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| − | 1) Let \( f,g:\mathbb{R}^n \rightarrow \mathbb{C} \) be continuous, bounded. Further, \( f,g \in L^1(\mathbb{R}^n) \). Define the convolution \(h:\mathbb{R}^n \rightarrow \mathbb{C} \) as
| + | Check out the document Lie-Gruppen, Bsp. 3.2 for part a) here: [[https://www.dropbox.com/sh/su8ja1eynb449nr/AABIiXtiB6SNdwjjWYtYN8Tua?dl=0]]. |
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| − | $$ h(x) = (f * g)(x) = \int_{\mathbb{R}^n } f(x-y) g(y) dy $$
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| − | == Part a) ==
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| − | Show that:
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| − | $$ (f*g)(x) = (g*f)(x) $$
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| − | === Solution part a) ===
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| − | $$ (f*g)(x) = \int_{\mathbb{R}^n } f(x-y) g(y) dy $$
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| − | $$ = \int_{-\infty}^{\infty} ... \int_{-\infty}^{\infty} f(x-y) g(y) dy_1 ... dy_n $$
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| − | The claim follows simply by substitution:
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| − | $$ t = x-y $$
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| − | $$ y = x-t $$
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| − | $$ dy_i = -dt_i $$
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| − | $$ \int_{\infty}^{-\infty} ... \int_{\infty}^{-\infty} f(t) g(x-t) (-dt_1) ... (-dt_n) $$
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| − | We can now change the orientation of the integral if we multiply (-1) n-times. These will cancel out all the n minus-signs from the \( dt_i\)`s. Thus we get:
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| − | $$ \int_{-\infty}^{\infty} ... \int_{-\infty}^{\infty} f(t) g(x-t) dt_1 ... dt_n $$
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| − | $$ = \int_{\mathbb{R}^n } f(t) g(x-t) dt = (g*f)(x) $$
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| − | This proves the claim. \( \square \)
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| − | == Part b) ==
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| − | Show that:
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| − | $$ \widehat{f*g}(k) = (2\pi)^{n/2} \hat f (k) \hat g (k) $$
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| − | === Solution part b) ===
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| − | The definition of the fourier coefficient is:
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| − | $$ \widehat{f*g}(k) = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } (f*g)(x) e^{-i<k,x>} dx $$
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| − | And with the defintion of the convolution:
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| − | $$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) dy \; e^{-i<k,x>} dx $$
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| − | Because \( e^{-i<k,x>} \) is independent of y, we can put it inside the inner integral.
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| − | $$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x>} dy \: dx $$
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| − | Now we can make a similar substitution as in part a)
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| − | $$ x = t+y $$
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| − | $$ t = x-y $$
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| − | $$ dx = dt $$
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| − | $$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(t) g(y) e^{-i<k,t+y>} dy \: dt $$
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| − | With the bilinearity of the scalar product we can write \(<k,t+y>\) as \(<k,t> + <k,y>\). We then get
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| − | $$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(t) g(y) e^{-i<k,t>-i<k,y>} dy \: dt $$
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| − | $$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(t) e^{-i<k,t>} g(y) e^{-i<k,y>} dy \: dt $$
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| − | As we can see, \( f(t) e^{-i<k,t>} \) is independent of y and \(g(y) e^{-i<k,y>}\) is independent of t. Therefore we can write it as two seperate integrals:
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| − | $$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } f(t) e^{-i<k,t>} dt \: \int_{\mathbb{R}^n } g(y) e^{-i<k,y>} dy $$
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| − | $$ = (2\pi)^{n/2} \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } f(t) e^{-i<k,t>} dt \: \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } g(y) e^{-i<k,y>} dy $$
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| − | $$ = (2\pi)^{n/2} \hat f (k) \hat g (k) \; \square $$
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| − | == Part c) ==
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| − | Let \( 0 < c < \infty\), and define \( F_c(x) = e^{-cx^{2}} \).
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| − | Compute \(\widehat{F_c}(k)\) by first showing that \(2c\phi'(k) + k\phi(k) = 0\), with \(\phi = \widehat{F_c}\), and then solving the differential equation.
