Suppose $f$ is holomorphic in a disk centered at the origin and $f$ satisfies the differential equation $$f'' = f.$$ Show that $f$ is of the form $$f(z)=A \sinh z + B \cosh z,$$ for suitable constants $A,B\in \mathbb C$.

Of course if $f$ were a function of one real variable, usual techniques to solve linear homogeneous differential equations (characteristic equations and so on) would quickly solve the problem. But using the same train of thought here, $f$ is a function of two real variables ($z = x + iy$), and so the same techniques cannot be applied directly. This is really a complex analysis problem.

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    $\begingroup$ It might be easier to show that $f(z)=Ce^z + De^{-z}$ for some $C,D$, then use the formula for $\cosh$ and $\sinh$ to figure out what $A,B$ should be. $\endgroup$ Feb 17, 2014 at 17:40
  • $\begingroup$ Is this in a differential equation course? Then perhaps there are some "existence and uniqueness" theorems. Was there a similar worked-out example, say the D.E. $f'' = -f$ just before this? $\endgroup$
    – GEdgar
    Feb 17, 2014 at 18:02
  • $\begingroup$ This is a complex analysis course. Unfortunately, there has been no mention of ODE's before this problem (it was taken from a different book). $\endgroup$
    – user129482
    Feb 17, 2014 at 18:14
  • $\begingroup$ Why can this not be trivially done by recognizing that $f(z) = A\cosh{z} + B \sinh{z}$ is the general solution to the differential equation $y'' - y = 0$? $\endgroup$
    – Ayesha
    Feb 25, 2014 at 1:57
  • $\begingroup$ Is it possible that this might have something to do with analytic continuation? $\endgroup$ Dec 4, 2015 at 7:00

4 Answers 4


We shall use a method identical to the one used for ODEs with a real variable, namely, integration after multiplying with a suitable integrating factor. As the equation is second order, we first rewrite the equation in a way that allows us to integrate it.

Adding $\,\,f'$ on both sides of the equation we obtain $$f''+f'=f'+f.$$ Thus setting $\,g=f+f'$, we get that $g'=g$, and hence multiplying by the integrating factor $\,\mathrm{e}^{-z}$ we obtain $$ 0=\mathrm{e}^{-z}\big(g'(z)-g(z)\big)=\big(\mathrm{e}^{-z}g(z)\big)', $$ which implies that $\mathrm{e}^{-z}g(z)$ is a constant analytic function (given that the domain of $f$ is open and connected), i.e., $\,g(z)=c\,\mathrm{e}^{z}$, for some $c\in\mathbb C$. Consequently $$ f'(z)+f(z)=c\,\mathrm{e}^{z}, $$ which when multiplied by $\mathrm{e}^{z}$ provides that $$ \mathrm{e}^{z}\big(f'(z)+f(z)\big)=c\,\mathrm{e}^{2z}, $$ or $$ \big(\mathrm{e}^{z}f(z)\big)' =c\,\mathrm{e}^{2z} =\frac{c}{2}\big(\mathrm{e}^{2z}\big)', $$ and hence $$ \mathrm{e}^{z}f(z)=\frac{c}{2}\mathrm{e}^{2z}+c', $$ for some $c'\in\mathbb C$, or $$ f(z)=\frac{c}{2}\mathrm{e}^{z}+c'\mathrm{e}^{-z}=a\cosh z+b\sinh z, $$ where $a+b=c \quad\text{and}\quad a-b=2c'.$

Conversely, every function of the form $$ f(z)=a\cosh z+b\sinh z,\quad a,b\in\mathbb C, $$ satisfies the equation $f''=f$.

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    $\begingroup$ Very nice, thanks for the detailed explanation. I think I was missing the key observation that, setting $g = f + f'$ yields $g' = g$. $\endgroup$
    – user129482
    Feb 17, 2014 at 18:18

Let : $$g=f-f(0)\cosh-f'(0)\sinh$$

This function is holomorphic in the same disc $D=D(0,r)$ as $f$.

You clearly have $g(0)=g'(0)=0$ and $g''=g$, so by induction $g^{(k)}(0)=0$ for all $k\geqslant 0$, and consequently $$g(z)=\sum_{k\geqslant 0} \frac{g^{(k)}(0)}{k!}z^k=0$$

for all $z$ in a disc $D(0,r')$ with $0<r'\leqslant r$.

Hence $g=0$ on the domain $D$, i.e. $f=f(0)\cosh+f'(0)\sinh$ on $D$.


