# Initial Value Problems with Delta Functions

• May 10th 2008, 08:20 PM
Chris L T521
Initial Value Problems with Delta Functions
I'm teaching myself Delta Functions and I'm stuck on this initial value problem. If you guys can help me out, I'd appreciate it! :D

Solve the initial value problem : $x^{//}+9x=\delta (t-3\pi)+cos(3t); x(0)=x^{/}(0)=0$.

Also, I don't get how to do this problem:

Apply Duhamel's principle to write an integral formula for the solution of the initial value problem $x^{//}+4x^{/}+8x=f(t); x(0)=x^{/}(0)=0$
• May 10th 2008, 08:36 PM
mr fantastic
Quote:

Originally Posted by Chris L T521
I'm teaching myself Delta Functions and I'm stuck on this initial value problem. If you guys can help me out, I'd appreciate it! :D

Solve the initial value problem : $x^{//}+9x=\delta (t-3\pi)+cos(3t); x(0)=x^{/}(0)=0$.

[snip]

I'd try solving it using Laplace Transforms - something that, if you don't know much about, you might consider learning more as a pre-requisite to solving differential equations involving generalised functions such as the Dirac Delta function .......

(Although I'm sure I saw something somewhere that suggested you loved Laplace transforms ....?)
• May 10th 2008, 09:13 PM
mr fantastic
Quote:

Originally Posted by Chris L T521
[snip]
Also, I don't get how to do this problem:

Apply Duhamel's principle to write an integral formula for the solution of the initial value problem $x^{//}+4x^{/}+8x=f(t); x(0)=x^{/}(0)=0$

Isn't Duhamel's Principle used to solve certain types of partial differential equations .......?

Or do you mean Duhamel's Convolution Principle - used to invert a Laplace transform (by using a convolution integral) when using Laplace Transform to solve IVP’s with general forcing functions?
• May 10th 2008, 09:20 PM
Chris L T521
I'm looking in my book, and it appears to be Duhamel's Convolution Principle (I see $\tau$ 's in there, so I think may have to do with convolution...)
• May 10th 2008, 10:57 PM
TheEmptySet
Quote:

Originally Posted by Chris L T521
Apply Duhamel's principle to write an integral formula for the solution of the initial value problem $x''+4x'+8x=f(t); x(0)=x'(0)=0$

I don't know what Duhamel's principle, but it can be solved using the Laplace transform and the convolution theorem.

Assuming that the Laplace transform of f(t) exists we get

$s^2X+4sX+8X=F(s) \iff X=\frac{F(s)}{s^2+4s+8}$

$X=\frac{F(s)}{(s+2)^2+4}$

The Covolution theorem states that

$\mathcal{L}^{-1}(F(s)G(s))=\int_{0}^{t}f(t-\tau)g(\tau)d\tau$

so $G(s)=\frac{1}{(s+2)^2+4}$

The inverse transform of the s-axis translation is

$g(t)=\frac{1}{2}e^{-2t}\sin(2t)$

Now using the above and the convolution theorem we get

$x(t)=\mathcal{L}^{-1}\frac{F(s)}{(s+2)^2+4}=\int_{0}^{t}f(t-\tau)(\frac{1}{2}e^{-2\tau}\sin(2\tau))d\tau$

I hope this helps.
• May 10th 2008, 11:15 PM
TheEmptySet
Quote:

Originally Posted by Chris L T521
I'm teaching myself Delta Functions and I'm stuck on this initial value problem. If you guys can help me out, I'd appreciate it! :D

Solve the initial value problem : $x^{//}+9x=\delta (t-3\pi)+cos(3t); x(0)=x^{/}(0)=0$.

Taking the Laplace transform of both sides gives

$s^2X+9X=e^{3s\pi }+\frac{s}{s^2+9}$

Solving for X gives

$X=\frac{e^{3s\pi}}{s^2+9}+\frac{s}{(s^2+9)^2}$

To avoid using partial fractions on the 2nd note that

$\frac{s}{(s^2+9)^2}=\frac{1}{6}(-1)\frac{d}{ds}\left( \frac{3}{s^2+9}\right)$

We can now use the derivative of a transform theorem backwards

$\mathcal{L}(t^nf(t))=(-1)^n\frac{d^n}{ds^n}F(s)$

so we get the inverse transform is $\frac{1}{6}t\sin(3t)$

so finally our solution is

$x(t)=\frac{1}{3}\mathcal{U}(t-3\pi)\sin(3(t-3\pi)+\frac{1}{6}t\sin(3t)$