Problem 27
Question
We use $$U(x, s)=c_{1} e^{-\sqrt{s+h} x}+c_{2} e^{\sqrt{s+h} x}$$ The condition \(\lim _{x \rightarrow \infty} u(x, t)=0\) implies \(\lim _{x \rightarrow \infty} U(x, s)=0,\) so we take \(c_{2}=0 .\) Therefore $$U(x, s)=c_{1} e^{-\sqrt{s+h} x}$$ The Laplace transform of \(u(0, t)=u_{0}\) is \(U(0, s)=u_{0} / s\) and so $$U(x, s)=u_{0} \frac{e^{-\sqrt{s+h} x}}{s}$$ and $$u(x, t)=u_{0} \mathscr{L}^{-1}\left\\{\frac{e^{-\sqrt{s+h} x}}{s}\right\\}=u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s} e^{-\sqrt{s+h} x}\right\\}$$ From the first translation theorem, $$\mathscr{L}^{-1}\left\\{e^{-\sqrt{s+h} x}\right\\}=e^{-h t} \mathscr{L}^{-1}\left\\{e^{-x \sqrt{s}}\right\\}=e^{-h t} \frac{x}{2 \sqrt{\pi t^{3}}} e^{-x^{2} / 4 t}$$ Thus, from the convolution theorem we obtain $$u(x, s)=\frac{u_{0} x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{e^{-h \tau-x^{2} / 4 \tau}}{\tau^{3 / 2}} d \tau$$
Step-by-Step Solution
Verified Answer
The solution for \( u(x, t) \) is \( \frac{u_{0} x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{e^{-h \tau - x^{2}/4 \tau}}{\tau^{3 / 2}} d\tau \).
1Step 1: Setting up the problem
We begin with the expression for \( U(x, s) \) as given: \( U(x, s) = c_{1} e^{-\sqrt{s+h} x} + c_{2} e^{\sqrt{s+h} x} \). The condition \( \lim_{x \rightarrow \infty} U(x, s) = 0 \) implies that \( c_{2} = 0 \), leaving us with \( U(x, s) = c_{1} e^{-\sqrt{s+h} x} \).
2Step 2: Apply initial condition at x=0
The Laplace transform of \( u(0, t) = u_{0} \) gives us \( U(0, s) = \frac{u_{0}}{s} \). By substituting \( x = 0 \) in \( U(x, s) = c_{1} e^{-\sqrt{s+h} x} \), we get \( U(0, s) = c_{1} = \frac{u_{0}}{s} \).
3Step 3: Substitute and rearrange
Substitute \( c_{1} = \frac{u_{0}}{s} \) back into the expression for \( U(x, s) \) to get \( U(x, s) = \frac{u_{0}}{s} e^{-\sqrt{s+h} x} \).
4Step 4: Inverse Laplace Transform
We are given \( u(x, t) = u_{0} \mathscr{L}^{-1}\left\{ \frac{1}{s} e^{-\sqrt{s+h} x} \right\} \). Use the inverse Laplace transform, particularly the first translation theorem, to derive that \( \mathscr{L}^{-1}\left\{ e^{-\sqrt{s+h} x} \right\} = e^{-h t} \mathscr{L}^{-1}\left\{ e^{-x \sqrt{s}} \right\} \).
5Step 5: Applying the known Laplace Transform Formula
The inverse Laplace transform of \( e^{-x \sqrt{s}} \) is known as \( \frac{x}{2\sqrt{\pi t^3}} e^{-x^2 / 4t} \). Therefore, we have \( \mathscr{L}^{-1}\left\{ e^{-x \sqrt{s}} \right\} = \frac{x}{2\sqrt{\pi t^3}} e^{-x^2 / 4t} \).
6Step 6: Final Solution Using Convolution Theorem
Using the convolution theorem, we integrate over \( \tau \) to get the final expression: \( u(x, t) = \frac{u_{0} x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{e^{-h \tau - x^{2}/4 \tau}}{\tau^{3 / 2}} d\tau \) as the solution for \( u(x, t) \).
Key Concepts
Inverse Laplace TransformConvolution TheoremTranslation TheoremDifferential Equations
Inverse Laplace Transform
The Inverse Laplace Transform is a method used to convert functions in the complex frequency domain back into the time domain. Whenever we have taken the Laplace Transform of a function to make solving differential equations easier, the inverse process allows us to revert to the function in its original domain.
This process is particularly crucial when dealing with complex systems and stability analysis.
