0
$\begingroup$

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

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

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!

$\endgroup$
1
  • $\begingroup$ Are we assuming $\lambda > 0$? $\endgroup$ Apr 15 at 13:25

1 Answer 1

1
$\begingroup$

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 $K$.

$\endgroup$
2
  • $\begingroup$ 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
    Apr 15 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 $\exp[(a + bi)z] - \exp[(a - bi)z] = 2i\exp(az)\sin(bz)$. $\endgroup$ Apr 16 at 3:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.