Fourier Transforms

ecause of the convolution structure. The transformed equation becomesf̂k2 Af̂k − ĥk 0and this is a quadratic equation that can be solved to givef̂k − A2 A22 ĥk .Both roots are important. It is sometimes possible to deduce what the solution should be bylooking at the structure. Write the two solutions asf̂1k − A 2 A22 ĥk , f̂2k − A 2 − A22 ĥk .If ĥk is absolutely integrable, then ĥk → 0ask → and sof̂1k → 0,f̂2k → −A.Thus if we require f̂k to be absolutely integrable, then only f̂1k is acceptable. If A 0,then both solutions can be important and an example has been previously given.ExampleUse the FT to solve the integral equation−fx − t ft dt exp − x22and check your answer by direct substitution.Outline of solution onlyStep 1: From (3.2.7), the integral is the convolution f ∗ fx. Take the FT of the equationand use (3.2.8) to obtainf̂k f̂k F exp − x 22 f̂k F exp − x22 gk sayStep 2: Invert transform using (3.2.9)fx 12−gk e −ikx dkStep 3: Determine fx, using contour integration if necessary. The solution should be17