# homogeneous differential equation with constant coefficients

#### confuseddove

Hi, I need some help solving this homogeneous differential equation with constant coefficients for initial condition;

y''''-4y=0
and
y(0)=1
y(1)=0
y(-1)=0
y(inf)=0

Well I get as far as

y=C1*e^(x*sqrt(2)) + C2*e^(-x*sqrt(2)) + C3*cos(x*sqrt(2)) + C4*sin(x*sqrt(2)),

but I don't know how to solve for initial contritions and would appreciate any help.

Well I think that

c1+c2+c3=1,

but that is as far as I get.

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#### johnsomeone

It seems to me that $$\displaystyle \lim_{x \to \infty} y(x) = 0$$ implies that $$\displaystyle c_1 = c_3 = c_4 = 0$$.

That's because, by combining $$\displaystyle c_3 \cos(\sqrt{2} x) + c_4 \sin(\sqrt{2} x)$$, y(x) is the sum of three functions, one of which goes to infinity (unless its coefficient is 0), one of which goes to 0, and one of which is periodic (period = $$\displaystyle \sqrt{2} \pi$$).

Thus $$\displaystyle \lim_{x \to \infty} y(x) = 0$$ forces $$\displaystyle c_1 = 0$$, and so $$\displaystyle 0 = \lim_{x \to \infty} y(x) = \lim_{x \to \infty} \left(c_3 \cos(\sqrt{2} x) + c_4 \sin(\sqrt{2} x)\right)$$.

But the only way a periodic function has limit 0 at infinity is if it's identically 0, and therefore $$\displaystyle c_3 = c_4 = 0$$.

Thus the condition $$\displaystyle \lim_{x \to \infty} y(x) = 0$$ implies that $$\displaystyle y(x) = c_2 e^{-\sqrt{2} x}$$.

But from there, $$\displaystyle y(1) = 0$$ implies that $$\displaystyle c_2 = 0$$, and so $$\displaystyle y$$ must be the constant function $$\displaystyle y(x) = 0$$. But that's impossible if $$\displaystyle y(0) = 1$$.

Thus it looks to me like there is no solution to that ODE.

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#### confuseddove

Thus it looks to me like there is no solution to that ODE.
It appears to me as well that there is no solution - trick question... or it might just be a typographical error...

#### HallsofIvy

MHF Helper
Hi, I need some help solving this homogeneous differential equation with constant coefficients for initial condition;

y''''-4y=0
and
y(0)=1
y(1)=0
y(-1)=0
y(inf)=0

Well I get as far as

y=C1*e^(x*sqrt(2)) + C2*e^(-x*sqrt(2)) + C3*cos(x*sqrt(2)) + C4*sin(x*sqrt(2)),

but I don't know how to solve for initial contritions and would appreciate any help.

Well I think that

c1+c2+c3=1,

but that is as far as I get.
Okay, f(0)= 1 is C1+ C2+ C3= 1 (don't switch from "C" to "c"- that's confusing).

The second condition is that y(1)= 0. Setting x= 1 in y=C1*e^(x*sqrt(2)) + C2*e^(-x*sqrt(2)) + C3*cos(x*sqrt(2)) + C4*sin(x*sqrt(2)) gives C1e^(sqrt(2))+ C2e^(-sqrt(2))+ C3 cos(sqrt(2)+ C4 sin(sqrt(2))= 0

The third condition is that y(-1)= 0. Setting x= -1 in y=C1*e^(x*sqrt(2)) + C2*e^(-x*sqrt(2)) + C3*cos(x*sqrt(2)) + C4*sin(x*sqrt(2)) gives C1e^(-sqrt(2))+ C2e^(sqrt(2))+ C3 cos(-sqrt(2)+ C4 sin(-sqrt(2))= 0
Since cosine is an"even" function and sine is odd, that is the same as
C1e^(-sqrt(2))+ C2e^(sqrt(2))+ C3 cos(sqrt(2)- C4 sin(sqrt(2))= 0

The fourth condition is that y(infinity)= 0. That means, immediately that C1 must be 0 since that first term would go to infinity or negative infinity if C1 were positive or negative. Adding the second and third equations will eliminate C4 leaving an equation in C2 and C3. Since C1 is 0, the first equation is C2+ C3= 1 so C3= 1- C2. Replace C3 in the result of adding the second and third equations by that to get a single equation in C2 and solve that for C2.