For instance, in our exercise, we apply the inverse Laplace transform to find the time-dependent function from its Laplace transform representation. It's apparent in the step where we find \( u(x, t) = u_{0} \mathscr{L}^{-1}\left\{ \frac{1}{s} e^{-\sqrt{s+h} x} \right\} \). We utilize known inverse transforms like \( \mathscr{L}^{-1}\left\{ e^{-x \sqrt{s}} \right\} \), making this approach systematic and efficient.
This process is particularly crucial when dealing with complex systems and stability analysis.
For instance, in our exercise, we apply the inverse Laplace transform to find the time-dependent function from its Laplace transform representation. It's apparent in the step where we find \( u(x, t) = u_{0} \mathscr{L}^{-1}\left\{ \frac{1}{s} e^{-\sqrt{s+h} x} \right\} \). We utilize known inverse transforms like \( \mathscr{L}^{-1}\left\{ e^{-x \sqrt{s}} \right\} \), making this approach systematic and efficient.
Convolution Theorem
The Convolution Theorem is an essential tool in the realm of transforms, including Laplace Transforms. It describes a way to transform the convolution of two functions from the time domain into simple multiplication in the Laplace domain. This greatly simplifies operations involving convolution, which can be complicated to calculate directly.
In practice, convolution allows us to integrate over a parameter to include the effects of distributed processes over time or space.
A straightforward example from our problem is obtaining the final form of \( u(x, t) \) by utilizing the convolution theorem. We integrated over \( \tau \), which effectively folded one function over another, thereby concluding with a neatly integrated expression for \( u(x, t) \). The result, \( \frac{u_{0} x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{e^{-h \tau - x^{2}/4 \tau}}{\tau^{3 / 2}} d\tau \), beautifully illustrates how convolution simplifies otherwise complex calculations.
In practice, convolution allows us to integrate over a parameter to include the effects of distributed processes over time or space.
A straightforward example from our problem is obtaining the final form of \( u(x, t) \) by utilizing the convolution theorem. We integrated over \( \tau \), which effectively folded one function over another, thereby concluding with a neatly integrated expression for \( u(x, t) \). The result, \( \frac{u_{0} x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{e^{-h \tau - x^{2}/4 \tau}}{\tau^{3 / 2}} d\tau \), beautifully illustrates how convolution simplifies otherwise complex calculations.
Translation Theorem
The Translation Theorem, also known as the shifting theorem in Laplace transforms, allows us to shift functions in the time domain by adjusting their Laplace Transform. When applicable, it provides an elegant way to handle exponential shifts in s, which corresponds to time-domain translations or delays.
This theorem is prominently used in our exercise to handle the term \( e^{-\sqrt{s+h} x} \). By using the translation theorem, we break down this expression into simple parts which can be inversely transformed easily. This theorem simplifies handling equations that involve shifts, particularly beneficial when tackling differential equations or control systems with delays or shifts in time.
This theorem is prominently used in our exercise to handle the term \( e^{-\sqrt{s+h} x} \). By using the translation theorem, we break down this expression into simple parts which can be inversely transformed easily. This theorem simplifies handling equations that involve shifts, particularly beneficial when tackling differential equations or control systems with delays or shifts in time.
Differential Equations
Differential equations play a significant role in modeling various dynamic systems, whether they be mechanical, electrical, hydraulic, or thermal aspects of natural and engineered systems. These equations relate a function with its derivatives, representing phenomena such as change, motion, or diffusivity.
In the context of our example problem, differential equations govern the behavior of a system modeled by the function \( u(x, t) \). By applying Laplace transforms to these equations, they become algebraic equations. Once solved, inverse transforms revert them back to their original domain, allowing us to find solutions like \( u(x, t) \). This method provides a powerful mechanism to solve complex differential equations step by step.
Successful application of these concepts, from setting up equations to using transforms, aids in understanding and solving problems within mathematics and physics efficiently.
In the context of our example problem, differential equations govern the behavior of a system modeled by the function \( u(x, t) \). By applying Laplace transforms to these equations, they become algebraic equations. Once solved, inverse transforms revert them back to their original domain, allowing us to find solutions like \( u(x, t) \). This method provides a powerful mechanism to solve complex differential equations step by step.
Successful application of these concepts, from setting up equations to using transforms, aids in understanding and solving problems within mathematics and physics efficiently.
Other exercises in this chapter
Problem 24
since erf \((0)=0\) and \(\lim _{x \rightarrow \infty}\) erf \((x)=1,\) we have $$\lim _{t \rightarrow \infty} u(x, t)=50[\text { erf }(0)-\text { erf }(0)]=0$$
View solution Problem 26
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View solution Problem 30
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View solution Problem 23
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