Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

can someone kindly help me with these few questions? :)

Find $L{e^tf(t)}$ in terms of f*(s) and state a range of s which this is defined.

I couldn't figure this out. You use the definition but i then get $e^t(1-s)f(t)$ and then i dont know what to do....

Using the Convolution Theorem, find the function f(t) satisfying the equation

$$f(t) = \int_0^t e^u f(t-u)\mathrm du + e^t$$

I know how to take the LT of both sides. Im not sure how to figure out the LT of the integral though. Ive for f(s) = something + 1/(s-1) at the moment.

share|improve this question
    
Hint: let $u = s - 1$ –  badatmath May 21 '11 at 20:31
    
user4645: So did you intend $f(t) = \int_0^t {ue^u f(t - u)du} + e^t $? –  Shai Covo May 23 '11 at 13:43

2 Answers 2

up vote 1 down vote accepted

Here is a detailed solution to the modified -- substantially more challenging -- problem (see the OP's comments below the previous answer; in particular, it is stated there that this is not homework).

To find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation $$ f(t) = \int_0^t {u e^u f(t - u)\,du} + e^t, \;\; t \geq 0, $$ begin, as in the previous answer, by writing $$ \hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1, $$ where this time $\hat \varphi$ is the Laplace transform of the convolution $$ (f*te^t )(t) = \int_0^t {ue^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)(t-u)e^{t - u}\,du}\bigg). $$ By the convolution theorem, $$ \hat \varphi(s) = \hat f(s) \frac{1}{{(s - 1)^2 }}. $$ It is worth noting that the term $1/(s-1)^2$ can be derived as follows, recalling that an exponential random variable with density function $\lambda e^{-\lambda t}$, $t \geq 0$, has mean equal to $1/\lambda$ (here $\lambda = s -1 > 0$): $$ \int_0^\infty {e^{ - st} te^t \,dt} = \frac{1}{{s - 1}}\int_0^\infty {t(s - 1)e^{ - (s - 1)t} \,dt} = \frac{1}{{(s - 1)^2 }}. $$ Solving for $\hat f(s)$ (using the above expression for $\hat \varphi(s)$) gives $$ \hat f(s) = \frac{{s - 1}}{{s^2 - 2s}} = \frac{{(s - 2) + 1}}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{s}\frac{1}{{s - 2}}. $$ Assuming that $s > 2$, it follows by inversion (and the Convolution Theorem) that $$ f(t) = 1 + (1 * e^{2t})(t),\;\; t \geq 0. $$ (Indeed, note that $\int_0^\infty {e^{ - st} 1\,dt} = \frac{1}{s}$ and $\int_0^\infty {e^{ - st} e^{2t} \,dt} = \frac{1}{{s - 2}}$.) Finally, from $$ (1 * e^{2t})(t) = \int_0^t {e^{2u} 1\,du} \,\bigg( = \int_0^t {1e^{2(t - u)} \,du} \bigg) , $$ it follows that $$ f(t) = 1 + \frac{{e^{2t} - 1}}{2} = \frac{{e^{2t} + 1}}{2},\;\; t \geq 0. $$ Indeed, this $f$ satisfies the original equation; that is, as one can easily verify, it holds $$ \frac{{e^{2t} + 1}}{2} = \int_0^t {ue^u \frac{{e^{2(t - u)} + 1}}{2}\,du} + e^t . $$

EDIT (in response to the OP's comment below). While inverting $1/s$ gives $1$ and inverting $1/(s-2)$ gives $e^{2t}$, inverting $1/(s(s-2))$ does not give the product of $e^{2t}$ and $1$; rather, by the Convolution Theorem, it gives the convolution of $e^{2t}$ and $1$. Since $(1 * e^{2t})(t) = \frac{{e^{2t} - 1}}{2}\,( = \int_0^t {e^{2u} \,du} )$, $$ f(t) = 1 + \frac{{e^{2t} - 1}}{2} = \frac{{e{}^{2t} + 1}}{2} $$ (which was verified by substitution into the original equation). However, as the OP observed, the solution can be obtained more elementarily by splitting $\hat f(s)$ into partial fractions. Specifically, $$ \hat f(s) = \frac{{s - 1}}{{s^2 - 2s}} = \frac{1}{{2s}} + \frac{1}{{2(s - 2)}}, $$ from which it follows, by inversion, that $$ f(t) = \frac{1}{2} + \frac{1}{2}e^{2t} = \frac{{e^{2t} + 1}}{2}. $$