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| − | === '''Solution''' ===
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| − | We prove that \(2c\widehat{F_c}'(k) + k\widehat{F_c}(k) = 0\). First we try to bring \(\widehat{F_c}'(k)\) into a form that is related to \(\widehat{F_c}(k)\):
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| − | We show two ways to prove that we can interchange \(\frac{d}{dk}\) and \(\int_{-\infty}^{\infty}\). For the exam I suggest you use the second way for it is shorter:
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| − | ''Proof 1'':
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| − | '''Dominated convergence theorem''':
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| − | If \( 1) f_n\) is a sequence of integrable functions with \(\lim\limits_{n \rightarrow \infty}{f_n(x)} = f(x) \)
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| − | \( 2) \exists g\) an integrable function with \(| f_n(x) | \le g(x) \forall n \in \mathbb{N}, x \in \mathbb{R} \)
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| − | \( \Rightarrow f\) is integrable and \(\lim\limits_{n \rightarrow \infty}{\int_{\mathbb{R}} f_n(x) dx} = \int_{\mathbb{R}} \lim\limits_{n \rightarrow \infty}{f_n(x)} dx = \int_{\mathbb{R}} f(x) dx \)
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| − | $$ \frac{d}{dk} \widehat{F_c}(k) = \lim\limits_{h \rightarrow 0}{\frac{\widehat{F_c}(k+h)-\widehat{F_c}(k)}{h}} = \lim\limits_{n \rightarrow \infty}{\frac{\widehat{F_c}(k+h_n)-\widehat{F_c}(k)}{h_n}} , \lim\limits_{n \rightarrow \infty}{h_n} = 0 $$
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| − | Let \(l(k) := e^{-ikx} , l'(k) = -ix e^{-ikx} \)
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| − | $$ \frac{\widehat{F_c}(k+h_n)-\widehat{F_c}(k)}{h_n} = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \frac{e^{-i(k+h_n)x}-e^{-ikx}}{h_n} e^{-cx^2} dx = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \frac{l(k+h_n)-l(k)}{h_n} e^{-cx^2} dx $$
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| − | Define \(f_n = \frac{l(k+h_n)-l(k)}{h_n} e^{-cx^2} = l'(k+\xi) e^{-cx^2} \) for \( \xi \in (0,h_n) \) (by the mean value theorem)
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| − | Now, let us check the conditions of the dominated convergence theorem:
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| − | $$ 1) \lim\limits_{n \rightarrow \infty}{f_n(x)} = l'(k) e^{-cx^2} = -ix e^{-ikx} e^{-cx^2} =: f(x) $$
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| − | $$ 2) | f_n (x) | = | l'(k+\xi) e^{-cx^2} | = | -ix e^{-i(k+\xi)x} e^{-cx^2} | \le |x| e^{-cx^2} = g(x) $$
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| − | $$ \Rightarrow \lim\limits_{n \rightarrow \infty}{\int_{\mathbb{R}} f_n(x) dx} = \int_{\mathbb{R}} \lim\limits_{n \rightarrow \infty}{f_n(x)} dx = \int_{\mathbb{R}} f(x) dx = \int_{\mathbb{R}} -ix e^{-ikx} e^{-cx^2} dx $$
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| − | $$ \Rightarrow \frac{d}{dk} \widehat{F_c}(k) = \lim\limits_{n \rightarrow \infty}{\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \frac{e^{-i(k+h_n)x}-e^{-ikx}}{h_n} e^{-cx^2} dx} = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} -ix e^{-ikx} e^{-cx^2} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx \square $$
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| − | ''Proof 2'':
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| − | We want to show that \(\frac{d}{dk}(\sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty} e^{-cx^{2}}e^{-ikx}dx) = \sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty} \frac{d}{dk}(e^{-cx^{2}}e^{-ikx})dx\). The lecture notes tell us that \(\widehat{xψ}(k)=i\frac{d}{dk}\widehat{ψ}(k)\). Now, taking a close look at the proposition yields:
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| − | $$ LHS = \frac{d}{dk} \widehat{e^{−cx^{2}}}(k) $$
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| − | $$ RHS = \sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty} -ixe^{-cx^{2}}e^{-ikx}dx = -i\widehat{xe^{-cx^{2}}} \square $$
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| − | Now let us start with the calculation:
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| − | $$ \frac{d}{dk} \widehat{F_c}(k) = \frac{d}{dk} (\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x)e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) \frac{d}{dk} e^{-ikx} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx $$
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| − | Now we use partial integration to rearrange the integral. We integrate \(xe^{-cx^{2}}\) and derive \(e^{-ikx}\). We get:
