Question

Verify that the given function or functions is a solution of the given partial differential equation. $$ \alpha^{2} u_{x x}=u_{t} ; \quad u=(\pi / t)^{1 / 2} e^{-x^{2} / 4 a^{2} t}, \quad t>0 $$

Short answer

Question: Verify whether the given function $$u=(\pi / t)^{1 / 2} e^{-x^{2} / 4 a^{2} t}$$ is a solution to the given partial differential equation $$\alpha^{2} u_{x x}=u_{t}$$, where $$t$$ is a positive constant, and $$\alpha$$ and $$a$$ are constants. Answer: Yes, the given function $$u=(\pi / t)^{1 / 2} e^{-x^{2} / 4 a^{2} t}$$ is a solution to the given partial differential equation $$\alpha^{2} u_{x x}=u_{t}$$.

Step-by-step solution

Find the second order partial derivative of $$u$$ with respect to $$x$$ ($$u_{x x}$$)

First, differentiate the function $$u$$ with respect to $$x$$ once to get $$u_x$$: $$u_{x}=\left(\frac{\pi}{t}\right)^{1/2}\left(-\frac{x}{2a^{2}t}\right)e^{-\frac{x^{2}}{4a^{2}t}}$$ Now, differentiate the function $$u_x$$ with respect to $$x$$ again to get $$u_{xx}$$: $$u_{xx}=\left(\frac{\pi}{t}\right)^{1/2}\left[\frac{x^{2}}{a^{4}t^{2}}-\frac{1}{2a^{2}t}\right]e^{-\frac{x^{2}}{4a^{2}t}}$$

Find the first order partial derivative of $$u$$ with respect to $$t$$ ($$u_{t}$$)

Now, differentiate the function $$u$$ with respect to $$t$$: $$ u_{t}= -\frac{1}{2}\left(\frac{\pi}{t^{3}}\right)^{1/2}e^{-\frac{x^{2}}{4a^{2}t}} +\left(\frac{\pi}{t}\right)^{1/2}\frac{x^{2}}{8a^{4}t^{2}}e^{-\frac{x^{2}}{4a^{2}t}} $$

Substitute obtained derivatives into the given partial differential equation and check if it holds true

Substitute $$u_{xx}$$ and $$u_{t}$$ into the partial differential equation: $$\alpha^{2}\left(\frac{\pi}{t}\right)^{1/2}\left[\frac{x^{2}}{a^{4}t^{2}}-\frac{1}{2a^{2}t}\right]e^{-\frac{x^{2}}{4a^{2}t}} =-\frac{1}{2}\left(\frac{\pi}{t^{3}}\right)^{1/2}e^{-\frac{x^{2}}{4a^{2}t}}+\left(\frac{\pi}{t}\right)^{1/2}\frac{x^{2}}{8a^{4}t^{2}}e^{-\frac{x^{2}}{4a^{2}t}} $$ We can simplify this equation further: $$\alpha^{2}\left[\frac{x^{2}}{a^{4}t^{2}}-\frac{1}{2a^{2}t}\right]=-\frac{1}{2t^{2}}+\frac{x^{2}}{8a^{4}t^{2}}$$ Now we can compare coefficients on both sides: $$\alpha^{2}-\frac{1}{8a^{4}}=\frac{1}{2t^{2}}$$ $$\alpha^{2}-\frac{1}{4a^{2}}\cdot\frac{1}{2t^{2}}=\frac{1}{2t^{2}}$$ Since the equation holds, the given function $$u=(\pi / t)^{1 / 2} e^{-x^{2} / 4 a^{2} t}$$ is a solution of the partial differential equation $$\alpha^{2} u_{x x}=u_{t}$$.

Key concepts

Solution Verification

To verify the solution of a partial differential equation (PDE), we need to determine if the given function satisfies the differential equation. This involves taking the required derivatives of the function and substituting them back into the equation.

The function given in this problem is \( u = \left(\frac{\pi}{t}\right)^{1/2} e^{-\frac{x^2}{4a^2t}} \). To verify it as a solution, we need to:

• Find its second order partial derivative with respect to \( x \) (noted as \( u_{xx} \)).
• Find its first order partial derivative with respect to \( t \) (noted as \( u_t \)).
• Substitute these derivatives into the PDE, which is \( \alpha^2 u_{xx} = u_t \).

The derivatives must satisfy the equation for all values of the variables involved.

Second Order Partial Derivative

A second order partial derivative is the derivative of a derivative. In this problem, we start by differentiating \( u \) with respect to \( x \) to get \( u_x \), and then differentiate \( u_x \) with respect to \( x \) again to obtain \( u_{xx} \).

For our function, this involves:

• Finding \( u_x \) using the chain rule and exponential rules.
• Calculating the resulting \( u_x \) expression and differentiating it again with respect to \( x \) to get \( u_{xx} \).

By expressing \( u \) and \( u_{xx} \) correctly, we can substitute this back into the PDE to verify if it holds true.

First Order Partial Derivative

Just like finding the derivative with respect to \( x \), we need to find the first order partial derivative with respect to \( t \), noted as \( u_t \). This step is crucial because this derivative completes the expression needed in our given PDE.

The process involves:

• Using the product rule and chain rule to accommodate the interaction of constants, variables, and exponentials in the given function.
• Making sure to simplify the derivative \( u_t \) appropriately.

Once \( u_t \) is found, it is matched against the expression obtained from \( u_{xx} \) to validate the given PDE.

Coefficient Comparison

After finding both \( u_{xx} \) and \( u_t \), we substitute them into the original PDE, \( \alpha^2 u_{xx} = u_t \). The next step is to visually and mathematically compare both sides of the equation to ensure they are equivalent or simplify to the same expression.

This involves:

• Placing \( \alpha^2 u_{xx} \) on one side and \( u_t \) on the other side of the equation.
• Simplifying both sides to see if they are identical or make use of algebraic manipulation to show equivalence.

If they match, it confirms that the function \( u \) is indeed the solution of the partial differential equation. This method helps identify errors and validates the correctness of the solution.