueness.

Put $w=v-v_0$, adjoin $z_0=t$, and write $z=(z_0,w)\in {\mathbb{R}}^{N+1}$. The autonomous system is $z^{\mathrm{\prime }}=A(z)=(1,H(z_0,v_0+w))$, $z(0)=0$. A formal series $z={\sum }_{q\ge 1}{a}_q{t}^q$ has the unique recursion $(q+1)a_{q+1}=[t^q]A(z)$. This coefficient depends only on $a_1,\dots ,a_q$: each substituted series has zero constant term, so no positive-degree factor can use $a_{q+1}$. Its zeroth coordinate is exactly $t$.

With $L=N+1$, F1 supplies $M\ge 1,r>0$ majorising every component of $A$ by $M\mathrm{/}(1-{\sum }_{i=0}^N{Z}_i\mathrm{/}r)$. The scalar equation $W-\frac{L}{2r}{W}^2=Mt$, $W(0)=0$, has an analytic branch by F3: the derivative in $W$ at zero is one. Its coefficient recursion $W=Mt+\frac{L}{2r}{W}^2$ gives $[t]W=M$ and every subsequent coefficient nonnegative. Differentiating the identity yields $W^{\mathrm{\prime }}=M\mathrm{/}(1-LW\mathrm{/}r)$ near zero, since the denominator there is one. Solving this quadratic on the branch through zero gives $W=(r\mathrm{/}L)(1-\sqrt{1-2LMt\mathrm{/}r})$. The square root is the analytic branch with value one at zero, so a positive convergence neighborhood is explicit; its positive real endpoint $t_{\ast }=r\mathrm{/}(2LM)$ is singular, since differentiating $W-L{W}^2\mathrm{/}(2r)=Mt$ there would give $0=M$ if W extended analytically through it. No entire majorant is claimed.

Compare the recursion in step 1.1 with the vector whose L components are W. At degree zero the initial coefficients agree. If $\mathrm{\mid }[t^j]z_i\mathrm{\mid }\le [t^j]W$ for $j\le q$, F2 bounds $[t^q]A_i(z)$ by $[t^q]M\mathrm{/}(1-LW\mathrm{/}r)=(q+1)[t^{q+1}]W$. Dividing by $q+1>0$ gives the next coefficient bound. This induction proves $z_i\ll W$ at every degree.

Choose $s>0$ strictly inside the convergence radius of W. Its coefficients obey $[t^q]W\le C{s}^{-q}$, where $C={\sum }_q[t^q]W{s}^q<\mathrm{\infty }$. Step 2.1 transfers this bound to z. F4 and F2 show the series sums and its compositions differentiate as in the formal recursion on a smaller interval, so $z^{\mathrm{\prime }}-A(z)=0$. The recursion also forces the Taylor series of every analytic solution; hence two such germs coincide. Finally, for the original nonnegative H with $v_0=0$, the recursion $(q+1)[t^{q+1}]v=[t^q]H(t,v)$ consists solely of nonnegative sums and products from its zero initial coefficient, so every coefficient of v is nonnegative.

After relabeling coordinates take ${\mathrm{\partial }}_{z_{d+1}}\varphi (a)\mathrm{\ne }0$. The map $\mathrm{\Psi }(z)=(z_1-a_1,\dots ,z_d-a_d,\varphi (z))$ has determinant ${\mathrm{\partial }}_{z_{d+1}}\varphi (a)$. F1 supplies an analytic inverse $z=\mathrm{\Phi }(x,t)$, and the initial surface is exactly t=0.

For any derivative $D_z^{\alpha }u$ of order m, repeated chain rule shows the coefficient of ${\mathrm{\partial }}_t^m(u\circ \mathrm{\Phi })$ is $(d\varphi {)}^{\alpha }$: obtaining m derivatives on u requires that each differentiation hit u, while differentiation of a coordinate coefficient leaves lower order on u. All transformed jet expressions are analytic and linear in the top-order jets. Differentiating the transformed nonlinear P in the pure t-jet therefore gives ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=m}{P}_{u_{\alpha }}(d\varphi {)}^{\alpha }$, the principal symbol of its linearization. This agrees with F2 applied to that linearized operator.

At the specified compatible jet P=0, and step 2.1 makes its derivative in the selected scalar slot nonzero. F1 gives a unique analytic branch expressing this slot in terms of the remaining jets and (x,t). The remaining derivatives all have total order at most m and normal order strictly less than m. This proves the claimed solved form only near the selected compatible jet.