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| − | $$ \frac{-i}{\sqrt{2\pi}} ([-\frac{e^{-cx^{2}}}{2c}e^{-ikx}]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{e^{-cx^{2}}}{2c}ike^{-ikx} dx) $$
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| − | The first term vanishes. It follows:
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| − | $$ \frac{d}{dk} \widehat{F_c}(k) = -\frac{k}{2c\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}e^{-ikx} dx = -\frac{k}{2c} \widehat{F_c}(k) $$
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| − | This gives us:
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| − | $$ 2c\widehat{F_c}'(k) + k\widehat{F_c}(k) = -k\widehat{F_c}(k) + k\widehat{F_c}(k) = 0 $$
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| − | Now we solve the differential equation \(2c\phi'(k) + k\phi(k) = 0\) by seperation of variables:
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| − | $$ 2c\phi'(k) + k\phi(k) = 0 $$
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| − | $$ \phi' = -\frac{k}{2c}\phi $$
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| − | $$ \frac{d\phi}{dk} = -\frac{k}{2c}\phi $$
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| − | $$ \int \frac{d\phi}{\phi} = -\frac{1}{2c} \int k dk $$
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| − | $$ ln(\phi) = -\frac{k^{2}}{4c} + p $$
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| − | $$ e^{ln(\phi)} = e^{-\frac{k^{2}}{4c} + p} $$
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| − | $$ \phi = re^{-\frac{k^{2}}{4c}} $$
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| − | while \(p\) and \(r\) are integration constants and \(r = e^{p}\). To evaluate the constant \(r\) we first create a boundary condition:
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| − | $$ \widehat{F_c}(0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}} dx $$
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| − | With the substitution \(x = \frac{t}{\sqrt{c}}\) we get \(dx = \frac{dt}{\sqrt{c}}\) and thus:
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| − | $$ \widehat{F_c}(0) = \frac{1}{\sqrt{2\pi c}} \int_{-\infty}^{\infty} e^{-t^{2}} dt = \frac{1}{\sqrt{2\pi c}} \sqrt{\pi} = \frac{1}{\sqrt{2c}} $$
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| − | using the Gaussian Integral. It follows:
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| − | $$ \phi(0) = \widehat{F_c}(0) $$
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| − | $$ r = \frac{1}{\sqrt{2c}} $$
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| − | and thus \(\phi(k) = \widehat{F_c}(k) = \frac{1}{\sqrt{2c}}e^{-\frac{k^{2}}{4c}}\).
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| − | == Part d) ==
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| − | Show that \(F_a*F_b = αF_c\), where \(c = c(a,b)\) and \(α = α(a,b)\).
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| − | === '''Solution''' ===
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| − | We show two ways to solve this problem. For the exam I suggest you use the second way for it is shorter:
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| − | ''Calculation 1'':
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| − | $$ (F_a*F_b)(x) = \int_{-\infty}^{\infty} F_a(x-y)F_b(y) dy = \int_{-\infty}^{\infty} e^{-a(x-y)^{2}}e^{-by^{2}} dy = \int_{-\infty}^{\infty} e^{-a(x^{2}-2xy+y^{2})}e^{-by^{2}} dy = e^{-ax^{2}} \int_{-\infty}^{\infty} e^{2axy-(a+b)y^{2}}dy $$
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| − | Using Completing the square we get:
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| − | $$ 2axy-(a+b)y^{2} = -(a+b)(y^{2}-\frac{2axy}{a+b}) = -(a+b)((y-\frac{ax}{a+b})^{2}-\frac{a^{2}x^{2}}{(a+b)^{2}}) $$
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| − | Therefore:
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| − | $$ e^{-ax^{2}} \int_{-\infty}^{\infty} e^{2axy-(a+b)y^{2}}dy = e^{-ax^{2}} \int_{-\infty}^{\infty} e^{-(a+b)((y-\frac{ax}{a+b})^{2}}e^{\frac{a^{2}x^{2}}{a+b}}dy = e^{-ax^{2}}e^{\frac{a^{2}x^{2}}{a+b}} \int_{-\infty}^{\infty} e^{-(a+b)((y-\frac{ax}{a+b})^{2}}dy $$
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| − | Substituting \(s = y-\frac{ax}{a+b}\) and solving the Gaussian integral gives us:
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| − | $$ (F_a*F_b)(x) = e^{-(a-\frac{a^{2}}{a+b})x^{2}} \int_{-\infty}^{\infty} e^{-(a+b)z^{2}}dz = \sqrt{\frac{\pi}{a+b}}e^{-(a-\frac{a^{2}}{a+b})x^{2}} = αF_c(x) $$
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| − | with \(α(a,b) = \sqrt{\frac{\pi}{a+b}}\) and \(c(a,b) = a-\frac{a^{2}}{a+b}\). The equation is valid since \(c = a-\frac{a^{2}}{a+b} > 0 \). This is because \(\frac{a^{2}}{a+b} < a \) for \(a, b > 0\).
