Question

In each exercise, (a) Show by direct substitution that the linear combination of functions is a solution of the given homogeneous linear partial differential equation. (b) Determine values of the constants so that the linear combination satisfies the given supplementary condition. \(u(x, t)=c_{1}+c_{2}(x-t)+c_{3}(x+t), \quad u_{x x}-u_{t t}=0\) \(u(x, 0)=1+2 x, \quad u_{t}(x, 0)=0\)

Short answer

Question: Verify that the linear combination \(u(x, t) = c_{1} + c_{2}(x-t) + c_{3}(x+t)\) is a solution of the homogeneous linear partial differential equation \(u_{xx} - u_{tt} = 0\), and determine the values of the constants \(c_{1}, c_{2},\) and \(c_{3}\), given the supplementary condition \(u(x, 0) = 1 + 2x\) and \(u_t(x, 0) = 0\). Answer: The linear combination \(u(x, t)\) is a solution of the given homogeneous linear partial differential equation. The constants that satisfy the given supplementary condition are \(c_{1} = 1\), \(c_{2} = 1\), and \(c_{3} = 1\).

Step-by-step solution

Verify by Direct Substitution that the Linear Combination Satisfies the Homogeneous Linear PDE

First, we need to find \(u_{xx}\), \(u_{tt}\), and substitute them into the given equation to verify that the linear combination is indeed a solution of the homogeneous linear PDE. Given linear combination: \(u(x, t) = c_{1} + c_{2}(x-t) + c_{3}(x+t)\). Now, find the second-order partial derivatives: \(u_{xx} = \frac{\partial^2 u}{\partial x^2} = 0\) \(u_{tt} = \frac{\partial^2 u}{\partial t^2} = 0\) Now substitute them into the given equation: \(u_{xx} - u_{tt} = 0 - 0 = 0\) Thus the linear combination \(u(x, t)\) is a solution of the given homogeneous linear PDE.

Determine the values of the constants to satisfy the supplementary condition

Now we will find \(u(x, 0)\) and \(u_t(x, 0)\), and then solve for the constants \(c_{1}, c_{2},\) and \(c_{3}\) to satisfy the given supplementary condition. First, find \(u(x, 0)\) and \(u_t(x, 0)\): \(u(x, 0) = c_{1} + c_{2}(x-0) + c_{3}(x+0) = c_{1} + c_{2}x + c_{3}x\) \(u_t(x, 0) = \frac{\partial u}{\partial t}|_{t=0} = -c_{2} + c_{3}\) Now, we will use the given supplementary condition: \(u(x, 0) = 1 + 2x\) \(u_t(x, 0) = 0\) Compare coefficients: \(c_{1} = 1\) \(c_{2} + c_{3} = 2\) \(-c_{2} + c_{3} = 0\) Solving for the constants, we find: \(c_{1} = 1\) \(c_{2} = 1\) \(c_{3} = 1\) Therefore, the constants that satisfy the given supplementary condition are \(c_{1} = 1\), \(c_{2} = 1\), and \(c_{3} = 1\).

Key concepts

Homogeneous Linear PDE

A partial differential equation (PDE) is termed 'homogeneous' if all terms contain the unknown function or its derivatives. The equation has no additional terms or "sources" added.
This concept means that the equation can be set equal to zero on one side. In simpler terms, it balances itself without a need for an external "force" to keep the equation equal.
When dealing with linear PDEs, especially homogeneous ones, we're essentially focusing on finding combinations of functions that solve the equation naturally. Here, linearity refers to the highest degree of derivatives or functions involved being one.
For example, in the PDE \(u_{xx} - u_{tt} = 0\), we're looking for functions \(u(x, t)\) such that their derivatives equate beautifully to zero. This symmetry is vital for understanding and predicting phenomena like wave behavior, heat diffusion, and more.
Homogeneous linear PDEs are significant in mathematical physics and engineering as they often describe idealized physical systems.

Direct Substitution Method

The direct substitution method is a practical approach to verify if a given function is a solution to a PDE. It involves directly plugging a proposed solution into the differential equation and checking if the equation holds true.
This method saves time and computation effort, ensuring that we don't need to "reinvent the wheel" when an obvious solution already exists.
In our exercise, we needed to verify that the function \(u(x, t) = c_{1} + c_{2}(x-t) + c_{3}(x+t)\) satisfies the PDE \(u_{xx} - u_{tt} = 0\).
This verification involved calculating the second-order partial derivatives:

• \(u_{xx} = 0\)
• \(u_{tt} = 0\)

By substituting these into the PDE, we confirm the solution satisfies the equation because both terms sum to zero.
If the substitution holds, as seen here, we have successfully demonstrated that the given function is, indeed, a solution.

Supplementary Condition

Supplementary conditions are additional constraints or boundaries needed to find specific solutions for differential equations. These are vital, especially in physical applications, to match the general equation to specific situations.
In mathematics and physics, such conditions might represent initial or boundary conditions which describe the system's state at a particular time or space.
In our exercise, the supplementary conditions were given by \(u(x, 0) = 1 + 2x\) and \(u_t(x, 0) = 0\).
These conditions were critical in determining the constants \(c_1\), \(c_2\), and \(c_3\) so that the chosen solution for the PDE would satisfy the real-world boundary or initial requirements of the problem.
By comparing and equating coefficients from these conditions, we precisely determined the constant values needed to tailor the solution to the specific parameters the problem described.

Solution Verification

Solution verification is the final step to ensure everything checks out in the problem statement.
It's like double-checking the math work after balancing both sides of an equation.
In this exercise, after calculating the constants \(c_1 = 1\), \(c_2 = 1\), and \(c_3 = 1\), it's essential to re-substitute these values back into the assumed solution to confirm everything aligns with both the PDE and the supplementary conditions given.
This includes checking both the initial function \(u(x, t) = 1 + (x-t) + (x+t)\) simplifies correctly to meet the conditions \(u(x,0) = 1 + 2x\) and \(u_t(x,0) = 0\).
Successful verification guarantees that the solution is not only mathematically sound but also contextually appropriate for the physical scenario or theoretical framework it models.
In essence, this ensures no detail is overlooked and confirms the reliability of the solution obtained.