Solve The Heat Equation Given That The Boundary Condition u(0,t)=0 U(1,t)=100
Introduction
The heat equation is a fundamental partial differential equation (PDE) that describes how heat diffuses through a medium over time. It is a crucial equation in various fields, including physics, engineering, and mathematics. In this article, we will discuss how to solve the heat equation with given boundary conditions, specifically the conditions u(0,t) = 0 and u(1,t) = 100.
The Heat Equation
The heat equation is a linear PDE that describes the distribution of heat (or temperature) in a medium over time. It is given by the following equation:
∂u/∂t = α ∂²u/∂x²
where u(x,t) is the temperature at point x and time t, α is the thermal diffusivity of the medium, and ∂/∂t and ∂²/∂x² are the partial derivatives with respect to time and space, respectively.
Boundary Conditions
Boundary conditions are essential in solving the heat equation. They specify the temperature at the boundaries of the medium at any given time. In this case, we are given the following boundary conditions:
u(0,t) = 0 (at x = 0) u(1,t) = 100 (at x = 1)
These boundary conditions indicate that the temperature at the left boundary (x = 0) is always 0, and the temperature at the right boundary (x = 1) is always 100.
Separation of Variables
To solve the heat equation with the given boundary conditions, we will use the method of separation of variables. This method involves assuming that the solution can be expressed as a product of two functions, one depending only on x and the other only on t.
Let u(x,t) = X(x)T(t)
Substituting this expression into the heat equation, we get:
X(x)T'(t) = α X''(x)T(t)
Dividing both sides by α X(x)T(t), we get:
T'(t)/α T(t) = X''(x)/X(x)
Since the left-hand side depends only on t and the right-hand side depends only on x, both sides must be equal to a constant, say -λ.
Eigenvalue Problem
We now have the following equation:
X''(x) + λ X(x) = 0
This is an eigenvalue problem, where λ is the eigenvalue and X(x) is the corresponding eigenfunction. The boundary conditions u(0,t) = 0 and u(1,t) = 100 imply that X(0) = 0 and X(1) = 0.
Solving the Eigenvalue Problem
To solve the eigenvalue problem, we need to find the eigenvalues and eigenfunctions that satisfy the boundary conditions. We can do this by using the following trial solution:
X(x) = A sin(√λ x) + B cos(√λ x)
Applying the boundary condition X(0) = 0, we get:
B = 0
Applying the boundary condition X(1) = 0, we get:
A sin(√λ) = 0
Since A ≠ 0, we must have sin(√λ) = 0. This implies that √λ = nπ, where n is an integer.
Eigenvalues and Eigenfunctions
The eigenvalues and eigenfunctions are given by:
λn = n²π² Xn(x) = sin(nπ x)
Time-Dependent Solution
Now that we have the eigenvalues and eigenfunctions, we can find the time-dependent solution. We can do this by substituting the expression for Xn(x) into the equation for T(t).
Tn(t) = e^(-λn α t)
General Solution
The general solution is a superposition of the time-dependent solutions:
u(x,t) = ∑[A_n sin(nπ x) e^(-n²π² α t)]
Boundary Conditions
To find the coefficients A_n, we need to apply the boundary conditions. We can do this by substituting the expression for u(x,t) into the boundary conditions.
Applying the boundary condition u(0,t) = 0, we get:
∑[A_n sin(nπ x) e^(-n²π² α t)] = 0
This implies that A_n = 0 for all n.
Applying the boundary condition u(1,t) = 100, we get:
∑[A_n sin(nπ) e^(-n²π² α t)] = 100
This implies that A_n = 100/∫[sin(nπ) e^(-n²π² α t) dx]
Coefficients A_n
To find the coefficients A_n, we need to evaluate the integral ∫[sin(nπ) e^(-n²π² α t) dx].
Using the fact that sin(nπ) = 0 for all n, we get:
A_n = 0
However, this is not possible, since we know that A_n ≠ 0. This implies that our solution is not valid.
Alternative Solution
To find an alternative solution, we can try a different approach. We can assume that the solution is of the form:
u(x,t) = X(x)T(t) + v(x,t)
where v(x,t) is a particular solution that satisfies the boundary conditions.
Particular Solution
To find the particular solution v(x,t), we can use the following equation:
v(x,t) = 100 - u(x,t)
Substituting this expression into the heat equation, we get:
∂v/∂t = α ∂²v/∂x²
Solving the Heat Equation
To solve the heat equation, we can use the method of separation of variables. We can assume that the solution is of the form:
v(x,t) = X(x)T(t)
Substituting this expression into the heat equation, we get:
X(x)T'(t) = α X''(x)T(t)
Dividing both sides by α X(x)T(t), we get:
T'(t)/α T(t) = X''(x)/X(x)
Eigenvalue Problem
We now have the following equation:
X''(x) + λ X(x) = 0
This is an eigenvalue problem, where λ is the eigenvalue and X(x) is the corresponding eigenfunction. The boundary conditions v(0,t) = 100 and v(1,t) = 0 imply that X(0) = 100 and X(1) = 0.
Solving the Eigenvalue Problem
To solve the eigenvalue problem, we need to find the eigenvalues and eigenfunctions that satisfy the boundary conditions. We can do this by using the following trial solution:
X(x) = A sin(√λ x) + B cos(√λ x)
Applying the boundary condition X(0) = 100, we get:
B = 100
Applying the boundary condition X(1) = 0, we get:
A sin(√λ) + 100 cos(√λ) = 0
Eigenvalues and Eigenfunctions
The eigenvalues and eigenfunctions are given by:
λn = (n-1)²π² Xn(x) = sin((n-1)π x) + 100 cos((n-1)π x)/sin((n-1)π)
Time-Dependent Solution
Now that we have the eigenvalues and eigenfunctions, we can find the time-dependent solution. We can do this by substituting the expression for Xn(x) into the equation for T(t).
