ODE (ar laplace transform)

**cyw1984** · August 3rd 2007, 11:53 PM

where L[u(x,t)] = U(x,s) with respect to t
Given u(x,o)=x u(0.t)=o

It is an imhonogeous equation. It need to find out general soland particular sol.

if it is a honogeous equation, just very easy to do.Am I correct in finding general sol.?
what is the yp(particular sol) in order to find U(x,s)

Finally, How to do inverse laplace transform to find u(x,t).
Thank

**ThePerfectHacker** · August 4th 2007, 05:44 PM

Give us the partial differencial equation with the boundary and initial value problems. The way you have it just makes no sense.

**cyw1984** · August 5th 2007, 05:14 PM

Originally Posted by ThePerfectHacker

Give us the partial differencial equation with the boundary and initial value problems. The way you have it just makes no sense.

updated
bivp is u(x,0)=x u(0,t)=0

**ThePerfectHacker** · August 5th 2007, 05:40 PM

That does not help. Give me the original equation without taking the Laplace Transform. The one you have is after you taken the Laplace Transformation.

**cyw1984** · August 5th 2007, 06:17 PM

Originally Posted by ThePerfectHacker

That does not help. Give me the original equation without taking the Laplace Transform. The one you have is after you taken the Laplace Transformation.

O..sorry

The original equation is

Uxx= Ut (-inf < x < + inf , t>0)
u(x,0)=x u(0,t)=0

Use the foruier transform and laplace transform to solve this PDE.

Thank, The above one I have is after I taken the Laplace Transformation.

**ThePerfectHacker** · August 5th 2007, 07:05 PM

$\mathcal{L}\left\{ \frac{\partial^2 u}{\partial x^2} \right\} = \mathcal{L} \left\{ \frac{\partial^2 u}{\partial t} \right\}$

$\frac{d^2 U(x,s)}{dx^2} = sU(x,s) - u(x,0) = sU(x,s)-x$

$\frac{d^2 U(x,s)}{dx^2} - sU(x,s) = x$

$U(x,s) = Ae^{x\sqrt{s}}+Be^{-x\sqrt{s}} - \frac{x}{s}$

Now solve for $A,B$ with the conditions given.

**cyw1984** · August 5th 2007, 07:54 PM

Originally Posted by ThePerfectHacker

$\mathcal{L}\left\{ \frac{\partial^2 u}{\partial x^2} \right\} = \mathcal{L} \left\{ \frac{\partial^2 u}{\partial t} \right\}$

$\frac{d^2 U(x,s)}{dx^2} = sU(x,s) - u(x,0) = sU(x,s)-x$

$\frac{d^2 U(x,s)}{dx^2} - sU(x,s) = x$

$U(x,s) = Ae^{x\sqrt{s}}+Be^{-x\sqrt{s}} - \frac{x}{s}$

Now solve for $A,B$ with the conditions given.

Now I have solve A=-B
$A = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}$

$U(x,s) = \frac{L}{s} \frac{1}{e^{L\sqrt{s}}-e^{-L\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}$
Am I correct?
Then how to do inverse laplace transform. The U(x,s) is very complex.....

**ThePerfectHacker** · August 6th 2007, 05:30 AM

Originally Posted by cyw1984

Now I have solve A=-B
$A = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}$

$U(x,s) = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}$
Am I correct?
Then how to do inverse laplace transform. The U(x,s) is very complex.....

I just realized that perhaps we need another point. You provided u(0,t)=0. Is there supposed to be another one? Because you cannot solve for A and B yet because there is 1 equation and two unknowns, if you provide one more conditions then we have 2 equations and 2 unknowns and it should solve.

**cyw1984** · August 6th 2007, 08:03 AM

Originally Posted by ThePerfectHacker

I just realized that perhaps we need another point. You provided u(0,t)=0. Is there supposed to be another one? Because you cannot solve for A and B yet because there is 1 equation and two unknowns, if you provide one more conditions then we have 2 equations and 2 unknowns and it should solve.

now if u(0,t)=0 u(L,t) =0

$U(x,s) = \frac{L}{s} \frac{1}{e^{L\sqrt{s}}-e^{-L\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}$

Is my answer correct???
Thank
and how to do inverse laplace.it is so complex

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