1 person

#### johnsomeone

Claim: There exists no $$\displaystyle f \in C^4[-1, \infty)$$ of the form $$\displaystyle f(x) = c_1 e^{x \sqrt{2}} + c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2}), c-1, c_2, c_3, c_4 \in \mathbb{R}$$ satisfying $$\displaystyle f(0) = 1, f(1) = 0, \lim_{x \to \infty} f(x) = 0$$.

Note 1: The satisfaction of y'''' = y at any point implies that y is 4 times differentiable there, and so its third derivative there is continuous. Also, y is continuous there, since its first derivative exists there. But then y'''' = y says that y'''' is also continuous there... thus the $$\displaystyle C^4$$ statement. Moreover, given its form, f is obviously $$\displaystyle C^{\infty}$$ on (the interior of) its domain.

Note 2: The choice of a single interval, $$\displaystyle [-1, \infty)$$, was based on the initial conditions. If you permit a break in the domain, so that the domain isn't that entire interval, then it becomes possible to satisfy that differential equation with those initial conditions. That's because a break in the domain means that the restrictions imposed by the initial conditions at 0, 1, and -1 no longer will carry over and influence the function on the disconnected interval where it's going to infinity. If you permitted a break in the domain, you'd in effect have different initial value problems on the different intervals, and there would be no conflict harmonizing into a single function, because the intervals would be disconnected.

Note 3: The -1 left endpoint of the domain was chosen because of the conditions, but its role in proving that claim is to have x = 0 in the interior of the function's domain. That's useful because the claim hinges on f(0) = 1, and you'd want to avoid any complications arising from derivatives and differential equations at closed interval endpoints, which might've posed a concern had the stated domain been $$\displaystyle [0, \infty)$$.

Proof: Assume there is such a function $$\displaystyle f$$ with all of those properties.

Define: $$\displaystyle g(x) = c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2})$$. In other words $$\displaystyle g(x) = f(x) - c_1 e^{x \sqrt{2}}$$.

Claim: The function $$\displaystyle g$$ is bounded on the positive reals.

Since for $$\displaystyle x> 0$$ you have that $$\displaystyle 0 < e^{-x \sqrt{2}} < 1$$, you get, for $$\displaystyle x> 0$$, that:

$$\displaystyle | c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2}) |$$

$$\displaystyle \le | c_2 | \ | e^{-x \sqrt{2}} | + | c_3 | \ | \cos(x \sqrt{2}) | + | c_4 | \ | \sin(x \sqrt{2}) |$$

$$\displaystyle \le | c_2 | + | c_3 | + | c_4 |$$.

Thus the function $$\displaystyle g(x) = c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2})$$ is bounded (by $$\displaystyle |c_2|+|c_3|+|c_4|$$) on the positive reals.

Claim: $$\displaystyle c_1 = 0$$ (and so $$\displaystyle f = g$$).

Assume $$\displaystyle c_1 \ne 0$$. Then since $$\displaystyle \lim_{x \to \infty} e^{x \sqrt{2}} = \infty$$, you'd have that $$\displaystyle \lim_{x \to \infty} c_1 e^{x \sqrt{2}} = \pm \infty$$.

But then $$\displaystyle 0 = \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( c_1 e^{x \sqrt{2}} + g(x) \right)$$ would require that $$\displaystyle \lim_{x \to \infty} |g(x)| = \infty$$,

since $$\displaystyle g$$ would have to "blow up in the opposite way" to counter $$\displaystyle \lim_{x \to \infty} c_1 e^{x \sqrt{2}} = \pm \infty$$ in order for the sum $$\displaystyle c_1 e^{x \sqrt{2}} + g(x)$$ to go to 0 as x went to infinity.

But since $$\displaystyle g$$ is bounded on the positive reals, that can't hold. Thus $$\displaystyle c_1 \ne 0$$ produces a contradiction, so $$\displaystyle c_1 = 0$$.

