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!
Best Answer
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 $K$.