Initial Value Problems with Delta Functions

**Chris L T521** · May 10th 2008, 08:20 PM

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!

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 $x^{//}+4x^{/}+8x=f(t); x(0)=x^{/}(0)=0$

**mr fantastic** · May 10th 2008, 08:36 PM

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!

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 ....?)

**mr fantastic** · May 10th 2008, 09:13 PM

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?

**Chris L T521** · May 10th 2008, 09:20 PM

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...)

**TheEmptySet** · May 10th 2008, 10:57 PM

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.

**TheEmptySet** · May 10th 2008, 11:15 PM

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!

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

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

Thread: Initial Value Problems with Delta Functions

LinkBack

Thread Tools

Display

Initial Value Problems with Delta Functions

Similar Math Help Forum Discussions

Initial Value Problems

initial value problems

Initial Value Problems

initial value problems

Initial value problems

Search Tags