For normal data, step 1.1 gives an analytic parametrization X of the surface. The nonvanishing analytic gradient has an analytic positive length: apply F1 to $s^2-\mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2=0$ at its positive root. Thus $\nu =\mathrm{\nabla }\varphi \mathrm{/}\mathrm{\mid }\mathrm{\nabla }\varphi \mathrm{\mid }$ is analytic. The map $\mathrm{\Theta }(x,t)=X(x)+t\nu (x)$ has independent tangent columns and its unit normal column, hence invertible derivative, so F1 again gives an analytic inverse. By F3 and the fact that Theta is affine in t, ${\mathrm{\partial }}_t^j(u\circ \mathrm{\Theta })(x,0)=D^{j}u(X(x))[\nu (x),\dots ,\nu (x)]$. The conormal dt in these coordinates is proportional to dphi on the surface; homogeneity of the degree-m symbol preserves its nonvanishing. Thus this flattening handles exactly the stated Euclidean normal data.

For $0\le j<m$, differentiation gives ${\mathrm{\partial }}_t^{j}P={\sum }_{\mathrm{\ell }=j}^{m-1}{t}^{\mathrm{\ell }-j}{g}_{\mathrm{\ell }}(x)\mathrm{/}(\mathrm{\ell }-j)!$. Thus ${\mathrm{\partial }}_t^{j}P(0,x)=g_j(x)$ and ${\mathrm{\partial }}_t^{m}P=0$. Its tangential derivatives are obtained by replacing $g_{\mathrm{\ell }}$ with ${\mathrm{\partial }}_x^{\alpha }{g}_{\mathrm{\ell }}$ and are analytic by F1.

For each allowed slot $(\alpha ,j)$ put $P_{\alpha ,j}={\mathrm{\partial }}_x^{\alpha }{\mathrm{\partial }}_t^{j}P$. The transformed right-hand side is $\tilde{F}(t,x,(V_{\alpha ,j}))=F(t,x,(V_{\alpha ,j}+P_{\alpha ,j}(t,x)))$. This finite analytic substitution is made near the actual initial jet, so after translation of that centre F1 applies to zero-constant increments. It is analytic near $(0,0,0)$ and introduces no new derivative slot. Since ${\mathrm{\partial }}_t^{m}P=0$, ${\mathrm{\partial }}_t^{m}u=F$ is exactly ${\mathrm{\partial }}_t^{m}v=\tilde{F}$.

Step 1.1 gives ${\mathrm{\partial }}_t^{j}v(0,x)={\mathrm{\partial }}_t^{j}u(0,x)-g_j(x)$, so the old data hold precisely when all the new data vanish. Conversely adding P to any zero-data solution reverses step 2.1 and restores every old data function. Addition and subtraction of P are inverse maps on the solution germs.

Write $u={\sum }_{\mathrm{\ell }\ge 0}{a}_{\mathrm{\ell }}(x)t^{\mathrm{\ell }}$ with $a_{\mathrm{\ell }}\in \mathbb{R}[[x]]$. The initial conditions force $a_0=\cdots =a_{m-1}=0$. Consequently each allowed jet has a factor $t^{m-j}$ and in particular zero total constant term. F1 makes substitution into F a well-defined formal series in x,t.

For $q\ge 0$, equating coefficients gives $\frac{(q+m)!}{q!}{a}_{q+m}(x)=[t^q]F(t,x,({\mathrm{\partial }}_x^{\alpha }{\mathrm{\partial }}_t^{j}u))$. In the right side, a coefficient of t-degree at most q in an allowed jet uses only $a_{\mathrm{\ell }}$ with $\mathrm{\ell }-j\le q$, hence $\mathrm{\ell }\le q+m-1$. Tangential differentiation changes x-degrees but never t-degree. Replacing u by its truncation through that index therefore leaves the coefficient unchanged.

Starting at q=0, step 2.1 defines each new $a_{q+m}$ by division by the nonzero integer $(q+m)!\mathrm{/}q!$. Every right-side coefficient is thus eventually matched, proving existence of a formal solution. The same recursion forces every coefficient of any other formal solution, proving uniqueness. For m=1 and a finite vector, apply the same recursion simultaneously to all components; no next coefficient of any component occurs on the right. F2 supplies the analytic-data reduction.

Expanding each denominator of G gives a product of two multinomial series. A monomial of positive total degree q has coefficient $M{r}^{-q}{\rho }^{-\mathrm{\ell }}$ times two positive integer multinomial factors, where $\mathrm{\ell }$ is its t-degree. This is at least $M{r}^{-q}$. The constant coefficient is zero, matching F(0)=0. Thus F is majorised by G.

