# Fourier transform of differentials equation

hey there.. i really dont know how to start answering this question.. can someone please guide me to solve this question.. gabbagabbahey
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Since the question tells you to use Fourier transforms, you start by taking the FT of both sides of the differential equation...

Since the question tells you to use Fourier transforms, you start by taking the FT of both sides of the differential equation...
still blur... gabbagabbahey
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$$F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)][/itex] Call the FT of $f(x)$, $F[f(x)]=\tilde{f}(k)$. Call the FT of $u(x)$, $F[u(x)]=\tilde{u}(k)$. Your textbook likely tells you how to express the Fourier transform of $-\frac{d^2u}{dx^2}$ and $a^2u(x)$ in terms of $\tilde{u}(k)$....do exactly that i dont know how to proceed from here... [tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)]$$

$$F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+a^2F\left[u\right]=F[f(x)]$$

$$F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-(-k^2\tilde{u}(k,y))\right]+a^2F\left[u\right]=F[f(x)]$$

gabbagabbahey
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When you take the fourier transform of $-\frac{d^2u}{dx^2}$ (That's what the $F[\ldots]$ means) you get $k^2F$, or $k^2\tilde{u}(k)$.

This gives you an algebraic (instead of differential) equation you can solve for $\tilde{u}$. To find $u$, you then take the inverse FT.

This is a very basic application of using FTs to solve DEs and I have a hard time believing that your textbook doesn't provide a similar example. You should really be in the habit of studying the relevant section of your text/notes before attempting homework problems that are based off of it.

i got the general solution already.. howeverr.. how am i going to prove... When you take the fourier transform of $-\frac{d^2u}{dx^2}$ (That's what the $F[\ldots]$ means) you get $k^2F$, or $k^2\tilde{u}(k)$.

This gives you an algebraic (instead of differential) equation you can solve for $\tilde{u}$. To find $u$, you then take the inverse FT.

This is a very basic application of using FTs to solve DEs and I have a hard time believing that your textbook doesn't provide a similar example. You should really be in the habit of studying the relevant section of your text/notes before attempting homework problems that are based off of it.

i've done the inverse..

F^-1[g^(k)] = g(x) = 1/ √2∏ ∫g^(k)e^ikx dk
.............................. = 1/ √2∏ ∫ 1/ (k^2 + a^2)e^ikx dk
...............................= √2∏ [(e^-a|x|) / 2a]

gabbagabbahey
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i've done the inverse..

F^-1[g^(k)] = g(x) = 1/ √2∏ ∫g^(k)e^ikx dk
.............................. = 1/ √2∏ ∫ 1/ (k^2 + a^2)e^ikx dk
...............................= √2∏ [(e^-a|x|) / 2a]

$$\tilde{u}(k)\neq\frac{1}{k^2+a^2}$$

What happened to the FT of $f(x)$?

accidently double post

Last edited: ok.. this is my full solution...
am i got it right until here?

gabbagabbahey
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That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...

That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...

inverse both side...
i get.. is that my general solution?

gabbagabbahey
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Does $\frac{1}{2a}\int f(x) e^{-a|x|}dx$ really represent the convolution of $f$ and $g$?

That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...

Does $\frac{1}{2a}\int f(x) e^{-a|x|}dx$ really represent the convolution of $f$ and $g$?

by convolution theorem,
F^−1{F*G} = f ·g
correct?
so, i just substitute it..?
kinda confused ok.. i revised bout it already.. gabbagabbahey
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That's better! However, that's just part of the general solution. You can also add any function, $u_h$, which satisfies the corresponding homogeneous ODE, $-\frac{d^2 u_h}{dx^2}+a^2u_h=0$, to the $u(x)$ you've found, and the result will still satisfy your ODE. So, you need to find that homogeneous solution and add it to what you've found above in order to obtain the true general solution.

That's better! However, that's just part of the general solution. You can also add any function, $u_h$, which satisfies the corresponding homogeneous ODE, $-\frac{d^2 u_h}{dx^2}+a^2u_h=0$, to the $u(x)$ you've found, and the result will still satisfy your ODE. So, you need to find that homogeneous solution and add it to what you've found above in order to obtain the true general solution.

should i solve the integration u(x) = 1/2a ∫ f(x - y) e^-a|y| dy?

gabbagabbahey
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First worry about finding the homogeneous part of the general solution. Solve the Homogeneous equation.

First worry about finding the homogeneous part of the general solution. Solve the Homogeneous equation.

can u give me more clue.. i really can't get it..
which homogeneous equation i need to solve?

gabbagabbahey
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is this my general solution?

$y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt$

now.. to answer my question, u'' + a^2u = x^5
after solving everything, i manage to get this equation..

$$\displaystyle u(x)= ∫- ∞ to 0 1/2a(x-y)^{5}e^{ay}dy+ ∫0 to ∞ (x-y)^{5}e^{-ay}dy should i solve it to get the general solution? because, i need to do integration by parts 5 times to solve it..$$

gabbagabbahey
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is this my general solution?

$y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt$

$u(x)=\frac{1}{2a}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt$ is part of your general solution. You are missing the homogeneous part.

Once you find the true general solution, you need only determine whether the integrals involved in exist/converge for your two $f(x)$'s to know whether or not a solution exists.

this is my general solution?

$u(x)=c_1 e^{-ax}+c_2 e^{ax}+y(x)$