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I have the operator $$ A : -e^{-2ax} \frac{\partial}{\partial x} \left(e^{2ax}\frac{\partial}{\partial x}\right)\\ D_A = \left( v \in C^2[0,L] \quad | \quad v(0) = v(L) = 0 \right) $$ and I need to find the eigenfunction $p$. But first I have a question. How do I apply this operator to a function $u$? I see no $u$ in the formula. Is it a typo, should it be $\frac{\partial u }{\partial x}$ in the last partial derivative within the parenthesis? If this is the case, this is my attempt: $$Ap = \lambda p$$ $$-2ap' - p'' - \lambda p = 0$$ $$p'' + 2ap' + \lambda p = 0$$ $$p(x) = C_1 e^{-a + \sqrt{a^2 - \lambda}} + C_2e^{-a - \sqrt{a^2 - \lambda}}.$$ But this doesn't at all look like the correct answer $$e^{-ax}\sin{\frac{k\pi x}{L}}, k=1,2,3,...$$ What am I doing wrong and how do I find the right answer (without complicating things)?

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How do we apply an operator such as

$A = -e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial}{\partial x} \right ) \tag 0$

to a function $u$? Typically we work from right to left, taking the argument $u$ to occupy the first or right-most "empty slot"; bearing this in mind:

We start with the following calculation:

$Au = -e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial}{\partial x} \right )u = -e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial u}{\partial x} \right )$ $=-e^{-2ax} \left ( 2ae^{2ax} \dfrac{\partial u}{\partial x} + e^{2ax} \dfrac{\partial^2u}{\partial x_2} \right ) = -2a\dfrac{\partial u}{\partial x} - \dfrac{\partial u^2}{\partial x^2}; \tag 1$

if $\lambda$ is an eigenvalue of this operator, then

$-e^{-2ax} \dfrac{\partial}{\partial x} \left (e^{2ax} \dfrac{\partial}{\partial x} \right )u = \lambda u, \tag 2$

then, via (1),

$-2a\dfrac{\partial u}{\partial x} - \dfrac{\partial u^2}{\partial x^2} = \lambda u, \tag 3$

or

$\dfrac{\partial u^2}{\partial x^2} + 2a\dfrac{\partial u}{\partial x} + \lambda u = 0; \tag 4$

since this is a linear, constant coefficient, second order ordinary differential equation, we may find solutions of the general form $e^{\rho x}$ where $\rho$ is a root of the quadratic equation

$\rho^2 + 2a \rho + \lambda = 0; \tag 5$

we have

$\rho_\pm = \dfrac{-2a \pm \sqrt{4a^2 - 4 \lambda}}{2} = -a \pm \sqrt{a^2 - \lambda}; \tag 6$

thus we examine functions of the form

$u(x) = C_+ e^{\rho_+ x} + C_- e^{\rho_- x} \tag 7$

which satisfy

$u(0) = u(L) = 0; \tag 8$

we note that (7) and (8) yield

$C_+ + C_- = u(0) = 0, \tag 9$

so that

$C_+ = -C_-; \tag{10}$

we can thus take

$C_+ = C = -C_-, \tag{11}$

and write

$u(x) = C(e^{\rho_+ x} - e^{\rho_- x}); \tag{12}$

we next note that for $\lambda < a^2$ in (6), we obtain two distinct real $\rho_\pm$, and if we apply the condition

$u(x) = C(e^{\rho_+ L} - e^{\rho_- L}) = u(L) = 0, \tag{13}$

we see that with $C \ne 0$ it leads to

$e^{\rho_+ L} = e^{\rho_- L}, \tag{14}$

which is impossible if $\rho_+ \ne \rho_-$; therefore (4), (5) have no solutions in $D_A$ with $\lambda < a^2$; since solutions to (4) are eigenfunctions of the operator $A$, we see it has no eigenvalues $\lambda < a^2$. If $\lambda = a^2$, then we see via (6) that (5) has a repeated root $\rho = \rho_\pm = -a$; it is well-known the solutions in this case are $e^{\rho x} = e^{-ax}$ and $xe^{-ax}$; then