To compute ${\mathrm{\partial }}_x^{\alpha }{\mathrm{\partial }}_t^{q+1}{u}_j(0)$, apply ${\mathrm{\partial }}_x^{\alpha }{\mathrm{\partial }}_t^q$ to $F_j(t,x,u,D_{x}u)$. Iterating the product and chain rules produces a finite sum of products of derivatives of F at zero and derivatives ${\mathrm{\partial }}_x^{\beta }{\mathrm{\partial }}_t^{\mathrm{\ell }}{u}_k(0)$ with $\mathrm{\ell }\le q$, multiplied by nonnegative integers. Indeed a spatial differentiation only increases beta; a time differentiation can increase ell at most by one, and there are only q such differentiations. The inner arguments at the origin are zero, since u and its initial spatial gradient vanish. The identical polynomial computation for U evaluates derivatives of G at the same zero argument by its stipulated origin conditions.

For normal order zero, all derivatives of the zero trace of u vanish, whereas the corresponding trace derivatives of U are nonnegative. Suppose the comparison holds at every normal order at most q, for all tangential indices and all components. In each finite product from step 1.2 take absolute values, apply those bounds and the derivative bounds from step 1.1, and then sum. Nonnegative polynomial coefficients give the comparison at normal order q+1 and simultaneously its nonnegativity for U. Induction proves every derivative bound; division by the same positive factorials gives $u_j\ll U_j$.

Admissible choices exist: $\rho =\min (1,r\mathrm{/}(2dNM))$ gives $a\ge 1\mathrm{/}2$. Fix any rho allowed by the statement, so $a>0$. The analytic function $P(y,q)=aq-b{q}^2-M\mathrm{/}(1-y\mathrm{/}r)+M$ vanishes at $(0,0)$ and has $P_q(0,0)=a>0$. F1 gives g. Writing $g={\sum }_{k\ge 1}{c}_k{y}^k$ yields $c_k=(b\mathrm{/}a){\sum }_{i=1}^{k-1}{c}_i{c}_{k-i}+(M\mathrm{/}a)r^{-k}$. At k=1 the sum is empty and $c_1=M\mathrm{/}(ar)>0$; successive coefficients are nonnegative.

The function $g(\sigma \mathrm{/}\rho +Nv)$ is analytic near zero, and its power-series coefficients in sigma,v are nonnegative by the finite binomial expansion and step 1.1. F2 therefore gives an analytic v with nonnegative coefficients. Moreover $v^{\mathrm{\prime }}(0)=g(0)=0$. Choose a positive sigma-radius small enough that $\mathrm{\mid }\sigma \mathrm{/}\rho +Nv(\sigma )\mathrm{\mid }$ is inside the actual convergence radius of g and less than r, and that $\mathrm{\mid }b{v}^{\mathrm{\prime }}(\sigma )\mathrm{\mid }<1$. These inequalities hold by continuity and their zero origin values.

Put $q=v^{\mathrm{\prime }}(\sigma )$ and $y=\sigma \mathrm{/}\rho +Nv(\sigma )$. The algebraic identity of step 1.1 is equivalent to $(q+M)(1-bq)=M\mathrm{/}(1-y\mathrm{/}r)$, hence $q=M\mathrm{/}((1-y\mathrm{/}r)(1-bq))-M$ on the neighborhood of step 2.1. With $\sigma =t+\rho {\sum }_i{x}_i$ we have ${\mathrm{\partial }}_t{U}_j=q$, ${\mathrm{\partial }}_{x_i}{U}_j=\rho q$, ${\sum }_j{U}_j=Nv$, and ${\sum }_{i,j}{\mathrm{\partial }}_{x_i}{U}_j=dN\rho q$. Substitution gives exactly the system G of F3.

On a sufficiently small polydisc in (t,x), $\mathrm{\mid }t\mathrm{\mid }+\rho {\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }$ is smaller than the radius selected in step 2.1, so this substitution converges absolutely. The coefficients of U and of its trace are nonnegative because those of v and the linear form are nonnegative. At the origin, $U_j=v(0)=0$ and ${\mathrm{\partial }}_{x_i}{U}_j=\rho v^{\mathrm{\prime }}(0)=0$. Thus every hypothesis on U in F3 holds and U dominates the formal zero-data solution.

By F1 set $w=u-g(x)$ to obtain $w_t=H(t,x,w,D_{x}w)$ with zero data. Put $c=H(0,0,0,0)$ and $z=w-tc$. Its equation is $z_t=H(t,x,z+tc,D_{x}z)-c=:K(t,x,z,D_{x}z)$, where K is analytic and K(0)=0. Its initial data remain zero. All substitutions stay inside the original analytic neighborhood after a finite shrinking.

F2 gives a unique formal z. Choose a common positive-radius coefficient bound M,r for the finite analytic family K (equivalently bound the absolute series on a smaller polydisc). F4 constructs the convergent positive Goursat majorant for these constants; F3 gives $\mathrm{\mid }[t^q{x}^{\alpha }]z_j\mathrm{\mid }\le [t^q{x}^{\alpha }]U_j$. Choose a positive polyradius s strictly inside the convergence region of U. If $C={\max }_j{\sum }_{\beta }[Z^{\beta }]U_j{s}^{\beta }$, then every coefficient of z is at most $C{s}^{-\beta }$.

