# Laplace Transformations

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.

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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

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 $$f(t) = \frac{1}{2} + \frac{1}{2}e^{2t} = \frac{{e^{2t} + 1}}{2}.$$

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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.

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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