$u(x) = C_1 e^{-ax} + C_2 x e^{-ax} = (C_1 + C_2x) e^{-ax}, \tag{15}$

whence

$C_1 = u(0) = 0, \tag{16}$

and hence

$u(x) = C_2 x e^{-ax}, \tag{17}$

whence

$C_2 L e^{-aL} = u(L) = 0 \Longrightarrow C_2 = 0 \Longrightarrow u(x) = 0, \tag{18}$

eigenfunctions cannot be zero; thus we rule out the case $\lambda = a^2$.

We are left only with the possibility that

$\lambda > a^2; \tag{19}$

then we have

$\rho_\pm = -a \pm i \sqrt{\lambda - a^2}; \tag{20}$

formulas (7), (9) and (12) still hold so we still write

$u(x) = C(e^{\rho_+ x} - e^{\rho_- x}); \tag{21}$

with $\rho_\pm$ as in (20) we find

$e^{\rho_\pm x} = e^{-ax} e^{\pm \sqrt{\lambda - a^2} x} = e^{-ax}(\cos \sqrt{\lambda - a^2} x \pm i \sin \sqrt{\lambda - a^2} x), \tag{22}$

so that

$e^{\rho_+ x} - e^{\rho_- x} = 2i e^{-ax} \sin \sqrt{\lambda - a^2} x, \tag{23}$

and

$u(x) = 2C i e^{-ax} \sin \sqrt{\lambda - a^2} x; \tag{24}$

the criterion

$u(L) = 0 \tag{25}$

forces

$\sin \sqrt{\lambda - a^2} L = 0, \tag{26}$

or

$\sqrt{\lambda - a^2} L = k\pi, \; 0 \ne k \in \Bbb Z \tag{27}$

(we preclude $k = 0$ to enforce $u(x) \ne 0$); then

$\lambda = a^2 + \dfrac{k^2 \pi^2}{L^2}, \; 0 \ne k \in \Bbb Z, \tag{28}$

and the eigenfunctions may be taken to be

$e^{-ax} \sin \sqrt{\lambda - a^2} x = e^{-ax} \sin \dfrac{x k\pi}{L}, 0 \ne k \in \Bbb Z. \tag{29}$

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Hint: Your functions are $$ p(x) = C_1 e^{(-a + \sqrt{a^2 - \lambda})x} + C_2e^{(-a - \sqrt{a^2 - \lambda})x} $$ Consider how they behave when $\lambda > a^2$. Then consider what $\lambda$ and the $C_i$ need to be to satisfy the boundary conditions.

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How do I apply this operator to a function $u$?

Your $D_A$ looks like the function space of all 2 times differentiable functions on $[0,L]$. So this seems like a function of one variable.

And therefore these seem ordinary differential operators, no partial ones, in the definition of $A$.

Thus I believe $A$ applies like this $$ \begin{align} A u &= -e^{-2ax} \partial_x \left(e^{2ax} \partial_x u \right) \\ &= -e^{-2ax}\left(e^{2ax} 2a \partial_x u + e^{2ax} \partial_x^2 u\right) \\ &= - (\partial_x^2 u + 2a \partial_x u)\\ \end{align} $$ to $u \in D_A$.

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  • $\begingroup$ Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer $\endgroup$ – Heuristics Jul 25 '18 at 18:25
  • $\begingroup$ Then I am at a loss what kind of function a feasible $u$ is. $\endgroup$ – mvw Jul 25 '18 at 18:27
  • $\begingroup$ Is my method of finding the eigenfunction correct (although somewhat incomplete)? $\endgroup$ – Heuristics Jul 25 '18 at 18:28
  • $\begingroup$ And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function. $\endgroup$ – mvw Jul 25 '18 at 18:30
  • $\begingroup$ Have a look at Universal oscillator. $\endgroup$ – mvw Jul 25 '18 at 18:37

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