You can always turn a linear homogeneous differential equation with constant coefficients (such as the one you are asking about) into a linear system. Here, you have $f''-f=0$, so if we set $v=\begin{pmatrix}f\\ f'\end{pmatrix}$, we have that $v'=\begin{pmatrix}f'\\f''\end{pmatrix}=\begin{pmatrix}f'\\ f\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}v$.

The point of this is that you reduce solving the equation to pure linear algebra. First, call $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, and note that we can diagonalize it as $A=S\Lambda S^{-1}$, where $S=\begin{pmatrix}-1&1\\1&1\end{pmatrix}$ and $\Lambda=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$. (Note that the first column of $A$ is an eigenvector with eigenvalue $-1$, and the second is an eigenvector with eigenvalue $1$.)

The reason why we do this is that $S\Lambda S^{-1}v=v'$ iff $\Lambda S^{-1}v=S^{-1}v'=(S^{-1}v)'$ so that, letting $w=S^{-1}v$, we have that the original equation is equivalent to $\Lambda w=w'$. Now, if $w=\begin{pmatrix}g\\ h\end{pmatrix}$, then this is saying that $g'=-g$, and $h'=h$, so that $g(x)=ae^{-x}$ and $h(x)=be^{x}$ for some constants $a,b$. Since $\begin{pmatrix}f\\f'\end{pmatrix}=v=Sw$, it follows that $f$ is a linear combination of $e^x$ and $e^{-x}$.

The approach is perfectly general, and just as easy as long as the resulting $A$ is diagonalizable. If $A$ is not diagonalizable (which for the matrices we get here, happens precisely when $A$ has repeated eigenvalues), we can still solve the system this way, using now the Jordan form of $A$. Yiorgos's solution can be easily seen to be equivalent to what we are doing here. A nice book explaining all of this is volume II of Apostol's Calculus.

Let me briefly explain the connection. Note first that $A^2-I=0$, and $p(x)=x^2-1$ is the smallest non-zero (monic) polynomial that vanishes when applied to $A$ (it is the minimal polynomial of $A$). This is just the characteristic polynomial of $A$. In fact, given any linear homogeneous differential equation with constant coefficients, rewriting it as a system $Av=v'$ always results in an $A$ whose minimal polynomial is also the characteristic polynomial, and is just the characteristic polynomial of the equation: If we want to solve $f^{(n)}-a_{n-1}f^{(n-1)}-\dots-a_0f=0$, this equation polynomial equation is $x^n-a_{n-1}x^{n-1}-\dots-a_0$. This shows that the eigenvalues of $A$ (the roots of $p$) are closely related to the solutions of the differential equation.

To make the relation more transparent, write $D$ for the derivative operator: $Df=f'$. Then the equation is just $(D^n-a_{n-1}D^{n-1}-\dots-a_0I)f=0$, where $I$ is the identity operator, $If=f$. To solve this, we factor the polynomial in $D$, and solve. For example, starting with $f''-f=0$, we get $(D^2-I)f=0$, or $(D-I)(D+I)f=0$, so if $g=(D+I)f$, then $(D-I)g=0$, or $g'=g$, or $g=ae^x$. Then we need to solve $(D+I)f=g$, or $f'+f=ae^x$, so $f=be^{-x}+ce^x$.

In general, if the resulting polynomial has degree $n$ and $n$ different roots $\lambda_1,\dots,\lambda_n$, the procedure just sketched gives us that the solution has the form $b_1e^{\lambda_1 x}+\dots+b_ne^{\lambda_n x}$ for some constants $\lambda_1,\dots,\lambda_n$. The case where the roots are repeated is slightly more involved, and instead of constants, the $b_i$ are now polynomials in $x$ ($b_i$ of degree $k_i-1$ if $\lambda_i$ appears as a root $k_i$ times). This can either be proved directly, and then used to deduce the general form of the Jordan form of $A$, or can be proved starting with the Jordan form of $A$.


Standard operator factoring $$ e^{z}\frac{d}{dz}\left[e^{-2z}\frac{d}{dz}(e^{z}f)\right]=f''-f. $$

  • $\begingroup$ Yes! ... btw, did you want $f'' - f$, there appears to be some typo $\endgroup$
    – orangeskid
    Dec 4, 2015 at 12:41
  • $\begingroup$ @orangeskid : Thank you! I fixed that typo now. $\endgroup$ Dec 4, 2015 at 14:17

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