Laplace Transforms — Question 10

Question 10 – Solution

Strategy. Multiplication by time and differentiation in time correspond to operations in $s$ that must be applied in the correct order.

Step 1: Compare the two orders. Since $(tf)(0)=0$, applying the time-derivative rule after time multiplication gives $$\boxed{\mathcal L\{(tf)'\}=-sF'(s).}$$ In the opposite order, parameter differentiation applies to the entire transform of $f'$: $$\boxed{\mathcal L\{tf'\}=-\frac d{ds}[sF(s)-f(0)]=-F(s)-sF'(s).}$$ Their difference is $F(s)$, exactly as required by $(tf)'-tf'=f$. The initial value $f(0)$ is constant in $s$ and differentiates to zero.

Step 2: Check a concrete example. For $f=e^{-t}$, $F=1/(s+1)$ and $F'=-1/(s+1)^2$. The formulas predict $$\mathcal L\{(1-t)e^{-t}\}=\frac{s}{(s+1)^2},\qquad \mathcal L\{-te^{-t}\}=-\frac 1{(s+1)^2}.$$ Direct use of the exponential and $te^{-t}$ integrals verifies both, for $s>-1$. Their difference is $1/(s+1)=F$.

Step 3: Differentiate the complete expression twice. Multiplication by $t^2$ corresponds to two parameter derivatives with positive sign. Thus $$\mathcal L\{t^2f''\}=\frac{d^2}{ds^2}[s^2F-sf(0)-f'(0)] =\boxed{2F+4sF'+s^2F''.}$$ The first derivative of $s^2F$ is $2sF+s^2F'$; differentiating again produces both product-rule contributions. The initial terms are at most linear in $s$, so their second derivatives vanish.

Step 4: Diagnose and verify. The proposed $s^2F''$ omits $2F+4sF'$. For $f=e^{-t}$, the correct result simplifies to $$\frac 2{s+1}-\frac{4s}{(s+1)^2}+\frac{2s^2}{(s+1)^3} =\boxed{\frac 2{(s+1)^3}}.$$ Since $f''=e^{-t}$, direct integration of $t^2e^{-(s+1)t}$ gives this same value for $s>-1$. Differentiating only $F$ while treating its prefactors as constants is the source of the error.