(Technically, if $$\displaystyle g$$ is bounded by $$\displaystyle B$$ on the positive reals, and $$\displaystyle c_1 >0$$, use that $$\displaystyle c_1 e^{x \sqrt{2}}$$ is eventually always greater than $$\displaystyle 1 + B$$ to conclude that $$\displaystyle f(x) = c_1 e^{x \sqrt{2}} + g(x)$$ is eventually always greater than 1,

and so it's limit to infinity can't be 0. Likewise, if $$\displaystyle c_1 < 0$$, then $$\displaystyle c_1 e^{x \sqrt{2}}$$ is eventually always less than $$\displaystyle -1 - B$$, so that $$\displaystyle f(x) = c_1 e^{x \sqrt{2}} + g(x)$$ is eventually always less than -1, and so it's limit to infinity can't be 0.)

Claim: $$\displaystyle c_3 = c_4 = 0$$.

Since $$\displaystyle \lim_{x \to \infty} f(x) = 0$$, have that $$\displaystyle \lim_{n \to \infty} f(x_n) = 0$$ for any sequence $$\displaystyle \{x_n\}_{n = 1}^{\infty}$$ such that $$\displaystyle x_n \to \infty$$.

Consider the sequences $$\displaystyle s_n = \frac{2 \pi n}{\sqrt{2}}$$ and $$\displaystyle t_n = \frac{(\pi/2) + 2 \pi n}{\sqrt{2}}$$. Clearly both go to infinity.

But $$\displaystyle f(s_n) = f\left( \frac{2 \pi n}{\sqrt{2}} \right)$$

$$\displaystyle = c_2 e^{-\left( \frac{2 \pi n}{\sqrt{2}} \right) \sqrt{2}} + c_3 \cos\left(\sqrt{2} \left( \frac{2 \pi n}{\sqrt{2}} \right)\right) + c_4 \sin\left(\sqrt{2}\left( \frac{2 \pi n}{\sqrt{2}} \right)\right)$$

$$\displaystyle = c_2 e^{-2 \pi n} + c_3 \cos( 2 \pi n ) + c_4 \sin(2 \pi n) = c_2 e^{-2 \pi n} + c_3$$.

Thus $$\displaystyle 0 = \lim_{n \to \infty} f(s_n) = \lim_{n \to \infty} \left( c_2 e^{-2 \pi n} + c_3\right) = c_3$$.

Likewise $$\displaystyle f(t_n) = f\left( \frac{(\pi/2) + 2 \pi n}{\sqrt{2}} \right)$$

$$\displaystyle = c_2 e^{-(\pi/2) -2 \pi n} + c_3 \cos\left( (\pi/2) + 2 \pi n \right) + c_4 \sin\left((\pi/2) + 2 \pi n\right)$$

$$\displaystyle = c_2 e^{-(\pi/2)}e^{ -2 \pi n} + c_4$$,

and so $$\displaystyle 0 = \lim_{n \to \infty} f(t_n) = \lim_{n \to \infty} \left(c_2 e^{-(\pi/2)}e^{ -2 \pi n} + c_4 \right) = c_4$$.

Completion of proof

Have shown that $$\displaystyle \lim_{x \to \infty} f(x) = 0$$ implies that $$\displaystyle c_1 = c_3 = c_4 = 0$$, and so have that $$\displaystyle f(x) = c_2 e^{-x \sqrt{2}}$$.

But then using that f(1) = 0 gives: $$\displaystyle f(1) = 0 \Rightarrow c_2 e^{-\sqrt{2}} = 0 \Rightarrow c_2 = 0$$. Thus $$\displaystyle f$$ is the constant function 0.

But that contradicts f(0) = 1, and so the original assumption - that there exists some f meeting all those stated criteria - has produced a contradiction. Therefore no such f exists. QED.

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1 person

#### confuseddove

johnsomeone and HallsofIvy thank you for extremely in-depth explanation. It really gives insight of how this kind of problems should be attacked.

I learn mathematics from books & internet. Yes I know, I should attended classes when I had the chance (almost decade ago), now I have to invest 10 times more work and time to actually understand what is going on.