An integral that will be very useful when we solve the heat equation on the half...

 

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An integral that will be very useful when we solve the heat equation on the half line is known as the error function. For any constant a, we define the error function to be 2 erf (az) = ₁ raz which contains the integral of the Gaussian over the interval [0, az]. Oftentimes, our solution u(x, t) will contain these Gaussian integrals, but over different intervals, for example over the intervals [az, ∞] or [-az, az]. It is standard practice to manipulate these integrals into the form of the error function (where the integral is taken over [0, 2]). Let's practice this below. (a) We saw, when working with the heat equation on the whole line, that Le-² 2 ✓= 1.° e-8² r∞ ds. Before we begin working with error functions, first calculate the value of 2 ✓ S ds = √R. e ds. (Hint: you do not need to calculate this integral directly. Instead, keep in mind properties of the Gaussian function.) (b) Next, show that ds = 1- erf(z). (Hint: to show this result, start with the left hand side and consider splitting the integral over multiple intervals of the domain. Then simplify to obtain the right hand side.) (c) Lastly, show that 2 ²/7 (e-de-ed²); ds = 2 erf(z). An integral that will be very useful when we solve the heat equation on the half line is known as the error function. For any constant a, we define the error function to be 2 erf (az) = ₁ raz which contains the integral of the Gaussian over the interval [0, az]. Oftentimes, our solution u(x, t) will contain these Gaussian integrals, but over different intervals, for example over the intervals [az, ∞] or [-az, az]. It is standard practice to manipulate these integrals into the form of the error function (where the integral is taken over [0, 2]). Let's practice this below. (a) We saw, when working with the heat equation on the whole line, that Le-² 2 ✓= 1.° e-8² r∞ ds. Before we begin working with error functions, first calculate the value of 2 ✓ S ds = √R. e ds. (Hint: you do not need to calculate this integral directly. Instead, keep in mind properties of the Gaussian function.) (b) Next, show that ds = 1- erf(z). (Hint: to show this result, start with the left hand side and consider splitting the integral over multiple intervals of the domain. Then simplify to obtain the right hand side.) (c) Lastly, show that 2 ²/7 (e-de-ed²); ds = 2 erf(z).

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a We use the standard techniques for calculating this integral ... View the full answer