F5 makes z an analytic sum on a smaller polydisc and permits its spatial and time derivatives to be taken termwise. Shrink once more so the absolute sums of z and its spatial derivatives are inside the convergence radii of K. Expanding K then gives an absolutely convergent substitution, whose Taylor coefficients agree with the formal substitution in F2. Each coefficient of $z_t-K(t,x,z,D_{x}z)$ is zero by the recursion; hence this function is identically zero there. All coefficients are real, so the restriction is a real analytic solution. Adding tc and g reverses step 1.1 and supplies the prescribed trace.

Any other analytic solution undergoes the same subtraction in step 1.1; its Taylor series solves the identical zero-data formal problem. F2 forces equality of every coefficient with z. On a common smaller neighborhood both analytic functions equal their series and therefore coincide. Reversing the subtraction proves uniqueness of the original germ.

For $\mathrm{\mid }\beta \mathrm{\mid }\le m-2$ prescribe ${\mathrm{\partial }}_t{U}_{\beta }=U_{\beta +e_t}$. For $\mathrm{\mid }\beta \mathrm{\mid }=m-1$ other than $(m-1)e_t$, choose the least spatial i with ${\beta }_i>0$ and prescribe ${\mathrm{\partial }}_t{U}_{\beta }={\mathrm{\partial }}_i{U}_{\beta -e_i+e_t}$. For $\beta =(m-1)e_t$, prescribe ${\mathrm{\partial }}_t{U}_{\beta }=F$, replacing jets of order at most m-1 by U, and each allowed order-m jet $\gamma$ by ${\mathrm{\partial }}_i{U}_{\gamma -e_i}$ using the least spatial i with ${\gamma }_i>0$. Such i exists because the pure m-time jet is excluded. Thus every right side uses only U and first spatial derivatives of U.

The prescribed data are analytic derivatives of the g_j, and the finite right-hand side is analytic near their compatible initial jet. For $d\ge 1$, F2 supplies a unique analytic vector U. If $d=0$, there are no spatial derivatives or mixed jets, and F4 supplies that vector as an analytic ODE solution. Independently F1 supplies the scalar formal solution u for the original data. Formal differentiation shows that $D^{\beta }u$ solves every equation in step 1.1 and has initial trace ${\mathrm{\partial }}_x^{\alpha }{g}_j$. The first-order formal uniqueness clause of F1 therefore identifies the Taylor series of each analytic U_beta with $D^{\beta }u$.

In particular the Taylor series of U_0 is u. Differentiating its convergent series shows $D^{\beta }{U}_0$ and U_beta have identical Taylor series; hence they coincide on a smaller neighborhood. Substituting in the pure-time equation in step 1.1 yields the scalar equation, and its first m data follow from the traces of U_{je_t}. Conversely for any analytic scalar solution, F3 makes its actual derivative vector satisfy every equation and datum in step 1.1. These constructions are inverse, including m=1 when the sole unknown is U_0 and the pure-time equation is the original equation.

Use F1 to choose analytic coordinates carrying the surface to t=0. For Euclidean normal jets use the normal-line coordinates of F1, so the prescribed functions are exactly the coordinate t-jets. At a selected compatible nonlinear jet F1 gives the analytic implicit branch; otherwise the equation is already solved. Every right-hand derivative now has total order at most m and t-order below m, and the lower/mixed initial jets lie in its analytic neighborhood by the data-compatibility hypothesis.

F2 subtracts the polynomial of the prescribed coordinate jets. F3 constructs the finite first-order jet system with zero analytic data. Its right side is analytic near the zero initial value and spatial jet, because F2 makes the scalar right side analytic at its zero-data centre and F3 replaces each allowed highest jet by a first spatial derivative. If there is at least one spatial variable, F4 supplies an analytic vector solution; with no spatial variables, the system is a finite analytic ODE system and F5 supplies it. F3 then recovers derivative compatibility and proves that the zero-order component solves the scalar zero-data equation. Adding the polynomial restores every prescribed jet. All arguments are local, so shrink finitely until the solution jet stays in the selected branch neighborhood; continuity and its prescribed centre jet ensure this. Composing with the analytic inverse coordinates gives a solution to the original PDE with its original normal data.

Two analytic solutions with these data and jets in the selected branch pull back to solutions of the same solved coordinate equation. Subtracting the same polynomial and taking derivative vectors gives two solutions of the same analytic jet system supplied by F3. If there is at least one spatial variable, F4 makes those vector solutions equal; with no spatial variables, F5 does so. Their zero-order components are therefore equal. Adding the polynomial and composing back proves equality of the original germs. No comparison with a different nonlinear branch or with nonanalytic solutions is used.