share|improve this answer
    
I finally understood all the steps except the last few. Inverting 1/s gives 1 which i understood. Inverting 1/(s-2) gives e^2t which I also understood. Shouldn’t the answer just be f(t) = 1 + (e^2t x 1) = 1 + e^2t? Or because it’s a product you have to use the Convolution Theorem again? Could you not just split everything up into partial fractions i.e. split up the 1/s(s+1)? Quick question. You wrote s > 1 in line 2 (the new answer). Is the 'bottom s bit' always >0? –  user4645 May 24 '11 at 10:40
    
user4645, see the EDIT; as for the quick question, consider $\int_0^\infty {e^{ - st} e^{at} \,dt} = \int_0^\infty {e^{ - (s - a)t} \,dt} = \frac{1}{{s - a}}$, for $s > a$. –  Shai Covo May 24 '11 at 12:37
    
okay -completely understood and now im getting most of the Laplace Transformation questions correct!!! Thanks so much. Its really helped! –  user4645 May 25 '11 at 10:35
    
Thanks, glad to help. –  Shai Covo May 25 '11 at 11:44

The solution to the first question follows easily from the definition: Can you bring $\int_0^\infty {e^{ - st} e^t f(t)\,dt}$ into the form $\int_0^\infty {e^{ - s't} f(t) \,dt}$? (Consider the hint you were already given above.)

The solution to the second question follows easily from the Convolution Theorem. You want to find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation $$ f(t) = \int_0^t {e^u f(t - u)\,du} + e^t, \;\; t \geq 0. $$ Your approach is right. Taking Laplace transform on both sides gives $$ \hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1, $$ where $\hat \varphi$ is the Laplace transform of the convolution $$ (f*e^t )(t) = \int_0^t {e^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)e^{t - u}\,du}\bigg). $$ (Note that the upper limit of the integral is $t$ in order to satisfy $t-u \geq 0$.) The Convolution Theorem states that the Laplace transform of a convolution is the product of the Laplace transforms of the individual functions. Hence, $$ \hat \varphi(s) = \hat f(s) \frac{1}{{s - 1}}. $$ Now you can solve for $\hat f(s)$, and, in turn by inversion (assuming that $s > 2$), find $f(t)$.

Note: After you find $f(t)$, verify that it satisfies the original equation. This is very easily done; indeed $f(t)$ is only slightly different from $e^t$ (in form).

EDIT (completing the solution; see Remark below):

Solving for $\hat f(s)$ gives $$ \hat f(s) = \frac{1}{{s - 2}}. $$ Assuming that $s > 2$, it follows by inversion that $f(t)=e^{2t}$. Indeed, this $f$ satisfies the original equation; that is, $$ e^{2t} = \int_0^t {e^u e^{2(t - u)} \,du} + e^t ,\;\; t \geq 0. $$

Remark. In view of the OP's comments below, it appears that the factor $u$ was forgotten in the convolution term of the original equation. A solution to the modified problem has now been posted.

share|improve this answer
    
Thank you for your detailed answer. I understood all the steps and managed to complete it correctly. However, the Laplace transformation of that integral is still really confusing me. You have taken e^u f(t-u) and not ue^u f(t-u) like in the question. Why does the ‘u’ vanish? Is there a step by step way of getting the Laplace transformation of ue^t f(t-u)? I do not get how you changed it. –  user4645 May 22 '11 at 12:52
    
I know the definition – f(s)g(s) = integral of etc. So say we take f(t)=f(t-u) and g(t)=ue^u. so Lf(t)=f(s) and Lg(t)=L(ue^u)=1/s^2(s-1)? If so then I got as a final answer f(t) = e^2t. Is that correct? You said I could check it… But when I plugged it back in I got (–te^t)-(e^2t)!!! :( Btw these are exam question not homework questions – I always get stuck on this part! –  user4645 May 22 '11 at 12:54
    
user4645, the $u$ did not vanish; it does not appear in the original question asking to find the function $f(t)$ satisfying the equation $f(t) = \int_0^t {e^u f(t - u)du} + e^t $. The solution to this equation is $f(t)=e^{2t}$, as you can easily verify by substitution. Perhaps you forgot to add the $u$? –  Shai Covo May 22 '11 at 19:18
    
Anyway, see the new answer, corresponding to the modified version (with $u$ added). –  Shai Covo May 23 '11 at 1:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.