Eigenfunction Expansion

**dwsmith** · February 12th 2011, 01:22 PM

Find the basic function at $x_0=0$ for

$\varphi''+4\varphi'+\lambda\varphi=0$

Also, find the solution which satisfies $\varphi(0)=3, \ \varphi'(0)=-1$

$m^2+4m+\lambda=0$

$\displaystyle m=\frac{-4\pm\sqrt{16-4\lambda}}{2}=-2\pm\sqrt{4-\lambda}$

$\varphi=\exp\left(z(-2+\sqrt{4-\lambda})\right)$

What I don't understand is why the book multiplied through by x instead of of z = x + yi, because looking at the first solution, the imaginary constant is being used in relations to sqrt{4-\lambda}.

How was $\varphi_2$ acquired and is the coefficient 2 associated with sine in the first solutions due to solving for the conditions?

The solution to the problem is:

$\displaystyle\varphi_1=e^{-2x}\left[\cos(x\sqrt{4-\lambda})+\frac{2\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right]$

$\displaystyle\varphi_2=e^{-2x}\frac{\sin (x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}$

**dwsmith** · February 14th 2011, 12:34 PM

$\varphi=\exp\left(x(-2\pm\sqrt{4-\lambda})\right)\Rightarrow \varphi=C_1\exp(-2x+x\sqrt{4-\lambda})+C_2\exp(-2x-x\sqrt{4-\lambda})$

Then solve for the initial conditions?

What about taking into account if $\lambda>4 \ \ \ \text{or} \ \ \ \lambda\in\left[ [4,0)\cup (0,-\infty)\right]\text{?}$

When $\lambda=0$ , the DE is reduced to $\varphi''+4\varphi'=0$ , and then we have

$t^2+4t=0\Rightarrow t=0, -4$

$\varphi=C_1+C_2e^{-4x}$

$\varphi_1(0)=C_1+C_2=3$

$\displaystyle\varphi_1'(0)=C_2=\frac{1}{4}$

$\displaystyle\varphi_1=\frac{1}{4}\left(11+e^{-4x}\right)$

$\varphi_2(0)=C_1+C_2=-1$

$\displaystyle\varphi_2'(0)=C_2=-\frac{3}{4}$

$\displaystyle\varphi_2=-\frac{1}{4}\left(1+3e^{-4x}\right)$

$\varphi(x)=A\left[\frac{1}{4}\left(11+e^{-4x}\right)\right]+B\left[-\frac{1}{4}\left(1+3e^{-4x}\right)\right]$

**dwsmith** · February 14th 2011, 03:25 PM

Now, if $\lambda\in\left[ [4,0)\cup (0,-\infty)]$ , then we obtain:

$\displaystyle \sqrt{4-\lambda}\in\mathbb{R}, \ \ \ \sqrt{4-\lambda}=\omega$

$\displaystyle \varphi=C_1\exp(-2x+x\omega)+C_2\exp(-2x-x\omega)$

$\varphi(0)=C_1+C_2=3$

$\varphi'(0)=C_1(\omega-2)+C_2(-2-\omega)=-1$

$\displaystyle C_1=\frac{5+3\omega}{2\omega}, \ \ C_2=\frac{-5+3\omega}{2\omega}$

$\displaystyle\varphi_1(x)=\frac{5+3\omega}{2\omega }\exp(-2x+x\omega)+\frac{-5+3\omega}{2\omega}\exp(-2x-x\omega)$

So omega can't be 0; therefore, lambda can't equal 4.

$\lambda\in\left[ [4,0)\cup (0,-\infty)]\Rightarrow \lambda\in\left[ (4,0)\cup (0,-\infty)]$

$\varphi_2(0)=C_1+C_2=-1$

$\varphi_2'(0)=C_1(\omega-2)+C_2(-2-\omega)=3$

$\displaystyle C_1=\frac{1-\omega}{2\omega}, \ \ C_2=\frac{-1-\omega}{2\omega}$

$\displaystyle\varphi_2(x)=\frac{1-\omega}{2\omega}\exp(-2x+x\omega)+\frac{-1-\omega}{2\omega}\exp(-2x-x\omega)$

$\displaystyle\varphi(x)=A\left[\frac{5+3\omega}{2\omega}\exp(-2x+x\omega)+\frac{-5+3\omega}{2\omega}\exp(-2x-x\omega)\right]+B\left[\frac{1-\omega}{2\omega}\exp(-2x+x\omega)+\frac{-1-\omega}{2\omega}\exp(-2x-x\omega)\right]$

Correct?

**dwsmith** · February 14th 2011, 04:03 PM

I don't understand how the book obtained its solution which would be for $\lambda>4$

$\displaystyle\varphi_1=e^{-2x}\left[\cos(x\sqrt{4-\lambda})+\frac{2\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right]$

$\displaystyle\varphi_2=e^{-2x}\frac{\sin (x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}$

When I do the problem, I obtain:

$\varphi=e^{-2x}\left(C_1\cos(x\sqrt{4-\lambda})+C_2i\sin(x\sqrt{4-\lambda})\right)$

$\varphi(0)=C_1=3$

$\varphi=e^{-2x}\left(3\cos(x\sqrt{4-\lambda})+C_2i\sin(x\sqrt{4-\lambda})\right)$

$\varphi'=-2e^{-2x}\left(3\cos(x\sqrt{4-\lambda})+C_2i\sin(x\sqrt{4-\lambda})\right)+e^{-2x}\left(C_2i\sqrt{4-\lambda}\cos(x\sqrt{4-\lambda})-3\sqrt{4-\lambda}\sin(x\sqrt{4-\lambda})\right)$

$\displaystyle\varphi'(0)=-2\cdot 3+C_2i\sqrt{4-\lambda}=-1\Rightarrow C_2=\frac{-5i}{\sqrt{4-\lambda}}$

Before I attempt to solve $\varphi_2$ , why am I so off with $\varphi_1\text{?}$

$\displaystyle\varphi_1(x)=e^{-2x}\left(3\cos(x\sqrt{4-\lambda})-i\frac{5\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right)\neq e^{-2x}\left[\cos(x\sqrt{4-\lambda})+\frac{2\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right]$

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