Tn(t) = e^(-λn α t)
General Solution
The general solution is a superposition of the time-dependent solutions:
v(x,t) = ∑[A_n (sin((n-1)π x) + 100 cos((n-1)π x)/sin((n-1)π)) e^(-((n-1)²π² α t)]
Boundary Conditions
To find the coefficients A_n, we need to apply the boundary conditions. We can do this by substituting the expression for v(x,t) into the boundary conditions.
Applying the boundary condition v(0,t) = 100, we get:
∑[A_n (sin((n-1)π x) + 100 cos((n-1)π x)/sin((n-1)π)) e^(-((n-1)²π² α t)] = 100
This implies that A_n = 100/∫[sin((n-1)π x) + 100 cos((n-1)π x)/sin((n-1)π) e^(-((n-1)²π² α t) dx]
Coefficients A_n
To find the coefficients A_n, we need to evaluate the integral ∫[sin((n-1)π x) + 100 cos((n-1)π x)/sin((n-1)π) e^(-((n-1)²π² α t) dx].
Using the fact that sin((n-1)π) = 0 for all n, we get:
A_n = 100/∫[100 cos((n-1)π x)/sin((n-1)π) e^(-((n-1)²π² α t) dx]
Final Solution
The final solution is a
Introduction
In our previous article, we discussed how to solve the heat equation with given boundary conditions. We used the method of separation of variables to find the solution. In this article, we will answer some common questions related to the heat equation and its solution.
Q: What is the heat equation?
A: The heat equation is a fundamental partial differential equation (PDE) that describes how heat diffuses through a medium over time. It is given by the following equation:
∂u/∂t = α ∂²u/∂x²
where u(x,t) is the temperature at point x and time t, α is the thermal diffusivity of the medium, and ∂/∂t and ∂²/∂x² are the partial derivatives with respect to time and space, respectively.
Q: What are the boundary conditions?
A: The boundary conditions are essential in solving the heat equation. They specify the temperature at the boundaries of the medium at any given time. In this case, we are given the following boundary conditions:
u(0,t) = 0 (at x = 0) u(1,t) = 100 (at x = 1)
Q: How do we solve the heat equation?
A: We use the method of separation of variables to solve the heat equation. This method involves assuming that the solution can be expressed as a product of two functions, one depending only on x and the other only on t.
Let u(x,t) = X(x)T(t)
Substituting this expression into the heat equation, we get:
X(x)T'(t) = α X''(x)T(t)
Dividing both sides by α X(x)T(t), we get:
T'(t)/α T(t) = X''(x)/X(x)
Q: What is the eigenvalue problem?
A: The eigenvalue problem is a crucial step in solving the heat equation. It involves finding the eigenvalues and eigenfunctions that satisfy the boundary conditions.
We have the following equation:
X''(x) + λ X(x) = 0
This is an eigenvalue problem, where λ is the eigenvalue and X(x) is the corresponding eigenfunction. The boundary conditions u(0,t) = 0 and u(1,t) = 100 imply that X(0) = 0 and X(1) = 0.
Q: How do we find the eigenvalues and eigenfunctions?
A: We can find the eigenvalues and eigenfunctions by using the following trial solution:
X(x) = A sin(√λ x) + B cos(√λ x)
Applying the boundary condition X(0) = 0, we get:
B = 0
Applying the boundary condition X(1) = 0, we get:
A sin(√λ) = 0
Since A ≠ 0, we must have sin(√λ) = 0. This implies that √λ = nπ, where n is an integer.
Q: What is the time-dependent solution?
A: The time-dependent solution is given by:
Tn(t) = e^(-λn α t)
Q: What is the general solution?
A: The general solution is a superposition of the time-dependent solutions:
u(x,t) = ∑[A_n sin(nπ x) e^(-n²π² α t)]
Q: How do we find the coefficients A_n?
A: We can find the coefficients A_n by applying the boundary conditions. We can do this by substituting the expression for u(x,t) into the boundary conditions.
Applying the boundary condition u(0,t) = 0, we get:
∑[A_n sin(nπ x) e^(-n²π² α t)] = 0
This implies that A_n = 0 for all n.
Applying the boundary condition u(1,t) = 100, we get:
∑[A_n sin(nπ) e^(-n²π² α t)] = 100
This implies that A_n = 100/∫[sin(nπ) e^(-n²π² α t) dx]
Q: What is the final solution?
A: The final solution is a superposition of the time-dependent solutions:
u(x,t) = ∑[A_n (sin((n-1)π x) + 100 cos((n-1)π x)/sin((n-1)π)) e^(-((n-1)²π² α t)]
Conclusion
In this article, we answered some common questions related to the heat equation and its solution. We hope that this article has been helpful in understanding the heat equation and its solution.
References
- [1] "Heat Equation" by Wikipedia
- [2] "Separation of Variables" by MathWorld
- [3] "Eigenvalue Problem" by MathWorld
Further Reading
- [1] "Heat Transfer" by Wikipedia
- [2] "Partial Differential Equations" by MathWorld
- [3] "Numerical Methods for Partial Differential Equations" by MathWorld