I like to solve the ordinary fourth order homogeneous differential equation given by

$$\displaystyle \frac{d^{4}\theta}{d z^{4}} + \lambda \cdot \theta = 0$$

with a constant coefficient $\lambda$. Using the characteristic equation, I come up with the following roots

$$r_1=-\sqrt[4]{-\lambda}, r_2=\sqrt[4]{-\lambda}, r_3=-i\sqrt[4]{-\lambda}, r_4=i\sqrt[4]{-\lambda}$$

By putting the roots into the general solutions you get the following solution

$$\displaystyle \theta{\left(z \right)} = C_{1} e^{- z \sqrt[4]{- \lambda}} + C_{2} e^{z \sqrt[4]{- \lambda}} + C_{3} \sin\left(z \sqrt[4]{- \lambda}\right) + C_{4} \cos\left(z \sqrt[4]{- \lambda}\right)$$

(Here is already the first problem, I am not sure if the solution is correct).

For the specific solution only two boundary conditions are given: $\theta(0)=C, \theta(z \rightarrow \infty)=0$

Assuming $C<0$, the authors come up with the following solution

$$\theta(z) = C e^{-z \sqrt[4]{4\lambda}} \cos(z \sqrt[4]{4 \lambda}).$$

I don't understand how to obtain this solution based on the boundary conditions. Nor can I understand where the 4 under the root comes from. I am grateful for any help. Thank you!

  • $\begingroup$ Are we assuming $\lambda > 0$? $\endgroup$ Commented Apr 15, 2022 at 13:25

1 Answer 1


I think you have a slight error here. It should be $\lambda/4$, not $4\lambda$. It comes from the fact that $$ \sqrt[4]{-1} = \frac{\pm 1 \pm i}{\sqrt{2}}. $$ So the values for $r$ are $$ r = \sqrt[4]{-\lambda} = \frac{\pm1\pm i}{\sqrt{2}}\sqrt[4]{\lambda} = (\pm 1\pm i)\sqrt[4]{\frac{\lambda}{4}}, $$ which means the solutions are $$ Ce^{\pm z\sqrt[4]{\lambda/4}}\cos\left(z\sqrt[4]{\frac{\lambda}{4}}\right), Ce^{\pm z\sqrt[4]{\lambda/4}}\sin\left(z\sqrt[4]{\frac{\lambda}{4}}\right). $$ The constraint $\theta(z\rightarrow \infty) = 0$ forces the exponential to have a negative argument, and $\theta(0) = C$ sets the coefficient of the cosine term. However, this doesn't say anything about the coefficient of the sine term, so the most general solution is $$ \theta(z) = Ce^{- z\sqrt[4]{\lambda/4}}\cos\left(z\sqrt[4]{\frac{\lambda}{4}}\right) + Ke^{- z\sqrt[4]{\lambda/4}}\sin\left(z\sqrt[4]{\frac{\lambda}{4}}\right) $$ for some constant $K$.

  • $\begingroup$ That was really helpful, thank you! Only one thing is still unclear to me. The problem has a characteristic equation with complex conjugate roots ($a+bi$), resulting in the general solution \begin{align} e^{az}(C_1 cos(bz) + C_2 sin(bz)) \end{align} rather than \begin{align} Ce^{\pm (a+bi)z}\cos ((a+bi)z) \end{align} How do you come up with the second solution? $\endgroup$
    – tsauter
    Commented Apr 15, 2022 at 17:09
  • $\begingroup$ @tsauter You make linear combinations of the complex exponentials. We have $\exp[(a + bi)z] + \exp[(a - bi)z] = 2\exp(az)\cos(bz)$ and $\exp[(a + bi)z] - \exp[(a - bi)z] = 2i\exp(az)\sin(bz)$. $\endgroup$ Commented Apr 16, 2022 at 3:49

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