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 $$\alpha^{2} u_{x x}=u_{t}$$.