91Ó°ÊÓ

Use the Laplace transform to solve the boundary value problem $$ \begin{aligned} \frac{\partial^{2} u}{\partial t^{2}} &=\frac{\partial^{2} u}{\partial x^{2}}+\sin \pi x, \quad 00 \\ u(0, t) &=0, \quad u(1, t)=0, t>0 \\ u(x, 0) &=0, \quad \frac{\partial u}{\partial t}(x, 0)=0, x>0 \end{aligned} $$

Evaluate the inverse Laplace transform of the given function. $$ F(s)=\frac{e^{-s}}{s^{3 / 2}} $$

Solve the given initial value problem. $$ y^{\prime \prime}+2 y^{\prime}+y=3 \delta_{0}(t-2), \quad y(0)=1, y^{\prime}(0)=0 $$

Express the solution of the given initial value problem as a convolution. $$ y^{\prime \prime}+y^{\prime}+y=f(t), \quad y(0)=0, y^{\prime}(0)=0 $$

A Gaussian function. Use the definitions of the Laplace transform and the complementary error function to show that. $$ \mathcal{L}\left(\frac{1}{a \sqrt{\pi}} e^{-t^{3} / 4 a^{2}}\right)=e^{a^{2} s^{2}} \operatorname{erfc}(a s) $$

Solution of Bessel's equation via the Laplace transform. (s) Show that after applying the Laplace transform to Bessel's equation of order 0, we obtain $$-\frac{d}{d s}\left[y^{2} Y-s y(0)-y^{\prime}(0)\right]+s Y-y(0)-Y^{\prime}=0$$ (b) Simplify the equation in \(Y\) to obtain \(\left(s^{2}+1\right) Y^{\prime \prime}=-8 Y\). (c) Solve the first order ordinary differential equation in \(Y\), and obtain that \(Y=\) \(\frac{K}{\sqrt{8^{2}+1}}\), where \(K\) is a constant. (d) Use the result of Exercise 50 to identify your solution as \(K J_{0}(t)\). You may wonder why the method did not yield a second solution, as expected from a second order differeatial equation. The reason is that by applying the Laplace transform. we are assuming that the (unknown) solution and its first and second derivatives have a Laplace transform. Since the second solution of Bessel's equation, \(Y_{0}\), is known to behave like Int for \(t\) near zero (Section 4.7, Exercise 28), the term ty" behaves like a constant multiple of \(\frac{1}{t}\) and does not have a Laplace transform, which explains why the method did not capture this second solution.