# PDE Problem on Eigenvalues and the Heat Equation

Let $a$ be any fixed real number. Consider the eigenvalue system:

$$\begin{cases} X'' + \lambda X = 0, & 0 ≤ x ≤ 1\\[0.1cm] aX(0) = X(1)\\[0.1cm] aX'(0) = -X'(1) \end{cases}.$$

Prove that if $a = \pm 1$, then every real $\lambda$ is an eigenvalue.

Prove that if $a \not= 1$, then no real $\lambda$ is an eigenvalue.

Approach:

So I looked for solutions of the ODE in the form:

$$X(x) = e^{bx}$$

Where I need to solve for b. Differentiation of $e^{bx}$ just multiplies the function by b, so then $X(x)$ becomes

$(b^2 + λ)*e^{bx} = 0$. And since $e^{bx}$ cannot be zero,

$$b^2 + λ = 0$$ So, $b=\pm\sqrt{-λ}x$ This gives the general form of $X(x) = c_1e^{\sqrt{-\lambda}x} + c_2e^{-\sqrt{-\lambda}x}$

Where $X'(x) = c_1\sqrt{-\lambda}e^{\sqrt{-\lambda}x} + c_2\sqrt{-\lambda}e^{\sqrt{-\lambda}x}$ is the 2nd derivative. Plugging in the boundary conditions gives:

$X(0) = c_1+c_2$

$X'(0) = c_1-c_2$

$X(1) = c_1e^{\sqrt{-\lambda}} + c_2e^{-\sqrt{-\lambda}}$

$X'(1) = c_1\sqrt{-\lambda}e^{\sqrt{-\lambda}} - c_1\sqrt{-\lambda}e^{\sqrt{-\lambda}}$

I then tried solving for $a$

$a = \frac{X(1)}{X(0)} = \frac{c_1e^{\sqrt{-\lambda}} + c_2e^{-\sqrt{-\lambda}}}{c_1+c_2}$

$a = \frac{X'(1)}{X'(0)} = \frac{c_1\sqrt{-\lambda}e^{-\sqrt{-\lambda}} - c_2\sqrt{-\lambda}e^{-\sqrt{-\lambda}}}{c_1-c_2}$

• Welcome to MSE! What are your thoughts on the problem? – MisterRiemann Aug 31 '18 at 12:19
• For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Aug 31 '18 at 12:22
• You should first solve the differential equation (in dependence of $\lambda$). – amsmath Aug 31 '18 at 12:58
• @amsmath So I look for solutions of the form X(x) = e^(bx). Differentiation of e^(bx) just multiplies the function by b, so then X(x) becomes (b^2 + λ)*e^(bx) = 0. And since e^(bx) cannot be zero, I solve b^2 + λ = 0. Here is where I am stuck. – Meghan Kapur Aug 31 '18 at 13:05
• $b = \pm\sqrt{-\lambda}$. So, the general solution is $X(t) = ce^{\sqrt{-\lambda}t} + de^{-\sqrt{-\lambda}t}$. Be careful! You have to treat the case $\lambda = 0$ separately. – amsmath Aug 31 '18 at 13:22

## 1 Answer

Note that $b^2+\lambda=0$ has solutions $b=\pm\sqrt{-\lambda}$. So the general solution is $$X(x)=C_1e^{\sqrt{-\lambda}x}+C_2e^{-\sqrt{-\lambda}x}.$$ Using the initial conditions $aX(0) = X(1),aX'(0) = -X'(1)$, one has $$a(C_1+C_2)=C_1e^{\sqrt{-\lambda}}+C_2e^{-\sqrt{-\lambda}},\tag{1}$$ and $$a\sqrt{-\lambda}(C_1-C_2)=-\sqrt{-\lambda}(C_1e^{\sqrt{-\lambda}}-C_2e^{-\sqrt{-\lambda}}).\tag{2}$$ The second equation above can be simplified as $$a(C_1-C_2)=-C_1e^{\sqrt{-\lambda}}+C_2e^{-\sqrt{-\lambda}}. \tag{3}$$ Now (1)+(3) and (1)-(3) give $$aC_1=C_2e^{-\sqrt{-\lambda}x}, aC_2=C_1e^{\sqrt{-\lambda}x}$$ from which it is easy to see $$(a^2-1)C_1C_2=0.$$ Case 1: $a^2-1=0$ or $a=\pm1$. Then $X(x)=C_1e^{\sqrt{-\lambda}x}+C_2e^{-\sqrt{-\lambda}x}$ is an eigenfunction (at least one of $C_1,C_2$ is not zero) with eigenvalue $\lambda$.

Case 2: $a^2-1\neq 0$ or $a\not=\pm1$. Then from (4), one has $C_1C_2=0$. So one of $C_1,C_2$ is zero. Suppose $C_1=0$. Then (1) and (2) become $$aC_2=C_2e^{-\sqrt{-\lambda}},-aC_2=C_2e^{-\sqrt{-\lambda}}$$ from which one has $C_2e^{-\sqrt{-\lambda}}=0$ or $C_2=0$. Thus $X(x)=0$. Similarly $C_2=0$ implies $C_1=0$. Thus $\lambda$ is not an eigenvalue.