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| − | ''Calculation'' 2:
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| − | In 1.b), we showed that for \(f,g∈L^1(\mathbb{R}^n)\), the Fourier transform of the convolution is \(\widehat{f∗g}(k) = (2\pi)^{\frac{n}{2}}\widehat{f}(k)\widehat{g}(k)\). Here, we have \(n = 1\) and can thus write \(\widehat{F_a∗F_b}(k) = \sqrt{2\pi}\widehat{F_a}(k)\widehat{F_b}(k)\). We insert the Fourier transform of \(F_a\) and \(F_b\), which we found in 1.c):
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| − | $$ \widehat{F_a∗F_b}(k) = \sqrt{2\pi}(\frac{1}{\sqrt{2a}}e^{-\frac{k^{2}}{4a}}\frac{1}{\sqrt{2b}}e^{-\frac{k^{2}}{4b}}) = \sqrt{\frac{\pi}{2ab}}e^{-\frac{k^{2}}{4}(\frac{1}{a}+\frac{1}{b})} = \sqrt{\frac{\pi}{a+b}}\sqrt{\frac{1}{2\frac{ab}{a+b}}}e^{-\frac{k^{2}}{4}\frac{a+b}{ab}} = α\widehat{F_c} $$
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| − | with \(α(a,b) = \sqrt{\frac{\pi}{a+b}}\) and \(c(a,b) = a-\frac{a^{2}}{a+b}\). The equation is valid since \(c = \frac{ab}{a+b} > 0 \). This is because \(a, b > 0\).
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| − | Because of the linearity of the Fourier transform, we can conclude: \((F_a∗F_b)(x)=αF_c(x)\).
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| − | == Part e) ==
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| − | Let \(g \in L^{1}(\mathbb{R})\), \(u \in L^{1}(\mathbb{R}) \cap C^{2}(\mathbb{R})\). Find a solution of the differential equation
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| − | $$ u^{(2)}(x) = u(x) - 2g(x) $$
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| − | using Fourier transform.
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| − | ''Hint:'' consider the Fourier transform of \(f(x) = e^{-|x|}\), \(x \in \mathbb{R}\).
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| − | === '''Solution''' ===
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| − | We first calculate the Fourier transform of \(f(x) = e^{-|x|}\), \(x \in \mathbb{R}\):
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| − | $$ \widehat{f}(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-|x|}e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} (\int_{-\infty}^{0} e^{-|x|}e^{-ikx} dx + \int_{0}^{\infty} e^{-|x|}e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} (\int_{-\infty}^{0} e^{x}e^{-ikx} dx + \int_{0}^{\infty} e^{-x}e^{-ikx} dx) $$
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| − | $$ = \frac{1}{\sqrt{2\pi}} (\int_{-\infty}^{0} e^{x(1-ik)} dx + \int_{0}^{\infty} e^{-x(1+ik)} dx) = \frac{1}{\sqrt{2\pi}} ([\frac{e^{x(1-ik)}}{1-ik}]_{-\infty}^{0} + [\frac{e^{-x(1+ik)}}{-(1+ik)}]_{0}^{\infty}) = \frac{1}{\sqrt{2\pi}} (\frac{1}{1-ik} - \frac{1}{-(1+ik)}) $$
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| − | $$ = \frac{1}{\sqrt{2\pi}} \frac{1+ik+1-ik}{1-i^{2}k^{2}} = \frac{1}{\sqrt{2\pi}} \frac{2}{1+k^{2}} $$
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| − | Now to the differential equation:
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| − | $$ u^{(2)}(x) = u(x) - 2g(x) $$
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| − | $$ \widehat{u^{(2)}}(k) = \widehat{(u - 2g)}(k) $$
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| − | $$ \widehat{u^{(2)}}(k) = \widehat{u}(k) - 2\widehat{g}(k) $$
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| − | $$ i^{2}k^{2}\widehat{u}(k) = \widehat{u}(k) - 2\widehat{g}(k) $$
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| − | using \(\widehat{(\lambda\psi + \mu\varphi)} = \lambda\widehat{\psi} + \mu\widehat{\varphi}\) and \(\widehat{(\frac{\partial}{\partial x_j}\psi)}(k) = ik_j\widehat{\psi}(k)\) for \(\psi,\varphi \in L^{1}(\mathbb{R})\) and \(\lambda,\mu \in \mathbb{C}\).
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| − | $$ -k^{2}\widehat{u}(k) = \widehat{u}(k) - 2\widehat{g}(k) $$
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| − | $$ \widehat{u}(k) (1 + k^{2}) = 2\widehat{g}(k) $$
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| − | $$ \widehat{u}(k) = \frac{2}{1+k^{2}}\widehat{g}(k) = \frac{\sqrt{2\pi}}{\sqrt{2\pi}} \frac{2}{1+k^{2}}\widehat{g}(k) = \sqrt{2\pi}\widehat{f}(k)\widehat{g}(k) $$
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| − | using the identity for \(\widehat{f}(k)\) evaluated in the first part of the exercise. Applying part b) we get:
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| − | $$ \widehat{u}(k) = \sqrt{2\pi} \left( \widehat{f}(k)\widehat{g}(k) \right) = \widehat{f * g}(k) $$
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| − | and the rest is done in silence:
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| − | $$ u(x) = (f * g)(x) = \int_{\mathbb{R}} f(x - y) g(y) \, dy $$
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